MIT Open Course Ware Lecture Compliled

5.61 Fall 2007 Lecture #1 page 1

Quantum Mechanics – Historical Background

Physics in the Late 19th Century (prior to quantum mechanics (QM))

• Atoms are basic constituents of matter

• Newton’s Laws apply universally

• The world is deterministic

According to classical mechanics (CM):

Given initial positions r�

0 and velocities v

�

0 , and given all forces F

�

( )t �

all the future can be predicted!

�F d

�v �

�v �

F �

��

�a= m

�

�v

t

dv�

� =( )t = � � dt � = m dt �

�v t

0 m0

�r

( )t = ��r

�r0

� d�r

� t

r��vdt �

�v� d = =

�� dtt

0

Physics was complete except for a few decimal places !

• Newtonian mechanics explained macroscopic behavior of

matter -- planetary motion, fluid flow, elasticity, etc.

• Thermodynamics had its first two laws and most of their

consequences

• Basic statistical mechanics had been applied to chemical

systems

• Light was explained as an electromagnetic wave

MIT Open Courseware Compillation Source: http://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2007/lecture-notes/

Prepared By Kevin Boudreaux 1 of 227


� However there were several experiments that could not be explained

by classical physics and the accepted dogma !

• Blackbody radiation

• Photoelectric effect

• Discrete atomic spectra

• The electron as a subatomic particle

� Inescapable conclusions would result from these problems

• Atoms are not the most microscopic objects

• Newton’s laws do not apply to the microscopic world of the

electron

OUTCOME � New Rules!!!

Quantum Mechanics!

• Describes rules that apply to electrons in atoms and molecules

• Non-deterministic, probabilistic ! A new philosophy of nature

� Explains unsolved problems of late 19th century physics

� Explains bonding, structure, and reactivity in chemistry




The DEMISE of CLASSICAL PHYSICS

(a) Discovery of the Electron

In 1897 J.J. Thomson discovers the electron and measures (e m e

)(and inadvertently invents the cathode ray (TV) tube)

Faraday (1860’s – 1870’s) had already shown using electrochemistry that

amounts of electric current proportional to amounts of some substances could

be liberated in an electrolytic cell. The term “electron” was suggested as a

natural “unit” of electricity.

But Thomson experimentally observes electrons as particles with charge & mass.

y = displacement

induced by

deflector voltage �

Deflector plates Phosphor

Cathodescreen

Anode

Thomson found that results are independent of (1) cathode material

(2) residual gas composition

� “electron” is a distinct particle, present in all materials!

Classical mechanics � force on electron due to deflector voltage:

Fy

= �e (force starts at time t = 0 when electron enters region between plates)

dvy

dvy

= �� e �

= m dt

( F = ma ) �dt �

�m

e �

Integrating � vy

=��

�

m

e

e �

��t [Note v

y (t = 0) = 0]




Integrating again since vy

=dy

and y t = 0) = 0 � y =� e � �t

2

f

dt(

��

me

�

2

tf

= total time electron is between the plates (easily calculated)

� e �Set voltage �, calculate time t

f , measure displacement y �

��

� = 1x10

11 C/kg

m e

� e �Modern day value is

��

me �� = 1.758x10

11 C/kg

(b) 1909 Milliken oil drop experiment determines e, me separately

mist of micron-size

oil droplets in air

FGravitational force downward: g

Fg

= �Mg M = mass of droplet, g = gravitational constant

Frictional force upward due to air: Ff

Ff

= 6�r�v r = radius of droplet, � = air viscosity, v = droplet velocity

Since Ff � v , terminal velocity v

t is reached when forces balance

6�r�vt

= Mg � get droplet mass M = 6�r�vt

g

neNow use x-rays or �-rays to add some charge ( ) to the droplets

Voltage � across plates exerts Coulomb force Fc

= �ne on the charged droplet

x-rays n-F

cF

gF

f




Adjust voltage until drop stops falling: v = 0 � Ff

= 0, Fc

= Fg

�ne = Mg Determine ne = Mg �

Mulliken did this for lots of droplets i = 1,2,3,...

e eThey all had different charges (ni

) but all integer multiples of charge ( )

Determined elementary charge as e = 1.59x10�19

C

(very close to today’s value e = 1.602x10�19

C )

Combining values for (e m e m = 9.11x10�31

kge

) and ( ) �e

Hydrogen mass was known: mH

= 1.66x10�27

kg � electron is subatomic!!

(c) Where are the electrons? What’s the structure of the atom?

Angstrom (10-10 m) atomic size scale already inferred from gas kinetics

First “jellium” model didn’t last long

Au foil

(jelly)n+

e�

Rutherford backscattering experiment

He2+

He2+

(no electrons) 98%

undeflected

2%

scatter back

� (1) He2+ nucleus very small, << 10-10 m (Rutherford estimated 10-14 m)

(2) Au atoms are mostly empty!




Rutherford planetary model: classical mechanical model of atomic structure

Coulomb attraction plays the role of gravity

r

+Z

2m v

centripetal force Fc

= e

r

v Ze2

e�

Coulomb force FC

=4��

0r

2

m v2

Ze2

Ze2

for stable orbit e = � r =r 4��

0r

24��

0 m

ev

2

This is stable compared to separated electron & nucleus

1 � Ze2 � 1 Ze

2

E = K.E. + P.E. = mev

2 + �� = � < 0

2 4��0 r 2 4��

0r

BUT model not consistent with classical electrodynamics:

Accelerating charge emits radiation! (centripetal acceleration = v2/r)

And since light has energy, E must be getting more negative with time

� r must be getting smaller with time!

� Electron spirals into nucleus in ~ 10-10 s !

Also, as r decreases, v should increase

Frequency � of emitted light = frequency of rotation

� (Hz = cycles/s) =v (m/s)

2�r (m/cycle)

circumference of orbit

� atom should emit light at all frequencies – that is it should produce a

continuous spectrum




BUT emission from atoms was known to be discrete, not continuous!

For H:

n1 = 3 n1 = 2 n1 = 1 n2 = 3 4 5 6… n2 = 2 3 4…

10,000 30,000 50,000 100,000 (cm-1

)

For the H atom, Rydberg showed that the spectrum was consistent with the

simple formula:

� � 1 1 � (cm

-1 ) = R � �

� �

2n

1

2n

2

with n1

= 1, 2, 3, ... and n2

= n1

+ 1, n1

+ 2, n1

+ 3, ...

R = 1.097x105 cm

-1 (Rydberg constant)

n1 = 1 Lyman series

n1 = 2 Balmer series

n1 = 3 Paschen series

visible & UV lines well known

Summary: Rutherford’s model of the atom

(1) Is not stable relative to collapse of electron into nucleus

(2) Does not yield discrete emission lines,

(3) Does not explain the Rydberg formula




The DEMISE of CLASSICAL PHYSICS (cont’d)

(a) Blackbody radiation – When heated all objects emit light!

Classically: (1) Radiation from a blackbody is the result of electrons

oscillating with frequency �

Oscillating charged particle � antennae

(2) The electrons can oscillate (& radiate) equally well at any

frequency

� Rayleigh-Jeans Law for spectral density �(�), where intensity of emitted

light in frequency range from � to � + d� is I(�) ~ �(�)d�

d� = � � ,T ) d� =8�kT

�2 d� ��

2( 3

c

where d�(�,T) = density of radiative energy in frequency range

from � to � + d� at temperature T

k = Boltzmann’s constant [= R/NA (gas constant per molecule)]

c = speed of light

� � ��2 � divergence at high freq. Rayleigh- ( )

Jeans Law � lots of emission in UV, x-ray regions!

“UV catastrophe” – not observed in exp’t �(�)

Exp’t T3

T1 < T2 < T3T1

T2

�




Planck (~1900) � First “quantum” ideas

(1) The energy of the oscillator � frequency

E � �

(2) The energy �

(the # of oscillators n)

E � n�

an integral multiple of �

or E = nh�

constant

h� becomes a “quantum” of energy

Planck used statistical mechanics (5.62) to derive the expression for black body

radiation

d� = � � ,T( ) d� =8�h�

3

c3

1

eh� kT

�1

�

�

�

�d�

�

�(�)

� �3

1

eh� kT

�1

�

��

�

�

Fitting exp’t to model � h = 6.626 x 10-34 J-sec

Planck’s constant




(b) Photoelectric effect

Exp’t: UV light e-

K.E. = _mv2

metal

Classically

expect: K.E. K.E.

I �

Exp’t:

Opposite!

K.E. linear K.E.

I �0 �

threshold

Einstein (1905) proposed:

(1) Light is made up of energy “packets: “photons”

(2) The energy of a photon is proportional to the light frequency

E = h� h Planck’s constant




New model of photoelectric effect:

E

0

� = work function

of metal

K.E.

of e-

energy of

photon = h�

� K.E. = h� – � = h� – h�0 = h(� – �0)

K.E. � K.E. = h(� – �0)

�0 = �/h �

Comparing to exp’t, value of “h” matches the one found by Planck!

This was an extraordinary result !

Summary:

(1) Structure of atom can’t be explained classically

(2) Discrete atomic spectra and Rydberg’s formula can’t be explained

(3) Blackbody radiation can be “explained” by quantifying energy of

oscillators E = h�

(4) Photoelectric effect can be “explained” by quantifying energy of light

E = h�



� � �

�


The ATOM of NIELS BOHRNiels Bohr, a Danish physicist who established the Copenhagen school.

(a) Assumptions underlying the Bohr atom

(1) Atoms can exist in stable “states” without radiating. The states have

discrete energies En, n = 1, 2, 3,..., where n = 1 is the lowest energy

state (the most negative, relative to the dissociated atom at zero

energy), n = 2 is the next lowest energy state, etc. The number “n” is

an integer, a quantum number, that labels the state.

(2) Transitions between states can be made with the absorption or

�Eemission of a photon of frequency � where � = .

h

En1

h� h�or

Absorption Emission

En2

These two assumptions “explain” the discrete spectrum of atomic vapor

emission. Each line in the spectrum corresponds to a transition between two

particular levels. This is the birth of modern spectroscopy.

h(3) Angular momentum is quantized: � = n� where � =

2�

Angular momentum

L = �r � p �= L

�r

LFor circular motion:

L

�is constant if

�r and p

�are constant

l = mrv is a constant of the motion � �p = mv

Other useful properties

1



� � �


v� = (2�r ) � � = r �rot rot

velocity

(m/s) circumference frequency angular

(m/cycle) (cycles/s) frequency

(rad/s)

� � = mvr = mr2�

rot

Recall the moment of inertia I = �mir

i

2

i

� For our system I = mr2

� � = I�rot

Note: Linear motion vs. Circular motion

mass m � I moment of inertia

velocity v � �rot

angular velocity

momentum p = mv � � = I� angular momentum

Kinetic energy is often written in terms of momentum:

12

p2

1 m2r

2 v

2 �

2

K.E. = =K.E. = mv =2 2m 2 mr

22I

Introduce Bohr‛s quantization into the Rutherford‛s planetary model.

For a 1-electron atom with

a nucleus of charge +Ze+Ze

r e-

Ze2

n2

(4��0

)�

2

r =4��

0 mv

2 Z me

2 The radius is quantized!!� r =

(4��0

)�

2

� a0

the Bohr radius2

me

2




For H atom with n = 1, r = a 0 = 5.29x10-11 m = 0.529 Å (1 Å = 10-10 m)

Take Rutherford‛s energy and put in r,

1 Ze2

1 Z2me

4

E = � � E = � Energies are quantized!!!2 4��

0r

n

n2

8�0

2h

2

For H atom, emission spectrum

E E me4 � 1 1 �

-1 n2 � n1 = �� (cm ) =hc hc 8�

0

2h

3c ��

n1

2n

2

2

4

Rydberg formula ! with R =me

= 109,737 cm-1

8�0

2 h

3c

Measured value is 109,678 cm-1 (Slight difference due to model that gives

nucleus no motion at all, i.e. infinite mass.)

3




WAVE-PARTICLE DUALITY of LIGHT and MATTER

(A) Light (electromagnetic radiation)

Light as a wave

For now we neglect the polarization vector orientation

Propagating in x-direction:

�0

� 0x (const. t) or

t (const. x)

� ( x,t) = �0

sin ��2� �

( x � ct) � �

2�Define k = (“wavevector” magnitude)

�

At some fixed time, say t = 0, can write simply � x,t = 0) = �0

( )( sin kx

Superposition principle

sin(kx)

Constructive+ = 2sin(kx) interference

sin(kx)

(in phase)




sin(kx + �)

(out of phase by �/2)

This leads to many interference phenomena

Young‛s 2-slit experiment

D

�

�=

s

�� D =

s

Light as a particle

Compton exp‛t

If just a wave, expect light to scatter off electron

Experimentally:

e-x-rays (� � 1Å)

e-

� �

e-

��

Destructive+ = 0 interference

sin(kx)

light, �

sl

D

Interference

pattern

screen observed




The backscattered wave is red-shifted (� � > � ), i.e. less energy/photon.

hc hcE � = < = E

��

Energy (and momentum) transferred to the electron.

Need relativistic mechanics to solve

h� � h �p =

c ��=� �

for the light

Light is a particle with energy

h�momentum p =

c

Light can behave both as a wave and as a particle!!

Which aspect is observed depends on what is measured.

(B) Matter

Matter as particles � obvious from everyday experience

Matter as waves (deBroglie, 1929, Nobel Prize for his Ph.D. thesis!)

Same relationship between momentum and wavelength for light and for matter

h hp = � � = � de Broglie wavelength

� p

Amazing notion! But wavelength only observable for microscopic momentum

Diffraction

patterne- crystal




Consequences (I)

(1) on Bohr atom

+

If e- wave does not close on

itself, eventually destructive

interference will kill it!

+

If e- wave does close on

itself, then constructive

interference preserves it.

Criterion for stability:

nh nh2�r = n� = =

p mv

nhor mvr = = n�

2�

� � = n�

As Bohr had assumed angular momentum is quantized!




WAVE-PARTICLE DUALITY of MATTER

Consequences (II)

�Heisenberg Uncertainty Principle �x�p

x�

2

Consider diffraction through a single slit

D

�

�=

s

� � D =

s

light, �

s

l

peak-nullD

distance

x

Coming out of the slit, the electrons

D spread out to form a diffraction pattern

with width D.

x

v�

v�

xv�

This means the electrons must go through the slit with some range of

velocity components vx

x

Now consider a beam of electrons with de Broglie wavelength �. The slit

restricts the possible positions of the electrons in the x direction: at the slit,

the uncertainty in the electron x-position is

xs�=



�


�v D � �x � = =

v � s �x

�x�vx

= �v or �x�px

= � p

hBut from deBroglie � =

p

� �x�px

= h

So the position and momentum of a particle cannot both be determined with

arbitrary position! Knowing one quantity with high precision means that the

other must necessarily be imprecise!

The conventional statement of the Heisenberg Uncertainty Principle is

�x�p �x

2

(depends on how “uncertainty” is defined: 1/e half-width, FWHM, etc.)

Uncertainty can always be larger than � 2 , but not smaller.

Note that this sort of uncertainty is standard in classical wave mechanics. If

you focus a light beam or a water wavelet to a small spot size, at the focus

there is a wide range of propagation directions. What is new is the idea that

particles inherently show wavelike behavior, with similar consequences.

Implications for atomic structure

Apply Uncertainty Principle to e- in H atom

me = 9.11 x 10-31 kg �x � 10-10 m (1 Å)




� � 1�p

x � � �v � � x10

�6m s

2�xx

2m�x 2

Basically, if we know the e- is in the atom, then we can’t know its velocity at all!

Bohr had assumed the electron was a particle with a known position and velocity.

To complete the picture of atomic structure, the wavelike properties of the

electron had to be included.

So how do we properly represent where the particle is??

Schrödinger (1933 Nobel Prize)

A particle in a “stable” or time-independent state can be represented

mathematically as a wave, by a “wavefunction” �(x) (in 1-D) which is a solution to

the differential equation

Time-independent

Schrödinger

equation ��

2

2m

�2�

�x2

+ V x x x( )� ( ) = E� ( )

potential energy total energy

We cannot prove the Schrödinger equation. But we can motivate why it might be

reasonable.

�1

( x,t) = Asin (kx ��t) is a right-traveling wave.

A v

2�k = � = 2�� = v

�

Similarly, a left-traveling wave can be represented as

�2

( x,t) = Asin (kx + �t) .

Both are solutions to the wave equation




�2� ( x,t) 1 �

2� ( x,t)

=�x

2 v

2 �t

2

Further, the sum � ( x,t) = �1

( x,t) + �2

( x,t) of left and right traveling waves is

also a solution.

� ( x,t) = A��sin (kx ��t) + sin (kx + �t) ( )cos (�t)�

�= 2 Asin kx

This is a stationary wave or standing wave. Its peaks and nulls remain stationary.

At various times during a full cycle (2�/�):

) = 2 Asin kx� ( x,0 ( ) t = 0

2

2

� 1 2� ��

x,8 ��

= 2 A

� � sin kx( )

�t = �/4�

� 1 2� t = �/2�

nodes

� x,� 4

��

= 0

� 2

2

t = 3�/4�

� 3 2� ��

x,8 � ��

= 2 A

�� sin kx( )

�t = �/�

� 1 2� � x,

��= 2 A �1 ( )

� 2 ��

( )sin kx

As in a vibrating violin string, the node positions are independent of time. Only

the amplitude of the fixed waveform oscillates with time.

More generally, we can write wave equation solutions in the form

� x,t) = � ( )cos (�t)( x

In the particular case above, � x ( ) .( ) = 2 Asin kx




For the general case,

�2 �

�x

(2

x,t)v

1

2

�2�

�t

(2

x,t) �

v

�2

2

� ( x,t) = �k2� ( x,t) (� k = v)= =

Plugging in � x,t) = � ( )cos (�t)( x

�2� x � 2� �

2( )( ) = �

�x2

= �k2� x

� �� ( )x

�2� x

2

de Broglie relation � =h

p �

�x

( )= �

�

�

p

��

�� x( )

2

�2� x( )

��2

= p2� x( )

�x2

2But p ) = 2m ��E �V x �

�(assuming t-independent potential) = 2m(K.E. ( )

�2 �

2� x

+ V x x x� �( )

( )� ( ) = E� ( )2m �x

2

time-independent Schrödinger equation in one dimension

We now have the outline of:

• a physical picture involving wave and particle duality of light and matter !

• a quantitative theory allowing calculations of stable states and their

properties !




SOLUTIONS TO THE SCHRÖDINGER EQUATION

Free particle and the particle in a box

Schrödinger equation is a 2nd-order diff. eq.

x

� + V x x x�

2 �2� ( )

( )� ( ) = E� ( )2m �x

2

We can find two independent solutions �1

x x( ) and �2

( )The general solution is a linear combination

� x A�1

( ) + B�2

( )x( ) = x

A and B are then determined by boundary conditions on � x � � x( ) and ( ) .

Additionally, for physically reasonable solutions we require that � x( ) and

� � x be continuous function. ( )

(I) Free particle V(x) = 0

�2 �

2� x( )

� = E� x( )2m �x

2

2mE �2 k

2

or E =Define k2

=�

22m

2

V x E =p

� p2

= �2 k

2 �( ) = 0,

2m

h 2�de Broglie p = � k =

� �

p = �k




�2� x( )

= �k2� xThe wave eq. becomes ( )

�x2

with solutions � ( ) = ( ) + B sin kxx Acos kx ( )

Free particle � no boundary conditions

�2 k

2

� any A and B values are possible, any E = possible 2m

So any wavelike solution (traveling wave or standing wave) with any wavelength,

wavevector, momentum, and energy is possible.

(II) Particle in a box � �

V x x < 0, x > a( ) = � ( ) V x 0 � x � a) V(x) ( ) = 0 (

0 0 ax

Particle can’ xt be anywhere with V ( ) = �

� � ( x < 0, x > a) = 0

For 0 � x � a , Schrödinger equation is like that for free particle.

�2 �

2� x

�( )

= E� x2

( )2m �x

�2� x( )

= �k2� x with same definition ( )

�x2

k2

2mE or E =

�2 k

2

=�

22m




again with solutions � ( ) = Acos kx ( )x ( ) + B sin kx

But this time there are boundary conditions!

Continuity of � x � ( ) = � ( ) = 0 ( ) � 0 a

(i) � 0 ( ) + B sin 0 A = 0( ) = Acos 0 ( ) = 0 �

(ii) � ( ) = B sin kaa ( ) = 0

Can’t take B = 0 (no particle anywhere!)

Must have sin ka � ka = n� n = 1, 2, 3,... ( ) = 0

n�� k is not continuous but takes on discrete values k =

a

Thus integer evolves naturally !!

So solutions to the Schrödinger equation are

� (0 � x � a) = B sin ��

� n�

a

x

�

n = 1,2,3,...

These solutions describe different stable (time-independent or

“stationary”) states with energies

�2 k

2

E = �2m

n2 h

2

E =n 2

8ma

Energy is quantized!! And the states are labeled by a quantum number n

which is an integer.

Properties of the stationary states




(a) The energy spacing between successive states gets progressively

larger as n increases

�E

2

n+1� E

2 � h2

n=�

(n + 1) � n� �� 8ma

2

E

n+1� E

n= (2n + 1) E

1

E

1= h

28ma

2

� �

E

5= 25E

1

E

4= 16E

1

E

3= 9E

1

E

2= 4E

1

(b) The wavefunction � ( x) is sinusoidal, with the number of nodes

increased by one for each successive state

� 1

= Bsin � x a

� 2

= Bsin 2� x a

( ) (0 nodes)

� 3

= Bsin 3� x a

( ) (1 node)

� 4

= Bsin 4� x a

( ) (2 nodes)

( ) (3 nodes)

�

3= 2a 3

�

4= a 2

2a�=

�

1= 2a

� # nodes = n - 1 n

= 2a n

0 ax

(c) The energy spacings increase as the box size decreases.

1E � a

2




We’ve solved some simple quantum mechanics problems! The P-I-B model is a

good approximation for some important cases, e.g. pi-bonding electrons on

aromatics.

Electronic transitions shift to lower energies as molecular size increases !




QUANTUM MECHANICAL PARTICLE IN A BOX

Summary so far:

� �V ( x < 0, x > a) = � � ( x < 0, x > a) = 0

V (0 � x � a) = 0 � (0 � x � a) = B sin �

� n�

a

x

�

�n

V(x)

0 n2 h

2 n� 2a

a En

=8ma

2 k =

a � =

nn = 1,2,3,... 0 x

What is the “wavefunction” � x( ) ?

Max Born interpretation:

( )2

x x is a probability distribution or probability density

for the particle

� x = � * ( )� ( )

� � x dx is the probability of finding the particle in the interval ( )2

between x and x + dx

This is a profound change in the way we view nature!! We can only know the

probability of the result of a measurement – we can’t always know it with

certainty! Makes us re-think what is “deterministic” in nature.

Easy implication: Normalization of the wavefunction

� �x

2

� x dx = probability of finding particle in interval ( )2

x1

The total probability of finding the particle somewhere must be 1.

For a single particle in a box,




� a

( )2

�0

� ( )2

Normalization condition �� x dx = x dx = 1

a � n� x ��

0

B2

sin2

�� a dx = 1 � B =

2

a

�n

( )x =a

2 sin

��

� n�

a

x

�

�n = 1,2,3,... Normalized wavefunction

(2 a)sin2

4� x a Interpretation of � x( )

2

( ) based on measurement

Each measurement of the

(2 a)sin2 (3� x a ) position gives one result.

Many measurements give

a probability distribution

of outcomes.

(2 a)sin2 (2� x a )

(2 a)sin2 (� x a )

Expectation values or average values

For a discrete probability distribution

0.4

0.35

0.3

0.25

e.g. 0.2

0.15

0.1

0.05

0

2 4 6 8 10 12 14

�1

2

=

�2

2

=

�3

2

=()()

�4

2

=




<x> = average value of x

= 4(0.1) + 6(0.1) + 8(0.2) + 10(0.4) + 12(0.2)

= 4(P4) + 6(P6) + 8(P8) + 10(P10) + 12(P12)

where Px is probability that measurement yields value “x”

� x = � xPx

Now switch to continuous probability distribution

P � � ( )x

2

dx x

� � �

�

� x = �� x � ( )2

x dx

Similarly �

2x

2 � x dx= �� x ( )

2

Often written in “sandwich” form

�

x = � * x x dx�� ( ) x� ( )�

2x = �� * ( ) x

2� ( )dxx x

For particle in a box

2 a � n� x �x =

a �

0

x sin2

�� a �dx




aIntegrate by parts � x =

2

The average particle position is in the middle of the box.




VARIANCE, ROOT-MEAN SQUARE, OPERATORS,

EIGENFUNCTIONS, EIGENVALUES

� Deviation of ith measurement from average value <x>xi � x

xi � x � Average deviation from average value <x>

But for particle in a box, xi � x = 0

)2

� Square of deviation of ith measurement from average

value <x>

( xi � x

)2

( xi � x � �

2 � the Variance in x

x

Note )2 2

22( xi � x = x � x = �

x

The Root Mean Square (rms) or Standard Deviation is then

��

��

�2

1 2

2= x x

x � �

The uncertainty in the measurement of x, �x, is then defined as

�x = �x

�x

for particle in a box

a �

�x

2 = � * x

2� x dx � � * x x dx( ) x ( ) ( ) x� ( )

0 ��

2

� 2 � a � n� x � �� 2 � a � n� x � �= a ��

0

x2

sin2

a ��dx �

�� a ��

0

x sin2

a ��dx

��




Evaluate integral by parts

� 2 2 �� 2 �

� �2

=

a

3 �

a

)2

��

a

4 �x

2(n�

a2 �(n� )

2 ��

2 = � 2

x

4(n� )2

� 3

�

1 2

��x = � =x

2(a

n� ) �

�(n�

3

)2

� 2�

�

Note that deviation increases with a, and depends weakly on n.

Now suppose we want to test the Heisenberg Uncertainty Principle for the

particle in a box.

2 �1 2

2 2We need and p to get �p = � = �p � pp

p �

�

But do we write p = � * x x dx ?�� ( ) p� ( )

what do we put in here??

We need the concept of an OPERATOR

Af ˆ x x( ) = g ( )

operator acts on function to get a new function




d � x2 � � 2x �

e.g. dx �

�3 �� =

�� 3 ��

operator function new function




Special Case

If Af x xˆ ( ) = a f ( )

number (constant)

then f x ( ) is called an eigenfunction of the operator

and is the eigenvalue.

This is called an eigenvalue problem (as in linear algebra).

Quantum mechanics is full of operators and eigenvalue problems!!

e.g. Schrödinger’s equation:

� �2

d 2 �

�� + V x x = E � x( )� � ( ) ( )2m dx

2

� �

H operator Eigenfunction constant

(Hamiltonian)

or H� = E� with H x�

2 d

2

+ V x (in 1D) ( ) = � ( )2m dx

2

The Hamiltonian operator, acting on an eigenfunction, gives the energy.

i.e. the Hamiltonian is the energy operator

2

If V x then ( ) = 0 , E = K.E. =p

2m

( )2

� H =p

( )2 d

2

� p = ��2

2m dx2

means p ( ) i.e. the operator acts sequentially on the function ( )p2

( )ˆ p




p f x ( )ˆ p x ˆ ˆ x ��

= p g ( )p ( ) f �� ( ) �

��( )

2

( ) = ˆ ( ) = p pf ˆ x

� p p

��i�

dx �� i�

dx ��= ��

2

dx2

( )( ) =� d � � d � d

2

d� p = �i� Momentum operator (in 1D)

dx

p

for Particle in a Box

2 � � d 2

� � d

�� ( )x dx

�

2

2�

2 = p � p = �� * ( )

��i�

dx �� x dx �

�� * x

�� dx �

x ( )��

( ) �i�p

Note order is now very important! Operator acts only on the function to its

right.

� � d p = �� * ( )

��i�x

�� x dx( )

dx

a � 2

1 2

� n� x � d � 2 1 2

� n� x = �

0

�

��

a ��sin

�� a ��i�

dx �� a ��

sin �� a ��

��dx

= 0

a �� 2 � � n� x � � d � � d � �� 2 �

1 2

� n� x � 1 2

2p = �

0 ��

a ��sin

�� a �� i�

dx �� i�

dx �� a ��

sin �� a ��

�dx

� �

2�2

a

2 2

a

� n�

a

x

��

�

��

� n

a

� � � n�

a

x � 2

a

�2

��

� n

a

�

��

�

0

a

sin2

��

� n�

a

x �=

0

sin ��

sin ��

dx =��

dx

n2�

2 �

2

=2

a



�


n2 h

2 n

2 h

2

= = 2mNote p2

= 2mE as expected 4a

28ma

2

E = K.E. since V(x) = 0

�2

p =

n2�

2

2�

2

= (�p)2

a

� �x�p =2(

a

n� )

�(n�

3

)2

� 2��

n�

a

�

2

�

�

(n�

3

)2

� 2��

=

1 2 1 2

always > 1

� �x�p � as expected from Heisenberg Uncertainty Principle 2




THE POSTULATES OF QUANTUM MECHANICS (time-independent)

Postulate 1: The state of a system is completely described by a

wavefunction � (r,t) .

Postulate 2: All measurable quantities (observables) are described by

Hermitian linear operators.

Postulate 3: The only values that are obtained in a measurement of an

observable “A” are the eigenvalues “an” of the corresponding operator “ A”. The

measurement changes the state of the system to the eigenfunction of A with

eigenvalue an.

Postulate 4: If a system is described by a normalized wavefunction � ,

then the average value of an observable corresponding to A is

a = �� *A� d�

Implications and elaborations on Postulates

2

#1] (a) The physically relevant quantity is �

� * (r,t)� (r,t) = � (r,t)2

� probability density at time t

and position r

(b) � (r,t) must be normalized

�� *� d� = 1

(c) � (r,t) must be well behaved




(i) Single valued

(ii) � and � � continuous

(iii) Finite

#2] (a) Example: Particle in a box eigenfunctions of H

� n� x �H x x � x � x( )�

n ( ) = E

n ( ) ( ) =

��

�

a

2

�

�1 2

sin �� a �

n n

But if � is not an eigenfunction of the operator, then the statement is not

true.

e.g. � ( ) above with momentum operator xn

� ( ) = �i�d d

�� 2 �1 2

� n� x � p

n nx

dx � ( )x = �i�

dx �� a �

sin �� a � �

�n

� �

�� 2 �1 2

� n� x � �� p

n �� a �

sin �� a � �

�

(b) In order to create a Q.M. operator from a classical observable, use

d x = x and p

x = �i� and replace in classical expression.

dx

e.g.

K.E. =1

p2

=1

p p�

2 d

2

(1D) ( )( ) = �2m 2m 2m dx

2

�2 � �

2 �

2 �

2 �= �

2m ��x

2 +�y

2 +�z

2 � (3D)

Another 3D example: Angular momentum L = r � p




� d d �l

x = yp

z � zp

y = �i�

��y

dz � z

dy ��

� d d �l

y = zp

x � xp

z = �i� z

�� dx � x

dz ��

� d d �l

z = xp

y � yp

x = �i�

��x

dy � y

dx ��

(c) Linear means

A f x x Af ˆ x Ag x and

A cf x ��

= cA f x

�� ( ) + g ( )�� = ( ) + ˆ ( )�� ( ) �

� ( )��

(d) Hermitian means that

�� 1

�A�

2 d� = �� 2 ( A�

1 )�

d�

and implies that the eigenvalues of A are real. This is important!!

Observables should be represented as real numbers.

Proof: Take A� = a�

�� ( A� )d� = �� ( A� )

�

d�

��a�d� = �� (a� )

�

d�

� a = a *

true only if a is real

(e) Eigenfunctions of Hermitian operators are orthogonal

i.e. if A� = a � and A� = a �m m m n n n




then �� m

��

nd� = 0 if m � n

Proof:

�� m

�A� d� = �� n ( A� )

�

d�n m

an �� m

��

nd� = a

m

�

�� n �

m

�d�

� �� (a

n � a

m ) �� m �

nd� = 0

a � a �

�� d� = 0( n m ) � m n

= 0 if = 0 if n � m

n = m

Example: Particle in a box

As much + as - area

�Eigenfunctions

of

�1

�2

�3

�4

�1

��

2

�1

��

3

�1

��

4

0 a 0 ax x

In addition, if eigenfunctions of A are normalized, then they are orthonormal

�� m

��

nd� = �

mn

Krönecker delta




�mn

= ��1 if m = n (normalization)

�0 if m � n (orthogonality)

#3] If � is an eigenfunction of the operator, then it’s easy, e.g.

H�n

= En �

n measurement of energy yields value

But what if � is not an eigenfunction of the operator?

e.g. � could be a superposition of eigenfunctions

� = c1�

1 + c

2�

2

where A�1

= a1�

1 and A�

2 = a

2�

2

Then a measurement of A returns either a1

or a2 , with probability c

1

2 or c2

2

respectively, and making the measurement changes the state to either �1

or �2 .

#4] This connects to the expectation value

(i) If �n

is an eigenfunction of A , then A�n

= an �

n

a = �� n

�A�

nd� = a

n �� n

��

nd� = a

n

= a only value possible a n

measure

(probability c1

2 )

(probability c2

2 )

a1

a2

�2

�1

�




(ii) If � = c1�

1 + c

2�

2 as above

�� ˆ � (c

1�

1 + c

2�

2 )�

ˆ ( �1

+ c2�

2 = c

1

2 a

1 + c

2

2 a

2a = A� d� = A c

1) d�

c12 is the probability of measuring a

1

<a > = average of possible values weighted by their probabilities




PRINCIPLES OF QUANTUM MECHANICS (cont’d) ,

COMMUTATORS

Order counts when applying multiple operators!

e.g. A = px = ? and can we write px = xp ?

� operate on function to obtain .

Af ˆ x x � px x��i�

d �x x

d � d �

�f ( )

( ) = g ( ) ( ) f ( ) =� dx �

( ) f ( )

= �i�x dx

f x x

��i�x

dx � i� x( ) � i�f ( ) =

� A =��i�x

d �

� dx � i�

�= ( px)

Now try B = xp

Bf ˆ x x� d d �

x �i�x x

� B = �i�xd

( ) = ( )��i�

dx �

�f ( ) =

�

�

dx �f ( )

= xpdx

� xp � px

Define commutator

For two operators A and B ,

��

A, B��

= AB � BA = C

need not be zero!

e.g.

�� x, p�� = xp � px = i� � 0 !




Important general statements about commutators:

1) For operators that commute

��

A, B��

= 0

• it is possible to find a set of wavefunctions that are eigenfunctions

of both operators simultaneously.

e.g. can find wavefunctions �n

such that

A� = a � and B� = b �n n n n n n

• This means that we can know the exact values of both observables

A and B simultaneously (no uncertainty limitation).

2) For operators that do not commute

��

A, B�� 0

• it is not possible to find a set of wavefunctions that are

simultaneous eigenfunctions of both operators.

• This means that we cannot know the exact values of both

observables A and B simultaneously � uncertainty !

e.g. �� x, p� = i� � 0 � �x�p ��

2



5.61 Fall 2007 Lectures #12-15 page 1

THE HARMONIC OSCILLATOR

• Nearly any system near equilibrium can be approximated as a H.O.

• One of a handful of problems that can be solved exactly in quantum

mechanics

examples

m1 m2 B (magnetic

field) A diatomic molecule μ

(spin

magnetic

moment) E (electric

field)

Classical H.O.

m

X0

k

X

Hooke’s Law: f = �k X � X0

) � �kx ((restoring force)

d2 x d

2 x � k �

f = ma = m

dt2

= �kx �dt

2 +�� m�

x = 0




Solve diff. eq.: General solutions are sin and cos functions

x t Asin �t � =( ) = ( ) + Bcos (�t)k

m

or can also write as

x t( ) = C sin (�t + �)

where A and B or C and � are determined by the initial conditions.

e.g. x 0 v 0( ) = x0

( ) = 0

spring is stretched to position x0 and released at time t = 0.

Then

x ( )0 = A sin 0 ( ) = x0

�( ) + B cos 0 B = x0

v 0 = � cos 0( ) �� sin 0( ) = 0 � A = 0( ) =dx

dt x =0

So x t( ) = x0

cos (�t)

Mass and spring oscillate with frequency: � =k

m

and maximum displacement x0

from equilibrium when cos(�t)= ±1

Energy of H.O.

Kinetic energy � K

1 2

1 � dx �2

1 2 1 kx

0

2 sin

2 �tK = mv = m = m �� x

0 sin �t =

2 2 �� dt � 2

� ( )�� 2

( )

Potential energy � U

f x � U = � f x dx = kx dx = kx2

=( ) = �dU

� ( ) � ( )1 1

kx0

2 cos

2 (�t)dx 2 2




Total energy = K + U = E

E =1

kx0

2 ��sin

2 (�t) + cos2 (�t)��2

12

E = kx0

2

x(t)

x0(t)

0 t

-x0(t)

12

U K

kx0

E2

0 t

Most real systems near equilibrium can be approximated as H.O.

e.g. Diatomic molecular bond A B

X

U

X

X0 A + B separated atoms

equilibrium bond length




U X ( X � X0

)2

+1 d

3U

( X � X0

)3

+�( ) = U ( X0

) +dU 1 d

2U

( X � X0

) +dX 2 dX

2 3! dX

3

X = X0 X = X

0 X = X

0

Redefine x = X � X0

and U ( X = X0

) = U ( x = 0) = 0

( ) =dU 1 d

2U 1 d

3U

U x x + x2

+ x3

+�dx

x =02 dx

2

x =0 3! dx

3

x =0

U

x

real potential

H.O. approximation

dUAt eq. = 0

dx x =0

For small deviations from eq. x3

<< x2

� U x1 d

2U

2 �

1 kx

2x( ) �

2 dx2

2 x =0




Total energy of molecule in 1D m1 m2

XX1 X2XCOM

xrel M = m

1 + m

2 total mass

m1m

2μ = reduced mass m

1 + m

2

XCOM

m1 X

1 + m

2 X

2 COM position =

m1

+ m2

xrel

= X2 � X

1 � x relative position

1 � dX1 �

2

1 � dX2 �

2

1 � dXCOM

�2

+1 � dx �

2

K = = μ2

m1

��

dt �� +

2 m

2

��

dt ��

2 M

��

dt ��

2 �� dt ��

1U = kx

2

2

1 � dXCOM

�2

+1 � dx �

2

1E = K + U = μ

2 M

��

dt ��

2 �� dt ��+

2 kx

2

COM coordinate describes translational motion of the molecule

2

1 � dXCOM

�E =

trans 2

M

��

dt ��

QM description would be free particle or PIB with mass M

We’ll concentrate on relative motion (describes vibration)

1 � dx �2

1 = μE

vib 2 �� dt ��

+2

kx2

and solve this problem quantum mechanically.



�


THE QUANTUM MECHANICAL HARMONIC OSCILLATOR

H� x��

2 d

21 �

�� ( ) = E� ( )+ kx2

x x( ) = �� 2m dx

22

K U

Note: replace m with μ (reduced mass) if

Goal: Find eigenvalues En and eigenfunctions �n(x)

Rewrite as:

m1 m2

d2� x( )

+2m

dx2

�2

� �� E �

1

2 kx

2� � � x( ) = 0

This is not a constant, as it was for P-I-B,

so sin and cos functions won’t work.

TRY: f ( ) = e�� x

22 (gaussian function) x

d 2

f x2

2 �� x2

2 ( ) + �2

x( )

= ��e�� x 2

+ �2 x e = �� f x x

2 f ( )

dx2

d 2

f x( )+ � f x x

2 f xor rewriting, ( ) �� 2 ( ) = 0w

dx2

which matches our original diff. eq. if

2mE mk � = and �

2 =

�2

�2

� E =2

k

m




We have found one eigenvalue and eigenfunction

m

k 1 or � =Recall � =

2�

k

m

1 1� E = �� = h�

2 2

This turns out to be the lowest energy: the “ground” state

For the wavefunction, we need to normalize:

where N is the normalization constant � x x2

2( ) = Nf ( ) = Ne�� x

1 4

�

N 2 �� x

�� ( )x

2

dx = 1 � ��

e2

= 1 � N =

��

� �

�

� �

�0

( ) =

�� 1 4

�� x2

2x

� �e

x� �0

( )11 1

E0

= ��E0

= � = h�22 2

x

Note �0

x x( ) is symmetric. It is an even function: �0

( ) = �0

(�x)There are no nodes, & the most likely value for the oscillator displacement is 0.

So far we have just one eigenvalue and eigenstate. What about the others?



� �

� �


1�� x 2 =x h��

0 ( ) =

��

�� 1 4

2

� �e E

02

(2�1 2

x) e�� x

22

E1

=2

3 x h��

1 ( ) =

1

1

8

2 ��

��

� �

5�

2 x

��

1 4

1 4

(4�x2 � 2) e

�� x2

2 E

2 =( ) =

�� 2 h�

�� 2

8�3 2

x3 �12�

1 2 x) e

�� x 2 E

3 =

7 x h��

3 ( ) =

1

48 �� 2�

1 4

(

� km �with � =

�� 2 ��

These have the general form

1 2

�n

x( ) =1

2n n!( )

1/ 2

�

�

�

��

�

��

1 4

H n �

1 2 x( ) e

�� x2

2 n = 0,1,2,...

Normalization Gaussian

Hermite polynomial (pronounced “air-MEET”)

y even n = 0

H1

( ) = 2 y ( )

H0

( ) = 1 ( )y odd n = 1

H2

( )y = 4 y2 � 2 even (n = 2)

H3

( ) = 8y ( )y3 �12 y odd n = 3

H4

y4 � 48y even (( ) = 16 y

2 + 12 n = 4)




� x � x( )2

n ( )

n

�

�

�

� x E3

=3

( )7��

2

x E2

= 2

( )5��

2

x E1

= 1

( )3��

2

0 ( )x E =

��

0 2

� 1�Energies are E

n =

��n +

��h�

2

Note E increases linearly with n.

Energy levels are evenly spaced

En+1

� En

=�

��(n + 1) +

1

2

�

�h� �

�

��n +

1

2

�

�h� = h� regardless of n

1 There is a “zero-point” energy E

0 = h�

2

E = 0 is not allowed by the Heisenberg Uncertainty Principle.




Symmetry properties of �’s

�0,2,4,6,....

are even functions � �x) = � ( )( x

�1,3,5,7,....

are odd functions � �x) = �� ( )( x

Useful properties: (even) (even) = even

(odd) (odd) = even

(odd) (even) = odd

d (odd)= (even)

d (even)= (odd)

dx dx

��

(odd) dx = 0 ��

(even) dx = 2��

(even) dx �� 0

Just from symmetry:

� � � d ( )�

x n

= �� n ( ) x�

n ( )dx = 0x x p = �

�

��ih

��

n x dx = 0

n �� n dx

odd odd

Average displacement & average momentum = 0

IR spectroscopy H.O. selection rules

Intensity of vibrational absorption features

n’ = 1

Vibrational transition h�

n = 0

�+ �-




2

Intensity Inn�

�d μ �

�� n

�x�

n ' dx

dx

1) Dipole moment of molecule must change as molecule vibrates

HCl can absorb IR radiation, but N2, O2, H2 cannot.

2) Only transitions with n� = n ± 1 allowed (selection rule).

(Prove for homework.)

QUANTUM MECHANICAL HARMONIC OSCILLATOR &

TUNNELING

Classical turning points

Classical H.O.: Total energy ET

=1

kx0

2

2 EToscillates between K and U.

Maximum displacement x0 occurs when

all the energy is potential.

E

x -x0 x0

x0

=2E

T

k is the “classical turning point”

The classical oscillator with energy ET can never exceed this displacement, since

if it did it would have more potential energy than the total energy.




Quantum Mechanical Harmonic Oscillator.

x( )2

x( )2

x( )2

x( )2

x

�3

�2

At high n, probability �1

density begins to look

classical, peaking at turning

points. �

0

�12

x( )2

()

1

2 kx

2

x�3

( )2

�2

�1

�0

x( )2

x( )2

x( )2

Non-zero probability at x > x0!

Prob. of (x > x0, x < -x0): 1

�

2 �0

2 ( )x dx = 2 �

��

�

� 2 �� x

2

e dx �

�1 2 �

�1 2

�

=2

�1 2

�

e� y

2

dy = erfc 11

( )

“Complementary error function”

tabulated or calculated

numerically

Prob. of (x > x0, x < -x0) = erfc(1)

= 0.16

Significant probability!

x




The oscillator is “tunneling” into the classically forbidden region. This is a purely

QM phenomenon!

Tunneling is a general feature of QM systems, especially those with very low

mass like e- and H.

� x( ) ~ sin kx + �( )

E

V0

� x( ) ~ sin kx( )Finite

barrier � x( ) ~ e

�� x

x Even though the energy is less than the barrier height, the wavefunction is

nonzero within the barrier! So a particle on the left may escape or “tunnel” into

the right hand side.

d 2� ( )x �2m V(

0 � E)

( ) � �2� ( )or = � �� x x

dx2

��

2

��1

( )Solutions are of the form � ( ) = Be

� x� = �

�2m V

�

0

2

E �2

x with

� �

Note � � (V0 E)

1 2

and � � m

�

2

2m

1 2

If barrier is not too much higher then the energy and if the mass is light, then

tunneling is significant.

Important for protons (e.g. H-bond fluctuations, tautomerization)

� d 2 �

� x x( ) = E� ( )Inside barrier: + V0�

� � dx

2




Important for electrons (e.g. scanning tunneling microscopy)

Nonstationary states of the QM H.O.

System may be in a state other than an eigenstate, e.g.

� = c0�

0 + c

1�

1 with c

0 + c

1 = 1 (normalization), e.g.

2 2

c0

= c1

=2

Full time-dependent eigenstates can be written as

� x,t) = � x0 0

( ( ) e i�

0t

� x,t) = � x1 1

( ( ) e i�

1t

where

1

1 1 3 3��

0 = E

0 = ��

vib � �

0 = ��

vib � �

1 =�

vib ��

1 = E

1 = �

vib 2 2 2 2

System is then time-dependent:

� ( x,t) = e x e i�

1t

�1

x t x t x i�

0t

�0

( ) + ( ) = c0

( )�0

( ) + c1

( )�1

( )1

2

1

2

i�0t i�

1t

where c0

( )t = e c1

( )t = e1

2

1

2

What is probability density?

�� ( x,t)� ( x,t) =

1

2

��

�x

i�0t �

xi�

1t � ��

x� i�

0t

x� i�

1t �

0

( ) e +1

( ) e

0 ( ) e +

1 ( ) e

1 i(�1 ��

0 )t � i(�1 ��

0 )t 1 =

2 ��

0

�

0 +

1

�

1 +

1

�

0e +

0

�

1e

� =

2

��

0

2 +

1

2 + 2

0

1 cos (�

vibt)�

Probability density oscillates at the vibrational frequency!




�t = 0, � 4, � 2, 3� 4, �

� ( x,t)2

x

�1

x( )2

()

�0

x( )2

()

x 2�

0�

1 cos (�t)

�t = 0, � 4, � 2, 3� 4, �

What happens to the expectation value <x>?




�

��

x = �� ( x,t) x� ( x,t) dx

1 �

= ��

0

�x

i�0t

+1

� ( ) ei�

1t �

�

0 ( ) e

� i�0t

x� i�

1t

�dx( ) e x

x x +

1 ( ) e

2

� � � �

=1

�� 0

�x

0 dx + �� 1

�x

1dx + �� 1

�x

0 e

i(�1 ��

0 )t

dx + �� 0

�x

1e � i(�1

��0 )t

dx ��

2

<x>0 = 0 <x>1 = 0 = cos (�vib

t) ��

0 x

1dx

��

<x>(t) oscillates at the vibrational frequency, like the classical H.O.! �

Vibrational amplitude is � �0 x�

1dx

�

1 1

2 2

�0

x

� � �1

2 �

��

� �

4

(2�1 2

x) e� x 2

e( ) =�� 4

� x 2 �

1 ( )x =

1

� x�0

( ) =�

��

� �

4

xe� x

22

= (2� )1 2

xx �1

( )

�

)1 2 �

)1 2

2� dx = dx = x ( )t = (2� )

1 2

cos (�vib

t)�� 0 x�

1(2� �� 0

(2�

Relations among Hermite polynomials

Recall H.O. wavefunctions

1

2

�n

x1

1/2 ��

��

� ��

� 4

H (�1 2 x) e

� x 2 n = 0,1,2,... ( ) =

n

2n n!( )

Normalization Gaussian

Hermite polynomial



� �


y even n = 0

H1

( ) = 2 y ( )

H0

( ) = 1 ( )y odd n = 1

H2

( ) = 4 y ( )y2 2 even n = 2

H3

( ) = 8y ( )y3 12 y odd n = 3

H4

( ) = 16 y ( )y4 48y

2 + 12 even n = 4

Generating formula for all the Hn:

H y 1 ey

2

en

( ) = ( )n d

n

y2

dyn

A useful derivative formula is:

n ( )

= 1 2 yey

2

e y

2

+ 1 ey

2dH y d

n d

n+1

e y

2

= 2 yH ( ) H ( )( )n

( )n

dy dyn

dyn+1 n

yn+1

y

Another useful relation among the Hn’s is the recursion formula:

( ) 2 yH ( ) + 2nH ( ) = 0Hn+1

yn

yn1

y

Substituting 2 yH y ( ) above gives ( ) = H ( ) + 2nH n

yn+1 n1

y

dH y

dy = 2nH

n1 y

n ( )

( )

Use these relations to solve for momentum <p>(t)

p =�

�� ( x,t) p� ( x,t) dx��

=1 �

� � ( ) ei�

0t �

xi�

1t �

ˆ �0

( ) e� i�

0t

x� i�

1t � �� 0

x +1

( ) e p

x +1

( ) e

dx2

� � � �

=2 ��

0 p

0 �� 1 p

1 �� 1 p

0 �� 0 p

1 �

1 � �ˆ dx +

�ˆ dx +

�ˆ e

i(�1 ��

0 )t

dx +�

ˆ e � i(�1

��0 )t

dx�

<p>0 = 0 <p>1 = 0




1 1

2 �� 2d �

0 x

�� 4

(�x) e� x 2

= �1

xdx

( ) =�� 2 ��

( )

1 1

�� i(�1

��0

�� 2 i(�1

��0 )t

��

� � � 2 i�

vibt

� �� 1 p

0 e

)t

dx = i�� 2 �

e �� 1

1dx = i�

�� 2 �e

To solve integral ��

�0

�p�

1e � i(�1

��0 )t

dx use relations among Hn’s ��

d �

1 x

d �N1 H

1 (�1 2 x) e

� x2

2

= �

1 2 N

1

d �H1

( )y e y

22� �( ) =

dy � dx dx �

xwith y � �1 2

dy = �1 2

dx dx = �1 2

dy d

= �1 2

d

dx dy

�d ( ) = �

1 2 N

1

� dH

1 y

y2

2 yH

1 ( ) e

y2

2�

1 x ( ) e

dx ��

dy y

d H

1 y y y( ) = 2nH

0 ( ) = 2H

0 ( )

dy

yH1

( )y =1

2

1

2 H

2 y�

�2nH

0 y y �

= H

0 ( ) +( ) + H

2 ( ) y ( )

d

dx �

1 x �

�( ) e

y2

2 H

2 ( )y e

y2

2

�

= �1 2

N1 �� 1 1

( )�

( ) = �1 2

N1

H0

y2

1 �

0 x �

2 x

�( )

2N2 �N

0

� � d

�� 0

�p

1e � i(�1

��0 )t

dx = e � i(�1

��0 )t (�i�) �� 0

�

1dx

dx

� � �= e

� i(�1 ��

0 )t (�i�)�1 2 N

1

1

�� 0

�

0 dx �

1

�� 0

�

2 dx�

�N0

2N2 �

1

2

= e � i(�1

��0 )t (�i�)�1 2

N

N

0

1

��

� �

2 ��

�e� i�

vibt

= �i�

Finally




� 1 � 1

p ( )t =2

1

i��

��

2 �

� 2

(ei�

vibt

e i�

vibt )�� = �

�

��

2

�

�

2

sin (�vib

t) � �

Average momentum also oscillates at the vibrational frequency.



5.61 Fall 2007 12-15 Lecture Supplement Page 1

Harmonic Oscillator Energies and Wavefunctions

via Raising and Lowering Operators

We can rearrange the Schrödinger equation for the HO into an interesting form ... 2

+ ( m�x)2 � ��

� = 1

2m� p2

+ (m�x)2 �� = E�

1 � � � d �� 2m i dx

with

H =1

� p2+ (m�x)

2 �2m � �

which has the same form as

u2+ v2

= (iu + v)(iu + v) .

We now define two operators

a±�

1

2�m��ip + m� x( )

that operate on the test function f(x) to yield

(aa

+x

�� 1

(ip + m� x)(ip + m� x)��f ( )) f ( ) = x

2�m�

=1

�� p

2 + (m� x)

2 im� (xp px)�� f (x)

2�m�

( p2+ (m�x)

2 ) [x, p } f ( )= {2�

1

m� 2

i

�] x

aa

+=

1 ( p2+ (m�x)

2 ) +2

1=�

1

� 2

1H +

2�m�

Which leads to a new form of the Schrödinger equation in terms of a+ and a- …

H� = ��aa

+

1�2 �

�

If we reverse the order of the operators-- aa

+� a

+a-- we obtain …




H� = ��a

+a

+1�2 �

�

or

��

��a

±a�

±1

2 �� = E�

and the interesting relation

aa+ a

+a = [a a ] = 1

+

A CLAIM: If � satisfies the Schrödinger equation with energy E, then a+�

satisfies it with energy (E+�� ) !

H (a+� ) = ��

��a

+a

+2

1��

(a+� ) = ��

��a

+aa

++

2

1a

+ ��

= ��a+

�

� aa

++

1

2

�

� � = a

+

�

��

� a

+a

+ 1 +1

2

�

� � ��

�

� a

+a

+1

2

�

� + ��

� � � � = a+

� �

= a+ (H + �� )� = (E + �� )(a+� )

H (a+� ) = (E + �� )(a+

� )

Likewise, a-� satisfies the Schrödinger equation with energy (E-�� ) …

H (a� ) = ��

�aa

+

1

2

�

�(a� ) = ��

�

�aa

+a

1

2a

�

�� = a

��

�

�a

+a

1

2

�

��

= a ��

��aa+

12

1� ��

= a (H �� )� = a (E �� )�

�

H (a� ) = (E �� )(a� )

So, these are operators connecting states and if we can find one state then we can

use them to generate other wavefunctions and energies. In the parlance of the

trade the a± are known as LADDER operators or

a+ = RAISING and a- = LOWERING operators.

We know there is a bottom rung on the ladder �0 so that

a� 0 = 0




2�m�

1 � d �

��dx

+ m�x�� 0 = 0

Integrating this equation yields d� 0 m�

dx=

�x� 0

�d� 0 =

m�� xdx � ln� 0 =

m2�

hx2 + A0

� 0 �

�0x e 2� and E

0=

2

1��

E0 comes from plugging � 0 into H� = E� . We will perform the normalization

below.

Now that we are firmly planted on the bottom rung of the ladder, we can

utilize a+ repeatedly to obtain other wavefunctions, �n , and energies, En. That is,

( ) = A0

m� x2

x ( , with E =��n +

2

1�� n ( ) = An a

+ )nem

2��x2

n

Thus, for �1 we obtain

1 1

( ) = A1

� m� � 4 � 2m� � 2 m� x2

�1x

�� xe 2�

where you still have to determine the normalization constant A1.




Algebraic Normalization of the Wavefunctions: We can perform the normalization

algebraically. We know that

a+� n = cn� n+1 a� n = dn� n1

What are the proportionality factors cn and dn ? For any functions f (x) and g(x)� �

�� f�(a± )gdx = �� (a

�f )�g dx

here a�

is the Hermitian conjugate of a±

Proof:

f �(a±g) dx =

1

(2�m� )1 2

� d

dx+ m�x

�g dx

� �

��f ��

recall that

a± = 1

(2�m� )1 2 [�ip + mwx] = 1

(2�m� )1 2

� ��i

�

��+ mwx

� � =

1

(2�m� )1 2

d �

��

d

dx+ mwx

� � �

� �� i dx

Integrate by parts

f�(a±g) dx =

1dx =

� ��

�±�

dx

d+ mwx

�

�f�

��

�

gdx = ��

�(a�f )�g dx

��(2�mw)

So we can write

��

(a±� n ) *(a±� n )dx = ��

(a�a±� n ) *� ndx

� �

We now use

��a±a� ±

2

1�� n = En� n and En =

��n +

2

1��w

�

�� a

+a

±

1

2

1

2

� �

�� n

�

�� nn +=

��

and therefore

a+a� n = n� n

And

�� aa+

1

2

1

2

�

�� n

� �

�� n = ��

� � n +




aa+� n = (n + 1)� n

We can now calculate cn :

�

�� (a+� n ) * (a+� n )dx = cn2

� * dx =�

(aa+� n ) *� ndx = (n + 1)�

� n*� n dx�� n+1� n+1 ��

n+1cn =

The calculation for dn proceeds in a similar manner : � � � �

�� (a� n ) * (a� n )dx = dn2�� n

*1� n1dx = ��(a+a� n )

*� ndx = n�� n

*� n dx

dn = n

Thus we obtain the two normalization constants for the a±

a+� n = (n + 1)1 2� n+1 a� n = n1 2 � n1

HO Wavefunctions: Rearranging the equations to a slightly more useful form yields

n + 1

1==� n+1

( )1 2 a+� n

� n1

1

n1 2a� n

We can now use these equations to generate other wavefunctions. Thus, if we start

with � 0 we obtain:

n = 0

n = 1 � 2

1 1

2 2

n = 2 � 3

2

1

+ 1 3

1

�2

n = 3 � 4 = 1 1

3 + 1 4 � 3 �2So that � n is

� 1 =1

0 + 1( )1 2 a+� 0 = a+� 0

= a+� 1 = a+( )2� 0

= a+� 2 = a+3� 0

a+� 3 = a+4� 0

� n =1n!a+( )

n� 0




Orthogonality of the HO Wavefunctions: Recall the orthogonality condition for

two wave functions is

��

� n*� mdx = �nm

�

Using the a± operators we can show this condition also holds for the HO

wavefunctions. The proof is as follows.

�� m* (a+a )� n dx = n�� m

*� n dx

� (a�� m )�

(a�� m )dx = � (a+a�� m )�� n dx = m�� m

*� n dx

(n m) �� m*� n dx = 0

The trivial case occurs when n = m ; but when n � m then

�� m*� n dx = 0

Potential Energy of the Harmonic Oscillator: We can now use the a± operators to

perform some illustrative calculations. Consider the potential energy associated with

the HO.

1 1V =

2kx2

=2m� 2x2

and therefore 1

2m� 2x2V = =

1

2m� 2

�� n* x2

� ndx

First, we express x and p in terms of a± operators …

1 2 1 2

x =��

2m

�

� ��

(a+ + a ) p = i�� m

2

�

��

(a+ a )

and

x2 � � �)

2 � � �=�� 2m� �

(a++ a

=�� 2m� �

(a+a+ + a+a + aa+ + aa )




Dirac Notation: Before we evaluate this expression let’s introduce some new

notation that will make life simpler for us on future occasions. Instead of writing

the integral between ±� we use brackets to denote this integral. The first

half is called a “bra” and the second a “ket”. That is, “bra”-c-“ket” notation is

bra = and ket =

and for the probability density we would have an expression such as

�

�� m*� ndx = m n

where the present of �m* is understood. Using this notation, integrals for

x2x , p , and assume the form

�

�� m* x� ndx = m x n and

�* p� ndx =�� m m p n

1

2m� 2x2V = =

2

1m� 2

�� n* x2

� ndx =2

1m� 2 n x2 n

� � � 1

2m� 2 �

V = n a+a+ n + n a+a n + n aa+ n + n aa n

�� 2m� ��

yielding

V =h�4

n + n + 1�� =��

2n +

12

�

��

�

It is important to not get totally embroiled in the equations and neglect the

chemistry and physics. Accordingly, we should ask the question as to the physical

significance of this formula ?



5.61 Fall 2007 Lecture Summary 12-15, Supplement page 1

Ehrenfest’s Theroem

In the lecture notes for the harmonic oscillator we derived the

expressions for x (t) and (t) using standard approaches – integrals

involving Hermite polynomials (see pages 17 and 18, Lecture Summary 12-15).

The calculations are algebraically intensive, but showed that

px

x (t) and

(t) oscillate at the vibrational frequency. The results were as follows:px

x (t) = (2� )1 2

cos(� vibt ) =�

� 2μ

�

�

�

�

1 2

cos(� vibt )

and

1 � �� p (t) =

2 i�

� 2 �

1 2

(ei�vibt e i�vibt )��

= �� μ

2

�

� �

1 2

sin(� vibt )

The issue considered here is an approach to calculate x (t) and p (t) in a

more straightforward manner.

Classically, (we use m instead of μ since we are dealing with a free

particle) dx

p = mv = mdt

So, quantum mechanically we might expect

p (t) = md x (t)dt

.

But, is this expression valid ? We can show that in fact it is with the

following argument.

d x (t)dt

For our original expression was …

d x (t)

dt= d

dt

��

�

��

*x dx

��

=

�

�d * �

�d

dtdxx dx +

*xdt

��

Recall the time dependent Schrödinger equation is

�� 1i� = H� or = H�

�t �t i�Inserting these results into the expression above yields




d x (t)

dt= �

1 �

� (H )*x dx +

1 �

�*x(H )dx

i��

i��

=1 �

�* ( xH � Hx)dx =

i �

�* (Hx � xH )dx

i��

��

Evaluating the commutator (assuming that H is the HO Hamiltonian) we find

…

(Hx � xH ) f (x) =�

��

2

�

μ

2

dx

d 2

2+U(x)

�

��x fˆ (x) � x

�

��

2

�

μ

2

dx

d 2

2+U(x)

�

��f (x)

= �2�

2 df �2� i

px�

��= �

i�px2μ dx

= �μ �� μ

And therefore

d x (t)

dt=i��

* (Hx � xH )dx =

1��

* px dx

��

m��

Which is the result that is desired

px (t) = μd x (t)

dt

Thus, we can now obtain (t) without the lengthy calculation contained inpxthe HO lecture notes.

�(t) = μ

d x (t)

dt= μ

dt

d

��

��

2μ

�

�

1 2

cos(�t )�

�

�= �

�� μ�

2

�

�

1 2

sin(�t )px

px (t) = �μ��

2�

��

�

1 2

sin �t( )

which is the result with which we started initially.

The equations above are a specific illustration of a more general

result due to Paul Ehrenfest (an Austrian physicst who later resided in

Leiden, The Netherlands) and known as Ehrenfest’s Theorem. In particular,

for any dynamical variable F




d F (t)

dt

i �

= �* (HF � FH )dx

��

For further information see McQuarrie Problems 4-43 and 4-44, p 187-188.



� � �

5.61 Fall 2007 Separable Systems page 1

��

�� x r = ( x y z ) = i x + j y + k z

� d � � ∂ � � ∂ � � ∂ + j + kp =

i dx p = ( p x p y p z ) = i

i ∂x i ∂y i ∂y � x, p� i� � � � �� = �� x, p x �� = i� �y p ˆ, ˆ y � = i� �z, p z � = i�

ˆ p2 −�2 d2ˆ p2 −�2 ∂2 −�2 ∂2 −�2 ∂2

T = = T = = + + 2m 2m dx 2 2m 2m ∂x 2 2m ∂y 2 2m ∂y 2

ψ (x ) ψ (x y z , , )

O = �ψ * (x )O ψ (x ) dx O = �ψ * (x y z , , )O ψ (x y z , , ) dx dy dz

�� !��" �� "��#ˆ ˆ = ˆ ˆ p y ˆ = yp p p ˆ = ˆ ˆ etc .xy yx ˆ ˆ ˆ ˆ p p z z z x x z

$%��"�� &��'(� "�&� �� %�� "��xf ˆ (y) = f (y) ˆ p f (x) = f ( x) ˆ f * (z) p = ˆ * ( )x ˆ z pz ˆ x px f z etc .

�"� ��)� �� "�*�� +%�� '��)�� 2 � ∂2 ∂2 ∂2 � � − 2

+ 2

+ 2 � +V ( x, y, z)�ψ ( x, y, z) = Eψ ( x, y, z)

� 2m ∂x ∂y ∂z � �

∇2 �"� ��(��

� �22 �

� − ∇ +V ( x, y, z)�ψ ( x, y, z) = Eψ ( x, y, z)� 2m �

H = − �2

∇2 +V ( x, ˆ, ˆ)y z ,�)�(�� -.2m

ˆ ( , y z , ) = ψ ( x, , ) -. �"�*�� +%��Hψ x E y z

/��)� ��0�� &��'(��




�$ V ( x, y, z) = Vx ( x) +Vy ( y ) +Vz ( z)

� �2 ∂2 � � �2 ∂2 � � �2 ∂2 �( ) 2 x ( ) � 2 y ( ) � 2 z ( ) �ˆ , − x y + − +V ˆH x, y z = +V ˆ + − +V ˆ z

�"�� 2m ∂x � � 2m ∂y � � 2m ∂z �= H + H + H

x y z

� �"�*��1� �+# '��)��H + H + H �ψ x, y, z = Eψ x, ,� x y z � ( ) ( y z )

�"�� (%�� ) ψ (x, y, z )= ψ x (x)ψ

y (y )ψ z (z )

/�� &��'(��02"�� !� ��%)� �"�� "� 3. �%�� "� �� 3. ��

H ψ ( x) = E ψ ( x)x x x x

H y ψ y ( y ) = Ey ψ y ( y )

H ψ (z) = E ψ (z)z z z z

$�� )H

x ψ x ( x)ψ y ( y )ψ z (z) =ψ y ( y )ψ z (z) H x ψ x (x) =ψ y ( y )ψ z (z) Ex ψ x ( x)

E ψ ( ) ( ) ψ z= x x x ψ y y z ( )

�)� �� H �� H �y z

H ψ = E ψ

�H x + H y + H z � �ψ x (x)ψ y ( y)ψ z ( z)� = (Ex + Ey + Ez ) �ψ x ( x)ψ y ( y )ψ z (z )�� ˆ ˆ ˆ � � � � �

E = E + E + E x y z

�"%�� "� ,�)�(�� "�� "�� ( ��)� �"� ��%�� "� -.,�)�(�� 4%�� %�� "� ��%�� "� 3. ,�)�(�� "� ��%�� +%�&�(�� "�� 3. ��'(�)�#

�� 2�&��%�� )%(��(� �� "� �� H �� '(� ��




H = H + H + H #x y z

�"� -. ��%�� %��((� ��)�(�5�� "� 3. !�&��%�� )�(�5��

ψ x ψ y ψ z ψ x ψ y ψ z dx dy dz =

� x * ( ) x ( ) � y

* ( ) y ( ) � z * ( ) z ( )

�� x * ( ) y

* ( ) z * ( ) x ( ) y ( ) z ( )

ψ x ψ x dx ψ y ψ y dy ψ z ψ z dz = 1

3 3 3�� "�� ψ !� �� "�� %�� ψ n � �� H

xx

��"�� !��" �� %�� ψ ny �� H

y �� %�� ψ nz� �� H

z #

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ψ111 ( , , ) =ψ x1 (x)ψ y1 (y)ψ z1 (z) �� ψ 311 (x y z ) =ψ x3 (x)ψ y1 (y) z1 ( )x y z , , ψ z #

��ψ111 * ( x, y z , )ψ 311 ( , , ) dx dy dz x y z

� ψ * * y ψ * ( ) x ψ ( ) z�� x1 ( ) x ψ y1 ( ) z1 z ψ x3 ( ) y1 y ψ z1 ( ) dx dy dz

� ψ ( ) x ψ ( ) x dx ψ y ( ) dy ψ ( ) ( ) z� x1*

x3 � y1* ( ) ψ y1 y � z1

* z ψ z1 dz = 0

6 3 3��%�� "� ��%�� "��( �� )�(�5�� "��1�� "��)�(�� !� �%))��5� �"�� '� !��

ψn n n * ( , , )ψm m m ( , , ) =δn ,m δn ,m δn ,m�� x y z

x y z x y z

x y z dx dy dz x x y y z z

��

��1� �� (� �� -. �%'4�� ,��)�� ( �� 5#$%��"�� %)� �"� �� " �� # �"�� )��"�'� ��%�� )�(�� (� �� "� �� )�&�� "� �� !�(( '� �� ) � �� 5 '��%�� "� ��"�� "�� (�� "�� 5 /�� '�(�! ��"�0# �%(��)��(




,��)�� (� �� (�� )�� '�� "� &�'�� (��)�� )�(��%(��# $�� )�(�� ,78 )��"� �� "� 7,��"� � �� "� 798 ��" �� 5 �� "� ,9798 '��#

�� "�� "� ��( ��( �� &�� '�

V x, y, z = k x + k y + k z = V x +V y +V z( ) 1

x 2 1

y 2 1 2

x ( ) y ( ) z ( ) z2 2 2 ��!� '��%�� "� ��( �� %) �� (� � ��( �� 5 ��(� !�� (� !�� "�,�)�(�� !� �� %)

H = H + H + H x y z

p x 2 1 2H = + k x x x 2m 2

ˆ p y 2

1 2 H = + k y y y2m 2

H = p z

2

+ 1

k z ˆ2 z z 2m 2

!"�� " 3. ,�)�(�� '�� (� �%'4�� ,��)�� (!��" �"� �� /: � :� �� :50# �� "� ��%��'�&�� !� �� ))��(� !�� !� �(( �"� ��%�� &�(%��

1

−αxx 2

−α y y 2

�α �4 −αzz 2

n n x yn ( x, , ) = N H x nx (α x

1/2 x) e 2 N H y ny (α y

1/2 y ) e 2 z � z nz

( z 1/2 ) 2ψ y z N H α z e

z π �� 1 � � �� 1 � � �� 1 � �

Ex nz

= nx + ��ω x + ny + ��ω y � + nx + ��ω zn n y � 2 � �� 2 � � � 2 � ��

(mk x )1 2

kxα = ω = etc. x x � m

�� "�� "� -. ��'(�) !� "�&� - +%��%) �%)'�� "�� !�&��%�� (( �"�� )%(��%�(�#

.�� -.� �"�� %)'�� "�� "�� "�� "�� !� ��1�� 3.# 8�� )�(� �� "�� "�� -. �� '(� �� !� ��




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k = k = k ≡ k � ω = ω = ω ≡ ω x y z x y z

� 3 �� En n n = �ω nx + ny + nz + �x y z 2 ��"� ��%�� "�� E000 = 2

3 �ω # �"�� (� �� !�� "�� / nx = 0, ny = 0, nz = 0 0� �� # ,�!�&�� "�� "��!�� "� �� 5

2 �ω nx = 1, ny = 0, nz = 0 �nx = 0, ny = 1, nz = 0 �� nx = 0, ny = 0, nz = 1# � !� �� "�� !� �"�� (( �"�� '� �+%�(�

3�ω E000 = nondegenerate level

2 5�ω

E100 = E010 = E001 = 3-fold degenerate level 2

��#2� !�(( ��((� ��! �"�� !��" � ��%�� (�:�

� ;

�336 �633 �363 �=66 �6=6 �66= >9��(� ��

�366 �636 �663 -9��(� ��

�666 ��9��

�� "� !�&��%�� ψ100 ( x, , ) ≠ψ 010 (x, y z , ) ≠ψ 001 (x, y, )y z z

�"�� (�� # %�� !� )�:� %� � !�&��%�� "�� %) �� !� ��

ψ (x, y, z) = a ψ 010 x, , ) + bψ 001 x, y, z )( y z (




!"�� ' �� # �"�� %�� %� �"�� ψ �� (�� "� ,�)�(��? �� "��

Hψ = H (a ψ + bψ ) = a H ˆψ + b H ˆψ010 001 010 001

= a 5�2 ωψ 010 + b 5�

2 ωψ 001 = 5�

2 ω (a ψ 010 + bψ 001 )

= 5�ωψ2

�"�� ((%�� "� �)�� "�� #

�� "� �� +%�( �)�%�� %)�� "� !�(( �� ))��!��" �� 5# �� "� ��))�� @'��:��A� �#�# kx ≠ ky ≠ kz �"��

�"� 366� 636 �� 663 �� !�(( '��)� ��9��# �� "�� !� !�((%�%�((� �� "�� "� �� "�� '�� @(��A �� "�� "� �� "�&� '�� (��# �"� (�� (��%�� )�� ) �"� ��( &��!< �� "� �� (� �(��"�(� �� "�� "� �� (�&�(� )��"� (��: (�:�

�

�666

�636�366�663

�633�336 �363

�=66 �6=6 �66=

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k = k ≡ k ≠ k � ω = ω ≡ ω ≠ ω x y z x y z

�"�� "� �� '��)�

En n n = �ω (nx + ny +1) + �ω z �nz +

1 ��

x y z 2 �� !� ��




1 E000 = �ω+ �ω z nondegenerate level

2 1

E100 = E010 = 2�ω+ �ω z 2-fold degenerate level 2

3 E001 = �ω+ �ω z nondegenerate level

2




�� !� "�� #��

5

�'�

� V ( x, y, z) = Vx ( x) +Vy ( y ) +Vz (z)

V ( ) = 0 0 ≤ x ≤x a x

Vy ( ) y = 0 0 ≤ y ≤ b

V ( ) z = 0 0 ≤ z ≤ c z

V x ,V y ,V z = ∞ otherwise x ( ) y ( ) z ( )

�� "� '� 8%�� "� '�

− �2 �

∂2

2 +

∂2

2 +

∂2

2

��ψ ( x, y, z) = Eψ ( x, y, z) ψ (x, y, z)= 0

2m ∂x ∂y ∂z �

2� �� (� �� &��'(�� H = H x + H

y + H z !"��

H x � H

y � H z �� " 3. ��(� �� '� ,�)�(��# � �"� ��(%�� "�

-. �+%�� %�� "� 3. ��(%��

� ψ ( x , y , z ) = ψ x ( x )ψ

y ( y )ψ

z ( z )n n n

!"�� ) �"� 3. ��'(�) !� "�&� �"� ��(%��1

� 2 �2 � nx π x � �2 nx 2

ψ nx ( ) x = � sin � nx = 1,2,3,... En =

2x a � a � 8m a 1

� 2 �2 � ny π y � �2 n 2 yψ ny

( ) y = � sin � ny = 1,2,3,... En = 2y b � b � 8m b

1

ψ nz ( ) z =

� 2 ��2

sin � nz π z

�� nz = 1,2,3,... Enz

= �2 n

2 z 2

c � c � 8m c 2"�� "� �� ! � �%�� (( �"�� %��%) �%)'��




�2 � n 2 n 2 n 2 �

En n n = E + En + E = x +

y +

2 z ��x y z nx y nz 8m a 2 b2 c �

.��

.�� %� �� )�(�� "�� "� ��# �#�# �� B ' B � �� %� -9.'�

� E = h2

(n 2 + n 2 + n 2 )n n xx ynz 2 y z 8ma

3h2

E111 = 2

�� 8ma

1

2 ψ111 ( x, y, z) =ψ1 x ψ1 ( ) 1

� 83 ��

�π x

��

�π y

��

�π z

��( ) y ψ ( )z = sin sin sin

a � a � a � a �

�%�;

E = E = E = h2

(22 +12 +12 ) -9��(� ��211 121 112 28ma

�� "� !�&��%��

ψ (x, y, z)≠ ψ (x, y, z)≠ ψ (x, y, z)211 121 112

�

�=== ��9��

�==3 �=3= �3==-9��(� ��

�=33 �3=3 �33=��9��

�333




�� "� ��))�� @'��:��A� �#�# a ≠ b ≠ c �"�� "� �� (��#

h2 � 1 1 4 � h2 � 1 4 1 �E112 = 2 +

2 +

2 � ≠ E121 = 2 +

2 +

2 �8m a b c � 8m a b c �

� �� #��

1

� 8 �2 � nx π x � � ny π y � � nz π z �ψ n n n ( x, , ) = � sin �sin � �y z sin

x y z abc � a � b � c �

h2 � n 2 n 2 n 2 �

+ �� n = 1,2,3,... n = 1,2,3,... n = 1,2,3,... n n n x y z E x y z

= x 2

+ 2

y

2 z

8m a b c �



5.61 Fall 2007 Rigid Rotor page 1

$�� $��

7�� "� �� !� ��(�� R ��) �� "��

)=

)3

ω r + r ≡ R1 2

�3 m r = m r �� )�� /78�01 1 2 2

78� �=

�"�� !� ��(�� %(� '� �� (�� /�� !"��" �� !�1� '�(��:�� "�� )0 �� !� �%�(�� /�� !"��" �� !�1� '� (��:�� )�� )�(��%(�# 7(��((�� " �� "�� '�� "�� %(��)�)��%) L = I ω !"�� ω �� "� ��%(�� &�(�� "� )�)�� i i

�� Ii = mr i 2 �� "� ��(�# �� "�� "� 78�� "� �!� '�� )%��

"�&� �"� ��)� ��%(�� +%��# �"� �(��( ,�)�(�� "� ��(�� L 2 L 2 1 1 1

H = 1 + 2 = m r 2ω2 + m r 2ω2 = 2 2 2 m r + m r ω1 1 2 2 ( 1 1 2 2 )2I 2I 2 2 21 2

�� "��:�� "�� !� �� (�� !�%(� '� ��((� �� !��%(� �"��: �� %�� "� ��# 2� �� "�� !� �� "� ��&� )�)��

m m I = m

1 r 12 + m

2 r

22 = μr

02 μ = 1 2

m + m 1 2

!"�� "� �� +%�(�� !� "�&� �� "�� "�� !� ��(� ��)'�"�&�� (� ��(� !��" � �� μ �� R

��) �"� ��# �"%� !� "�&�1 2 L2

H = Iω = 2 2I

!"�� "� �� +%�(��!� "�&� �� "� ��%(��)�)��%) �� "�� &��(�� L = Iω # �"� ��'(�)�� ! ��)�(��(� ��%�� 39'�� '(�) !��" )�� μ#

�)�(��(�� !� "�&� � ��%� ��'4�� "�� "�(� ��

�

5

θ

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φ




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x / R y 0 �"�� %�� /�0# �"�� !� !�%(� !�� "� �'�&�

ˆ ˆ ˆ ˆ� ��)�� %��(� �� R R ≠ R R # �"�� "�� %�� %(� "�� "��x y y x

�� !��" +%��%) )��"�� C �� (( �"�� "�� +%��%) )��"��(�'�%� �"� '� ��!� �'�&� C '%� "�� &��"�� !��" ��)��# �"%�� !�!�(( �� "�� !"�(� (�� )�)��%) �� ))%�� !��" �� "��/ p p ˆ ˆ = p p ˆˆ 0 �"� ��)� !�(( �� '� ��%� �� %(�� )�)�� '��%�� "�� (��x y y x

ˆ ˆ ˆ�� / L L ˆ ≠ L L 0#x y y x

7(��((�� %(�� )�)��%) �� &�� '�L = r × p # �"�� )�� "� ��ˆ+%��%) �� "�%(� '� L = r × pˆ ˆ # �"�� &�� "�� "��

��)�� !"��" !� �� "� ��%(�� )�)��%) �� %�� "�� "� �"�� 7�� 5




i j kˆ ˆ × p = ( x, ˆ, ˆ) × ( p x , p y , ˆ z )L = r y z p = x y z

p x p y p z

( yp − zp ˆ y ) i ( − xp ˆ ˆ z ) j + ( xp ˆ ˆ − ˆ ˆ x )= ˆ ˆ z ˆ + zp ˆ x y yp k

ˆ ˆ ˆ zp ˆ xp − yp ˆ ˆ ∴ Lx = yp ˆ ˆ z − zp ˆ y Ly = ˆˆ x − xp ˆ ˆ z Ly = ˆ ˆ y x

��!� �� %�� %� �"�� "�� %(�� )�)��%) �� ))%��# $�� )�(�

+ L x L

y = +(yp z − zp y )(zp x − xp z ) = + yp z zp x − zp y zp + zp y xp z

− L y L

x = −(zp x − xp z )(yp z − zp y )= − zp x yp z + zp x zp − xp z zp y

ˆ ˆ� [Lx , Ly ]= yp x [p z , z]+ xp y [z, p z ]

= −i �yp + i�xp = i�L ∴x y z

��!� !� ��%(� �� !��: �%� �"� ��(��"�� L y , L

z �� L z , L

x �� #

,�!�&�� !� �� "�� %� ��!�� #� �� x → y, y → z,z → x # �"�� )%�� '� �"� �� '��%�� %� (�'�(�� "� � � �� 5 � �� ((� ��'�� !� �"�� (�'�( �%� � �� "��x → y, y → z,z → x � �"�� (( �"� )��"�)�� !�%(� !��: �%� � ��(� �"� ��)��!��" �"� (�� "%��(�� %�� "� �� )��# �"� ��(� �"�� !�)%�� '� ��%( �� "�� %� ��(�'�(�� &�� "� ��"�9"�� "��(�� %� ��# 7��(�� )%�� &� �"� "��!"�(� � ��)�(� ��"�� !� � �� /�#�#x ↔ y 0 !�(( ��&�� "� "�� %� �� &� %� �"� !�� !�� /�� 0# �"�� (��&�� &�� )�� '��%�� %�� "� !��: !� �� '� �� -� '%� !� )%�� '� ��%( �� (� �� (�# �� "� �%�%�� !� ��"�� "� ��%(� �� "� 5 � �� "�� "� ��%(�� '� ��(�� )%��# �� "�� !� ��

�L , L � = i L � ˆ� x y � z

⎯⎯⎯ L � ˆ x→y

→ ��L , ˆ �� = i L y z x

y→z z→x

zzy

zzx

pypx

pxpy

ˆˆˆˆ

ˆˆˆˆ

+

−




2 z zL L ��,

6

⎯⎯⎯ L � ˆ x→y

→ ��L z , ˆ

x �� = i L y y→z z→x

��%�� "� ��%(�� )�)��%) �� ))%�� "� �� "�� ))�� %��# �"%�� !� ��&� �� "� �)�� (%�� "�� % � �� & ��# �"� '�� !� �� "�� !�((9�� %(��)�)��%) ��%�� 3 � ��#

��%�� "� ��'(�) �� !� �� "�� "� ,�)�(��

H = L2 1 2 2 2 = (L + L + L ) #2I 2I x y z

�� (��&�(� �� "�! �"�� !"�(� �"� � � �� 5 ��%(�� )�)��%)�� ))%�� !��" ��" ��"�� "�� ))%�� !��" L2

�L ,L2 � = �L , L2 � + �L , L2 � + �� z � � z x � � z y � �

= �L , L2 � + �L , L2 � = �L ,L �L + L �L ,L � + �L ,L �L + L �L ,L �� z x � � z y � � z x � x x � z x � � z y � y y � z y �

=(− )yi L � L + L (x i L − � + Ly ) + (x )xi L L� y (y i L � x ) = 0

�� "�� !� )�:� %�� (�� )%�� "�� L y , L

2 �� = 0 ��

�L ,L2 � = 0 �� !�((# �� L z ��))%�� !��" L2 � �"� �!� �� "�� x �

��))�� %�� !� �� (: �'�%� � ��(� !��" � !�((9�� 59��)�� %(�� )�)��%) /��(( �� m0 �� !�((9�� ( ��%(��)�)��%) /��(( �� l0# �"%�� !"�� !� �� (��:�� "� ��%�� "�� ,�)�(�� !� �� (��:�� '� �!� +%��%)�%)'�� l �� m# �"�� )�:�� '��%�� !� �� !��" � -. ��) �"��!�%(� "�&� "�� "�� +%��%) �%)'�� '%� !�1&� ��! �� "� )�� "� �%�� "�� !"��" �� !� ��)��(# $�� %�" � =. ��)� !�� = +%��%) �%)'��# �� "� �'�&� ��(�� !� !�(( �� "��%(�� )�)��%) �� '� Yl

m � !��" m �� !��" �"� ��&�(%� ��L �� l �� !��" L2 #z

2� �� ! (�� !��" �"� ��(� ��%(� ��'(�) �� (&�� "� ��&�(%��(� �� Yl

m � �� "� �� # �� "�� "� ��&�(%�� "�




�� ,�)�(�� !�(( ��(� �� l '��%�� "� ,�)�(�� ( �� L2 # �� %((� ��(&� �"�� ( �+%�� )��&�� !��: �� "��( ��(�� # �� )�� '��:�� φ �� '� �"� ��(� ��(��&� �� "� z 5� �� !"�(� �"� &�� )�4�� +%��%))��"�� %�� θ �� "�� # 2�!�(( %�� "� (�� '%� '� ��%(�"�� +%�� :�� ) ��"�� %��%�� "�� )� ��&��? ,�� )�%��%( ��(�� "��( ��(��x ≡ r cos φ sin θ

y ≡ r sin φ sin θ

z ≡ r cos θ

2 1 ∂ 2 ∂ 1 ∂2 1 ∂ ∂ ∇ = r + + sin θ r 2 ∂r ∂r r 2 sin 2 θ ∂φ2 r 2 sin θ ∂θ ∂θ

�� )�:� �� !� �� "� ��%(�� )�)��%) �� "��( ��(�� !�((# �"�� )�� %�� %� �� '� ��(� ��%� ��(�� "� �"�� %(� �� !� !�(( )��(� �� "� ��%(��

� ∂ ∂ �L

x = − i� − sin φ − cot θ cos φ � ∂θ ∂φ �� ∂ ∂ �

L y = − i� cos φ − cot θ sin φ �

∂θ ∂φ �∂

L = − i�z ∂φ

L2 = L2 x + L2

Y + L2 z � L2 = − �2

� 1 ∂

�sin θ

∂ ��

+ 1

2

∂2

2 ��

�sin θ ∂θ ∂θ � sin θ ∂φ �� %(�� "� �"�*�� +%�� "� �� '��)��

−�2 � 1 ∂ � ∂ � 1 ∂2 � m m+ Y ( , ) = E Y (θ φ , ) sin θ � 2 2 � l θ φ l l 2I �sin θ ∂θ ∂θ � sin θ ∂φ �

φ

θ

��




!"�� !� "�&� �� "�� '��%�� "� �� (� �� '� ��"� �%�� "�� %� R� �"� �� !�&��%�� )%�� '� ��%�� θ �� φ ��(�#

�� %�� %�� "�� )� &�� (��'�� :� �"�� ((�! %�� "� ��&�(%�� (� +%��:(�# �"�� :� �� (��%� �� "� �� (�!�� )��%(�� !� %�� (&� �"� ,��)�� ((�� !�!�(( %�� "�) �� &� �"� ��&�(%�� "� �� (��%��# �"� ��%(� �� "��"� �� "� �� '��

�2

El = l (l +1) l = 0, 1, 2,... 2I

8�� %�� '�%� ��%(�� )�)��%) �� "�� :�� %(�� )�)��%)� !� %�� (��# � !"�(� ( �� "� (�� "�� (�� )�&�� %�� "� �%�(�%�� G �� "� �"�� (�� "� �� )�� )�(��%(�

�2

E = J ( J +1) J = 0, 1, 2,... J 2I �"%�� "� �� '��!�� "� �� (�&�(� �� !��" �� J� %�(�:�� "� "��)�� ((�� !"�� "�� !�� +%�((� ��

�2 �2

EJ +1 − EJ = ��( J +1)( J + 2) − J ( J +1)�� = ( J +1)2I I

$%��"�� " J !� "�&� )%(��(� ��'(� &�(%�� m /�(�� :��!� �� K ��)� �� 0:

m = 0, ±1, ± 2,..., ±J �"%�� " �� (�&�( �� /=JH309��(� ��# �"��((�� m ��(�� "��)�� "� ��%(�� )�)��%) �(�� "� 59��# $�� J� �"�� &�(%�� m ��(�� "� �� "� ��%(�� )�)��%)&�� %(� '� �� C �� (��&� m �"� ��%(�� )�)��%) ��)��(� �(�� H5< �� m �� 5�� "��%(�� )�)��%) �� "��( �� 5#�"��((�� !� :��! �"�� "� �� "� �� 1� �� "� ��# �"�� (�� "� �� "�� "� �� (� �� J� !"��" )��%�� "� (��"�� "� &�� #

5

�

= +m J

= −m J

0=m



5.61 Angular Momentum Page 1

'�� (��

�� L2 �� L ��))%�� "�� "�� ))�� %��# �"�� %�� z

� ��)�(� �)�� "� �� %(�� )�)��%) ��'(�)� C �"��)�� "� �((�!�� &�(%�� %(�� )�)��%) �� )� (�:� �"� �� "� �� &��(�'(� �� "� ��)# �"� �� "�� !� !�%(� (�:� �� :��!�� "� ��&�(%�� !��" �"�� %��# 2� !�(( �� "��&�(%�� L2 �� L

z '� α �� β� ��&�(� �� "��2 β ( , ) = α Y β , ) ˆ β θ φ ) = βY β ( , )L Y θ φ (θ φ L Y ( , θ φ α α z α α

$�� '��&�� !"�� ((�!� !� !�(( �)�� "� �� "� �� θ

�� φ �� "�� "� �'�&� �+%�� '��)�ˆ2 β β ˆ β βL Y = α Y L Y = βYα α z α α

�� &�� "� �� (�!�� /�� "� ��)�(�� "� ,��)�� ((��?0

L ≡ L ± iL ± x y

2"��" �� "� ��))%�� (��L , L � = 2�L �L , L � = ± �L �L , L2 � = 0� + − � z � z ± � ± � ± �

�"�� (�� (��&�(� �� &� %�� "� ��))%�� (�� !�1&��(�� &��

� ˆ ˆ ˆ � ˆ ˆ ˆ � ˆ ˆ ˆ � ˆ2 ˆ � =�Lx , Ly �� = i L� z �Ly , Lz �� = i L� x �Lz , Lx �� = i L� y �L , Lz � 0

$�� )�(��L , L � = �L , L � ± i �L , L �� z ± � � z x � � z y �

= i Ly ± i (− � x ) = ± � (Lx y )� i L ± iL

= ± �L ±

�"� �� (�!�� "�&� � ��%(�� "� ��&�(%� �� L z

ˆ ˆ β ˆ ˆ ˆ ˆ β ˆ ˆ β ˆ βL (L Y ) = (�L , L � + L L )Y = (±�L + L β )Y = (β ± �)(L Y )z ± α � z ± � ± z α ± ± α ± α

�"%�� L+ / L− 0 �� /(�!��0 �"� ��&�(%� �� L z'� � � "�� "� ��)��# ��

�"� �� (�!�� ))%�� !��" L2 �"�� "�� "� &�(%�� α �� !� �� !��

ˆ β ∝ Y β ±�L Y± α α

�� "� ��&�(%�� L z �� &��(� ��?

2"�� "� (�)�� "�� (�� &�(%��I ��(( �"�� "� "��)��((�� !� ��%�� "�� "�� !�� )��)%) ��&�(%� �� "� �� %(�




'� �� '� �%��&� ��(�� "� �� "� (�!�� #�"�� (�� )��)%) ��&�(%� �� "�� # �� "�� "��

2 2 2 2ˆ ˆ ˆ ˆ 0 x y x yL L L L+ = + ≥

�"�� %(� ��)�(� ��(�� "� �� "�� %J��:� �� '��&�'(� �� +%�� %J �� '��: � ��&� �%)'��# �� &� &�(%� �� "��&�� &�(%� �� L2

x �� L2 y !�%(� �)�(� �� )�� &�(%� �� L

x �� L y � !"��" ��

�)��'(� �� "�� ,��)��# �� !"�� !�%(� �� )��%(�� )�)��%) )��I 2� ��! ��(� �"� �'�&� �+%�� "� ��!�&��%�� Yα

β

0 ≤ �Yαβ* ( L2

x + L2 y )Yα

β = �Yαβ* (L2 − L2

z )Yαβ

= �Yαβ* (α − β 2 )Yα

β

= α − β 2

,�� β 2 ≤ α �� "�� − α ≤ β ≤ α # 2"��" )�� "�� "�� '��")� �)%) �� )��)%) &�(%�� "�� β �� :� �� &�� α# �� !� �� "��&�(%�� '� β)� �� β)�� &�(�� "�� (�� "��

ˆ βmax ˆ βmin L Y = 0 L Y = 0 #+ α − α

2� �� "�� %�� "�� :��!(�� )� �(��'�� :� ��: �� )�� "��(��"�� '��!�� α �� β)� /�� β)��0# $�� "��

ˆ ˆ β ˆ ˆ β� L L Y max = 0 L L Y min = 0− + α + − α

2� �� "�� (��(� �� )� �� L x �� L

x L − ( − L L ) Y max = 0 L + L + ( − L L ) Y min = 0� L + i L L i L L( ˆ2

xˆ2

yˆ

yˆ

xˆ

xˆ

y ) αβ ( ˆ2

xˆ2

yˆ

yˆ

xˆ

xˆ

y ) αβ

,�!�&�� "�� "� )�� &�� ) �� "� �� '��%�� !� ��1�:��! !"�� L

x �� L y ��&�� !"�� Yα

β # ,�!�&�� !� �� 9!�� "� ��)�� )� �� L2 �� L

z ˆ2 ˆ2 ˆ ˆ ˆ ˆ(L + L ± ( − L L ))i L Lx y y x x y

L2 − L2 −i L� ˆ z z

� �"�� !� "�&�� (L2 − L2

z − �L z )Yα

β max = 0 (L2 − L2 z + �L

z )Yαβmin = 0

� (α − β 2 − �β ) = 0 (α − β 2 + �β ) = 0 max max min min

�α = β max (β max + �) = βmin (βmin − �) � β = − β ≡ �lmax min




!"�� "� (�� (�� !� "�&� ��)�(� �� ! &��'(�� l� �"�� )��(��/�� "�� "�� "� %�� %(�� )�)��%)0# � ��)'�� "�� )��)%)�� )� �)%) &�(%�� !� "�&� �"�� −�l ≤ β ≤ �l # $%��"�� !� �� ) �"�(�!�� "� "��"�� &�(%� �� )�� '� �%��&� ��(�� "� �� (�� "�� "� �� '��!�� "� "��"�� (�!�� &�(%�� K�j − (−�j) = 2�l L )%�� '� �� )%(��(� �� # �"%�� l ��(� )%��"�� '� �� "�(�9��#

�%�� (( �"�� "�� !� ��(%�� /.�� m ≡ β / � 0

2 31 2 2

ˆ ( 1) 0, ,1, ,2... m ml lY l l Y l= + =2L �

��ˆ , 1... 1, m mz l lL Y m Y m l l l l= = − − + −�

!"�� !� "�&� ��(�� α !��" l �� β !��" m �� "�� Yαβ '��)�� Yl

m # �(�� "�� +%�� !� "�&� �� "�� 0 ≤ L2 = �2l (l +1) �)�(�� l ≥ 0 # �"�� "�

�%��)��( ��&�(%� �+%�� (( ��)� �� %(�� )�)��%)#

�� "�� "�� "�� ) !"�� !� ��! �� "� �� #�"�� !� "��

�2

E = J ( J +1) J = 0, 1, 2,... J 2I!"�� )�� "� +%��%) �%)'�� J �� "� �� +%�&�(�� "�+%��%) �%)'�� l �� '�&�# ,�� "� �� "� �� J CE ∝ J (J +1) 9 �� "� ��)� �� !� ��%�� %� ��&�� L2 # �"� �� 1 / 2I

��)�(� �� ) �"� �� "�� "� �� ,�)�(�� L2 / 2I ��"�� "��L2 # �"� ��( �� "�� J �� # �� %J)��"� �"��: �"�� )�� !� )�� )��:� �� %��&�� '�&� �� "�� l �"�%(� ��(� '� �� "�(� ��#,�!�&�� "�� # �"� �� '��%�� %� ��&�� '�&� ��&�(�� )�� # �"%�� !"�(� �� &�(%�� l )�� %(�� )�)��%) /�#�# �"�� 1� ��%� �� "� ��0 !� !�(( �� (�� "�� "�� "�� %(��)�)��%)# �� '(�� (�� "�&� �� %(�� )�)��%) !��"l = 1 # �"%�� !"�(� ��&��%�( ��)� )�� "�&� ��( �� "�2




�((�!�� &�(%�� l� ��%(�� )�)��%) �� (!�� '�� "� �'�&� ��&�(%��(��#



5.61 Spherical Harmonics page 1

�� ! � �� " �� ! � �� #

� ∂ ∂ �L

x = − i� �− sin φ − cot θ cos φ �� ∂θ ∂φ �� ∂ ∂ �

L y = − i� �cos φ − cot θ sin φ �

� ∂θ ∂φ �∂

L = − i�z ∂φ

2 ˆ2 ˆ2 ˆ2 2 2 � 1 ∂ � ∂ � 1 ∂2 L = Lx + LY + Lz L = − � � �sin θ � +

2 2 � sin θ ∂θ � ∂θ � sin θ ∂φ �

$� �� ! �� %��&�� '��

m L2 m mHY ˆ = Y = E Y l l l l2I

−�2 � 1 ∂ � ∂ � 1 ∂2 m m + Y , = E Y l , � �sin θ � 2 2 � l (θ φ ) l (θ φ )2I sin θ ∂θ � ∂θ � sin θ ∂φ �

�� l �� '�� L2 �� m �� '�� L z "

��(�� L2 �� L z

� �� l = 0, 12 ,1, 2

3 ,2,... m = − l, −l +1,..., l

)� �� l �� ! �� '�� " $�� '�� ! � ��

2 m 2 m 2 mL Y l = L Y l = � l (l +1)Yl

L Y m = �mY m z l l

�� ! �� ! Ylm ! ��

�� " $� �� *+ �� '�� " ��,�� - �� .�� /" �� #




Ylm (θ φ ) lm Pl

m im φ , = A (cos θ )e

�� Alm �� 0�� Plm ( x) ��

1��" �� 1�� #P0 (cos θ ) = 1 P1

0 (cos θ ) = cos θ0

P1 (cos θ ) = sin θ P0 (cos θ ) = 1 (3cos 2 θ −1)1 2 2

P1 (cos θ ) = 3cos θ sin θ P2 (cos θ ) = 3sin 2 θ2 2

�� % �� 2�� 0� �� #

• �� 0� �� θ �� φ"

Ylm (θ φ , ) ∝ f (θ ) g (φ )

�� 2��! �� 2�� 2�� θ �� 2�� φ#

H =−�2

�� 1 ∂

��sin θ

∂ ��

+ 1

2

∂2

2 �

≠ H θ + H

φ2I sin θ ∂θ � ∂θ � sin θ ∂φ �2� ��! �� (�� θ �� φ �� # � ��3� � �� (� �� θ �� φ �� ! 4�� x �� y �� -��" �� θ 5φ �� *+�� "

• $� �� L z #

m ∂ m ∂ m im φ m im φ mL Y = − i� Y = − i� A P (cos θ )e = �mA P (cos θ ) e = �mY z l l lm l lm l l∂φ ∂φ

�� L2 ! �� Ylm

2 m.�� / �� L Y l �� 4�� 2l (l +1)Yl

m "• )� �� 5�� l �� " ��

�� φ �� -5� �� 6 �� *π" )�� l

m ( , �� φ → φ + πY θ φ ) 2 7 � ��! �� φ �� *π# �3��




2φ φ π = im im im e e e

4�� " ��! ��

�� ! � �� # (θ φ ) = Y , + π )Yl

m , lm (θ φ 2

mlm lP (cos θ )eim (φ+2π ) m

lm lPA (cos θ ) eim φ = A

im φ im (φ+2π ) e = e

1 = eim 2π

m = an integer ��! m �� ! �� l � ��" 2� ��! �� m �� l! l �� " ��

�� l ��

�� " • ��

��# 8/ �� φ �� l */ ��l�� '�� l 9/ �� m ≠ 0 �� 2�� - ��Yl

m* = Yl −m " ��! �� ±m ��

Rlm = 1 (Y m + Yl

−m ) Ilm = 1 (Yl

m − Yl −m )2 l i 2

�� ! �� ! �� "

• �� ! �� 2�� " ��! l:6 �� ;�3! l:8 �� ; 3! l:* �� ;�3<"

• $� �� ! �� ! �� 2�� 2l+1-�� #

l m Y l m 's 2l + 1

6 6 Y 0 0 8

8 58! 6! 8 Y 1 −1 , Y

1 0 , Y

1 1 9

* 5*! 58! 6! 8! * Y 2 −2 , Y

2 −1 , Y

2 0 , Y

2 1 , Y

2 2 =

9 59! 5*! 58! 6! 8! *! 9 Y 3 −3 , Y

3 −2 , Y

3 −1 , Y

3 0 , Y

3 1 , Y

3 2 , Y

3 3 >



�


�� ! � �� - �� (�� " $� �� ! � �- �� "

��! �� 0�! �� 2�� #

( ) ( ), cos φθ φ θ= mm im

l lm lY A P e

( )2 2 2ˆ 1 0,1,2,3... = = + =�m m m l l lL Y L Y l l Y l

ˆ , 1,..., = = − − +�m m z l lL Y mY m l l l

�� ! "�# $�� #�%�� #&��%�

)� �� '��2 � 1 ∂ � ∂ � 1 ∂2

− � �sin θ � + 2 2 �Y (θ φ , ) = EY (θ φ , )

2I sin θ ∂θ � ∂θ � sin θ ∂φ �ˆ�� HY (θ φ , ) = EY (θ φ , ) �� " �� '��!

� ∂ � ∂ � 2IE 2 ∂2

�sin θ �sin θ � + 2

sin θ �Y ( , ) = − 2

Y (θ φ )θ φ , ∂θ � ∂θ � � � ∂φ

�� θ �� φ

)�3�� ! 4�� 9+ �� "

∴ �� Y (θ φ ) = Θ (θ )Φ (φ ), ��

2IE +�� β ≡

2 (note β ∝ E )

� ∂ � ∂ � 2 ∂2

�sin θ ∂θ

�sin θ ∂θ

� + β �Θ( ) ( ) φ = − ∂φ 2 ( ) Φ ( ) φsin θ θ Φ Θ θ

� � �

+�� Θ(θ )Φ (φ ) ��




sin θ ∂ � ∂ � 2 1 ∂2

Θ θ + β sin θ = − Φ φ�sin θ � ( ) ( ) Θ θ ∂θ ∂θ Φ φ ∂φ( ) � � ( ) 2

�� θ �� φ

%�� θ �� φ �� ! �� '�� '�� ≡ �*"

1 ∂22 φ = − Φ ( ) m

Φ φ( ) ∂φ 2 $

sin θ ∂ � ∂ � 2 2�� sin θ �Θ( ) + sin θ mθ β = Θ( ) θ � ∂ �θ ∂ θ

$$

?�� Φ (φ ) �� $

2∂ Φ (φ )= − m 2Φ φ( )

∂φ 2

%�� Φ (φ ) = A e im φ and A e −im φ m −m

@�� '��0��

Φ (φ + 2π ) = Φ (φ )

im (φ+2π ) im φ −im (φ+2π ) −im φ A e = A e and A e = A e m m −m −m

im (2π ) −im (2π )∴ e = 1 and e = 1

�� m = 0, ±1, ± 2, ± 3,....

� �� A��B '��

∴ Φ (φ ) = A e im φ m = 0, ±1, ± 2, ± 3,.... m



2π e m

" �� $$


2π

��0��# Φ φ Φ φ dφ = 1� ∗ ( ) ( ) 0

Φ ( ) = 1 im φ = 0, ±1, ± 2, φ ± 3,...

�� 3� ��( �� Θ(θ )sin θ ∂ ∂

Θ( ) θ ��

∂θ ��Θ( ) + sin θ m �sin θ �

θ β 2 = 2

θ ∂

dx C�� # x = cos θ Θ( ) θ = P ( ) x = dθ

−sin θ

%�� 0 ≤ θ ≤ π −1 ≤ x ≤ + 1

�� sin 2 θ = 1− cos 2 θ = 1− x 2

�� '�� 3� �'�� Θ#

sin θ d

��sin θ

dΘ �+ (β sin 2 θ − m 2 )Θ( ) θ = 0

dθ dθ �

�� 5 ��#

d2Θ dΘ sin 2 θ dθ 2 + sin θ cos θ

dθ + (β sin 2 θ − m 2 )Θ θ( ) = 0

�� x = cosθ �� Θ(θ)= P(x)"

dΘ=

dP dx = − sin θ

dP = − (1− x 2 )

1 2

1 2

dP dθ dx d θ dx dx

d2Θ=

d ��dΘ

� = ��

dx �

d ��−(1− x 2 ) dP

�

dθ 2 dθ dθ � dθ � dx dx �

� 2� x dP d P �

= − sin θ �2 1 2 dx

− (1− x 2 )1 2

dx 2 �� (1− x ) � �

2 = − x dP

+ (1− x 2 ) d P 2dx dx

%�� 3� �'��




1− x 2 d P − 2x 1− x 2 dP

+ β 1− x 2 − m 2 P ( ) = 0( )2

dx

2

2 ( ) dx ( ( ) ) x

+�� (1− x 2 ) �� '�� #

2 2

(1− x 2 ) d P − 2x dP

+ �β −

m P x = 0� � ( )

dx 2 dx � 1− x 2 ��

�� '�� (�� ! �� " �� ! Pl

m ( ) x " �� DmD�� '�� m2#

m m P

l (x)= Pl ( )cosθ

0 0P (cosθ)= 1 P (cosθ)= 1 (3cos2 θ − 1)0 2 2

0 1P (cosθ)= cosθ P (cosθ)= 3cosθ sinθ 1 2

P1 (cosθ)= sinθ P2 (cosθ)= 3sin2 θ 1 2

��"

1

2 m

�� 2l + 1� (l − m )!%� Θ θ A

lm l (cosθ) Alm

= �� 2 � l + m )!��

( )= P

(

�� A �� 0�� lm

2 2

π � Alm � P

l

m (cosθ) sinθdθ = 1 0 �

%� �� #

ψ lm (r

0,θ ,φ)= Y

lm (θ ,φ)= Θ

l

m (θ)Φ m (φ)

1

2

Ylm (θ ,φ)= �

�� 2l + 1

�� (l − m )!

P cosθ �� 4π � (l + m )!��

�l

m ( )eimφ

�� "




��#�%�� #&��%� �'&&�#(

Y m (θ ,φ)= Θ m (θ)Φ (φ)l l m

1

2�� 2l + 1� (l − m )! mmYl (θ ,φ)=

�� 4π �� (l + m )!

��

Pl (cosθ)eimφ

l = 0, 1, 2,... m = 0, ± 1, ± 2, ± 3,... ± l

mYl 3� �� Hψ = Eψ �� "

1

1 � 5 �2 Y 0 = Y2

0 = � � (3cos 2 θ −1)0 1 2 (4π ) �16 π �

1 1

2 2 0 � 3 � ±1 � 15 � ±iφY1 = � � cos θ Y2 = � � sin θ cos θ e � 4π � � 8π �

1 1

Y 1 = � 3 �2

iφ ±2 � 15 �22 ±2iφ

1 � � sin θ e Y2 = � � sin θe � 8π � � 32 π �

1

2−1 � 3 � −iφY1 = � � sin θ e

� 8π �

Ylm 3� �� # ��Y

lm′

′∗ (θ ,φ)Ylm (θ ,φ)sinθdθdφ = δ

ll ′δ

mm′

�1 if l = l′ �1 if m = m′ normalization E�&��(�� δ

ll ′= � δ

mm′= �

�0 if l ≠ l ′ �0 if m ≠ m′ orthogonality

��# % �� l → J �� '�� .�"�"# l (l +1) J (J +1) J = 0, 1, 2,... "/



5.61 Fall 2007 Lecture # page 1

SPHERICAL HARMONICS

(� ,�) = �l

� ( )Yl

m m ( )�m

�

1

!�2

m � m im�Y

l (� ,�) = �

��

� 2

4

l

�

+ 1 ((

l

l

�

+

m

m

))!��

Pl

(cos�)e

��

l = 0, 1, 2,... m = 0, ± 1, ± 2, ± 3,... ± l

Yl

m ’s are the eigenfunctions to H� = E� for the rigid rotor problem.

1

Y0

0 =

1

(4� )1 2

Y2

0 =

�

�

16

5

� ��

� 2

(3cos2 � �1)

1 1

Y1

0 =

�

�

4

3

� ��

� 2

cos� Y2

±1 =

�

�

8

15

� ��

� 2

sin� cos�e± i�

1 1

� 3 � 2 � 15 � 2

Y1

�1 =

� 8� �� 32� ��sin

2 �e

±2i�sin�e

i�Y

2

±2 =

1

2

Y1

1 =

� 3 � � i�

� 8� ��sin�e

m m�� mY

l ’s are orthonormal: ��Y

l � (� ,�)Yl

(� ,�)sin�d�d� = �ll ��

mm�

�1 if l = l � �1 if m = m� normalization Krönecker delta �

ll �= � �

mm�= �

�0 if l � l � �0 if m � m� orthogonality

ˆ m )Energies: (eigenvalues of HYl

m = E

lmY

l

Switch l � J conventional for molecular rotational quantum #

2IE Recall � = = l l ( + 1) � J J ( + 1) J = 0, 1, 2,...

�2

20



5.61 Fall 2007 page 2

�E

�2

EJ

= J J + 1( )2I

6�2

0 ±1 ±2 ±3Y

3, Y

3, Y

3, Y

3 (7x degenerate)J = 3 E

3 =

I

3�2

0 ±1 ±2J = 2 E2

= Y2

, Y2

, Y2

(5x degenerate)I

�2

0 0J = 1 E1

= Y1

, Y1

(2x degenerate)I

0J = 0 E0

= 0 Y0

(nondegenerate)

Degeneracy of each state gJ

= (2J + 1)from m = 0, ± 1, ± 2,..., ±J

Spacing between states � as J �

�2

�2

=EJ +1

� EJ

2IJ + 1 ) � J J + 1

I (J + 1)�

�( )(J + 2 ( )�� =

Transitions between rotational states can be observed through

spectroscopy, i.e. through absorption or emission of a photon

�+

�-

h� Absorption

EJ

�+

�-EJ+1

h�

�+

�-EJ

Emission

�+

�-EJ-1

or

Lecture #20



5.61 Fall 2007 page 3

J (ξ ⋅ μ)

Jdx

Molecules need a permanent dipole for rotational transitions.

Oscillating electric field grabs charges and torques the molecule.

2

dμStrength of transition I

JJ dx

electric field dipole moment

of light of rotor

Leads to selection rule for rotational transitions: J = ±1

Recall angular momentum is quantized (in units of ). Photon carries one quantum of angular momentum.

Conservation of angular momentum J = ±1 Angular momentum of molecule changes by 1 quantum upon absorption or emission of a photon.

Ephoton

= hphoton

= Erot

= EJ +1

EJ =

I

2

(J +1)photon

=4

h

2 I (J +1)

J J +1 J J +1 J J +1

Define

hB rotational constant (Hz)

82 I

and

hB rotational constant (cm-1)

82cI

(Hz) = 2B J +1)J J +1

(J J +1

(cm-1

) = 2B J +1)(

Lecture #20



5.61 Fall 2007 page 4

E

E3

= 12Bhc

E2

= 6Bhc

E1

= 2Bhc J = 1

J = 2

J = 3

�E0�1

= 2Bhc

�E2�3

= 6Bhc

�E1�2

= 4Bhc

This gives rise to a rigid rotor absorption spectrum with evenly spaced lines. J = 0

��0�1

�1�2

�2�3

�3�4

�4�5

�5�6

2B

Spacing between transitions is 2B (Hz) or 2B (cm-1

)

�J +1�J +2

��J �J +1

= 2B ��(J + 1) + 1�

�� 2B J + 1( ) = 2B

Use this to get microscopic structure of diatomic molecules directly from

the absorption spectrum!

Get B directly from the separation between lines in the spectrum.

Use its value to determine the bond length r0!

Lecture #20



5.61 Fall 2007 page 5

h 2

m1m

22B =

4�2cI

I = μr0

μ =m

1 + m

2

1 1

� � � � h 2 h 2

(B in cm-1

) or r0

= (B in Hz) � r0

= ��

��

��

�� 8�

28�

2 BμcBμ

Lecture #20




HYDROGEN ATOM

Schrodinger equation in 3D spherical polar coordinates:

�2 1 � �

2 � 1 � � � 1 �

2 ��

2μ� r

2 �r ��r

�r ��+

r2 sin� ��

sin��

+r

2 sin

2 � ��

2

�� (r,� ,�) + U (r,� ,�)� (r,� ,�) = E� (r,� ,�)

�Ze2

with Coulomb potential U (r) =4��

0r

Rewrite as

2��

2 � � � �

2 U r

�r, ,�) + L

2 r, ,�) = 0

��+ 2μr �

( ) � E �� ( (� �r ��

r�r �

function of r only function of �,� only

r is separable � is separable

Angular momentum: solutions are spherical harmonic wavefunctions

m� r,� ,�) = R r � ,�( ( )Y ( )

l

m mwith L

2Y (� ,�) = �

2l l ( + 1)Y (� ,�) l = 0,1,2,...

l l

Radial equation for the H atom:

��

�

�

�

�

��

�dR r �

2l l + 1�

2 d ( )

+( )

( ) � E ( ) = 02

r + U r R r

2μr2

dr dr 2μr2

Solutions R r( ) are the H atom radial wavefunctions

Simplest case: l = 0 yields solution

( ) = 2

��

�

a

Z

0 ��

�3 2

e�Zr a

0R r exponential decay away from nucleus




with

E = �Z2e

28��

0a

0 lowest energy eigenvalue

�μe2 Bohr radius a

0 � �

0h

2

General case: solutions are products of (exponential) x (polynomial)

Energy eigenvalues:

E =�Z

2e

2

=�Z

2μe

4

n = 1,2,3,... 8��

0a

0n

28�

0

2h

2n

2

Radial eigenfunctions:

� �1 2

l +3 2

Rnl

( ) = �� (n � l �1)!

3 ��

�

�

na

2Z

0 �

�l �Z r na

0 L2l +1

�

� 2

na

Zr

0 �

� r r e

n+ l

�2n � (n + l)!��

2l +1where L (2Zr na

0 ) are the associated Laguerre functions, the first few of which are:

n+ l

n = 1 l = 0 L1

1 = �1

�1

n = 2 l = 0 L2

= �2! 2 � Zr

��

� a0

l = 1 L3

3 = �3!

�n = 3 l = 0 = �3! 3 � 2Zr + 2Z

2 2 �2

L1

3 � a

0

r

9a0 �

�3

l = 1 L4

= �4! 4 � 2Zr �

� 3a0 �

l = 2 L5

5 = �5!

Normalization:

Spherical harmonics �2�

d��

d sin Yl

m* ( ,�)Yl

m ( ,�) = 1 0 0

*Radial wavefunctions �0

�

dr r2 R

nl ( ) R

nl ( ) = 1r r




TOTAL HYDROGEN ATOM WAVEFUNCTIONS

m� r,� ,�) = R r � ,�

nlm (

nl ( )Y

l ( )

principle quantum number n = 1,2,3,...

angular momentum quantum number l = 0,1,2,...,n �1

magnetic quantum number m = 0,±1,±2,...,±l

ENERGY depends on n: E = �Z 2e

2 8��

0a

0n

2

ORBITAL ANGULAR MOMENTUM depends on l: L = � l l ( + 1)

ANGULAR MOMENTUM Z-COMPONENT depends on m: Lz

= m�

Total H atom wavefunctions are normalized and orthogonal:

2� � �*

�0

d� �0

sin�d� �0

r2dr �

nlm (r,� ,�)�

n ' l ' m ' (r,� ,�) = �

nn '�

ll '�

mm '

msince components R r � ,�) are normalized and orthogonal. ( )Y (nl l

Lowest few total H atom wavefunctions, for n = 1 and 2 (with � = Zr a0 ):




1

�

� Z �3/ 2

�� n = 1 l = 0 m = 0 �

100 =

�

a0 � e =�

1s

� Z �3/ 2

�� / 2 n = 2 l = 0 m = 0 �

200 =

�

a � (2 �� )e =�

2s

1

1

32� 0

� Z �3/ 2

l = 1 m = 0 �210

=32� �

a

0 � �e

�� / 2 cos� =�

2 pz

�3/ 2

Z ��e

�� / 2 sin� e

± i�l = 1 m = ±1 �

21±1 =

1

64� �

a0 �

or the alternate linear combinations

� Z �3/ 2

=�

a � �e

�� / 2 sin� cos� =

1

2 (� +� )

21+1 21�1�

2 px

1

32� 0

� Z �3/ 2

�2 p

y

1

32� �

a0 � �e

�� / 2 sin� sin� ==

1

2i

� ��(21+1 21�1

)

The value of l is denoted by a letter: l = 0,1,2,3...

s,p,d,f orbitals

The value of m is denoted by a letter for l = 1: m = 0, ± 1 linear combinations

pz , p

x ,p

y orbitals

HYDROGEN ATOM ENERGIES

Potential energy of two electrons separated by the Bohr radius:

U = e2

4��0a

0_ one “atomic unit” (a.u.) of energy.

H atom energies: E = �Z 2e

2 8��

0a

0n

2 = �Z

2 2 n

2 a.u.

n En (a.u.)

1 -1/2

2 -1/8

3 -1/18

4 -1/32

5 -1/50

0




H atom energies &transitions E E/hc n (a.u.) (cm-1)

0 0

4 -1/32 -6,855

3 -1/18-12,187

-1/82

-1/2 -109,6801

H atom emission spectraLyman series Balmer Paschen

�(A)

DEGENERACIES OF H ATOM ENERGY LEVELS

As n increases, the degeneracy of the level increases.

What is the degeneracy gn of each level as a function of n?

Does this help understand the periodic table?

�(1012Hz)




SHAPES AND SYMMETRIES OF THE ORBITALS

S ORBITALS �r / 2a

03�

1s = (�a

0 )�1 2

e�r / a

0 �2s

= (32�a0

3 )(2 � r a0 )e

l = 0 spherically symmetric

n - l - 1 = 0 radial nodes n - l - 1 = 1

l = 0 angular nodes l = 0

n - 1 = 0 total nodes n - 1 = 1

Electron probability density given by � (r,� ,�)2

Probability that a 1s electron lies between r and r + dr of the nucleus:

2� � �1

� d� �0

d� sin�*

r,� ,�)�1s

(r,� ,�)r2dr = 4� �a

0

3 ) e�2r / a

0 r2dr�

1s ( (

0

P ORBITALS: wavefunctions

Not spherically symmetric: depend on � ,�

m = 0 case: �210

=�2 p

z

= (32�a0

3 )�1/ 2

(r a0 )e

�r / 2a0 cos

�2 p

z

independent of � symmetric about z axis

radial nodes n - l - 1 = 0 (note difference from 2s: Rnl

( ) depends on l as well as n)r

angular nodes l = 1

total nodes n - 1 = 1

xy nodal plane – zero amplitude at nucleus

�2 p

x

= (32�a0

3 )�1/ 2

(r a )e�r / 2a

0 sin� cos�0

m = ±1 case: Linear combinations give �1/ 2

�2 p

y

= (32�a0

3 ) (r a0 )e

�r / 2a0 sin� sin�

Equivalent probability distributions




0

0.1

0 5 10 15 20 25 300

0.1

0 5 10 15 20 25 300

0.1

0.2

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0 5 10 15 20 25 300

0.1

0.2

0 5 10 15 20 25 300

0.1

0.2

1s

2s

2p

r2R

nl

2 r( ) a

0 ()

3s

3p

3d

H atom radial probability densities

0 5 10 15 20 25 30

s = r/a0




MAGNETIC FIELD EFFECTS

Electron orbital angular momentum (circulating charge) magnetic moment

Magnetic field B applied along z axis interacts with μ:

Potential energy

Include in potential part of Hamiltonian operator:

H = H0 +

eB z

L

2mz

e

H atom wavefunctions are eigenfunctions of both H0

and L z

operators

eigenfunctions of new H operator.

Energy eigenvalues are the sums

Z 2e

2 eB

E = + z m

80a

0n

22m

e

Energy depends on magnetic quantum number m when a magnetic field is applied.

2p orbitals: m = -1,0,+1 states have different energies

Splitting proportional to applied field Bz.

U = −μ•Β = −μzΒz= eBz

2me

Lz

μ = −e

2me

L




Applied magnetic field

No magnetic field

2p 1s Emission spectra�

m = +1

2p m = 0

m = -1

Energy

m = 01s

One line Three lines

Complex functions �21�1

and �21+1

are eigenfunctions of L z

with eigenvalues ±m� .

�2 p

x

and �2 p

y

are eigenfunctions of H0

but not of L z � no longer energy

eigenfunctions once magnetic field is applied.



5.61 Physical Chemistry 23-Electron Spin page 1

��

��

��

��!"�� #$�� %�� & � �'�� (��)*�� & �� '�� +�� & �� ,�� -��

.��/ �� "�� *� ��+�� +�� +� �� 0 �� 0 1�%231��

%�� #$�� %� � �� *�� 4 ��4�� +� �� + �� ,�� ,��4 ��+ �� 4 +� ��

�� +'�&�� $�� & �� +� ��

�� 5 � ��6� ��'�� ' �� ,�� 7 �� '��

��

8�� 9�� : = ��

: = � � (� + �) � = �� ,�� ≤ � − �: = ��-

� = )4 ±�4 ±�4�4 ±�




�� 9��

� ≡ ��

� = � � (� + �) = �

� = �� ,�� =

� = � �

; �

� �

- �

� = ±� � �5�& ��

2 m 2 m( , ) = l (l +1)� Yl (θ φ) l = 0,1,2,... L Yl θ φ , n for H atom

L Y m , Y m (θ φ(θ φ) = m� , ) m = 0, ±1, ±2,... ± n for H atom z l l

S 2α = s (s +1)�2α S 2β = s (s +1)�2β s =1

always 2

αS α = m �α m =1 β 1

S β = m �β m = −z s s z s s2 2

�� &�� α and β �� &�� & �� ,�� +�� <

α ≡ "spin up" β ≡ "spin down"

�� &��

� * � *α αdσ = β βdσ = 1 σ ≡ spin variable

* *�α βdσ = � β αdσ = 0

�� '� �� 3�'��4 �� &�� 4 � � ��




Electron orbital magnetic moment Electron spin magnetic moment

e eμμμμL = − L μμμμ s = − gS

2m 2me e

e� e�= − l (l +1) ≡ −β0 l (l +1) μμμμ s = − g s (s +1) = −β0 g s (s +1)μμμμL 2m 2me e

e e� e e�μLz

= − Lz = − m = −β0m μSz= − gSz = − gms = −β0 gms ≈ ±β02m 2m 2m 2me e e e

g ≡ "electronic g factor" = 2.002322

=�� +'�&�� %�=2�: �� %23 ��#� �� -�� +'�&�� -��

(r, , , ) =ψ (r,θ φ α σ, ) ( ) or ψ r,θ φ, ) β (σ )Ψ θ φ σ (

�� &�� +'�&��

1 2 1 2

Ψ 1 =��

Z 3 �� e−Zrα Ψ 1 =

��

Z 3 �� e−Zrβ

100 100−2 � π � 2 � π �

+� � �� -�� 3�� ,�� + !!

��



5.61 Physical Chemistry 24 Pauli Spin Matrices Page 1

Pauli Spin MatricesIt is a bit awkward to picture the wavefunctions for electron spin because – the electron isn’t spinning in normal 3D space, but in some internal dimension

that is “rolled up” inside the electron. We have invented abstract states “α” and “β” that represent the two possible orientations of the electron spin, but because there isn’t a classical analog for spin we can’t draw “α” and “β” wavefunctions. This situation comes up frequently in chemistry. We will often deal with molecules that are large and involve many atoms, each of which has a nucleus and many electrons … and it will simply be impossible for

us to picture the wavefunction describing all the particles at once. Visualizing oneparticle has been hard enough! In these situations, it is most useful to have an abstract way of manipulating operators and wavefunctions

without looking explicitly at what the wavefunction or operator looks like in real space. The wonderful tool that we use to do this is called Matrix Mechanics (as opposed to the wave mechanics we have been using so far). We will use the simple example of spin to illustrate how matrix mechanics works.

The basic idea is that we can write any electron spin state as a linear combination of the two states α and β:

ψ ≡ cαα + cβ β

Note that, for now, we are ignoring the spatial part of the electron wavefunction (e.g. the angular and radial parts of ψ). You might ask how we

can be sure that every state can be written in this fashion. To assure yourself that this is true, note that for this state, the probability of finding

22the electron with spin “up” (“down”) is cα ( cβ ). If there was a state that

could not be written in this fashion, there would have to be some other spin state, γ, so that

ψ ≡ cαα + cβ β + cγ γ

2

In this case, however, there would be a probability cγ of observing the

electron with spin γ – which we know experimentally is impossible, as the electron only has two observable spin states.

The basic idea of matrix mechanics is then to replace the wavefunction with a vector:



1


⎛ cα ⎞ ψ ≡ cαα + cβ β → ψψψψ ≡ ⎜ ⎟

⎝ cβ ⎠

Note that this is not a vector in physical (x,y,z) space but just a convenient

way to arrange the coefficients that define ψ. In particular, this is a nice way to put a wavefunction into a computer, as computers are very adept at dealing with vectors.

Now, our goal is to translate everything that we might want to do with the wavefunction ψ into something we can do to the vector ψψψψ . By going through

this stepbystep, we arrive at a few rules

Integrals are replaced with dot products. We note that the overlap between any two wavefunctions can be written as a modified dot product

between the vectors. For example, if φ ≡ dαα + dβ β then:

1 0 0 1

∫ * *

∫ * *

∫ * *

∫ * *

∫ * φ ψ dσ = dα cα α α dσ + dα cβ α β dσ + dβ cα β α dσ + dβ cβ β β dσ

* * = dα cα + dβ cβ

⎛ c ⎞ = (d *

d * )⎜ α ⎟α β

⎝ cβ ⎠

≡ φφφφ† iψψψψ

Where, on the last line, we defined the adjoint of a vector: †

⎛ x1 ⎞ * *

⎜ ⎟ ≡ ( x1 x2 ) . ⎝ x2 ⎠

Thus, complex conjugation of the wavefunction is replaced by taking the adjoint of a vector. Note that we must take the transpose of the vector

and not just its complex conjugate. Otherwise when we took the overlap we would have (column) x (column) and the number of rows and columns would not match.

These rules lead us to natural definitions of normalization and orthogonality

of the wavefunctions in terms of the vectors:

2

∫ * † ( α *

β * )⎜⎛

c

cα ⎟⎞

ψ ψ dσ = 1 → ψψψψ iψψψψ = c c = cα + cβ = 1

⎝ β ⎠

and

2




⎛ c ⎞ φ ψ dσ = 0 → φ ψφφφ ψψψ = (d d )⎜ ⎟ = d c + d c = 0i∫ *

α *

β *

c α

α *

α β *

β ⎝ β ⎠

Finally, we’d like to be able to act operators on our states in matrix mechanics, so that we can compute average values, solve eigenvalue equations, etc. To understand how we should represent operators, we note

that in wave mechanics operators turn a wavefunction into another wavefunction. For example, the momentum operator takes one wavefunction and returns a new wavefunction that is the derivative of the original one:

x ⎯⎯ xψ ( ) p→ dψ

( )dx

Thus, in order for operators to have the analogous behavior in matrix

mechanics, operators must turn vectors into vectors. As it turns out this is the most basic property of a matrix: it turns vectors into vectors. Thus we have our final rule: Operators are represented by matrices.

As an illustration, the S z operator will be represented by a 2x2 matrix in

spin space: ⎛? ?⎞

S z → Sz ≡ ⎜ ⎟

⎝? ?⎠ We have yet to determine what the elements of this matrix are (hence the question marks) but we can see that this has all the right properties: vectors

are mapped into vectors:

⎛? ?⎞⎛ cα ⎞ ⎛ cα '⎞ S

z ψ =ψ ′ → ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝? ?⎠⎝ cβ ⎠ ⎝ cβ ' ⎠

and average values are mapped into numbers

∫ψ * S

z ψ dσ → (cα *

cβ * )⎛

? ? ⎟⎞⎜⎛

c

cα ⎟⎞

⎜ ⎝? ?⎠⎝ β ⎠

Now, the one challenge is that we have to determine which matrix belongs to

a given operator – that is, we need to fill in the question marks above. As an exercise, we will show how this is done by working out the matrix

representations of S2 , S

z , S x and S

y .

We’ll start with S2 . We know how S

2 acts on the α and β wavefunctions: 3 3

S2α = �2α S

2β = �2β 4 4



�


Now represent S2 as a matrix with unknown elements.

⎛ c d ⎞ S

2 = ⎜ ⎟⎝ e f ⎠

In wave mechanics, operating S2 on α gives us an eigenvalue back, because α

is and eigenfunction of S2 (with eigenvalue

4

3 �2 ). Translating this into matrix

mechanics, when we multiply the matrix S2 times the vector α

� , we should

get the same eigenvalue back times α :

S2α =

3 �

2α → S2α �

= 3 �

2α �

4 4

In this way, instead of talking about eigenfunctions, in matrix mechanics we

talk about eigenvectors and we say that α � is an eigenvector of S

2 .

We can use this information to obtain the unknown constants in the matrix

S2 :

2 3 2 2 � 3 2 � ⎛ c d ⎞ ⎛ 1 ⎞ 2 ⎛ 1 ⎞ S α = � α → S α = � α ⇒ ⎜ ⎟ ⎜ ⎟ =

4

3 � ⎜ ⎟4 4 ⎝ e f ⎠ ⎝ 0⎠ ⎝ 0⎠

⎛ c ⎞ ⎛ 3 �2 ⎞

⇒ ⎜ ⎟ = ⎜ 4

⎟ e⎝ ⎠ ⎝ 0 ⎠

so c = 3 �

2 e = 0

4

Operating on β

3 � 3 2 � ⎛ c d ⎞⎛ 0⎞ 3 2 ⎛ 0⎞ S2β = �2β → S2α = � α ⇒ ⎜ ⎟⎜ ⎟ = 4 � ⎜ ⎟

4 4 ⎝ e f ⎠⎝ 1⎠ ⎝ 1 ⎠

⎛ d ⎞ ⎛ 0 ⎞ ⇒ ⎜ ⎟ = ⎜ 3 2 ⎟

⎝ f ⎠ ⎝ 4 � ⎠

Thus: d = 0 f = 3 �

2

4

So for S2 we have

S2 = 3 �

2 ⎛⎜

1 0⎞⎟

4 ⎝ 0 1 ⎠

We can derive Sz in a similar manner. We know its eigenstates as well:

S α = � α S β = − � β z 2 z 2

So that the matrix must satisfy:



�

�

�

�


� ⎛ c d ⎞⎛ 1 ⎞ ⎛ 1 ⎞ S

z α = α → ⎜ ⎟ ⎜ ⎟ = 2 ⎜ ⎟

2 ⎝ e f ⎠⎝ 0⎠ ⎝ 0⎠

⎛ c ⎞ ⎛ 2

� ⎞ ⇒ ⎜ ⎟ = ⎜ ⎟ ⎝ e⎠ ⎝ 0⎠

which gives us

c = e = 0 2

Analogous operations on β give

� ⎛ c d ⎞ ⎛ 0⎞ ⎛ 0⎞ S

z β = − β → ⎜ ⎟ ⎜ ⎟ = − 2 ⎜ ⎟

2 ⎝ e f ⎠ ⎝ 1 ⎠ ⎝ 1⎠

⎛ d ⎞ ⎛ 0 ⎞ ⇒ ⎜ ⎟ = ⎜ � ⎟

⎝ f ⎠ ⎝ − 2 ⎠

d = 0 f = − 2

Compiling these results, we arrive at the matrix representation of S : z

� ⎛ 1 0 ⎞ Sz = ⎜ ⎟

2 ⎝ 0 −1⎠

Now, we need to obtain Sx and S

y , which turns out to be a bit more tricky.

We remember from our operator derivation of angular momentum that we can rewrite the S and S in terms of raising and lowering operators:

x y

1 1 S

x = (S+ + S- ) S

y = (S+ − S - )

2 2i where we know that

S β = c α S α = 0 and S α = c β S β = 0+ + + − − −

where c+ and c are constants to be determined. Therefore for the raising

operator we have

S β = c α →⎛ c d ⎞⎛ 0⎞

= ⎛ c+ ⎞

S α = 0 → ⎛ c d ⎞⎛ 1⎞

= ⎛ 0⎞

+ + ⎜ ⎟⎜ ⎟ ⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ e f ⎠⎝ 1⎠ ⎝ 0 ⎠ ⎝ e f ⎠⎝ 0⎠ ⎝ 0⎠

⇒ d = c f = 0 ⇒ c = 0 e = 0+

And for the lowering operator

⎛ c d ⎞⎛ 1 ⎞ ⎛ 0 ⎞ ⎛ c d ⎞⎛ 0⎞ ⎛ 0⎞ S

−α = c−β → ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ S −β = 0 → ⎜ ⎟⎜ ⎟ = ⎜ ⎟

⎝ e f ⎠⎝ 0⎠ ⎝ c− ⎠ ⎝ e f ⎠⎝ 1⎠ ⎝ 0⎠ ⇒ c = 0 e = c ⇒ d = 0 f = 0

−

Thus

⎛ 0 1⎞ ⎛ 0 0⎞ S

+ = c+ ⎜ ⎟ S-

= c− ⎜ ⎟⎝ 0 0⎠ ⎝ 1 0⎠



= �


Therefore we find for S and S x y

1 ⎛ 0 c ⎞ 1 ⎛ 0 −ic ⎞ S

x = 1

2 (S+ + S− ) = ⎜ +

⎟ S y =

2

1

i (S+ − S− ) = ⎜ +

⎟2 ⎝ c− 0 ⎠ 2 ⎝ ic − 0 ⎠

Thus, we just need to determine the constants: c+ and c. We do this by

using two bits of information. First, we note that Sx and S

y are observable,

so they must be Hermitian. Thus, *

1 1* ⎛ 0 c ⎞⎛ 0⎞ ⎛ ⎛ 0 c ⎞⎛ 1⎞⎞

∫α * S x β dσ = (∫ β * S

x αdσ ) → (1 0)⎜ 1 c 2

0

+

⎟ ⎜ 1 ⎟ = ⎜ (0 1)⎜ 1 c

2

0

+

⎟⎜ 0⎟⎟

⎝ 2 − ⎠⎝ ⎠ ⎝ ⎝ 2 − ⎠⎝ ⎠⎠ * ⇒ c = c+ −

This result reduces our work to finding one constant (c+) in terms of which we have

1 ⎛ 0 c+ ⎞ 1 ⎛ 0 −ic + ⎞ S

x = ⎜ * ⎟ S y = ⎜ * ⎟

2 ⎝ c + 0 ⎠ 2 ⎝ ic + 0 ⎠

Note that S and S have the interesting property that they are equal to x y

their adjoints: †

1 ⎛ 0 c ⎞ 1 ⎛ 0 c ⎞ S

x † = ⎜ *

+

⎟ = ⎜ *

+

⎟ = S x

2 ⎝ c+ 0 ⎠ 2 ⎝ c+ 0 ⎠ †

1 ⎛ 0 −ic ⎞ 1 ⎛ 0 −ic ⎞ S

y † = ⎜ *

+

⎟ = ⎜ *

+

⎟ = S y

2 ⎝ ic + 0 ⎠ 2 ⎝ ic + 0 ⎠

This property turns out to be true in general: Hermitian operators are represented by matrices that are equal to their own adjoint. These selfadjoint matrices are typically called Hermitian matrices for this reason, and

the adjoint operation is sometimes called Hermitian conjugation.

To determine the remaining constant, we use the fact that S2

= S x

2 + S

y

2 + S

z

2 .

Plugging in our matrix representations for Sx , S

y , Sz and S

2 we find:

3�2 ⎛ 1 0⎞ �2 ⎛ 1 0 ⎞ ⎛ 1 0 ⎞ 1 ⎛ 0 c+ ⎞ ⎛ 0 c+ ⎞ 1 ⎛ 0 −ic + ⎞ ⎛ 0 −ic + ⎞ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + ⎜ * ⎟ ⎜ * ⎟ + ⎜ * ⎟ ⎜ * ⎟

4 ⎝ 0 1⎠ 4 ⎝ 0 −1⎠ ⎝ 0 −1⎠ 4 ⎝ c+ 0 ⎠ ⎝ c+ 0 ⎠ 4 ⎝ ic + 0 ⎠ ⎝ ic + 0 ⎠ 2 2

�2 ⎛ 1 0⎞ c ⎛ 1 0⎞ c ⎛ 1 0⎞+ + = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟

4 ⎝ 0 1⎠ 4 ⎝ 0 1 ⎠ 4 ⎝ 0 1⎠ 2 2

3 2 �2 c+ c+⇒ � = + +

4 4 4 4 2 2⇒ c+

⇒ c = �+




where on the last line, we have made an arbitrary choice of the sign of c+ . Thus we arrive at the final expressions for S and S

x y

� ⎛ 0 1 ⎞ � ⎛ 0 −i⎞ S

x = ⎜ ⎟ S y = ⎜ ⎟

2 ⎝ 1 0⎠ 2 ⎝ i 0 ⎠ In summary, then, the matrix representations of our spin operators are:

� ⎛ 0 1 ⎞ � ⎛ 0 −i ⎞ � ⎛1 0 ⎞ 2 3 2 ⎛1 0⎞ S

x = ⎜ ⎟ S y = ⎜ ⎟ S

z = ⎜ ⎟ S = � ⎜ ⎟2 ⎝ 1 0⎠ 2 ⎝ i 0 ⎠ 2 ⎝ 0 −1⎠ 4 ⎝ 0 1 ⎠

Given our representations of Sx , S

y and Sz now we can work out any property

we like by doing some linear algebra. For example, we can work out the commutator of S and S :

x y

� ⎛ 0 1⎞ � ⎛ 0 −i ⎞ � ⎛ 0 −i ⎞ � ⎛ 0 1⎞⎡S ,S ⎤ = S S − S S =⎣ x y ⎦ x y y x ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ⎜ ⎟

2 ⎝ 1 0⎠ 2 ⎝ i 0 ⎠ 2 ⎝ i 0 ⎠ 2 ⎝ 1 0⎠

�2 ⎡⎛ 0 1 ⎞⎛ 0 −i ⎞ ⎛ 0 −i⎞ ⎛ 0 1 ⎞⎤

= ⎢⎜ ⎟⎜ ⎟ − ⎜ ⎟ ⎜ ⎟⎥ 4 ⎣⎝ 1 0⎠⎝ i 0 ⎠ ⎝ i 0 ⎠ ⎝ 1 0⎠⎦

�2 ⎡⎛ i 0 ⎞ ⎛ −i 0⎞⎤ �2 ⎛ i 0 ⎞

= ⎢⎜ ⎟ − ⎜ ⎟⎥ = ⎜ ⎟4 ⎣⎝ 0 −i ⎠ ⎝ 0 i ⎠⎦ 2 ⎝ 0 −i⎠

⇒ ⎣⎡S x ,S y ⎦⎤ = i�S z

It is comforting to see that the matrices that represent the angular momentum operators obey the commutation relations for angular momentum!

It is similarly possible to work out the eigenvalues of S2 , S

z and S x by

computing the eigenvalues of Sx , S

y and Sz ; average values can be obtained

by taking (row)*x(matrix)x(column) products. We are thus in a position to compute anything we want for these operators. As it turns out, any operator

in this 2x2 space can be written as a linear combination of Sx , S

y , Sz and S

2 .

So, in some sense, we now have all the information we could possibly want about spin½ systems.

In honor of Pauli, it is conventional to define the dimensionless versions of

Sx , S

y , Sz (i.e. matrices without the leading factor of � / 2 ) as Pauli Spin

matrices:

⎛ 0 1⎞ ⎛ 0 −i⎞ ⎛ 1 0 ⎞ σ x ≡ ⎜ ⎟ σ y ≡ ⎜ ⎟ σ z ≡ ⎜ ⎟

⎝1 0⎠ ⎝ i 0 ⎠ ⎝0 −1⎠

One of the reasons the Pauli spin matrices are useful (or equivalently the matrix representations of S

x , Sy and S

z ) is that they are the simplest




possible place to practice our skills at using matrix mechanics. We will see later on that more complicated systems can also be described by matrixvector operations. However, in these cases the matrices will have many more elements – sometimes even infinite numbers of elements! Therefore, it is best to get some practice on these simple 2x2 matrices before taking on

the more complicated cases.



5.61 Physical Chemistry 25 Helium Atom page 1

HELIUM ATOM

Now that we have treated the Hydrogen like atoms in some detail, we now

proceed to discuss the nextsimplest system: the Helium atom. In this situation, we have tow electrons – with coordinates

R

z r1 and r2 – orbiting a nucleus with charge Z = 2

located at the point R. Now, for the hydrogen atom we were able to ignore the motion of the nucleus

r2 by transforming to the center of mass. We then

obtained a Schrödinger equation for a single y effective particle – with a reduced mass that was very close to the electron mass – orbiting the

x origin. It turns out to be fairly difficult to r1 transform to the center of mass when dealing with

three particles, as is the case for Helium. However, because the nucleus is

much more massive than either of the two electrons (MNuc ≈ 7000 mel) it is a very good z approximation to assume that the nucleus sits at

the center of mass of the atom. In this approximate set of COM coordinates, then, R=0 r2

and the electron coordinates r1 and r2 measure y

the between each electron and the nucleus. Further, we feel justified in separating the

motion of the nucleus (which will roughly x r1correspond to rigidly translating the COM of the atom) from the relative d the electrons orbiting

the nucleus within the COM frame. Thus, in what follows, we focus only on the motion of the electrons and ignore the motion of the nucleus.

We will treat the quantum mechanics of multiple particles (1,2,3…) in much the same way as we described multiple dimensions. We will invent operators r1 , r2 ,

r3 , … and associated momentum operators p1 , p2 , p3 …. The operators for a

given particle (i) will be assumed to commute with all operators associated with any other particle (j):

[r1, p 2 ] = [p

2 , r3 ] = [r 2 , r3 ] = [p

1, p3 ] = ... ≡ 0




Meanwhile, operators belonging to the same particle will obey the normal commutation relations. Position and momentum along a given axis do not

commute:

⎡ x , p ⎤ = i� ⎡ y , p ⎤ = i� ⎡z , p ⎤ = i�⎣ 1 x1 ⎦ ⎣ 1 y1 ⎦ ⎣ 1 z1 ⎦

while all components belonging to different axes commute: ˆ ˆ = ˆ ˆ p ˆ y p ˆ p = ˆ p .x y y x y = ˆ ˆ p p ˆ etc 1 1 1 1 z1 1 1 z1 z1 x1 x1 z1

As you can already see, one of the biggest challenges to treating multiple electrons is the explosion in the number of variables required!

In terms of these operators, we can quickly write down the Hamiltonian for the Helium atom:

Kinetic Energy NucleusElectron 1 Of Electron 1 ElectronElectron

Attraction Repulsion

p2 p2 Ze 2 1 Ze 2 1 e2 1H ≡ 1 + 2 − − +

2m 2me 4πε 0

r1

4πε 0

r2

4πε 0 r −re

1 2 Kinetic Energy NucleusElectron 2 Of Electron 2 Attraction

This Hamiltonian looks very intimidating, mainly because of all the constants (e, me,

ε0, etc.) that appear in the equation. These constants result from our decision to use SI units (meter, gram, second) to express our lengths, masses and energies. This approach is quite awkward when the typical mass we’re dealing with is 1028

grams, the typical distance is 1010 meters and the typical energy unit is 1018

Joules. It is therefore much simpler to work everything out in what are called atomic units. In this system of units we choose our unit of mass to be equal to the

electron mass, me, our unit of charge to be equal to the electron charge, e, and our unit of angular momentum to be � . Further, we choose to work in electrostatic units, so that the permittivity of free space (4π ε0) is also 1. The result of these

four choices is twofold. First of all, the Hamiltonian of the Helium atom (and indeed of any atom or molecule) simplifies greatly because all the constants are

unity and can be omitted in writing the equations:




p2 p2 Z − Z 1

H ≡ 1 + 2 − + 2 2 r

1 r2 r − r

1 2

∇2 ∇2 Z Z 1= − 1 − 2 − − +

2 2 r1

r2 r

1 − r

2

This will greatly simplify our algebra in the future and we will typically use atomicunits from this point forward. The second benefit of this choice of units is that

now our units of energy, mass, distance, angular momentum, etc. are of theappropriate size for dealing with atoms and molecules:

Atomic units and their SI equivalents

Quantity Natural unit SI equivalent

Electron mass m = 1 9.11x10−31

kg

Charge e = 1 1.06x10−19

C

Angular momentum � = 1 1.05x10−34

J ⋅ s

Permittivity κ0

= 4πε 0

= 1 1.11x10−10

C2 ⋅ J-1 ⋅ m-1

Length κ0�

2 me

2 = a = 1 (bohr) 5.29x10−11

m0

(Bohr radius)

Energy me 4 κ

0

2�

2 = e 2 κ

0a

0 = 1 (hartree) 4.36x10

−18 J = 27.2 eV

(twice the ionization energy of H)

Time κ 2�3 me

4 = 1 2.42x10−17

s0

(period of an electron in the first Bohr orbit)

Speed e 2 κ

0� = 1 2.19x10

6 kg

(speed of an electron in the first Bohr orbit)Electric potential me

3 κ0

2�

2 = e 2 κ

0a

0 = 1 27.21 V

(potential of an electron in the first Bohr orbit)

Magnetic dipole moment e� m = 1 1.85x10−23

J ⋅ T-1

(twice a Bohr magneton)

Thus, so long as we work consistently in atomic units, we will tend to findenergies that are of order unity, distances that are of order unity, momenta

that are of order unity … without having to keep track of so much scientificnotation.




Once Schrodinger had solved the Hydrogen atom, it was generally believed that the solution of the Helium atom would follow not long afterward. However,

Scientists have tried for decades to solve this three body problem without succeeding. Very, very accurate approximations were developed, but no exact solutions were found. As it turns out, even with the simplifications described

above it is impossible to determine the eigenstates of the Helium atom. This situation is common in chemistry, as most of the problems we are interested in cannot be solved exactly. We therefore must resort to making approximations, as

we will do throughout this course.

NonInteracting Electron Approiximation For Helium, the first thing we notice is that the Hamiltonian becomes separable if we neglect the electronelectron repulsion term:

∇2 ∇2 Z Z ∇2

Z ∇2 Z1 ˆ ˆH = 1 + 2 − − = − + 2 − ≡ H + Hind 2 2 r r

12 r 2 r 1 2

1 2 2

Electron One Electron Two

Thus, if we neglect the interaction between the electrons, the Hamiltonian reduces to the sum of two hydrogenic atom Hamiltonians (with Z=2). Based on our experience with separable Hamiltonians in the past (Hamiltonians Add �

Wavefunctions Multiply � Energies Add), we can immediately write down the correct form for the eigenfunctions and eigenvalues of this Hamiltonian:

Ψn l m s ;n l m s (r1,σ1;r

2,σ 2 ) =ψn l m s (r1

,σ1 )ψn l m s (r2,σ 2 )

11 1 1 2 2 2 2 11 1 1 2 2 2 2

2 2Z Z

E = E + E = − − n n n n 2 21 2 1 2 2n1 2n2

where, in the second line, we have made use of atomic units ( � =me=e=4π ε0=1) to simplify our energy expression. Thus, by neglecting the electron repulsion (an approximation) we move from a problem that is impossible to solve to one that we

can already solve easily.

However, the outstanding question is: how good is this approximation? The easiest way to test this is to look at the ground state. According to the Pauli exclusion principle, we can’t put two electrons in the same state. However, as we learned in

freshman chemistry, we can put one electron in 1s with spin up and the other in 1s with spin down:

Ψ1sα;1 sβ (r1

,σ1;r2,σ 2 ) =ψ

100 α (r1,σ1 )ψ100 β (r2

,σ 2 )




This wavefunction has an energy 2 2

Z Z 2E11 = − − = − Z =-4 a.u. = − 108.8 eV

2 2

How good is this result? Well, we can determine the ground state energy of Helium by removing one electron to create He+ and then removing the second to create He2+ . As it turns out, the first electron takes 24.2 eV to remove and the

second takes 54.4 eV to remove, which means the correct ground state energy is 78.8 eV. So our noninteracting electron picture is off by 30 eV, which is a lot of energy. To give you an idea of how much energy that is, you should note that a

typical covalent chemical bond is worth about 5 eV of energy. So totally neglecting the electron interaction is not a very good approximation.

Independent Electron Approximation So how can we go about improving this approximation? Well, first we note that the

product wavefunction described above is not antisymmetric. To test antisymmetry, all we have to do is recall that, since the electrons are identical, the labels “1” and “2” are arbitrary. By swapping these labels we can’t possibly change the outcome

of any measurement. Upon interchanging the labels “1” and “2”, an antisymmetric wavefunction will give the same wavefunction back times a minus sign. However, our proposed wavefunction does not do this:

Interchange

1 and 2 ψ (r ,σ )ψ (r ,σ ) ⎯⎯⎯⎯→ψ (r ,σ )ψ (r ,σ )1sα 1 1 1sβ 2 2 1sα 2 2 1sβ 1 1

≠ − ψ (r ,σ )ψ (r ,σ )1sα 1 1 1sβ 2 2

This is a problem, because we said that all electron wavefunctions should be antisymmetric under exchange. We can fix this problem by taking the ““

combination of the wavefunction we proposed and its exchange partner:

Ψ1sα;1 sβ (r1,σ

1; r2,σ

2 ) ≡ 1 (ψ (r ,σ )ψ (r ,σ ) −ψ (r ,σ )ψ (r ,σ ))1sα 1 1 1sβ 2 2 1sα 2 2 1sβ 1 1

2

where the leading factor of 1/√2 ensures that the new wavefunction is normalized. We check that this is antisymmetric:

σ σΨ1sα ;1 sβ (r

1,

1;r

2,

2 ) ≡ 1 (ψ1sα (r

1,σ

1 )ψ1sβ (r2,σ

2 ) −ψ1sα (r

2,σ

2 )ψ1sβ (r1,σ

1 ))2

1↔2 1 ⎯⎯⎯→ (ψ1sα (r

2,σ

2 )ψ1sβ (r1,σ

1 ) −ψ1sα (r

1,σ

1 )ψ1sβ (r2,σ

2 ))2

1 = − (ψ1sα (r

1,σ

1 )ψ1sβ (r2,σ

2 ) −ψ1sα (r

2,σ

2 )ψ1sβ (r1,σ

1 ))2

= −Ψ 1sα ;1 sβ (r

1,σ

1;r

2,σ

2 )




Does this new wavefunction give a better energy? As it turns out, this change by itself does nothing for the energy prediction. The new wavefunction is a linear

combination of two degenerate eigenstates of the independent electron Hamiltonian. As we learned before, any sum of degenerate eigenstates is also an eigenstate with the same eigenvalue. So, while quantum mechanics says we have to

make the wavefunction antisymmetric, antisymmetry by itself does not affect our energy for Helium.

The simplest way for us to improve our estimate of the helium ground state energy is to consider not the eigenvalues of our approximate Hamiltonian with our

approximate eigenfunctions, but instead look at the average energy of our approximate function with the exact Hamiltonian. That is to say, a better approximation to the energy can be got from

H = ∫Ψ*

1sα ;1 sβ H Ψ

1sα ;1 sβ dτ

1dτ

2

where dτ1 = dr1 dσ1 and similarly for dτ2. We refer to this picture as an independent electron approximation. Within the wavefunction the electrons

behave as if they do not interact, because we have retained the separable form. However, in computing the energy, we fold these interactions back in in an approximate way by computing the average energy including the interaction.

We can simplify the average energy pretty quickly:

⎛ ⎞ ⎟ Ψ dτ1dτ2∫Ψ*

1sα ;1 sβ ⎜⎜ H1 + H2 +

r −

1

r ⎟ 1sα ;1 sβ 1 2⎝ ⎠

⎛ ⎞ ⇒ ∫Ψ*

1sα ;1 sβ ⎜⎜ −2 + − 2 +

r −

1

r ⎟ Ψ dτ dτ ⎟ 1sα ;1 sβ 1 2

1 2⎝ ⎠

Ψ dτ dτ⇒ −4∫Ψ*

1sα ;1 sβΨ

1sα ;1 sβ dτ1dτ2 + ∫Ψ*

1sα ;1 sβ r1 −

1

r2 1sα ;1 sβ 1 2

2 1 Ψ

∫ r1

1sα

−

;1

r sβ

2

⇒ −4 + dτ1dτ2

We thus have for the average energy:2

Ψ 1sα ;1 sβ

H = −4 + dτ dτ1 2∫ r − r

1 2



2


The first term is simply the noninteracting electron energy from above. The second term is the average value of the electronelectron repulsion. Because the

integrand is strictly positive (as one would expect for electron repulsion) this new term will only raise the average energy, which is a good thing, since our noninteracting picture gave an energy that was 30 eV too low! We can further expand

the repulsion term based on the antisymmetric form of the wavefunction. First, we note that we can factorize the antisymmetric wavefunction into a space part times a spin part:

1 (ψ1sα (r1,σ1 )ψ1sβ (r2 ,σ2 ) −ψ1sα (r2 )ψ1sβ (r1,σ1 ))2

⇒ 1 (ψ1s (r1 )α (σ1 )ψ1s (r2 ) β (σ2 ) −ψ1s (r2 )α (σ2 )ψ1s (r1 ) β (σ1 ))2

⇒ ψ (r )ψ (r ) 1 (α (σ ) β (σ ) − β (σ )α (σ ))1s 1 1s 2 1 2 1 2 2

Ψ space (r1, r2 ) Ψspin (σ1,σ2 )

With these definitions, it is easy to verify that the space and spin wavefunctions are individually normalized. Note, in the absence of a magnetic field, you will always be able to write the eigenfunctions of the Hamiltonian in this form because

H is separable into a space and spin part

H = H + H space spin

With the spin part being (trivially) zero. As a result, eigenfunctions of H will always be products of a space part and a spin part as above. With this space/spin

separation in hand, we can simplify the repulsion term: 2 2 2 2

Ψ Ψ Ψ space spin Ψ space (r1, r2 ) Ψ 1sα ;1 sβ

dτ1dτ2 = ∫ dr1dr2dσ1dσ2 = ∫∫ r1 − r2 r1 − r2 r1 − r spin (σ1,σ2 )

2

dr1dr2dσ1dσ

2

2 Ψ space (r1, r2 ) = Ψspin (σ1,σ2 ) dσ dσ ∫ r − r1 2

dr1dr21 2∫21

Ψ space (r1, r2 ) = ∫ dr1dr2

r1 − r2

The evaluation of this 6 dimensional integral is very tedious (cf. McWeeny

problems 839 and 840) but the result is that 2

Ψ space dr1dr2 = =∫ r1 − r2

5Z a.u. = +34 eV

5

8 4




Adding this average repulsion term to our noninteracting energy gives a ground state estimate of 108.8 eV +34 eV= 74.8 eV, which is only 4 eV off of the correct

energy. The energy is still not very close, but at least we are making progress.

As we have seen already, the independent electron picture is not all that accurate

for describing atoms. However, chemists are very pragmatic and realize that the ease of solving noninteracting problems is extremely valuable and as we will soon see, it gives us a picture (molecular orbital theory) that allows us to describe a

wide range of chemistry. Therefore, chemists are extremely reluctant to abandon an independent particle picture. Instead, a great deal of work has gone into making

more accurate models based on independent particles – either by making more sophisticated corrections like the one above or by coming up with a different noninteracting Hamiltonian that gives us a better independent particle model. We will

spend the next several lectures discussing the qualitative features of electronic structure and bonding that come out of this picture.



5.61 Physical Chemistry Lecture #26 page 1

TWO ELECTRONS: EXCITED STATES

In the last lecture, we learned that the independent particle model gives a reasonable

description of the ground state energy of the Helium atom. Before moving on to talk about manyelectron atoms, it is important to point out that we can describe many more properties of the system using the same type of approximation. By using the

same independent particle prescription we can come up with wavefunctions for excited states and determine their energies, their dependence on electron spin, etc. by examining the wavefunctions themselves. That is to say, there is much we can determine from simply looking at Ψ without doing any significant computation.

We will use the excited state 1s2s configuration of Helium as an example. For the ground state we had:

Ψspace (r1,r

2) × Ψspin (σ1

,σ2)

α σ β σ − β σ α σ ⇒ ψ (r )ψ (r ) 1

2 ⎜⎝

⎛ ( 1) ( 2 ) ( 1) ( 2 ) ⎟⎠⎞

1s 1 1s 21s

In constructing excited states it is useful to extend the stick diagrams we have used before to describe electronic configurations. Then there are four different configurations we can come up with:

Ψspace (r1,r

2) × Ψspin (σ1

,σ2)

2s ? ?

1s

2s ? ?

1s

2s ? ?

1s

2s ? ?

1s




Where the question marks indicate that we need to determine the space and spin wavefunctions that correspond to these stick diagrams. It is fairly easy to come up

with a reasonable guess for each configuration. For example, in the first case we might write down a wavefunction like:

ψ r σ ψ r σ =ψ r ψ r α σ α σ 1sα ( 1 1) 2sα ( 2 2 ) 1s ( 1) 2s ( 2 ) ( 1) ( 2 )

However, we immediately note that this wavefunction is not antisymmetric. We can perform the same trick as before to make an antisymmetric wavefunction out this:

1 ⎛ψ r ψ r α σ α σ −ψ r ψ r α σ α σ ⎞⇒ ⎜ 1s ( 1) 2s ( 2 ) ( 1) ( 2 ) 1s ( 2 ) 2s ( 1) ( 2 ) ( 1) ⎟2 ⎝ ⎠

⇒ 1 ⎛ψ r ψ r −ψ r ψ r ⎞ α σ α σ ⎜ 1 ( 1) 2s ( 2 ) 1s ( 2 ) 2s ( 1)⎟ ( 1) ( 2 )2 ⎝ s ⎠

Ψ Ψspace spin

Applying the same principle to the 1s↑2s↓ configuration gives us a bit of trouble:

⇒ ⎜⎛

⎝ ψ ) ) β σ ) ) ( ( ⎞

⎠ Ψspace Ψψ

1s (r1) 2s (r2 α σ ( 1 ( 2

−ψ1s (r2

ψ2s (r1)α σ

2 ) β σ 1) ⎟ ≠ spin

Hence, the pure ↑↓ configuration can’t be separated in terms of a space part and a spin part. We find a similar result for 1s↓2s↑:

⇒ 1 ψ r ψ r β σ α σ −ψ r ψ r β σ α σ ≠ Ψ Ψspace 2 ⎜⎝

⎛ 1s ( 1) 2s ( 2 ) ( 1) ( 2 ) 1s ( 2 ) 2s ( 1) ( 2 ) ( 1) ⎟

⎠

⎞ spin

Since we know the wavefunction should separate, we have a problem. The solution

comes in realizing that for an open shell configuration like this one, the 1s↑2s↓ and 1s↓2s↑ states are degenerate eigenstates and so we can make any linear combinations of them we like and we’ll still obtain an eigenstate. If we make the “+” and ““ combinations of 1s↑2s↓ and 1s↓2s↑ we obtain:

( 1s (r 1)ψ

2s (r 2

α σ 1) β ( 2 ) −ψ

1s r 2 )ψ

2s r 1

α σ 2 ) β (σ

1)) ±⇒ ψ ) ( σ ( ( ) ( ψ r ) ( α σ − r ψ σ ) ( ( ( ) ψ (r β σ ) ( ) ψ ( ) ( ) ( r β α σ ))1s 1 2s 2 1 2 1s 2 2s 1 2 1

ψ r r α σ σ ) ( ⇒ ( ( ) ψ (r ) �ψ (r )ψ ( ) ) ( ( ) ( β σ ) ± β ( α σ ))1s 1 2s 2 1s 2 2s 1 1 2 2 1

1 2

Ψ Ψ space spin

which separates nicely. Performing similar manipulations for the ↓↓ configuration and

taking care to make sure that all our spatial and spin wavefunctions are individually normalized allows us to complete the table we set out for the 1s2s excited states:




Ψspace (r1,r

2) × Ψspin (σ1

,σ2)

2s 1 ψ α σ ⇒ ( 1s (r1 )ψ2s (r2 ) −ψ1s (r2 )ψ2s (r1 )) × ( 1 )α σ ( 2 )2

1s

2s 1 1(ψ1s (r1 )ψ2s (r2 ) +ψ1s (r2 )ψ2s (r1 )) × ( ( 1 β σ ( − ( 1 α σ ( )α σ ) 2 ) β σ ) 2 )1s 2 2

and 1 12s α σ β σ ((ψ1s (r1 )ψ2s (r2 ) −ψ1s (r2 )ψ2s (r1 )) × ( ( 1 ) ( 2 ) + β σ 1 )α σ ( 2 ))2 2

1s

2s ⇒ 1 ψ r ψ r −ψ r ψ r × β σ β σ ( 1s ( 1 ) 2s ( 2 ) 1s ( 2 ) 2s ( 1 )) ( 1 ) ( 2 )2

1s

We notice several things about these wavefunctions: • While the overall wavefunction is always antisymmetric by construction, the

spatial part can be either antisymmetric (cases 1, 3 and 4) or symmetric (case 2). This effect is compensated for in the spin part, which can also be antisymmetric (case 2) or symmetric (cases 1,3 and 4). The resulting

wavefunction always has a symmetric part times an antisymmetric part, resulting in an antisymmetric wavefunction.

• The spin part of Case 2 is exactly the same as the spin part of the ground state

of the helium atom. Thus, just as we thought of the electrons in the ground state as being “paired”, we say the electrons in Case 2 are paired.

• The spatial parts of three of the states above (cases 1,3 and 4) are the same. Case 2 has a different spatial part. Because the Hamiltonian only depends on spatial variables and not spin, we immediately conclude that 1,3 and 4 will be

degenerate states – even when we take into account the electronelectron interaction. State 2, however, will generally have a different energy once we account for interactions. In common spectroscopic parlance the three

degenerate states are called a triplet and the unique state is called a singlet. Further, because these states arise from degenerate spin states, they are called singlet and triplet spin states.




Energies of Singlet and Triplet States As we showed above, we expect the singlet and triplet states to have different

energies once electron repulsion is taken into account. Which one will be lower? To decide this, we note that the triplet spatial wavefunction is zero when the two electrons are at the same position:

r1 = r2 ⇒ ΨT = 1 (ψ1s (r1 )ψ2s (r1 ) −ψ1s (r1 )ψ2s (r1 )) = 0 2

whereas the singlet wavefunction is nonzero:

r1 = r ⇒ ΨS = 1 (ψ1s (r1 )ψ2s (r1 ) +ψ1s (r1 )ψ2s (r1 )) = 2 ψ1s (r1 )ψ2s (r1 ) ≠ 02 2

Because the electrons repel each other more when they are close to one another, we

therefore expect the singlet to have more electronelectron repulsion and a higher energy. This rule turns out to hold quite generally and is called Hund’s rule: for degenerate noninteracting states, the configuration with highest spin multiplicity

lies lowest in energy. Hund actually has three rules (of which this is the first) concerning the ordering of degenerate noninteracting states. The others apply only to atoms and will not be discussed here, but see McQuarrie Section 9.119.12 for

more on this topic.

So we expect the triplet to be lower. How much lower? To answer this question, we

have to compute the average energies of the singlet and triplet wavefunctions. Recall that the spin part never matters for the energy:

∫ Ψ * H Ψd d r σ = ∫ Ψ space

* Ψspin * H Ψ space Ψspin d d r σ = ∫ Ψspin

* Ψspin dσ ∫ Ψ space * H Ψ space dr

= ∫ Ψ space * H Ψ space dr

1

The influence of the spin wavefunction is only indirect: if the spin part is antisymmetric (e.g. singlet) then the spatial part must be symmetric and vice versa. To simplify our algebra, it is convenient to create the obvious shorthand notation:

ΨT = (ψ (r )ψ (r ) −ψ (r )ψ (r )) ≡ ( s s − s s 12 1s 1 2s 2 1s 2 2s 1

12

1 2 2 1 ) 1 1 1 2 s s ΨS = 2 (ψ1s (r1 )ψ2s (r2 ) +ψ1s (r2 )ψ2s (r1 )) ≡

2 ( s s + 2 1 )

where we just need to remember that the first function in a product will be the one that has electron “1” while the second will have electron “2”. Proceeding then:




⎛ ⎞ E / = ∫ΨS T /

* H ΨS T dr1dr = 1 ˆ ˆ 1 ⎟

⎟ (1 2 s ± 2 1 ) dr1dr2( )

( ) (

( ) (

1 2

1

1

1

1 2 2 1 2

1 1 1 1 2 2 1 2

2 2

5 1 1 1 2 2 1

2 2

s H H⎜±∫

1 2

s s s

s s s s

s s s s s s

⎜⎝

⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠

+ + −

± − − + −

= − + ± −

∫

∫

2

2

2

r r

r r

r r

s s s S T / 2

⎠

= 1 2 s ± 2 1 ) dr drs s s 1 2

± 2 1 ) drs s 1dr2

On the second line, we have used the fact that both ΨS and ΨT are eigenstates of the independent particle Hamiltonian (by construction) and on the third line, we have

taken the independent particle energy outside the integral because ΨS and ΨT are normalized. Thus, we see that the average energy takes on the familiar form of

(noninteracting energy)+(interactions). The interaction term can be simplified further:

1 1

2 ∫ ( s s ± 2 1 ) r1 − r2

1 2 s s ( s s ± 2 1 ) dr dr21 2 s s 1

1 1 1 1 1 ⇒ 1 2 ss 1 2 s −1 2 ss s 2 1 − 2 1 s s s s 1 2 s + 2 1 s s s 2 1 r1d 2s s d r∫2 r1 − r2 r1 − r2 r1 − r2 r1 − r2

We note that the first and last terms are the same if we just interchange the dummy integration variables:

1s (r1 ) 2s (r2 ) dr1dr2 ⎯⎯⎯→∫1s (r2 ) 2s (r1 ) 1s (r ) 2s (r ) dr dr∫1s (r1 ) 2s (r2 ) r1 −

1

r2

1↔2

r2

1

− r1

2 1 2 1

= ∫ 2s (r1 )1s (r2 ) 1

2s (r1 )1s (r2 ) dr1dr2 ≡ J12

r1 − r2

Meanwhile the second and third terms are also the same:

∫1s (r1 ) 2s (r2 ) r −

1

r

1↔2 1 2s (r1 )1s (r2 ) dr1dr2 ⎯⎯⎯→∫1s (r2 ) 2s (r1 )

r − r 2s (r )1s (r ) dr dr2 1 2 1

1 2 2 1

= ∫ 2s (r1 )1s (r2 ) r1 −

1

r2

1s (r2 ) 2s (r1 ) dr1dr2 ≡ K12

These integrals are called Coulomb(J) and exchange(K) integrals, respectively. Both are positive numbers (because they arise from electron repulsion) and it can be rigorously proved that J>K always (i.e. no matter what functional form the 1s and 2s

wavefunctions have). Thus, in terms of J and K the energies of the singlet and triplet states become:

5 + J ± KES T = −

12 12 / 2

Thus we see that, as expected, the singlet state is higher in energy than the triplet. In fact, we can even give a numerical estimate for the splitting by evaluating K12. Plugging in the forms of the 1s and 2s orbitals of helium and doing the integrals, we obtain K12=32/729=1.2 eV and a splitting of 2K12=2.4 eV. The latter is quite a bit larger than the experimental singlettriplet splitting in helium, which comes out to




only .8 eV. Once again, we see the independent particle model gives us a qualitatively correct picture (i.e. the sign of the splitting is correct and of the right order of

magnitude) but we fail to obtain quantitative results. We therefore arrive at the following qualitative picture of the 1s2s excited state of Helium:

It is interesting to note that the exchange interaction results from the fact that the

electrons are indistinguishable. Notice that, if we had not antisymmetrized our wavefunctions, the spatial part would have just been a direct product 1s2s instead of the symmetric/antisymmetric 1s2s±2s1s combinations we obtained for the singlet and

triplet. In the former case, the electrons are being treated as distinguishable (e.g. electron “1” is always 1s while electron “2” is always 2s) and the exchange term disappears from the interaction:

1 11 2 s s 1 r2

distinguishable → s

1

2 ∫ ( s s ± 2 1 ) r1 − r2

1 2 s s ( s s ± 2 1 ) dr d ⎯⎯⎯⎯⎯⎯ ∫1 2 s r1 − r2

1 2 sd r1d 2 = J12 s r

Clearly exchange – which arises from the cross terms on the left – is absent on the right. Thus, the K integrals only arise when we have terms in the wavefunction where two electrons have exchanged places. Hence the name “exchange.” It is important to

note that, next to the Pauli exclusion principle, this is the biggest impact that antisymmetry has on chemistry.

There was a lot of interest in class about how we reconcile the fact that, in other chemistry courses you’ve been taught that there are only two spin states for a pair of

electrons: ↑↓ and ↑↑. The former represented the singlet state and the latter the triplet state. You referred to the singlet electrons as being “paired” and the triplets as being “unpaired.” However, how do these strange spin states we’ve derived connect

with the “paired” and “unpaired” ideas? To answer this question, we first of all we should note that neither of the antiparallel states we’ve derived is strictly ↑↓. Instead, they look like ↑↓±↓↑, with ↑↓-↓↑ being the singlet and ↑↓+↓↑ being part of the

triplet. The idea that the singlet state is ↑↓ is a white lie that we tell in order to simplify our arguments: as long as the subtle difference between ↑↓ and ↑↓-↓↑ isn’t important, we can get away with explaining much (though not all) chemistry by treating

the singlet state as ↑↓.




In the more precise picture we’ve derived here, it the spin part of the wavefunction determines whether the electrons are paired or not. An electron pair has the

characteristic spin part αβ−βα. That is to say, paired electrons form a singlet. Spin

parts that look like αα, αβ+βα, or ββ are unpaired triplet configurations. As we have

seen above, pairing two electrons raises the energy through the exchange integral. In some situations, this is called the “pairing energy.” The counterintuitive thing that we have to relearn is that αβ+βα does not describe an electron pair. In every way it

behaves like αα or ββ : the energies are the same and (as you will show on the

homework) the eigenvalues of S2 are the same. This idea really does not fit into the total

simple qualitative picture of triplet states being ↑↑, but it is nonetheless true.

The fact that there are three elements of the triplet state is not a coincidence. As ˆ2 2you will show, the eigenvalues of Stotal for the triplet states are all 2� , which is

ˆ2consistent with a total spin of S=1, because the eigenvalues of Stotal would then be

�2S (S + 1) = 2�2 . This picture is also

consistent with the idea that, if we add two spins with s=½ parallel to one

another we should get a total S of S= ½+½=1. Given this picture, we note that

the three triplet states would then correspond to the three possible zprojections of spin. That is to say the

three triplet states should have MS=+1,0 and1, respectively. This gives us at least some qualitative picture of what the αβ+βα state means and why it

corresponds to unpaired electrons. In the αβ+βα state the spins are oriented parallel to each other, but they are both oriented perpendicular to the z axis, so that on average you will always find one spinup and one spindown along z. This is a very

simple example of the addition of angular momentum, a topic which is covered in much greater depth in McQuarrie.

x

z

y

1= +S

M

1= −S

M

0= S

M




MANY ELECTRON ATOMSThus far, we have learned that the independent particle model (IPM) gives a qualitatively correct picture of the eigenstates of the helium atom. What about

atoms with more than two electrons, such as lithium or carbon? As it turns out, the IPM is capable of giving a realistic picture of atomic structure in essentially an analogous fashion to the helium case. To begin with, we set up our coordinates so that

the nucleus is at the origin and the N electrons are at positions r1, r2, r3, …rN. In terms of these variables, we can quickly write down the many-electron Hamiltonian (in atomic

units):

ˆ 1 N 2

N Z N N 1 H = − ∑∇ j − ∑ + ∑∑

2 j=1 j=1 r j i=1 j>i r

i − r j

Kinetic Energy Electron-Nuclear Electron-Electron

Attraction Repulsion

Thus, the Hamiltonian has the same three sources of energy as in the two electron case, but the sheer number of electrons makes the algebra more complicated. As before, we note that we can make the Hamiltonian separable if we neglect the

electron-electron repulsion:

⎛ ⎞ H = −

1 N ∑ ∇2 −

N ∑

Z =

N ∑ ⎜ − 1 ∇2 −

Z ⎟ ≡ N ∑ h

NI 2 j r ⎜ 2 j r ⎟ i j = 1 j = 1 j j = 1⎝ j ⎠ j = 1

where each of the independent Hamiltonians h i describes a single electron in the field

of a nucleus of charge +Z. Based on our experience with separable Hamiltonians, we

can immediately write down the eigenstates of this Hamiltonian as products with energies given as sums of the independent electron energies:

Ψ = ψ 1 ψ 2 ψ 3 ... ψ N E = E + E + E + ... + E

Where (1) is a shorthand for (r1,σ1) and k ≡ {n , l , m , s } specifies all the quantum

k1

( ) k2

( ) k3

( ) kN ( ) k

1 k

2 k

3 kN

i i i i i

numbers for a given hydrogen atom eigenstate. Of course, there is a problem with

these eigenstates: they are not antisymmetric. For the Helium atom, we fixed this by making an explicitly antisymmetric combination of two degenerate product states:

ψ (1) ψ (2)1 1ψ 1 ψ 2 −ψ 2 ψ 1 =( 1sα ( ) 1sβ ( ) 1sα ( ) 1sβ ( ) ) ψ

1

1

s

sα

β ( ) 1 ψ1

1

s

sα

β ( ) 22 2




where on the right we have noted that this antisymmetric product can also be written as a determinant of a 2x2 matrix. As it turns out, it is straightforward to extend this

idea to generate an N particle antisymetric state by computing an NxN determinant called a Slater Determinant:

ψ (1) ψ (1) ψ (1) � ψ (1)k k k k1 2 3 N

k1

( ) k2

( ) k3

( ) kN ( ) ψ 2 ψ 2 ψ 2 � ψ 2

Ψ(1,2,..., N ) = ψ 3 ψ 3 ψ 3 � ψ 3k ( ) k ( ) k ( ) k ( ) 1 2 3 N

� � � � �

k1

( N ) ψ k

2 ( N ) ψ

k3

( N ) � ψ kN

( N )

1

!N

ψ

As you can imagine, the algebra required to compute integrals involving Slater

determinants is extremely difficult. It is therefore most important that you realize several things about these states so that you can avoid unnecessary algebra:

• A Slater determinant corresponds to a single stick diagram. This is easy to

see by example:

2px 1sα (1) 1sβ (1) 2sα (1) 2 p α (1)x

sα ( ) 1 β ( ) 2sα 2 2 p α ( ) 1 2 s 2 ( ) 2 x2s ⇒ Ψ (1, 2,3, 4 ) = 1sα 3 1sβ 3 2sα 3 2 p α 3

( ) ( ) ( ) x ( ) ( ) ( ) ( ) x ( )

1sα 4 1sβ 4 2sα 4 2p α 41s

It should be clear that we can extend this idea to associate a determinant with an arbitrary stick diagram. Further, recall that for the excited states of

helium we had a problem writing certain stick diagrams as a (space)x(spin) product and had to make linear combinations of certain states to force things to separate. Because of the direct correspondence of stick diagrams and

Slater determinants, the same pitfall arises here: Slater determinants sometimes may not be representable as a space)x(spin) product, in which case a linear combination of Slater determinants must be used instead. This generally only happens for systems with unpaired electrons, like the 1s↑2s↓ configuration of helium or the …2px↑2py↓ configuration of carbon.

• A Slater determinant is anitsymmetric upon exchange of any two electrons.

We recall that if we take a matrix and interchange two its rows, the determinant changes sign. Thus, interchanging 1 and 2 above, for example:




ψ (2) ψ (2) ψ (2) � ψ (2)k k k k1 2 3 N

ψ 1 ψ 1 ψ 1 � ψ 1( ) k ( ) ( ) k ( ) k k1 2 3 N

Ψ (2,1,..., N ) = N 1

! ψ

1 ( )

2 ( ) ψ

33 �

N ( ) 3 ψ 3 ( ) ψ 3

k k k k

� � � � �

ψ ( N ) ψ ( N ) ψ ( N ) � ψ ( N )k k k k1 2 3 N

ψ (1) ψ (1) ψ (1) � ψ (1)k k k k1 2 3 N

ψ 1

( ) 2

( ) ψ 3

2 � N

( )2 ψ 2 ( ) ψ 2k k k kInterchange

Rows 1&2 1= −Ψ (1,2,..., N )3 ψ 3 ( ) ψ 3⎯⎯⎯⎯⎯⎯→ − ψ ( ) ( ) ψ 3 � ( ) k k k k

1 2 3 NN ! � � � � �

ψ ( N ) ψ ( N ) ψ ( N ) � ψ ( N )k k k k1 2 3 N

A similar argument applies to any pair of indices, and so the Slater determinant antisymmetric under any i↔j interchange.

• The determinant is zero if the same orbital appears twice. We recall that if we take a matrix and interchange two of its columns, the determinant also

changes sign. Assuming k1 =k3 above:

ψ (1) ψk (1) ψ (1) � ψ

k (1)k k1 2 1 N

ψ 1

( ) 2

( ) ψ 1

2 � N

( ) 2 ψ 2 ( ) ψ 2k k k k

Ψ (1,2,..., N ) = 1 ψ ( ) ( ) ψ 3 � ( ) 3 ψ 3 ( ) ψ 3

k k k k1 2 1 NN ! � � � � �

ψ ( N ) ψk ( N ) ψ ( N ) � ψ

k ( N )k k1 2 1 N

ψ (1) ψ (1) ψ (1) � ψ (1)k k k k1 2 1 N

ψ 1

( ) 2

( ) ψ 1

2 � N

( ) 2 ψ 2 ( ) ψ 2k k k kInterchange

Columns 1&3 1= −Ψ (1,2,..., N )⎯⎯⎯⎯⎯⎯⎯→ − ψ 3 ψ 3 ψ 3 � ψ 3( ) k ( ) ( ) k ( ) k k

1 2 1 NN ! � � � � �

ψ ( N ) ψ ( N ) ψ ( N ) � ψ ( N )k k k k1 2 1 N

The only way a number can be equal to its opposite is if it is zero. Thus, in enforcing antisymmetry the determinant also enforces the Pauli exclusion

principle. Thus, Slater determinants immediately lead us to the aufbau




principle we learned in Freshman chemistry: because we cannot place two electrons in orbitals with the same quantum numbers, we are forced to fill up

orbitals with successively higher energies as we add more electrons: 1s↑, 1s↓, 2s↑, 2s↓, …. This filling up is dictated by the fact that the wavefunction is antisymmetric, which, in turn, results from the fact that the electron is spin-½. As it turns out, all half-integer spin particles (Fermions) have antisymmetric wavefunctions and all integer spin particles (Bosons) have symmetric wavefunctions. Imagine how different life would be if electrons were Bosons

instead of Fermions! • Ψ is normalized if the orbitals ψ are normalized and two determinants are

ki

orthogonal if they differ in any single orbital. These two facts are relatively tedious to prove, but are useful in practice.

• The noninteracting energy of a Slater determinant is the energy of the

orbitals that make it up. Just as was the case for helium, our antisymmetric wavefunction is a linear combination of degenerate non-interacting eigenstates:

2 ψk 3 ψ N )Ψ (1,2,..., N ) = 1

(ψ k1

( ) 1 ψ k2

( ) ψ k3

( ) 3 � kN

( N ) −ψ k1

( ) 2 ψ 2

( ) 1 ψ k3

( ) � kN

(N !

( ) ψ k ( ) 2 ψ

k ( N −ψ ( ) ψ k 3 ψ ( ) �ψ

k ( N ) ...) +ψ k 3 1 ψ

k ( ) � ) k 1 ( ) k 2 1 2 3 N 1 2 3 N

Thus the determinant itself is an eigenstate of the non-interacting Hamiltonian

with the same eigenvalue as each state in the sum: E = E + E + E + ... + EΨ k k k k

1 2 3 N

We have only completed half of the independent particle picture at this point. We have the noninteracting energy; what remains is the computation of the average

energy. To compute this, we would need to do a 2N-dimensional integral involving a Slater Determinant on the left, the Hamiltonian in the middle (including all interaction terms) and a Slater Determinant on the right. The book-keeping is quite tedious, but

can be worked out in the most general case. The result is that the energy breaks down into terms that we already recognize:

Noninteracting Average Energy Repulsion

N N H = ∑E

i + ∑J� ij − K�

ij i=1 i< j

ψ 2 1 ψ 2ψ ( ) ( ) dr dr dσ dσJ� ij ≡ ∫∫ ψk *

i ( ) 1

k *

j ( ) 1

k k 1 2 1 2i jr − r

1 2

K� ≡ ψ * 1 ψ * 2 1 ψ 2 ψ 1 dr dr dσ dσij ∫∫ k ( ) k ( ) ( ) k ( ) 2 2k 1 1

i ji j r − r1 2




This energy expression has a nice intuitive feel to it: we get contributions from the interaction of each electron orbital with the nucleus (first term) as well as from the mutual repulsion of each pair of orbitals (second term). The repulsion has the

characteristic “Coulomb minus Exchange” form because of the antisymmetry of the Slater determinant, which leads to a minus sign every time we exchange 1 and 2. For example, when we expand the determinant in terms of product functions, we will have

a term like: X ≡ψ

k1 (1)ψ

k2 (2)ψ

k3 (?)�ψ

kN (?)

we will have a corresponding term with 1↔2 with a minus sign: Y ≡ψ

k1 (2)ψ

k2 (1)ψ

k3 (?)�ψ

kN (?)

When we compute the average repulsion we will have:

1 ≈ ∫∫ ( X − Y ... ) 1

( X − Y ... ) dr1dr2 ... r r12 12

The +XX and +YY terms will give us Coulomb integrals and the cross terms -XY and

-YX will give us exchange integrals. We note that, as defined, the Coulomb and Exchange terms are both positive, so that the exchange integral always reduces the repulsion energy for a Slater determinant. These arguments do not constitute a proof

of the energy expression above – they are merely intended to give you a general feeling that this expression is plausible. Deriving the average energy actually takes quite a bit of time and careful bookkeeping of different permutations (e.g. 1↔2↔4

versus 4↔2↔3 …). If you want to delve deeper into these kinds of things, we recommend you take 5.73, which covers much of the material here in greater depth.

In any case, most of the time we need not worry about the details of how the average energy expression is derived for a general determinant. Usually what we want to do is

use the formula to compute something interesting rather than re-derive it. Toward this end we note that the Coulomb and exchange integrals above involve integration over spin variables, which we can be done trivially. Basically, the integration over σ1

(σ2) will give unity if the left and right functions for electron 1 (electron 2) have the same spin, and zero otherwise. For the Coulomb integral, the left and right functions are the same, so the spin integration always gives 1 and we can get rid of the spin

integration:

* * 1 J ψ 1 ψ 2 ψ 1 ψ 2 dr dr= ∫∫ ( ) ( ) ( ) ( ) ij i j i j 1 2

r − r 1 2




where here we use the shorthand i ≡ {n , m , l } to represent all the spatial quantum i i i

numbers for the ith orbital, and similarly for the jth one. For the exchange integrals, the left and right functions are different, and so the exchange integral is only nonzero if the ith orbital and the jth orbital have the same spin part:

⎧ * * 1 ψ 2 2 1 =ψ ( ) ψ ( ) dr dr if s s⎪∫∫ ψ

i ( ) 1 j ( )

i j 1 2 i jK = ⎨ r − r

1 2ij ⎪ ⎩ 0 Otherwise

In practice (for example on a Problem Set) it is useful to begin with the general expression for the independent particle energy and then integrate out the spin part to get a final expression that only involves the spatial integrals J and K rather than

the space-spin integrals J� and K� . For example, say we are interested in the 1s↑1s↓2s↑ configuration of lithium. The energy is given by:

ˆ

1

= + −∑ ∑ = <

= + + + − + − + −

� �

� � � � � �

N N H E J K

i ij ij i i j

E E E J K J K J K

J1s1s J1s2s K1s2s J1s2s 0

1s 1s 2s 1sα;1 sβ 1sα;1 sβ 1sα;2 sα 1sα; 2 sα 1sβ ;2 sα 1sβ ;2 sα = 2E + E + J + 2J − K

1 1 s s 1 2 1s 2s s s 1 2 s s

Thus, in going from space-spin integrations to space integrations, we acquire a few factors of 2, but the formula still looks similar to the expressions we’ve seen before.

The independent particle energy expression is extremely powerful: it allows us to make rough predictions of things like atomic ionization energies and excitation

energies once we have the Coulomb and exchange integrals. The independent particle energy also gives us a rigorous explanation of another freshman chemistry effect: shielding. As we know, while 2s and 2p are degenerate for the hydrogen atom they

are not degenerate as far as the order of filling up orbitals. 2s comes first and 2p comes second. As you know, physically this arises because the 2s orbital sees an

effective nuclear charge that is bigger than 2p because it is shielded less by the 1s orbital. It turns out that the IPM has a simple means of explaining this result. If we look at the 1s↑1s↓2px↑ configuration of lithium and go through exactly the same

manipulations as we did above for 1s↑1s↓2s↑ we get:

H = 2E + E + J + 2J − K 1 1 s p 1 2 1s 2 px s s 1 2 x s px

taking the difference between 1s↑1s↓2px↑ and 1s↑1s↓2s↑ gives:

0




ΔE = E − E = + 2 1 1

2 1 1

E

E J

+ J + 2J − K − 1s 22p 1s 22s

2E 1s p

x s s s px 1 2 p

x 1 2 s

+ + + 2J − K )(2E 1s s s s s s 1 2 s1 2 s

= 2J − 2J + K − K ≠ 0 1 2 s p

x 1 2 s s 1 2 x s s 1 2 s p

Thus, while 1s↑1s↓2px↑ and 1s↑1s↓2s↑ give the same energies if the electrons do not interact, they give different energies once we include the average interaction. Further, it is clear from the above expression that the relevant interactions are

between the 1s and 2p orbitals on the one hand and the 1s and 2s orbitals on the other. Using our physical argument of shielding, we assert that the 1s and 2s orbitals will repel each other less than the 1s and 2p orbitals, leading to a net stabilization of

1s↑1s↓2s↑ relative to 1s↑1s↓2px↑. We explore this idea in more detail on the problem set.



1 5.61 Physical Chemistry Lecture #28

MOLECULAR ORBITAL THEORY PART I At this point, we have nearly completed our crashcourse introduction to

quantum mechanics and we’re finally ready to deal with molecules. Hooray! To begin with, we are going to treat what is absolutely the simplest

molecule we can imagine: H+ . This simple molecule will allow us to work 2

out the basic ideas of what will become molecular orbital (MO) theory.

We set up our coordinate

system as shown at right, with the electron positioned at r, and the two nuclei

positioned at points RA and RB, at a distance R from one HA another. The Hamiltonian is RA

easy to write down: H2

H = − 1 ∇2 −∇2

A −∇2

B − 1 − 1 + 1

ˆ2 r 2MA

2MB R − r R − r R − R

B A BA

e

HB RB

r

rArB

R + Coordinates

Electron HA HB e HA e HB HA HB

Kinetic Kinetic Kinetic Attraction Attraction Repulsion Energy Energy Energy

Now, just as was the case for atoms, we would like a picture where we can separate the electronic motion from the nuclear motion. For helium, we

did this by noting that the nucleus was much heavier than the electrons and so we could approximate the center of mass coordinates of the system by placing the nucleus at the origin. For molecules, we will make a

similar approximation, called the BornOppenheimer approximation. Here, we note again that the nuclei are much heavier than the electrons.

As a result, they will move much more slowly than the light electrons. Thus, from the point of view of the electrons, the nuclei are almost sitting still and so the moving electrons see a static field that arises from fixed nuclei. A useful analogy here is that of gnats flying around on the back of an elephant. The elephant may be moving, but from the gnats’ point of view, the elephant is always more or less sitting still. The

electrons are like the gnats and the nucleus is like the elephant.




The result is that, if we are interested in the electrons, we can to a good approximation fix the nuclear positions – RA and RB – and just look at the

motion of the electrons in a molecule. This is the BOppenheimer approximation, which is sometimes also called the clampednucleus approximation, for obvious reasons. Once the nuclei are clamped, we can

make two simplifications of our Hamiltonian. First, we can neglect the kinetic energies of the nuclei because they are not moving. Second, because

the nuclei are fixed, we can replace the operators R A and R

B with the

numbers RA and RB. Thus, our Hamiltonian reduces to

ˆ rH (R ,R ) = − ∇2

− 1 1 1

− + el A B 2 R − r R − r R − R

A B A B

where the last term is now just a number – the electrostatic repulsion between two protons at a fixed separation. The second and third terms

depends only on the position of the electron, r, and not its momentum, so we immediately identify those as a potential and write:

ˆ ( ) ∇2 r RA , RB ( ) 1

H R ,R = − +V r + el A B 2 eff R − R

A B

This Hamiltonian now only contains operators for the electrons (hence the subscript “el”), but the eigenvalues of this Hamiltonian depend on the

distance, R, between the two nuclei. For example, the figure below shows the difference between the effective potentials the electron “feels” when the nuclei are close together versus far apart:

R Small R Large

Veff(r)

Likewise, because the electron feels a different potential at each bonddistance R, the wavefunction will also depend on R. In the same limits as

above, we will have:

R Small R Large

ψel(r)




Finally, because the electron eigenfunction, ψel, depends on R then the eigenenergy of the electron, Eel(R), will also depend on the bond length.

Mechanically, then, what we have to do is solve for the electronic eigenstates, ψel, and their associated eigenvalues, Eel(R), at many different fixed values of R. The way that these eigenvalues change with R will tell us

about how the energy of the molecule changes as we stretch or shrink the bond. This is the central idea of the BornOppenheimer approximation, and it is really very fundamental to how chemists think about molecules. We

think about classical pointlike nuclei clamped at various different positions, with the quantum mechanical electrons whizzing about and gluing the nuclei together. When the nuclei move, the energy of the system changes because

the energies of the electronic orbitals change as well. There are certain situations where this approximation breaks down, but for the most part the BornOppenheimer picture provides an extremely useful and accurate way to

think about chemistry.

How are we going to solve for these eigenstates? It should be clear that looking for exact solutions is going to lead to problems in general. Even for H2

+ the solutions are extremely complicated and for anything more complex

than H2+exact solutions are impossible. So we have to resort to

approximations again. The first thing we note is that if we look closely at our intuitive picture of the H2

+ eigenstates above, we recognize that these

molecular eigenstates look very much like the sum of the 1s atomic orbitals for the two hydrogen atoms. That is, we note that to a good approximation we should be able to write:

ψ el (r ) ≈ c11sA (r ) + c

21sB (r )

where c1 and c2 are constants. In the common jargon, the function on the

left is called a molecular orbital (MO), whereas the functions on the right are called atomic orbitals (AOs). If we write our MOs as sums of AOs, we are using what is called the linear combination of atomic orbitals (LCAO)

approach. The challenge, in general, is to determine the “best” choices for c1

and c2 – for H2+ it looks like the best choice for the ground state will be

c1=c2. But how can we be sure this is really the best we can do? And what about if we want something other than the ground state? Or if we want to describe a more complicated molecule like HeH+2?




THE VARIATIONAL PRINCIPLE

In order to make further progress, we will use the Variational Principle to predict a better estimate of the ground state energy. This method is very general and its use in physical chemistry is widespread. Assume you

have a Hamiltonian (such as the Helium atom) but you don’t know the ground state energy E0 and or ground state eigenfunction φ0.

Hφ = E φ ⇒ H = ∫φ 0*Hφ

0 dτ = ∫φ

0*E

0 φ

0 dτ = E

00 0 0

Now, assume we have a guess, ψ , at the ground state wavefunction, which we

will call the trial wavefunction. Compute the average value of the energy for the trial wavefunction:

∫ψ *Hψ dτ * Êavg = = ∫ψ Hψ dτ (if ψ normalized)

∫ * dτψ ψ

the Variational Theorem tells us that Eavg

≥ E0 for any choice of the trial

function ψ! This make physical sense, because the ground state energy is, by definition, the lowest possible energy, so it would be nonsense for the average energy to be lower.

SIDEBAR: PROOF OF VARIATIONAL THEOREM Expand ψ (assumed normalized) as linear combination of the unknown

eigenstates, φ n , of the Hamiltonian:

ψ = ∑an φ

n

n

Note that in practice you will not know these eigenstates. The important

point is that no matter what function you choose you can always expand it in

terms of the infinite set of orthonormal eigenstates of H .

∫ * dτ = ∑an am ∫φn * md = ∑an amδmn = ∑ n

2ψ ψ * φ τ * a = 1

, , nn m n m

Eavg = ∫ψ *Hψ dτ = ∑an * am ∫φn *H φmdτ = ∑an

* am ∫φn *Emφmdτ n m , ,n m

= ∑an * amEmδmn = ∑ an

2 En

, nn m

Now, subtracting the ground state energy from the average




E = 1• E = ∑ an

2 E

0 0 0 n

⇒ Eavg − E0

= ∑ an

2 En − ∑ an

2 E

0 = ∑ an

2

(E − E ) ≥ 0 since E ≥ En 0 n 0 n n n

Where, in the last line we have noted that the sum of terms that are non

negative must also be nonnegative. It is important to note that the equals sign is only obtained if an=0 for all states that have En>E0. In this situation, ψ actually is the ground state of the system (or at least one

ground state, if the ground state is degenerate).

The variational method does two things for us. First, it gives us a means of

comparing two different wavefunctions and telling which one is closer to the ground state – the one that has a lower average energy is the better approximation. Second, if we include parameters in our wavefunction

variation gives us a means of optimizing the parameters in the following way. Assume that ψ depends on some parameters c – such as is the case for our

LCAO wavefunction above. We’ll put the parameters in brackets ψ[c] – in order to differentiate them from things like positions that are inside of

parenthesis ψ(r).Then the average energy will depend on these parameters:

ψ c * Hψ c dτ

Eavg ( ) c =

∫ψ [ ] [ ] c dτ

∫ [ ] [ ]

c *ψ

Note that, using the variational principle, the best parameters will be the ones that minimize the energy. Thus, the best parameters will satisfy

* ψ c Hψ c dτ

= = 0 ∂E

ave ( ) c ∂ ∫ [ ] ˆ [ ] *

i ψ c τ∂ci

∂c ∫ [ ] c ψ [ ] d

Thus, we can solve for the optimal parameters without knowing anything

about the exact eigenstates!

Let us apply this in the particular case of our LCAOMO treatment of H2+.

Our trial wavefunction is: ψ [c] = c 1s + c 1s

el 1 A 2 B




where c=(c1 c2). We want to determine the best values of c1 and c2 and the variational theorem tells us we need to minimize the average energy

to find the right values. First, we compute the average energy. The numerator gives:

∫ψ el

* H el ψ

el dτ = ∫ (c

11s

A + c

21s

B ) *

H (c11s

A + c

21s

B ) dτ

= c * c 1s H 1s d τ + c

* c 1s H 1s d τ + c

* c 1s H 1s d τ + c

* c 1s H 1s d τ

1 1 ∫ A el A 1 2 ∫ A el B 2 1 ∫ B el A 2 2 ∫ B el B

≡H11 ≡H12 ≡H21 ≡H22

* * * * ≡ c H c + c H c + c H c + c H c1 11 1 1 12 2 2 21 1 2 22 2

while the normalization integral gives: *

∫ψ el

*ψel

dτ = ∫ (c11s

A + c

21s

B ) (c11s

A + c

21s

B ) dτ

= c c 1s 1s d τ + c c 1s 1s d τ + c c 1s 1s d τ + c c 1s 1s d τ1

*

1 ∫ A A 1

*

2 ∫ A B 2

*

1 ∫ B A 2

*

2 ∫ B B

≡S11 ≡S12 ≡S21 ≡S22

* * * * ≡ c S c + c S c + c S c + c S c1 11 1 1 12 2 2 21 1 2 22 2

So that the average energy takes the form: * * * *

c H c + c H c + c H c + c H c E

avg ( ) c = 1

*

11 1 1

*

12 2 2

*

21 1 2

*

22 2

c S c + c S c + c S c + c S c1 11 1 1 12 2 2 21 1 2 22 2

We note that there are some simplifications we could have made to this

formula: for example, since our 1s functions are normalized S11=S22=1. By not making these simplifications, our final expressions will be a little more general and that will help us use them in more situations.

Now, we want to minimize this average energy with respect to c1 and c2.

Taking the derivative with respect to c1 and setting it equal to zero [Note: when dealing with complex numbers and taking derivatives one must treat variables and their complex conjugates as independent variables. Thus d/dc1

* has no effect on c1 ]: ∂E * *

c H + c Havg 1 11 2 21 = 0 = * * * * ∂c c S c + c S c + c S c + c S c

1 1 11 1 1 12 2 2 21 1 2 22 2

* * * *

− c

1 H

11 c

1 + c

1 H

12 c

2 + c

2 H

21 c

1 + c

2 H

22 c

2 (c * S + c

* S )2 1 11 2 21

c S c + c S c + c S c + c S c( 1

*

11 1 1

*

12 2 2

*

21 1 2

*

22 2 )




* * * *

⇒ 0 = (c * H + c

* H ) −

c1

H11

c1

+ c1

H12

c2

+ c2

H21

c1

+ c2

H22

c2 (c

* S + c

* S )1 11 2 21 * * * * 1 11 2 21

c S c + c S c + c S c + c S c1 11 1 1 12 2 2 21 1 2 22 2

⇒ 0 = (c1

* H

11 + c

2

* H

21 ) − Eavg (c1

* S

11 + c

2

* S

21 ) Applying the same procedure to c2:

∂E c * H + c

* Havg

= 0 = 1 12 2 22

* * * *∂c c S c + c S c + c S c + c S c2 1 11 1 1 12 2 2 21 1 2 22 2

* * * *

− c

1 H

11 c

1 + c

1 H

12 c

2 + c

2 H

21 c

1 + c

2 H

22 c

2

2 (c1

* S

12 + c

2

* S

22 )(c * S c + c * S c + c * S c + c * S c )1 11 1 1 12 2 2 21 1 2 22 2

* * * *

⇒ 0 = (c * H + c

* H ) −

c1

H11

c1

+ c1

H12

c2

+ c2

H21

c1

+ c2

H22

c2 (c

* S + c

* S )1 12 2 22 * * * * 1 12 2 22

c S c + c S c + c S c + c S c1 11 1 1 12 2 2 21 1 2 22 2

⇒ 0 = (c1

* H

12 + c

2

* H

22 ) − Eavg (c1

* S

12 + c

2

* S

22 ) We notice that the expressions above look strikingly like matrixvector operations. We can make this explicit if we define the Hamiltonian matrix:

⎛ H11 H12 ⎞ ⎜⎛ ∫1sAH

el 1sAdτ ∫1sAH el 1sBdτ

⎟⎞

H ≡ ⎜ ⎟ ≡ ⎝ H21 H22 ⎠ ⎜⎜

⎝ ∫1sBH el 1sAdτ ∫1sBH

el 1sBdτ ⎟⎟⎠

and the Overlap matrix:

⎛ S11 S12 ⎞ ⎜⎛ ∫1sA1sAdτ ∫1sA1sBdτ

⎟⎞

S ≡ ≡⎜ S S

⎟ ⎜ ⎟⎝ 21 22 ⎠ ⎜⎝ ∫1s 1s dτ ∫1s 1s dτ ⎟

⎠B A B B

Then the best values of c1 and c2 satisfy the matrix eigenvalue equation:

(c1

* c2

* ) ⎜⎛

H

H11

H

H12

⎟⎞

= Eavg (c1

* c2

* ) ⎜⎛

S

S11

S

S12

⎟⎞

⎝ 21 22 ⎠ ⎝ 21 22 ⎠

Which means that:

∂E avg † † = 0 ⇔ c iH = E c iS Eq. 1 ∂c

avg

This equation doesn’t look so familiar yet, so we need to massage it a bit. * First, it turns out that if we had taken the derivatives with respect to c1

* and c2 instead of c1 and c2, we would have gotten a slightly different equation:

⎛ H11 H

12 ⎞ ⎛ c1 ⎞ ⎛ S11 S

12 ⎞ ⎛ c1 ⎞ ⎜ ⎟ ⎜ ⎟ = Eavg ⎜ ⎟ ⎜ ⎟ ⎝ H21

H22 ⎠ ⎝ c2 ⎠ ⎝ S21

S22 ⎠ ⎝ c2 ⎠

or




∂E avg 0 ⇔ H c = E S c Eq. 2= i i

∂c * avg

* * Taking the derivatives with respect to c1 and c2 is mathematically equivalent to taking the derivatives with respect to c1 and c2 [because we

can’t change c1 without changing its complex conjugate, and vice versa]. Thus, the two matrix equations (Eqs. 1&2) above are precisely equivalent, and the second version is a little more familiar. We can make it even more

familiar if we think about the limit where 1sA and 1sB are orthogonal (e.g. when the atoms are very, very far apart). Then we would have for the Overlap matrix:

⎛ ∫1sA1sAdτ ∫1sA1sBdτ ⎞ ⎛ 1 0⎞⎜ ⎟S ≡ = ⎜ ⎟ = 1 ⎜ 1s 1s dτ 1s 1s dτ ⎟ ⎝ ⎠⎜∫ B A ∫ B B

⎟ 0 1 ⎝ ⎠

Thus, in an orthonormal basis the overlap matrix is just the identity matrix and we can write Eq. 2 as:

∂E avg 0 ⇔ H c = E= i c

∂c * avg

Now this equation is in a form where we certainly recognize it: this is an

eigenvalue equation. Because of its close relationship with the standard eigenvalue equation, Eq. 2 is usually called a Generalized Eigenvalue Equation.

In any case, we see the quite powerful result that the Variational theorem

allows us to turn operator algebra into matrix algebra. Looking for the lowest energy LCAO state is equivalent to looking for the lowest eigenvalue of the Hamiltonian matrix H. Further, looking for the best c1 and c2 is

equivalent to finding the lowest eigenvector of H.

Let’s go ahead and apply what we’ve learned to obtain the MO coefficients c1

and c2 for H2+. At this point we make use of several simplifications. The

offdiagonal matrix elements of H are identical because the Hamiltonian is Hermitian and the orbitals are real:

* ˆ * ˆ ˆ∫1sAHel 1sBdτ = (∫1sB Hel 1sAdτ ) = ∫1sBHel 1sAdτ ≡ V

12

Meanwhile, the diagonal elements are also equal, but for a different reason. The diagonal elements are the average energies of the states 1sA and 1sB. If

these energies were different, it would imply that having the electron on one




side of H2+ was more energetically favorable than having it on the other,

which would be very puzzling. So we conclude

∫1s H 1s dτ = ∫1s H 1s dτ ≡ εA el A B el B

Finally, we remember that 1sA and 1sB are normalized, so that

∫1sA1sAdτ = ∫1sB1sBdτ = 1

and because the 1s orbitals are real, the offdiagonal elements of S are also

the same:

S12 = ∫1sA1sBdτ = ∫1sB1sAdτ = S21 .

Incorporating all these simplifications gives us the generalized eigenvalue equation:

⎛ ε V12 ⎞ ⎛ c1 ⎞ ⎛ 1 S

12 ⎞ ⎛ c1 ⎞ ⎜ ⎟ ⎜ ⎟ = Eavg ⎜ ⎟ ⎜ ⎟ ⎝V12

ε ⎠ ⎝ c2 ⎠ ⎝ S21 1 ⎠ ⎝ c2 ⎠

All your favorite mathematical programs (Matlab, Mathematica, Maple, MathCad…) are capable of solving for the generalized eigenvalues and

eigenvectors, and for more complicated cases we suggest you use them. However, this case is simple enough that we can solve it by guessand test. Based on our physical intuition above, we guess that the correct eigenvector

will have c1=c2. Plugging this in, we find:

⎛ ε V12 ⎞ ⎛ c1 ⎞ ⎛ 1 S

12 ⎞ ⎛ c1 ⎞ ⎜ ⎟ ⎜ ⎟ = Eavg ⎜ ⎟ ⎜ ⎟ ⎝V12

ε ⎠ ⎝ c1 ⎠ ⎝ S21 1 ⎠ ⎝ c1 ⎠

⎛ (ε + V12 ) c1 ⎞ ⎛ (1 + S ) c1

⎞12 ⇒

⎝⎜

(ε + V12 ) c1 ⎠

⎟ = Eavg

⎝⎜

(1 + S12 ) c1 ⎠

⎟

ε + V ⇒ Eavg = 12 ≡ Eσ

1 + S12

Thus, our guess is correct and one of the eigenvectors of this matrix has c1=c2. This eigenvector is the σbonding state of H2

+, and we can write down

the associated orbital as: ψ σ = c 1s + c 1s = c 1s + c 1s ∝ 1s + 1s

el 1 A 2 B 1 A 1 B A B

where in the last expression we have noted that c1 is just a normalization

constant. In freshman chemistry, we taught you that the σbonding orbital existed, and this is where it comes from.

We can also get the σ∗ antibonding orbital from the variational procedure. Since the matrix is a 2x2 it has two unique eigenvalues: the lowest one

(which we just found above) is bonding and the other is antibonding. We can



5.61 Physical Chemistry Lecture #28 10

again guess the form of the antibonding eigenvector, since we know it has the characteristic shape +/, so that we guess the solution is c1=c2:

⎛ ε V12 ⎞ ⎛ c

1 ⎞ ⎛ 1 S12 ⎞ ⎛ c

1 ⎞ ⎜V ε ⎟ ⎜ −c

⎟ = Eavg ⎜ S 1

⎟ ⎜ −c

⎟ ⎝ 12 ⎠ ⎝ 1 ⎠ ⎝ 21 ⎠ ⎝ 1 ⎠

⎛ (ε − V ) c ⎞ ⎛ (1 − S ) c ⎞12 1 12 1⇒ ⎝⎜

− (ε − V12 ) c1 ⎠

⎟ = Eavg

⎝⎜

− (1 − S12 ) c1 ⎠

⎟

ε − V ⇒ Eavg = 12 = Eσ *

1 − S12

so, indeed the other eigenvector

has c1=c2. The corresponding antibonding orbital is given by: ψσ(r)

*ψ σ = c 1s + c 1s = c 1s − c 1s ∝ 1s − 1sel 1 A 2 B 1 A 1 B A B

where we note again that c1 is

just a normalization constant.Given these forms for thebonding and antibonding orbitals, ψσ(r)we can draw a simple picture forthe H2

+ MOs (see right).

We can incorporate the energies obtained above into a simple MO diagram ofH2

+:ε − V

Eσ∗ = 12

1 − S

Eσ = 12

12 1 S

ε +

+

V

12

E1sA =ε E1sB =ε

On the left and right, we draw the energies of the atomic orbitals (1sA and1sB) that make up our molecular orbitals (σ and σ*) in the center. We note




that when the atoms come together the energy of the bonding and antibonding orbitals are shifted by different amounts:

ε − V12

ε − V12

ε (1 − S12

) εS12

− V12 Eσ * − E1s = − ε = − =

1 − S 1 − S 1 − S 1 − S12 12 12 12

E1s − Eσ *

= ε −ε + V12 =

ε (1 + S12 ) −ε + V12 =

ε S12 − V12

1 + S 1 + S 1 + S 1 + S12 12 12 12

Now, S12 is the overlap between two 1s orbitals. Since these orbitals are never negative, S12 must be a positive number. Thus, the first denominator

is greater than the second, from which we conclude εS − V εS − V

Eσ* − E1s = 12 12 > 12 12 = E1s − Eσ* 1 − S 1 + S12 12

Thus, the antibonding orbital is destabilized more than the bonding orbital is stabilized. This conclusion need not hold for all diatomic molecules, but it is a good rule of thumb. This effect is called overlap repulsion. Note that in

the special case where the overlap between the orbitals is negligible, S12

goes to zero and the two orbitals are shifted by equal amounts. However, when is S12 nonzero there are two effects that shift the energies: the

physical interaction between the atoms and the fact that the 1sA and 1sB orbitals are not orthogonal. When we diagonalize H, we account for both these effects, and the orthogonality constraint pushes the orbitals upwards

in energy.




MOLECULAR ORBITAL THEORY PART II

For the simple case of the oneelectron bond in H2+ we have seen that using

the LCAO principle together with the variational principle led to a recipe for computing some approximate orbitals for a system that would be very

difficult to solve analytically. To generalize this to the more interesting case of many electrons, we take our direction from our experience with the independent particle model (IPM) applied to atoms and we build up

antisymmetrized wavefunctions out of the molecular orbitals. This is the basic idea behind molecular orbital theory – there are many variations on the central theme, but the same steps are always applied. Rather than go step

bystep and deal with H2 and then Li2 and then LiH … we will instead begin by stating the general rules for applying MO theory to any system and then proceed to show some illustrations of how this works out in practice.

1) Define a basis of atomic orbitals For H2

+ our atomic orbital basis was simple: we used the 1s functions

from both hydrogen atoms and wrote our molecular orbitals as linear combinations of our basis functions:

ψ = c 1s + c 1s1 A 2 B

Note that the AO basis determines the dimension of our MO vector and also determines the quality of our result – if we had chosen the 3p orbitals instead of the 1s orbitals, our results for H2

+ would have been

very wrong!

For more complicated systems, we will require a more extensive AO basis.

For example, in O2 we might want to include all the 2s and 2p orbitals on both oxygens, in which case our MOs would take the form

ψ = c12s

A + c22 p

xA + c32 p

yA + c42 p

zA + c52s

B + c62 p

xB + c72 p

yB + c82 p

zB

Meanwhile, for methane we might want to include the 1s functions on all four hydrogens and the 2s and 2p functions on carbon:

ψ = c11s

1 + c

21s

2 + c

31s

3 + c

41s

4 + c

52s + c

62 p

x + c72 p

y + c82 p

z

In the general case, we will write: N

AO ψ = ∑ciφi i=1

and represent our MOs by column vectors:




⎛ c1 ⎞ ⎜ ⎟

ψ�

= ⎜ c

2 ⎟ ⎜ ... ⎟ ⎜ ⎟ ⎝ cN ⎠

We note that for the sake of accuracy it is never a bad idea to include

more AO functions than you might think necessary – more AO functions will always lead to more accurate results. The price is that the more accurate computations also tend to be more complicated and time

consuming. To illustrate, note that we could have chosen to write the H2+

MOs as linear combinations of four functions – the 1s and 2s states on each atom:

ψ = c 1s + c 1s + c 2s + c 2s1 A 2 B 3 A 4 B

Now, when we use the variational principle to get the coefficients of the lowest MO , c0, we are guaranteed that there is no set of coefficients that will give us a lower energy. This is the foundation of the variational

method. Note that one possible set of coefficients is c3=c4=0, in which case our 4function expansion reduces to the 2function expansion above.

Thus, the variationally optimal 4function MO will always have an energy less than or equal to the optimal 2function MO. As a result, the expansion with four functions allows the approximate MO to get closer to

the ground state energy. This makes sense, as the four AO expansion has more flexibility than the constrained two AO expansion used previously. The reason we didn’t use the four function expansion from

the beginning is that all the algebra is twice as difficult when we use four functions as two: the vectors are twice as long, the matrices are twice as big…. At least for a first try, it is generally good to start with the

smallest conceivable set of AOs for performing a calculation. If higher accuracy is required, a longer expansion can be tried. 2) Compute the relevant matrix representations

For H2+ we had to compute two matrices – the Hamiltonian and the

overlap, which were both 2by2 by virtue of the two AO basis functions:

⎛ H11 H12 ⎞ ⎛⎜ ∫1sAH

el 1sAdτ ∫1sAH el 1sBdτ ⎞

⎟H ≡ ⎜ ⎟ ≡ ⎝ H21 H22 ⎠ ⎜⎜

⎝ ∫1sBH el 1sAdτ ∫1sBH

el 1sBdτ ⎟⎟⎠

⎛ S11 S12 ⎞ ⎜⎛ ∫1s 1s dτ ∫1s 1s dτ

⎟⎞

A A A B S ≡ ≡⎜

S S ⎟ ⎜ ⎟⎝ 21 22 ⎠ ⎜⎝ ∫1sB1sAdτ ∫1sB1sBdτ ⎟

⎠




In the general case, the Hamiltonian and overlap become NbyNmatrices of the form:

⎛ ∫φ1

AO H φ1

AO ∫φ1

AO H φ2

AO ∫φ1

AO H φNAO ⎞

⎛ H11 H12 ... H1N ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ AO ˆ AO AO ˆ AO AO ˆ AO ⎟ ⎜ H21 H22 ... H2N ⎟ ⎜ ∫φ2 Hφ1 ∫φ2 Hφ2 ∫φ2 HφN ⎟H ≡ ≡⎜ ... ... ... ... ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ HN1 HN 2 ... H NN ⎠ ⎜

⎝ ∫φNAO

H φ1

AO ∫φN

AO H φ2

AO ∫φN

AO H φN

AO ⎟⎠

⎛ ∫φ1

AO φ1

AO ∫φ1

AO φ2

AO ∫φ1

AO φNAO ⎞

⎛ S11 S12 ... S1N ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ AO AO AO AO AO AO ⎟ ⎜ S21 S22 ... S2N ⎟ ⎜ ∫φ2 φ1 ∫φ2 φ2 ∫φ2 φN ⎟S ≡ ≡⎜ ... ... ... ... ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ SN1 SN 2 ... SNN ⎠ ⎜

⎝ ∫φNAO φ1

AO ∫φN

AO φ2

AO ∫φN

AO φNAO ⎟⎠

This step is where much of the hard work is done in most MO

calculations. Not only can the integrals between the different AO functions be very tricky to work out, there are a lot of them to be computed – N2 of them, to be exact! This hard work is best done in an

automated fashion by a computer, and in practice we will usually give you explicit values for the matrix elements for this step. However, it is important for you to realize what the matrix elements mean. The

diagonal elements of H represent the average energies of putting electrons in each AO and the offdiagonal terms tell us how strongly coupled one AO is to another. The diagonal elements of S are

normalization integrals and the offdiagonal terms tell us how much spatial overlap there is between the different AOs

3) Solve the generalized eigenvalue problem For every MO problem, the central step is determining the MOs, which always involves solving the generalized eigenvalue problem:

α αH c i = Eα S c i

The eigenvalues from this equation are the MO energies. The

eigenvectors are the coefficients of the molecular orbitals, written as sums of AOs:

N

ψ α ( ) = ∑ci αφi

AO ( ) rr i=1

In general, we will obtain N molecular orbitals out of N atomic orbitals. This step is precisely the same as what we did for H2

+, just generalized to the Norbital case. We note that for anything larger than a 2by2, it




is usually best to ask a computer to solve the generalized eigenvalue problem for you. 4) Occupy the orbitals according to a stick diagram

At this point, we must depart from the H2+ model and begin to account

for the fact that we have multiple electrons. To do so, we follow the prescription of the independent particle model and build a Slater

determinant out of our orbitals. However, whereas for atoms we built the determinant out of atomic orbitals, for molecules we will build the determinant out of molecular orbitals:

( ) 1 1ψ ↑ ( ) 1

1ψ ↓ ... ( ) 1

Nψ ↓

Ψ ≡ ( ) 1 2ψ ↑ ( ) 1

2ψ ↓ ... ( ) 2

Nψ ↓

... ... ... ...

( )1 Nψ ↑ ( )1

Nψ ↓ ... ( )N

Nψ ↓

As was the case for atoms, it is much easier to reason in terms of stick diagrams, rather than write out all of the orbitals in determinant form.

So, for example, we would associate a stick diagram like this

ψ2

ψ1

with a determinant:

ψ 1 ψ 1 ψ 1 ψ 1

1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( )

1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( )

ψ 2 ψ 2 ψ 2 ψ 2 Ψ (1, 2,3, 4 ) ≡

1↑ ψ 1↓ 3

2↑ ( ) ψ 2↓ 3ψ ( ) 3 ( ) ψ 3 ( )

1↑ ( ) 1↓ ( ) 2↑ ( ) 2↓ ( ) ψ 4 ψ 4 ψ 4 ψ 4

But all the information we would need is contained in the stick diagram and, of course, the MOs.

5) Compute the energy There are a variety of ways to compute the energy once the MOs have

been obtained. The simplest is to use the noninteracting particle picture we used for atoms. Here, the energy of N electrons is just given by the sum of the energies of the N orbitals that are occupied:



� * *


N

E = ∑Ei

i=1

A more accurate way is to use the independent particle model to add an average electronelectron repulsion to the energy:

N N E = H = ∑Ei +∑J� ij − K� ij

i=1 i< j

Where now the Coulomb and exchange integrals use molecular orbitals rather than atomic orbitals:

Jij ≡ ∫∫ ψ i (1)ψ j (2) r −

1

r ψ i (1)ψ j (2) dr

1dr

2dσ

1dσ

2

1 2

K� ij ≡ ∫∫ ψ i * (1)ψ *

j (2) r −

1

r ψ i (2)ψ j (1) dr

1dr

2dσ

1dσ

2

1 2

In fact, as we will see later on, there are even more elaborate ways to obtain the energy from an MO calculation. When we work things out

by hand, the noninteracting picture is easiest and we will usually work in that approximation when dealing with MOs.

Diatomic molecules As a first application of MO theory, it is useful to consider firstrow

diatomic molecules (B2, C2, N2,O2, CO,CN, NO, etc.), which actually map rather nicely on to an MO picture. We’ll go stepby step for the generic “AB” diatomic to show how this fits into the MO theory framework.

1) Define a basis of atomic orbitals. To begin with, one would consider a set consisting of 10 atomic orbitals – 5 on A and 5 on B:

ψ = c11sA + c

21sB + c

32sA + c

42sB + c

52 pzA + c

62 pzB + c

72 pyA + c

82 pyB + c

92 pxA + c

10 2 pxB

However, for all the diatomics above, the 1s orbitals on both atoms will

be doubly occupied. Since we will primarily be interested in comparing the MO descriptions of different diatomics the eternally occupied 1s orbital will have no qualitative effect on our comparisons. It is therefore

customary to remove the 1s orbitals from the expansion: ψ = c

12sA + c

22sB + c

32 pzA + c

42 pzB + c

52 pyA + c

62 pyB + c

72 pxA + c

82 pxB

The latter approximation is referred to as the valence electron or

frozen core approximation. The advantage is that it reduces the length of our vectors from 10 to 8.




2) Compute the Matrix Representations. Here, we have the rather daunting task of computing two 8by8 matrices. As mentioned above, we won’t be concerned in this class about filling in precise values for matrix

elements here. However, we will be very interested in obtaining the proper shape of the matrix by determining which matrix elements are zero and which are not. The Hamiltonian takes the shape:

s Hs s Hs s Hp s Hp s Hp s Hp s Hp s Hp ⎛ ∫ A ˆ

A ∫ A ˆ

B ∫ A ˆ

zA ∫ A ˆ

zB ∫ A ˆ

yA ∫ A ˆ

yB ∫ A ˆ

xA ∫ A ˆ

xB ⎞

⎜ ⎟ ⎜ ˆ s Hs ˆ s Hp s Hp ˆ s Hp s ˆ s Hp ˆ s ˆ ⎟ ⎜ ∫ sB Hs A ∫ B B ∫ B

ˆ zA ∫ B zB ∫ B

ˆ yA ∫ B Hp yB ∫ B xA ∫ B Hp xB ⎟

⎜ ⎟ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ⎜ ∫ pzA Hs A ∫ pzA Hs B ∫ pzA Hp zA ∫ pzA Hp zB ∫ pzA Hp yA ∫ pzA Hp yB ∫ pzA Hp xA ∫ pzA Hp xB ⎟

⎜ ⎟ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ⎜ ∫ pzB Hs A ∫ pzB Hs B ∫ pzB Hp zA ∫ pzB Hp zB ∫ pzB Hp yA ∫ pzB Hp yB ∫ pzB Hp xA ∫ pzB Hp xB ⎟

H ≡ ⎜ ⎟ ˆ⎜ pyA Hs ˆ

A pyA ˆ

B pyA Hp zA pyA ˆ

zB pyA Hp yA pyA ˆ

yB pyA Hp xA pyA ˆ

xB ⎜ ∫ ∫ Hs ∫ ˆ ∫ Hp ∫ ˆ ∫ Hp ∫ ∫ Hp ⎟⎟

⎜ ⎟ p Hs p Hs p Hp p Hp p Hp p Hp p Hp p Hp ⎜ ∫ yB

ˆ A ∫ yB

ˆ B ∫ yB

ˆ zA ∫ yB

ˆ zB ∫ yB

ˆ yA ∫ yB

ˆ yB ∫ yB

ˆ xA ∫ yB

ˆ xB ⎟

⎜ ⎟ Hs ˆ Hp ˆ Hp ˆ Hp ⎜ ∫ pxA Hs ˆ

A ∫ pxA ˆ

B ∫ pxA Hp zA ∫ pxA ˆ

zB ∫ pxA Hp yA ∫ pxA ˆ

yB ∫ pxA Hp xA ∫ pxA ˆ

xB ⎟ ⎜ ⎟ ⎜ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ⎟⎝ ∫ pxB Hs A ∫ pxB Hs B ∫ pxB Hp zA ∫ pxB Hp zB ∫ pxB Hp yA ∫ pxB Hp yB ∫ pxB Hp xA ∫ pxB Hp xB ⎠

We assume, for simplicity, that the ABbondlies along the zaxis. Then it is relatively easy

A B

yto see that the molecule is symmetric upon

reflection around the x and y axes. As aresult, the Hamiltonian for AB is also z

symmetric (even) with respect to reflection

about x and y. Similarly, the s and p orbitals x

all have definite reflection symmetries:

Hamiltonian s

pz

py

px

X Reflection Y Reflection + + + +

+ + +

+

Further, we note that if we perform an integral, if the integrand is odd

with respect to reflection about either x or y the integrand will be zero.




This knocks out a bunch of integrals for us. For example, looking atreflection about x:

∫ / Hp

/ = (+)( + − = − ⇒ 0s

A B ˆ

xA B )( )

∫ /ˆ

/ = (+)( +)( −) = − ⇒ 0p

zA B Hp xA B

pyA B /

Hp /

= (+)( +)( −) = − ⇒ 0∫ ˆ xA B

∫ / Hp

/ = ( )( +)( −) = + ⇒ ≠ 0p

xA B ˆ

xA B −

We have analogous expressions for reflection about y zero several more

integrals. The result is that the Hamiltonian simplifies to

s Hs s Hs s Hp s Hp 0 0 0 0⎛ ∫ A ˆ

A ∫ A ˆ

B ∫ A ˆ

zA ∫ A ˆ

zB ⎞

⎜ ⎟ ⎜ ˆ Hs ˆ s Hp s ˆ 0 0 0

⎟ ⎜ ∫ sB Hs A ∫ sB B ∫ B

ˆ zA ∫ B Hp zB 0 ⎟

⎜ ⎟ˆ ˆ ˆ ˆ⎜ ∫ pzA Hs A ∫ pzA Hs B ∫ pzA Hp zA ∫ pzA Hp zB 0 0 0 0 ⎟

⎜ ⎟ ⎜ ∫ pzB Hs ˆ

A ∫ pzB ˆ

B ∫ pzB Hp zA ∫ pzB ˆ

zB 0 0 0 0Hs ˆ Hp ⎟ H ≡ ⎜ ⎟

0 0 0 0 p Hp p Hp 0 0⎜ ∫ yA ˆ

yA ∫ yA ˆ

yB ⎟

⎜ ⎟ ⎜ ⎟ ⎜ 0 0 0 ∫ pyB Hp ˆ

yA ∫ pyB ˆ

yB 0 0 ⎟0 Hp

⎜ ⎟ˆ ˆ⎜ 0 0 0 0 0 0 ∫ pxA Hp xA ∫ pxA Hp xB ⎟

⎜ ⎟ ⎜ 0 0 0 0 0 0 ∫ pxB Hp ˆ

xA ∫ pxB ˆ

xB ⎟⎝ Hp

⎠

which we write: ⎛ H11 H12 H13 H14 0 0 0 0 ⎞ ⎜ ⎟ ⎜ H21 H22 H23 H24 0 0 0 0 ⎟ ⎜ H31 H32 H33 H34 0 0 0 0 ⎟ ⎜ ⎟

H ≡ ⎜ H41 H42 H43 H44 0 0 0 0 ⎟ ⎜ 0 0 0 0 H H 0 0 ⎟ ⎜

55 56 ⎟

⎜ 0 0 0 0 H65 H66 0 0 ⎟ ⎜ 0 0 0 0 0 0 H77 H78

⎟ ⎜ ⎟⎜ ⎟⎝ 0 0 0 0 0 0 H87 H88 ⎠

It is easy to show that the overlap matrix has the same overall shape




⎛ S11 S12 S13 S14 0 0 0 0 ⎞ ⎜ ⎟ ⎜ S21 S22 S23 S24 0 0 0 0 ⎟ ⎜ S31 S32 S33 S34 0 0 0 0 ⎟ ⎜ ⎟ ⎜ S41 S42 S43 S44 0 0 0 0 ⎟S ≡ ⎜ 0 0 0 0 S S 0 0 ⎟ ⎜

55 56 ⎟

⎜ 0 0 0 0 S65 S66 0 0 ⎟ ⎜ 0 0 0 0 0 0 S77 S78

⎟ ⎜ ⎟⎜ 0 0 0 0 0 0 S S ⎟⎝ 87 88 ⎠

There are some additional symmetries in these matrices but the reflection symmetry properties are the most important.

3) Solve the generalized eigenvalue problem. This part would be impossible if we hadn’t simplified our matrices above. However, with the simplifications, it is clear that our matrices are block diagonal. For

example:

sA sB pzA pzB pyA pyB pxA pxB

⎛ H11 H12 H13 H14 0 0 0 0 ⎞ sA

⎜ ⎟ ⎜ H21 H22 H23 H24 0 0 0 0 ⎟

sB

⎜ H31 H32 H33 H34 0 0 0 0 ⎟ pzA

⎜ ⎟ ⎜ H41 H42 H43 H44 0 0 0 0 ⎟ pzB

H ≡ ⎜ 0 0 0 0 H55 H56 0 0 ⎟ pyA

⎜ ⎟ ⎜ 0 0 0 0 H65 H66 0 0 ⎟ pyB

77 78

87 88

⎟ ⎟

H ⎟⎠

H

H H

pxA ⎜ 0 0 0 0 0 0 ⎜⎜ pxB ⎝ 0 0 0 0 0 0

And similarly for the overlap matrix. The nice thing about block diagonal matrices is you can reduce a large eigenvalue to several smaller ones. In

this case, our matrices break down into a 4by4 block (sA, sB, pzA, pzB) a 2by2 block (pyA,pyB) and another 2by2 block (pxA,pxB). All the rest of the matrix is zero. As a result, we can decompose the above 8by8 into

three separate eigenvalue problems: A) The first eigenvalue problem to be solved is a 4by4:




⎛ H H H H ⎞ ⎛ c1

α ⎞ ⎛ S S S S ⎞ ⎛ c1

α ⎞ ⎜

11 12 13 14

⎟ ⎜ α ⎟ ⎜

11 12 13 14

⎟ ⎜ α ⎟

⎜ H21 H22 H23 H24 ⎟ ⎜ c2 ⎟ = Eα ⎜ S21 S22 S23 S24 ⎟ ⎜ c2 ⎟

⎜ H31 H32 H33 H34 ⎟ ⎜⎜

c3

α ⎟⎟ ⎜ S31 S32 S33 S34

⎟ ⎜⎜

c3

α ⎟⎟

⎜ ⎟ ⎜ ⎟ ⎝ H41 H42 H43 H44 ⎠ ⎜⎝ c4

α ⎟⎠ ⎝ S41 S42 S43 S44 ⎠ ⎜⎝ c4

α ⎟⎠

which will give us four molecular orbitals that can be written as linear combinations of the first four AOs (sA, sB, pzA, pzB)

ψ α = c1

α 2s

A + c2

α 2s

B + c3

α 2 p

zA + c4

α 2 p

zB

Because these orbitals are symmetric with respect to reflection about both x and y, they will look something like the H2

+ bonding and antibonding orbitals, and so they are referred to as σorbitals. For

example, we can make the +/ combinations of the 2s orbitals to obtain one bonding orbital and one antibonding:

+

σ1orbital

σ1∗orbital

A B

A B

A B

we can make the similar linear combinations of the 2pz orbitals to obtain:

σ2∗orbital

A B

A B

A B

+

σ2orbital

where we label the upper orbital σ* because of the nodes between the

nuclei, whereas the σ orbital has no nodes between the nuclei. Note




that these +/ combinations are just to illustrate what the orbitals

will look like; in order to get the actual molecular orbitals we would need to diagonalize the 4by4 and get the eigenvectors. However, if we do that for a molecule like N2 we actually get orbitals that look

strikingly similar to the ones above:

σ2*

σ2

σ1*

σ1

B) The second eigenvalue problem to be solved is a 2by2:

⎛ H55 H

56 ⎞ ⎛ c5

α ⎞ α ⎛ S55

S56 ⎞ ⎛ c5

α ⎞ ⎜ ⎟ ⎜ ⎟ = E ⎜ ⎟ ⎜ ⎟ ⎝ H65

H66 ⎠ ⎜⎝ c6

α ⎟⎠ ⎝ S65

S66 ⎠ ⎜⎝ c6

α ⎟⎠

which will give us two molecular orbitals that can be written as linear

combinations of the next two AOs (pyA,pyB): ψ α = c

5

α 2 pyA + c

6

α 2 pyB

These orbitals get ““ signs upon reflection about y, so we designate them πy orbitals. We can again make the +/ combinations to get an idea what these orbitals look like:




πyorbital

A B

+

A B

A B

πy * orbital

Here again, the use of +/ combinations is only for illustration

purposes. Unless we have a homonuclear diatomic, the coefficients will not be ±1. However, for a molecule like N2, the orbitals look strikingly similar again:

πy

πy *

C) The last eigenvalue problem is also 2by2:

⎛ H77 H

78 ⎞ ⎛ c7

α ⎞ α ⎛ S77

S78 ⎞ ⎛ c7

α ⎞ ⎜ ⎟ ⎜⎜ α

⎟⎟

= E ⎜ ⎟ ⎜⎜ α ⎟⎟⎝ H87

H88 ⎠ ⎝ c8 ⎠ ⎝ S87

S88 ⎠ ⎝ c8 ⎠

which will give us two molecular orbitals that can be written as linear combinations of the last two AOs (pxA,pxB):

ψ α = c5

α 2 p

xA + c6

α 2 p

xB

These orbitals get ““ signs upon reflection about x, so we designate them πx orbitals. The qualitative picture of the πx orbitals is the same




as for the πy orbitals above, expect that the πx orbitals come out of

the page. 4) Occupy the orbitals based on a stick diagram. The most important

thing here is to know the energetic ordering of the orbitals. This would

come out of actually evaluating the nonzero matrix elements in matrices above and then solving the generalized eigenvalue problem, which is tedious to do by hand. As a general rule however, there are only two

commonly found MO diagrams for diatomics: σ2∗ σ2∗

πx*,πy

* πx* ,πy

*

σ2

σ1

σ1∗

σ2

πx,πyπx,πy

σ1

σ1∗ Versus

Hence, the only question is whether the second σbonding orbital is above or below the πbonding orbitals. In practice, the σorbital (which has significant pz character) is stabilized as you move from left to right along

the periodic table, with the σorbital being less stable for atoms to the left of and including nitrogen and more stable for atoms to the right of N. Once we have the orbital energy diagram in hand, we can assign the

electrons based on stick diagrams. For example, for CO we have 10 valence electrons and we predict a stick diagram for the ground state like the one at left below .Meanwhile for NO, which has 11 valence

electrons, we have the stick diagram shown on the right.




σ1

σ1∗

σ2

πx,πy

πx *,πy

*

σ2∗

σ1

σ1∗

σ2

πx,πy

πx *,πy

*

σ2∗

CO MO diagram NO MO diagram

We note one important feature we get directly out of the stick

diagrams: the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). For example, in CO the HOMO is a πbonding orbital, whereas the LUMO is a π* antibonding

orbital. Meanwhile, for NO the HOMO and LUMO are both π* orbitals. These orbitals determine reactivity in a crude fashion, as when electrons are taken out of the molecule, they are removed from

the HOMO, and when electrons are added, they are added to the LUMO.

5) Compute the energy. Here we can say very little about diatomics,

because we don’t even know the orbital energies exactly, making it difficult to predict the energies of the whole molecule. If we knew

the orbital energies, the total energy for CO, for example, would be: ECO=2Eσ1+2Eσ1∗+2Eπx+2Eπy+2Eσ2

As we don’t know these orbital energies, we cannot evaluate the

accuracy of this independent electron model for diatomics. However, the bond order is a useful descriptor that correlates very well with the MO energy. For a diatomic, the bond order is simply:

((# Bonding Electrons)(# Antibonding Electrons))/2 The factor of two reflects the requirement of two electrons for forming a bond. A higher bond order implies a stronger bond and a

lower bond order a weaker bond. Thus, MO theory predicts CO will have a stronger bond (bond order 3) than NO (bond order 2.5), which




is experimentally verifiable: the bond energy in NO is 6.5 eV, while

the bond energy in CO is 11.1 eV.

There are a number of other successful predictions of MO theory for

diatomics: O2 is correctly predicted to be a spin triplet ground state, CO is correctly predicted to be slightly more stable than N2, the highest occupied molecular orbital for C2 is predicted to be degenerate …. Overall, given its

basis on the independent particle model, MO theory predicts a surprisingly large array of chemical features correctly.




HÜCKEL MOLECULAR ORBITAL THEORY

In general, the vast majority polyatomic molecules can be thought of as consisting of a collection of twoelectron bonds between pairs of atoms. So the qualitative picture of σ and πbonding and antibonding orbitals that we

developed for a diatomic like CO can be carried over give a qualitative starting point for describing the C=O bond in acetone, for example. One place where this qualitative picture is extremely useful is in dealing with

conjugated systems – that is, molecules that contain a series of alternating double/single bonds in their Lewis structure like 1,3,5hexatriene:

Now, you may have been taught in previous courses that because there areother resonance structures you can draw for this molecule, such as:

that it is better to think of the molecule as having a series of bonds of order 1 ½ rather than 2/1/2/1/… MO theory actually predicts this

behavior, and this prediction is one of the great successes of MO theory as a descriptor of chemistry. In this lecture, we show how even a

very simple MO approximation describes conjugated systems.

Conjugated molecules of tend to be planar, so that we can place all the atoms

in the xy plane. Thus, the molecule will have reflection symmetry about the zaxis:

z

Now, for diatomics, we had reflection symmetry about x and y and this gave

rise to πx and πy orbitals that were odd with respect to reflection and σ

orbitals that were even. In the same way, for planar conjugated systems the orbitals will separate into σ orbitals that are even with respect to reflection




and πz orbitals that are odd with respect to reflection about z. These πz

orbitals will be linear combinations of the pz orbitals on each carbon atom:

z

In trying to understand the chemistry of these compounds, it makes sense to focus our attention on these πz orbitals and ignore the σ orbitals. The πz

orbitals turn out to be the highest occupied orbitals, with the σ orbitals

being more strongly bound. Thus, the forming and breaking of bonds – as implied by our resonance structures – will be easier if we talk about making and breaking π bonds rather than σ. Thus, at a basic level, we can ignore the

existence of the σorbitals and deal only with the πorbitals in a qualitative MO theory of conjugated systems. This is the basic approximation of

Hückel theory, which can be outlined in the standard 5 steps of MO theory:

1) Define a basis of atomic orbitals. Here, since we are only interested

in the πz orbitals, we will be able to write out MOs as linear combinations of the pz orbitals. If we assume there are N carbon atoms, each contributes a pz orbital and we can write the µ th MOs as:

N

π µ = ∑ci µ

pz

i

i=1

2) Compute the relevant matrix representations. Hückel makes some radical approximations at this step that make the algebra much

simpler without changing the qualitative answer. We have to compute two matrices, H and S which will involve integrals between pz orbitals on different carbon atoms:

H p H p d = p d τij = ∫ z

i ˆ z

j τ Sij ∫ pz

i z

j

The first approximation we make is that the pz orbitals are orthonormal. This means that:

⎧1 i = j S

ij = ⎨ ⎩0 i ≠ j




Equivalently, this means S is the identity matrix, which reduces our generalized eigenvalue problem to a normal eigenvalue problem

i α = EαS c i µ ⇒ H c µ = Eµcµ

H c i

The second approximation we make is to assume that any Hamiltonian

integrals vanish if they involve atoms i,j that are not nearest neighbors. This makes some sense, because when the pz orbitals are far apart they will have very little spatial overlap, leading to an

integrand that is nearly zero everywhere. We note also that the diagonal (i=j) terms must all be the same because they involve the average energy of an electron in a carbon pz orbital:

H = p H p dτ ≡ αii ∫ z

i ˆ z

i

Because it describes the energy of an electron on a single carbon, α is often called the onsite energy. Meanwhile, for any two nearest

neighbors, the matrix element will also be assumed to be constant:

H = p H p dτ ≡β i,j neigbors ij ∫ z

i z

j

This last approximation is good as long as the CC bond lengths in the molecule are all nearly equal. If there is significant bond length alternation (e.g. single/double/single…) then this approximation can be

relaxed to allow β to depend on the CC bond distance. As we will see, β allows us to describe the electron delocalization that comes from multiple resonance structures and hence it is often called a resonance

integral. There is some debate about what the “right” values for the α, β parameters are, but one good choice is α=11.2 eV and β=.7 eV.

3) Solve the generalized eigenvalue problem. Here, we almost always

need to use a computer. But because the matrices are so simple, we can usually find the eigenvalues and eigenvectors very quickly.

4) Occupy the orbitals according to a stick diagram. At this stage, we

note that from our N pz orbitals we will obtain N π orbitals. Further, each carbon atom has one free valence electron to contribute, for a total of N electrons that will need to be accounted for (assuming the

molecule is neutral). Accounting for spin, then, there will be N/2 occupied molecular orbitals and N/2 unoccupied ones. For the ground

state, we of course occupy the lowest energy orbitals. 5) Compute the energy. Being a very approximate form of MO theory,

Hückel uses the noninteracting electron energy expression:




N

Etot = ∑Ei i=1

where Ei are the MO eigenvalues determined in the third step.

To illustrate how we apply Hückel in practice, let’s work out the energy of benzene as an example.

1

2

35

6

4

1) Each of the MOs is a linear combination of 6 pz orbitals

⎛ cµ ⎞ ⎜

1

µ ⎟ ⎜ c2 ⎟

6 ⎜ cµ ⎟ ψ µ = ∑c

µ pz

i → cµ = ⎜ 3

µ ⎟i i=1 ⎜ c4 ⎟

⎜ cµ ⎟ ⎜ 5 ⎟⎜ µ ⎟⎝ c6 ⎠

2) It is relatively easy to work out the Hamiltonian. It is a 6by6 matrix. The first rule implies that every diagonal element is α:

⎛α ⎞ ⎜ ⎟ ⎜ α ⎟ ⎜ α ⎟

H = ⎜ ⎟ ⎜ α ⎟ ⎜ α ⎟ ⎜ ⎟⎜ α ⎟⎝ ⎠

The only other nonzero terms will be between neighbors: 12, 23, 34, 45, 56 and 61. All these elements are equal to β:

⎛α β β ⎞ ⎜ ⎟ ⎜ β α β ⎟ ⎜ β α β ⎟

H = ⎜ ⎟ ⎜ β α β ⎟ ⎜ β α β ⎟ ⎜ ⎟⎜ ⎟⎝ β β α ⎠

All the rest of the elements involve nonnearest neighbors and so are zero:




⎛α β 0 0 0 β ⎞ ⎜ ⎟ ⎜ β α β 0 0 0 ⎟ ⎜ 0 β α β 0 0 ⎟

H = ⎜ ⎟ ⎜ 0 0 β α β 0 ⎟ ⎜ 0 0 0 β α β ⎟ ⎜ ⎟⎜ ⎟⎝ β 0 0 0 β α ⎠

3) Finding the eigenvalues of H is easy with a computer. We find 4 distinct energies:

E6=α−2β

E4=E5=α−β

E2=E3=α+β

E1=α+2β

The lowest and highest energies are nondegenerate. The second/third and fourth/fifth energies are degenerate with one another. With a little more work we can get the eigenvectors. They are:

⎛ +1⎞ ⎛ +1⎞ ⎛ +1⎞ ⎛ +1⎞ ⎛ +1⎞ ⎛ +1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ −1⎟ ⎜ −2⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ +2⎟ ⎜ +1⎟

1 ⎜ +1⎟ 1 ⎜ +1⎟ 1 ⎜ −1⎟ 1 ⎜ −1⎟ 1 ⎜ +1⎟ 1 ⎜ +1⎟ c 6 = ⎜ ⎟ c 5 = ⎜ ⎟ c 4 = ⎜ ⎟ c 3 = ⎜ ⎟ c 2 = ⎜ ⎟ c 1 = ⎜ ⎟

6 ⎜ −1⎟ 12 ⎜ +1⎟ 4 ⎜ +1⎟ 4 ⎜ −1⎟ 12 ⎜ −1⎟ 6 ⎜ +1⎟ ⎜ +1⎟ ⎜ −2⎟ ⎜ 0 ⎟ ⎜ 0 ⎟ ⎜ −2⎟ ⎜ +1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ −1⎠ ⎝ +1⎠ ⎝ −1⎠ ⎝ +1⎠ ⎝ −1⎠ ⎝ +1⎠

The pictures at the bottom illustrate the MOs by denting positive (negative) lobes by circles whose size corresponds to the weight of that particular pz

orbital in the MO. The resulting phase pattern is very reminiscent of a

particle on a ring, where we saw that the ground state had no nodes, the first and second excited states were degenerate (sine and cosine) and had

one node, the third and fourth were degenerate with two nodes. The one




difference is that, in benzene the fifth excited state is the only one withthree nodes, and it is nondegenerate.4) There are 6 π electrons in benzene, so we doubly occupy the first 3 MOs:

E6=α−2β

E4=E5=α−β

E2=E3=α+β

E1=α+2β

5) The Hückel energy of benzene is then: E = 2E + 2E + 2E = 6α + 8β

1 2 3

Now, we get to the interesting part. What does this tell us about the bonding in benzene? Well, first we note that benzene is somewhat more

stable than a typical system with three double bonds would be. If we do Hückel theory for ethylene, we find that a single ethylene double bond has an energy

EC=C = 2α + 2β

Thus, if benzene simply had three double bonds, we would expect it to have a total energy of

E = 3EC=C = 6α + 6β

which is off by 2β. We recall that β is negative, so that the ππππelectrons in

benzene are more stable than a collection of three double bonds. We call this aromatic stabilization, and Hückel theory predicts a similar stabilization of other cyclic conjugated systems with 4N+2 electrons. This energetic

stabilization explains in part why benzene is so unreactive as compared to other unsaturated hydrocarbons.

We can go one step further in our analysis and look at the bond order. In Hückel theory the bond order can be defined as:

occ µ µ

Oij ≡ ∑c

i cj µ=1

This definition incorporates the idea that, if molecular orbital µ has a bond

between the ith and jth carbons, then the coefficients of the MO on those carbons should both have the same sign (e.g. we have pz

i + pzj). If the orbital




is antibonding between i and j, the coefficients should have opposite signs(e.g. we have pz

i pzj). The summand above reflects this because

ci µ c

µ j > 0 if c

i µ , c

µ j have same sign

ci µ c

µ j < 0 if c

i µ , c

µ j have opposite sign

Thus the formula gives a positive contribution for bonding orbitals and a

negative contribution for antibonding. The summation over the occupied orbitals just sums up the bonding or antibonding contributions from all the occupied MOs for the particular ijpair of carbons to get the total bond

order. Note that, in this summation, a doubly occupied orbital will appear twice. Applying this formula to the 12 bond in benzene, we find that:

O ≡ 2cµ=1

cµ=1 + 2c

µ=2 c

µ=2 + 2cµ=3

cµ=3

12 1 2 1 2 1 2

⎛ +1 ⎞ ⎛ +1 ⎞ ⎛ +1 ⎞ ⎛ +2 ⎞ ⎛ +1 ⎞ ⎛ 0 ⎞ = 2 ⎜ ⎟ × ⎜ ⎟ + 2 ⎜ ⎟ × ⎜ ⎟ + 2 ⎜ ⎟ × ⎜ ⎟

⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 12 ⎠ ⎝ 12 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ 1 2 2

= 2 + 2 = 6 12 3

Thus, the C1 and C2 formally appear to share 2/3 of a πbond [Recall that we are omitting the σorbitals, so the total bond order would be 1 2/3 including

the σ bonds]. We can repeat the same procedure for each CC bond in benzene and we will find the same result: there are 6 equivalent πbonds, each of order 2/3. This gives us great confidence in drawing the Lewis structure we all learned in freshman chemistry:

You might have expected this to give a bond order of 1/2 for each CC πbond rather than 2/3. The extra 1/6 of a bond per carbon comes directly

from the aromatic stabilization: because the molecule is more stable than three isolated πbonds by 2β, this effectively adds another πbond to the

system, which gets distributed equally among all six carbons, resulting in an increased bond order. This effect can be confirmed experimentally, as benzene has slightly shorter CC bonds than nonaromatic conjugated

systems, indicating a higher bond order between the carbons.

Just as we can use simple MO theory to describe resonance structures and

aromatic stabilization, we can also use it to describe crystal field and ligand field states in transition metal compounds and the sp, sp2 and sp3 hybrid




orbitals that arise in directional bonding. These results not only mean MO theory is a useful tool – in practice these discoveries have led to MO theory becoming part of the way chemists think about molecules.




MODERN ELECTRONIC STRUCTURE THEORY At this point, we have more or less exhausted the list of electronic

structure problems we can solve by hand. If we were limited to solving problems manually, there would be a lot of chemistry we wouldn’t be able to explain! Fortunately, the advent of fast personal computers allows chemists

to routinely use more accurate models of molecular electronic structure. These types of calculations typically play a significant role in interpreting experimental results: calculations can be used to assign spectra, evaluate

reaction mechanisms and predict structures of molecules. In this way computation is complementary to experiment: when the two agree we have confidence that our interpretation is correct.

The basic idea of electronic structure theory is that, within the Born Oppenheimer approximation, we can fix the M nuclei in our molecule at some

positions RI. Then, we are left with the Hamiltonian for the electrons moving in the effective field created by the nuclei:

N N M N

H ≡ − 1

2 ∑∇i

2-∑∑

ZI +∑

1 Eq. 1

i=1 i=1 I =1 r − R i< j r − ri I i j

Where the first term is the kinetic energy of all N electrons, the second

term is the attraction between the electrons and nuclei and the third is the pairwise repulsion between all the electrons. The central aim of electronic

structure theory is to find all the eigenfunctions of this Hamiltonian. As we have seen, the eigenvalues we get will depend on our choice of

the positions of the nuclei – Eel(R1,R2,R3,…RM). As was the case with diatomics, these

energies will tell us how stable the molecule is with a given configuration of the nuclei {RI} –

if Eel is very low, the molecule will

R1 be very stable, while if Eel is high, the molecule will be unstable in

that configuration. The energy Eel(R1,R2,R3,…RM) is called the potential energy surface, and it contains a wealth of information, as illustrated in the picture at above. We can determine the equilibrium configuration of the

Equilibrium

Conformation

Unstable

intermediate

Reaction

Barrier

R2

Eel(R1,R2)




molecule by looking for the minimum energy point on the potential energy surface. We can find metastable intermediate states by looking for local minima – i.e. minima that are not the lowest possible energy states, but which

are separated from all other minima by energy barriers. In both of these cases, we are interested in points where ∇E

el = 0 . Further, the potential

surface can tell us about the activation energies between different minima

and the pathways that are required to get from the “reactant” state to the “product” state.

Solving the electronic Schrödinger also gives us the electronic wavefunctions Ψel(r1,r2,r3,…rN), which allow us to compute all kinds of electronic properties – average positions, momenta, uncertainties, etc – as

we have already seen for atoms.

We note that while the Hamiltonian above will have many, many eigenstates,

in most cases we will only be interested in the lowest eigenstate – the electronic ground state. The basic reason for this is that in stable molecules, the lowest excited states are usually several eV above the ground

state and therefore not very important in chemical reactions where the available energy is usually only tenths of an eV. In cases where multiple

electronic states are important, the Hamiltonian above will give us separate potential surfaces E1el, E

2el,

Erxn

σ* potential E3el … and separate wavefunctions surface Ψ1

el, Ψ2el, Ψ 3el. The different

potential surfaces will tell us about

the favored conformations of the molecules in the different electronic σ potential

states. We have already seen this surface

for H2+. When we solved for the

electronic states, we got two eigenstates: σ and σ*. If we put the electron in the σ orbital, the molecule

was bound and had a potential surface like the lower surface at right. However, if we put the electron in the σ∗ orbital the molecule was not bound

and we got the upper surface.




So, at the very least our task is clear cut: solve for the eigenstates of Eq. 1. Unfortunately, this task is also impossible in practice, even on a computer. Thus, over the years chemists have developed a vast array of sophisticated

techniques that allow us to at least approximate these solutions to within a tolerable degree of accuracy. Learning all the details of these approximations would require a course unto itself: the derivations of the

individual approximations are exceedingly complex, and the sheer number of different approximations that can be made is quite impressive. These detailed derivations teach us a lot about what molecules and properties we

should expect our approximations to work for and how we should think about improving our calculations in cases where the theory fails. However, the thing that has really brought computational chemistry into the mainstream is

the fact that one does not have to understand every nuance of a method in order to know how to use it successfully. It suffices to have a simple,

qualitative understanding of how each method works and when it can be applied. Then, coupling that knowledge with a little technical proficiency at using commercial chemistry software packages allows us to run fairly

sophisticated calculations on our desktop computer. The next two lectures are intended to prepare us to run these types of calculations.

First, we note that nearly all the popular approximations are still based on MO theory – MO theory on steroids in some cases, but MO theory nonetheless. Thus, there are still 5 steps in the calculation

1) Choose an Atomic Orbital Basis 2) Build the Relevant Matrices 3) Solve the Eigenvalue Problem

4) Occupy the orbitals based on a stick diagram 5) Compute the energy

In a typical calculation, the computer automatically handles steps 24 automatically – we don’t have to tell it anything at all. It is sometimes helpful to know that the computer is doing these things (e.g. The calculation

crashed my computer. What was it doing when it crashed? Oh, it was trying to solve the eigenvalue problem.) but we essentially never have to do them ourselves. This leaves two steps (1 and 5) that require some input from us

for the calculation to run properly

Choosing an Atomic Orbital Basis




The first point here is that for real electronic structure calculations, you typically use a basis set that is much larger than your intuition might tell you. For example, for H2

+ we guessed that we would get a decent result if

we wrote: ψ = c

11sA + c

21s

B

A basis of this type would be called a minimal basis set, because it is the smallest one that is even close to right. In a more accurate calculation, you

might use a basis that looks more like: ψ = c

11s

A + c

21s

B + c

32s

A + c

42s

B + c

52 p

xA + c

62 p

xB + c

72 p

yA

+ c82 p

yA + c

92 p

zA + c

10 2 p

zB + c

11 3s

A + c

12 3s

B

The reason we use such extended basis sets arises from a point that was discussed earlier concerning MO theory. Because our results are variational, a bigger basis always gets us a lower energy, and therefore a longer AO

expansion always gets us closer to the ground state energy. In the worst case, the unimportant basis functions will just have coefficients that are

very near zero. While such extended basis sets would be a significant problem if we were working things out by hand, computers have no problem dealing with expansions involving even 10,000 basis functions.

The second important point about the atomic orbitals we use is that they are not hydrogenic orbitals. The reason for this is that the twoelectron

integrals involving hydrogenic orbitals cannot all be worked out analytically, making it very difficult to complete Step 2. Instead of hydrogenic orbitals –

which decay like e −r – we will use Gaussian orbitals that decay like e

−α r 2

. Gaussians do not look very much like

hydrogenic orbitals – they don’t have a cusp at r=0 and they decay much too fast at

large distances. About the only good thing about them is that they have a mximum at r=0 and decay. These differences between

Gaussians and hydogenic orbitals are not a problem, though, because we use extended basis sets as emphasized above. Basically,

given enough Gaussians, you can expand anything you like – including a hydrogenic orbital, as shown in the picture at right. So




while using Gaussians may mean we have to use a few extra AOs, if we use enough of them we should be able to get the same answer.

So we plan to use relatively large Gaussian basis sets for our calculations. How exactly do we choose those basis sets? Thankfully, a significant amount of trialanderror research has distilled the choices down to a few key basis

set features. Depending on the problem at hand and the accuracy desired we only need to consider three aspects of the AO basis.

Single, Double, Triple, Quadruple Zeta Basis Sets As we have already discussed for MO theory of diatomics, the smallest basis we can think of for describing bonding would include all the valence orbitals

of each atom involved. Thus, for H we had 1 sfunction, for C there were 2 sfunctions and one set of p’s. Similarly, for sulfur we would have needed 3

sfunctions and 2 p’s …. A basis of this size is called a minimal or single zeta basis. The term “single zeta” refers to the fact that we have only a single set of the valence functions (Note: single valence might seem like a more

appropriate name, but history made a different choice). The most important way to expand the basis is to include more than a single set of valence functions. Thus, in a double zeta (DZ) basis set, one would include 2 s

functions for H, 3 s and 2 pfunctions for C and 4 s and 3 pfunctions for S. Qualitatively, we think of these basis functions as coming from increasing the n quantum number: the first s function on each atom is something like 1s,

the second something like 2s, the third like 3s …. Of course, since we are using Gaussians, they’re not exactly 1s, 2s, 3s … but that’s the basic idea. Going one step further, a triple zeta (TZ) basis would have: H=3s, C=4s3p,

S=5s4p. For Quadruple zeta (QZ): H=4s, C=5s4p, S=6s5p and so on for 5Z, 6Z, 7Z. Thus, one has:

H,He LiNe NaAr Names Minimal 1s 2s1p 3s2p STO3G DZ 2s 3s2p 4s3p 321G,631G, D95V

TZ 3s 4s3p 5s4p 6311G,TZV

Unfortunately, the commonly used names for basis sets follow somewhat

uneven conventions. The basic problem is that many different folks develop basis sets and each group has their own naming conventions. At the end of the table above, we’ve listed a few names of commonly used SZ,DZ and TZ

basis sets. There aren’t any commonly used QZ basis sets, because once




your basis is that large, it is best to start including polarization functions (see below).

Polarization Basis Functions Note that no matter how high you go in the DZ, TZ, QZ hierarchy, you will never, for example, get a pfunction on hydrogen or a dfunction on carbon.

These functions tend to be important for describing polarization of the electrons; at a qualitative level, the pfunctions aren’t as flexible in their angular parts and it’s hard to get them to “point” in as many directions as d

functions. Thus, particularly when dealing with directional bonding in molecules, it can be important to include some of these higher angular momentum functions in your AO basis. In this situation the basis set is said

to contain some “polarization” functions. The general nomenclature of polarization functions is to add the letter “P” to a basis set with a single set

of polarization functions, and “2P” to a basis with two sets. Thus, a DZP basis would have: 2s1p on hydrogen, 3s2p1d on carbon and 4s3p1d on sulfur. A TZP basis set would have 3s1p on hydrogen, 4s3p1d on carbon and 5s4p1d

on sulfur.

H,He LiNe NaAr Names

DZP 2s1p 3s2p1d 4s3p1d 631G(d,p), D95V TZP 3s1p 4s3p1d 5s4p1d 6311G(d,p),TZVP

We note that in practice it is possible to mixandmatch different numbers of polarization functions with different levels of zeta basis sets. The nomenclature here is to put (xxx,yyy) after the name of the basis set. “xxx”

specifies the number and type of polarization functions to be placed on Hydrogen atoms and “yyy” specifies the number and type of polarization

functions to be placed on nonhydrogen atoms. Thus, we would have, for example:

H,He LiNe NaAr 6311G(2df,p) 3s1p 4s3p2d1f 5s4p2d1f

Diffuse Functions Occasionally, and particularly when dealing with anions, the SZ/DZ/TZ/… hierarchy converges very slowly. For anions, this is because the extra

electron is only very weakly bound, and therefore spends a lot of time far




from the nucleus. It is therefore best to include a few basis functions that decay very slowly to describe this extra electron. Functions of this type

are called “diffuse” functions. They are still Gaussians ( e −α r

2

), but the value of α is very, very small causing the atomic orbital to decay slowly. Similar to

the situation for polarization functions, diffuse functions can be added in a mixandmatch way to standard basis sets. Here, the notation “+” or “aug“

is added to a basis set to show that diffuse functions have been added. Thus, we have basis sets like 321++G, 631+G(d,p), augTZP.

Aside: Transition Metals Those of you interested in inorganic chemistry will note that no transition metals appear in the tables above. This is not because there aren’t basis

sets for transition metals – it is just more complicated to compare different transition metal basis sets. First, we note that many of the basis sets above are defined for transition metals. Thus, for example, a 631G(d,p) basis on

iron is 5s4p2d1f while a TZV basis for iron is 6s5p3d. The reason we didn’t include this above is that the idea of “valence” for a transition metal is a subject of debate: is the valence and s and d function? An s a p and a d?

Hence, depending on who put the basis set together, there will be some variation in the number of functions. However, one still expects the same ordering in terms of quality: TZ will be better than DZ, DZP will be better

than a minimal basis, etc. Thus, you can freely use the above basis sets for all the elements between K and Kr without significant modification. Extending the above table for specific basis sets gives:

KCa ScZn GaKr

321G 5s4p 5s4p2d 5s4p1d 631G(d,p) 5s4p1d 5s4p2d1f N/A 6311G(d,p) 8s7p2d N/A 8s7p3d

TZV 6s3p 6s3p2d 6s5p2d

Things also become more complicated when dealing with second row

transition metals. Here, relativistic effects become important, because the Schrödinger equation predicts that the 1s electrons in these atoms are actually moving at a fair fraction of the speed of light. Under these

circumstances, the Schrödinger equation is not strictly correct and we need




to start considering corrections for relativistic effects. The most efficient way to incorporate the relativity is to use an effective core potential (ECP). An effective core potential removes the core electrons from the problem

and replaces them with an effective potential that the valence electrons feel. This potential reflects the combined interaction with the nucleus and the (relativistic) core electrons. Thus, for an ECP we specify both how many

core electrons we want to neglect and how many basis functions we want to use to describe the valence electrons. For example, one popular double zeta ECP is the LANL2DZ basis. As an example, for ruthenium LANL2DZ

replaces the 28 core electrons (1s22s22p63s23p63d10=Argon) with an effective potential and uses a 3s3p2d basis to describe the valence orbitals. Thus, for the second transition series we have (using

[CoreSize]/ValenceBasisSize as our shorthand): YCd HfHg

LANL2DZ [Argon]/3s3p2d N/A SDD [Argon]/8s7p6d [Kr4d104f14]/8s7p6d

As one progresses further up the periodic table, fewer and fewer basis sets are available, simply because less is known about their chemistry.

This is just a very brief overview of what basis sets are available and how

good each one is expected to be. The general idea of using basis sets is to use larger and larger basis sets until the quantity we are computing stops changing. This is based on the idea that we are really using the AO

expansion to approximate the exact solution of the Schrödinger equation. If we had an infinite basis, then we would get the exact answer, but with a large and finite basis we should be able to get close enough. Note, however,

that the calculation will typically take more time in a larger basis than in a smaller one. Thus, we really want to make the basis just big enough to get the answer right, but no larger.

Computing the Energy

For simple MO theory, we used the noninteracting (NI) electron model for the energy:




N N

ENI

= E = ∑∫ψ i ( ) 1 H ( ) τ∑ ψi

1 di

i=1 i=1

Where, on the right hand side we have noted that we can write the NI energy as a sum of integrals involving the orbitals. We already know from

looking at atoms that this isn’t going to be good enough to get us really accurate answers; the electronelectron interaction is just too important. In real calculations, one must choose a method for computing the energy

from among several choices.

The Hartree Fock (HF) Approximation

The HartreeFock method uses the IPM energy expression we’ve already encountered:

N

EIPM

= ∑Ei

+ ∑ J� ij

− K� ij

i=1

N

Ei

= ∑∫ i ( ) ˆψi ( ) dψ 1 H 1 τ

i=1

J� ≡ ∫∫ ψ i * ( ) * ( )

r1

− 1

r2

ψ i ( ) 1 ψ ( ) 2 dr dr2dσ dσ

21 ψ 2ij j j 1 1

K� ≡ ψ * 1 ψ * 2 1 ψ i 2 ψ 1 dr

1dr dσ

1dσij ∫∫ i ( ) j ( )

r1

− r2

( ) j ( ) 2 2

Since the energy contains the average repulsion, we expect our results will be more accurate. However, there is an ambiguity in this expression. The IPM energy above is correct for a determinant constructed out of any set of

orbitals {ψ i } and the energy will be different depending on the orbitals we

choose. For example, we could have chosen a different set of orbitals, {ψ ' } ,i

and gotten a different energy: N

E ' NI

= ∑E ' i

+ ∑ J� ' ij − K� '

ij

i=1

How do we choose the best set of orbitals then? HartreeFock uses the

variational principle to determine the optimal orbitals. That is, in HF we find the set of orbitals that minimize the independent particle energy. These orbitals will be different from the noninteracting orbitals because they will

take into account the average electronelectron repulsion terms in the Hamiltonian. Thus, effects like shielding that we have discussed

qualitatively will be incorporated into the shapes of the orbitals. This will tend to lead to slightly more spread out orbitals and will also occasionally



5.61 Physical Chemistry Lecture #32

10

change the ordering of different orbitals (e.g. σ might shift below π once

interactions are included).

Now, the molecular orbitals (and hence the energy) are determined by their

coefficients. Finding the best orbitals is thus equivalent to finding the best coefficients. Mathematically, then, we want to find the orbitals that make the derivative of the IPM energy zero:

∂E ∂ N IPM � α

= α ∑E

i + ∑ J�

ij − K

ij = 0

∂ci

∂ci i=1

In order to satisfy this condition, one typically resorts to an iterative

procedure, where steps 25 of our MO procedure are performed repeatedly:

1) Choose an AO Basis

1’) Guess an IPM Hamiltonian Heff

2) Build Heff, S matrices

Choose 3) Solve the Eigenvalue Problem

Better

Heff 4) Occupy Lowest Orbitals

dE 5) Compute E,

dc

dE = 0? Done

No dc Yes

Here, HF makes use of the fact that defining an IPM Hamiltonian, Heff, completely determines the molecular orbital coefficients, c. Thus, the most

convenient way to change the orbitals is actually to change the Hamiltonian that generates the orbitals. The calculation converges when we find the molecular orbitals that give us the lowest possible energy, because then dE

= 0 . These iterations are called selfconsistent field (SCF) iterations dc and the effective Hamiltonian Heff is often called the Fock operator, in honor of one of the developers of the HartreeFock approximation.

Generally Hartree Fock is not very accurate, but it is quite fast. On a decent computer, you can run a Hartree Fock calculation on several hundred

atoms quite easily, and the results are at least reasonable.




Density Functional Theory (DFT) Here, we still use a Slater determinant to describe the electrons. Hence, the things we want to optimize are still the MO coefficients c α . However, we

use a different prescription for the energy – one that is entirely based on the electron density. For a single determinant, the electron density, ρ(r) is just the probability of finding an electron at the point r. In terms of the

occupied orbitals, the electron density for a Slater Determinant is:

N 2

ρ r ψ r( ) = ∑ α ( ) Eq. 2 α =1

2

This has a nice interpretation: ψi

r( ) is the probability of finding an

electron in orbital i at a point r. So the formula above tells us that for a

determinant the probability of finding an electron at a point r is just the sum of the probabilities of finding it in one of the orbitals at that point.

There is a deep theorem (the HohenbergKohn Theorem) that states:

There exists a functional Ev[ρ] such that, given the ground state density, ρ0, Ev[ρ0]=E0 where E0 is the exact ground state energy. Further, for any density, ρ’, that is not the ground state density, Ev[ρ’]>E0.

This result is rather remarkable. While solving the Schrödinger Equation

required a very complicated 3N dimensional wavefunction Ψel(R1, R2,…RN), this theorem tells us we only need to know the density which is a 3D

function! – and we can get the exact ground state energy. Further, if we don’t know the density, the second part of this theorem gives us a simple way to find it: just look for the density that minimizes the functional Ev.

The unfortunate point is that we don’t know the form of the functional Ev. We can prove it exists, but we can’t construct it. However, from a

pragmatic point of view, we do have very good approximations to Ev, and the basic idea is to choose an approximate (but perhaps very, very good) form for Ev and then minimize the energy as a function of the density. That is, we

dE vlook for the point where = 0 . Based on Eq. 2 above, we see that ρ just

dρ




depends on the MOs and hence on the MO coefficients, so once again we are dE

vlooking for the set of MO coefficients such that = 0 . Given the dc

similarity between DFT and HF, it is not surprising that DFT is also solved by self consistent field iterations. In fact, in a standard electronic

structure code, DFT and HF are performed in exactly the same manner (see flow chart above). The only change is the way one computes the energy and dE

. dc

Now, as alluded to above, there exist good approximations (note the plural) to Ev. Just as was the case with approximate AO basis sets, these

approximate energy expressions have strange abbreviations. We won’t go into the fine differences between different DFT energy expressions here. I’ll simply note that roughly, the quality of the different functionals is

expected to follow: LSDA < PBE ≈ BLYP < PBE0 ≈ B3LYP

Thus, LSDA is typically the worst DFT approximation and B3LYP is typically

among the best. I should mention that this is just a rule of thumb; unlike the case for basis sets where we were approaching a welldefined limit, here

we are trying various uncontrolled approximations to an unknown functional. Experience shows us that B3LYP is usually the best, but this need not always be the case.

Finally, we note that the speed of a DFT calculation is about the same as Hartree Fock – both involve selfconsistent field iterations to determine the

best set of orbitals, and so both take about the same amount of computer time. However, for the same amount of effort, you can get quite accurate results. As a rule of thumb, with B3LYP you can get energies correct to

within 3 kcal/mol and distances correct to within .01 Å.

PostHartree Fock Calculations

Here, the idea is to employ wavefunctions that are more flexible than a Slater determinant. This can be done by adding up various combinations of Slater determinants, by adding terms that explicitly correlate pairs of

electrons (e.g. functions that depend on r1 and r2 simulataneously) and a variety of other creative techniques. These approaches are all aimed at




incorporating the correlation between electrons – i.e. the fact that

electrons tend to spend more time far apart from one another as opposed to close together. This correlation reduces the average repulsion employed in HF and brings us closer to the true ground state energy. In each case, one

usually does a HartreeFock calculation first and then includes the correlation afterward, leading to the heading of “postHartree Fock” methods. Once again, there are a number of acronyms, and we merely assert

that the quality of the results goes approximately like: HF < CASSCF < MP2 < CCSD < MP4 < CCSD(T)

Here, the ordering is rigorous: we have something solid that we are tyring to

approximate and going from left to right we are making better and better approximations. On the scale above, DFT typically gives results of about

MP2 quality.

As a general rule, postHF calculations are much, much more expensive than

HF or DFT and also require bigger basis sets: whereas HF might converge with a DZ or TZ basis, a postHF calculation might require QZ or even 5Z. Hence, they should only be attempted for relatively small molecules where

high accuracy is required

Combining what we have learned, then, the approximations we can make fit

nicely into a twodimensional parameter space:

321G TZVP

HF

CASSCF

DFT

MP2

CCSD

CCSD(T)

Method

STO3G 631G(d,p) 6311G+(2df,p)

Exact

Answer

Basis

Feasible Calculations




On the one axis, we have the basis sets we can choose from. On the other,

we have the different methods for approximating the energy. The get close to the exact answer, we need to employ a large basis set and an accurate energy method. Unfortunately, both increasing the size of the basis and

improving the method tend to slow our calculations down. Given that we don’t want to wait years and years to find out the result of a calculation, modern computers therefore limit how accurate our answers can be (as illustrated

with the red line above). As we become experts at what is and is not feasible with current computing power, we become better able to get good accuracy for a variety of problems.




SPECTROSCOPY: PROBING MOLECULES WITH LIGHT In practice, even for systems that are very complex and poorly

characterized, we would like to be able to probe molecules and find out as much about the system as we can so that we can understand reactivity, structure, bonding, etc. One of the most powerful tools for interrogating

molecules is spectroscopy. Here, we tickle the system with electromagnetic radiation (i.e. light) and see how the molecules respond. The motivation for this is that different molecules respond to light in different ways. Thus, if

we are creative in the ways that we probe the system with light, we can hope to find a unique spectral fingerprint that will differentiate one molecule from all other possibilities. Thus, in order to understand how spectroscopy

works, we need to answer the question: how do electromagnetic waves interact with matter?

The Dipole Approximation An electromagnetic wave of wavelength λ, produces an electric field, E(r,t), and a magnetic field, B(r,t), of the form:

E(r,t)=E0 cos(k·r – ωt) B(r,t)=B0 cos(k·r – ωt)

Where ω=2πν is the angular frequency of the wave, the wavevector k has a

magnitude 2π/λ and k (the direction the wave propagates) is perpendicular to E0 and B0. Further, the electric and magnetic fields are related:

E0· B0=0 |E0|=c|B0|

Thus, the electric and magnetic fields are orthogonal and the magnetic field is a factor of c (the

speed of light, which is 1/137 in atomic units) smaller than the electric field. Thus we obtain a

picture like the one at right, where the electric and magnetic fields

oscillate transverse to the direction of propagation.

Now, in chemistry we typically deal with the part of the spectrum from ultraviolet (λ≈100 nm) to radio waves (λ≈10 m)1. Meanwhile, a typical molecule

There are a few examples of spectroscopic measurements in the XRay region. In these

cases, the wavelength can be very small and the dipole approximation is not valid.

http://www.monos.leidenuniv.nl

1




is about 1 nm in size. Let us assume that the molecule is sitting at the origin. Then, in the 1 nm3 volume occupied by the molecule we have:

k·r ≈ |k| |r| ≈ 2p/(100 nm) 1 nm = .06

Where we have assumed UV radiation (longer wavelengths would lead to even smaller values for k·r). Thus, k·r is a small number and we can expand the electric and magnetic fields in a power series in k·r:

E(r,t)≈E0 [cos(k·0-ωt)+O(k·r)]≈E0 cos(ωt)

B(r,t)≈B0 [cos(k·0-ωt)+O(k·r)]≈B0 cos(ωt)

Where we are neglecting terms of order at most a few percent. Thus, in most chemical situations, we can think of light as applying two time dependent fields: an oscillating, uniform electric field (top) and a

uniform, oscillating magnetic field (bottom). This approximation is called the Dipole approximation – specifically when applied to the electric (magnetic) field it is called the electric (magnetic) dipole approximation. If

we were to retain the next term in the expansion, we would have what is called the quadrupole approximation. The only time it is advisable to go to higher orders in the expansion is if the dipole contribution is exactly zero as

happens, for example, due to symmetry in some cases. In this situation, even though the quadrupole contributions may be small, they are certainly large

compared to zero and would need to be computed.

The Interaction Hamiltonian How do these oscillating electric and magnetic fields couple to the molecule?

Well, for a system interacting with a uniform electric field E(t) the interaction energy is

ˆ ˆ îHE (t ) = − µµµµiE (t ) = −e r E (t )

where µµµµ is the electric dipole moment of the system. Thus, uniform electric

fields interact with molecular dipole moments.

Similarly, the magnetic field couples to the magnetic dipole moment, m.

Magnetic moments arise from circulating currents and are therefore proportional to angular momentum – larger angular momentum means higher circulating currents and larger magnetic moments. If we assume that all the

angular momentum in our system comes from the intrinsic spin angular momentum, I=(Ix , Iy ,Iz), then the magnetic moment is strictly proportional to

I. For example, for a particle with charge q and mass m then

ˆ q g ˆHB ( )t = −m Bˆ i ( )t = − I B ( )ti

2m




where g is a phenomenological factor (creatively called the “gfactor”) that takes into account the internal structure of the particle containing the intrinsic spin – for an electron ge=2.0023, while for a proton gp=5.5857.

So now suppose that we have a molecule we are interested in, and it has a

Hamiltonian, H0 , when the field is off. Then, when the field is on, the

Hamiltonian will be

H (t ) = H + H (t ) + H (t )0 E B

Actually, in most cases, the simultaneous effects of electric and magnetic fields are not important and we will consider one or the other:

H (t ) ≡ H + H (t ) H (t ) ≡ H (t ) or H (t ) .0 1 1 E B

Thus, in the presence of light, a molecule feels a timedependent

Hamiltonian. This situation is quite different with what we have discussed so far. Previously, our Hamiltonian has been time independent and our job

has simply reduced to finding the eigenstates of H . Now, we have a Hamiltonian that varies with time, meaning that the energy eigenvalues and

eigenstates of H also change with time. What can we say that is meaningful about a system that is constantly changing?

Time Dependent Eigenstates

As it turns out, the best way to think about this problem is to think about

the eigenstates of H0 . When the field is off, each of these eigenstates

evolves by just getting a phase factor: iE t �

H φ n (t ) = E φ

n (t ) ⇒ φ (t ) = e − n / φ

n (0)0 n n

Thus, things like the probability density do not change because multiplying by the complex conjugate wipes out the phase factor:

φ n ( ) t

2

= {e −iE t /�φ

n ( ) 0 } * e −iE t /�φ

n ( ) 0 =eiE t /�φ

n ( ) 0 * e −iE t /�φ

n ( ) 0 = φ n ( )

2 n n n n 0

Thus, when considering measurable quantities (which always involve complex conjugates) the eigenstates of the Hamiltonian appear not to change with time. However, when the field is on the eigenstates will change with

time. In particular, we will be interested in the rate at which the field induces transitions between an initial eigenstate φi and a final state φf.




To work out these rates, we first work out the time dependence of some arbitrary state, ψ(t). We can expand ψ(t) as a linear combination of the eigenstates:

ψ (t ) = ∑c (t )φ (t )n n

n

where cn(t) are the coefficients to be determined. Next, we plug this into the TDSE:

i�ψ� (t ) = Hψ (t )

⇒ i� ∂ ∑c

n ( ) t φ n ( ) t = H ∑c

n ( ) t φ n ( ) t

∂t n n

⇒ i�∑c� n ( ) t φ

n ( ) t + cn ( ) t φ�

n ( ) t = ∑cn ( ) t (H

0 + H

1 ( ) t )φ n ( ) t

n n

⇒ i�∑c� n ( ) t φ

n ( ) t − iE

n cn ( ) t φ

n ( ) t = ∑cn ( ) t (E

n + H

1 ( ) t )φ n ( ) t

n � n

⇒ i�∑c� n ( ) t φ

n ( ) t + n

n

E c ∑ n ( ) t φ n ( ) t = ∑c

n ( ) t (En

+ H1 ( ) t )φ (t )n

n n

⇒ i�∑c� ( ) t φ ( ) t = ∑c ( ) t H ( ) t φ n ( ) t

n n n 1

n n

Next, we multiply both sides by the final state we are interested in (φf*) and

then integrate over all space. On the left hand side, we get:

i�∫φ f

* (t ) ∑c� n (t )φ

n (t ) dτ = i�∑c� n (t )∫ n (t ) dτ = i�c�

f (t ) n n

δnf

Meanwhile, on the right we get:

t t H ˆ φ t c φ t t dτ∫φ f

* ( ) ∑cn ( ) 1 (t ) n ( ) dτ = ∑ n (t )∫ f

* ( ) H1 ( )φ n (t )

n n

Combining terms gives:

⇒ i�c� (t ) = ∑∫φ * (t ) H (t )φ (t ) dτ c (t ) Eq. 1f f 1 n n

n

Up to this point, we haven’t used the form of H1 at all. We note that we can

rewrite the lightmatter interaction as:

H1 (t ) = V cos (ωt )

where, for electric fields V î 0

and for magnetic fields ˆ ≡ − q g î .≡ −er E V I B2m

In either case, we can rewrite the cosine in terms of complex exponentials:

( )*

f tφ φ




H (t ) = V 1 (eiωt + e −iωt )1 2

Plugging this into Eq.1 above gives:

i�c� f (t ) = ∑∫φ f

* (t ) 12 V (e

iωt + e −iωt )φ

n (t ) dτ cn (t )

n

= ∑∫φ f

* ( ) 0 eiE t /�

2 V (e

iωt + e −iωt ) e

−iE t /�φ n

0 dτ cn

tf 1 n ( ) ( ) n

= ∑∫φ * ( ) 0 1

2 Vφ ( ) 0 dτ (e

( n −E f −�ω ) / � + e

−i E n −E f +�ω ) / ) c ( ) t −i E t ( t �

f n n

n

= ∑ 2 V

fn (e ( −E −�ω ) / �

+ e −i E −E +�ω ) / ) cn ( ) t

−i E t ( t �1 n f n f

n

Tickling the Molecule With Light To this point we haven’t made any approximations to the time evolution. We now make some assumptions that allow us to focus on one particular i→f

transition. We make two physical assumptions: 1) The molecule starts in a particular eigenstate, φi, at t=0. This sets

the initial conditions for our coefficients: only the coefficient of state i can be nonzero initially:

c (0) = 0 if n ≠ i c (0) = 1 n i

It is easy to verify that this choice gives the desired initial state: ψ (0) = ∑c (0)φ (0) = 0 + 0 + ...1 iφ (0) + 0.... = φ (0)n n i i

n

2) The interaction only has a small effect on the dynamics. This is

certainly an approximation, and it will not always be true. We can certainly guarantee its validity in one limit: if we reduce the intensity of our light source sufficiently, we will reduce the

strength of the electric and magnetic fields to the point where the influence of the light is small. As we turn up the intensity, there may be additional effects that will come into play, and we will come

back to this possibility later on. However, if we take this assumption at face value, we can assume on the right hand side




that the coefficient, cn, of a state other than φi will be much smaller than ci for all times:

c (t ) � c (t ) if n ≠ i c (t ) ≈ 1 n i i

Where, in the second equality we have noted that if all the other coefficients are tiny, ci must be approximately 1 if we want our

state to stay normalized.

These two assumptions lead to an equation for the coefficients of the form: i( En −E −�ω ) / � −i E n −E f +�ω ) / t �

i�c� ( ) t = ∑ 12 V

fn (e − f t

+ e ( ) cn ( ) t

f

n

⇒ i�c� ( ) t = V (e ( E −E −�ω ) / �

+ e −i E −E +�ω ) / ) c ( ) t

−i t ( t �1 i f i f

f 2 fi i

( i −E f −�ω ) / � −i E i −E f +�ω ) / t � = 1 V (e

−i E t + e

( )2 fi

Now we can integrate this new equation to obtain c (t ) :f

T

( ) 2 fi ∫ (−i E i t ( i t )i�c T = 1 V e

( −E f −�ω ) / � + e

−i E −E f +�ω ) / � dt

f

0

⇒ cf ( ) =

fi

∫ (e ( i −E f −�ω ) / �

+ e −i E i −E f +�ω ) / ) Eq. 2

V T −i E t ( t �

T dt 2i�

0

Now, this formula for cf (T ) is only approximate, because of assumption 2).

If we wanted to improve our result for, we could plug our approximate final expression (Eq. 2) back in on the RHS of Eq. 1 and then integrate the

equation again. This would lead to a better approximate solution for cf (t ) .

Most importantly, while our approximate solution is linear the interaction

matrix element, Vfi , after plugging the result back in, we would get terms

that were quadratic in Vfi . By assumption 2) above, these quadratic terms

will be much smaller than the linear ones we have retained above and so we feel safe in neglecting them. For these reasons, assumption 2) is known as the linear response approximation.

We now make the final rearrangement: we recall that we are interested in

the probability of finding the system in the state f. This is given by c Tf ( ) :

2




2 2T

2 V ( ) =

fi

2 ∫ (e ( −E −�ω ) / �

+ e −i E −E +�ω ) / ) dt

−i E i f t ( i f t � P T = c T

f ( ) f 4�

0

Fermi’s Golden Rule

Now, usually our experiments take a long time from the point of view of electromagnetic waves. In a single second a light wave will oscillate billions of times. Thus, our observations are likely to correspond to the longtime

limit of the above expression: 2 2

TVfi

T →∞ ∫ (e ( i −E f −�ω ) / �

+ e −i E i −E f +�ω ) / ) dt

−i E t ( t � P = lim

f 24�

0

and in fact, we are usually not interested in probabilities, but rates, which are probabilities per unit time:

2 2TV

Wfi

= fi

2

1 ∫ (e

( i −E f −�ω ) / � + e

−i E i −E f +�ω ) / ) dt −i E t ( t �

lim 4� T →∞ T

0

This integral looks very difficult. However, it is easy to work out with pictures because it is almost always zero. Note that both the real and imaginary parts of the integrand oscillate. Thus, we will be computing the

integral of something that looks like:

Thus, as long as the integrand oscillates, the positive regions will cancel out the negative ones and the integral will be zero. There only two situations

where the integrand is not oscillatory: Ei

− Ef

− �ω = 0 (in which case the

first term is unity) and Ei

− Ef

+ �ω = 0 (in which case the second term is

unity). We can therefore write 2

V W

fi ∝

fi

2 ⎣⎡δ (E

i − E

f − �ω) + δ (E

i − E

f + �ω)⎦⎤ 4�

where δ(x) is a function that is defined to be nonzero only when x=0. This result is called Fermi’s golden rule. It gives us a way of predicting the rate

of any i→f transition in any molecule induced by an electromagnetic field of




arbitrary frequency coming from any direction. This formula – as well as generalizations that relax the electric dipole and linear response approximations – is probably the single most important relationship in terms

of how chemists think about spectroscopy, and so we will dwell a bit on the interpretation of the various terms.

On the one hand, the probability of an i→f transition is proportional to 22

V = ∫φ f

* Vφ

idτ

fi

Thus, if the matrix element of the interaction operator V between the

initial and final states is zero, then the transition never happens. This is called a selection rule, and a transition that does not occur because of a selection rule is said to be forbidden. For example, in the case of the

electric field, 2 22 2

V = ∫φ f

* µµµµ iE0φ

idτ = E

0 i∫φ f

* µµµµφidτ = E

0 iµµµµ

fifi

Thus, for molecules interacting with electric fields, the transition i→f is forbidden unless the matrix element of the dipole operator between i&f is

nonzero. Meanwhile, in the case of a magnetic field, 2 2

22 q g q gV = ∫φ

f

* m Bi

0φ

idτ = B

0 i∫φ f

* Iφ

idτ = B iI

fi 0 fi2m 2m

Thus, a magnetic field can only induce an i→f transition if the matrix element of one of the spin angular momentum operators is nonzero between the initial and final states. Selection rules of this type are extremely

important in determining which transitions will and will not appear in our spectra.

Ef Ei

The second thing we note about Fermi’s Golden Rule is that it enforces energy

�ω �ωconservation. We note that the energy carried by a photon is �ω . The δfunction portion is only nonzero if E

f − E

i = �ω

EfEi (second term) or E

i − E

f = �ω (first

term). Thus, the transition only occurs Ef

− Ei

= �ω Ef

− Ei

= −�ω

if the energy difference between the two states exactly matches the energy of the photon we are sending in. This is depicted in the picture at right.




The way these terms are interpreted are as follows: in the first case, the light increases the energy in the system by exactly one photon worth of energy. Here, we think of a photon being absorbed by the molecule and

exciting the system. In the second case, the light reduces the energy of the system by exactly one photon worth of energy. Here, we think of the molecule emitting a photon and relaxing to a lower energy state. The fact

that photon emission by a molecule can be induced by light is called stimulated emission, and is the principle on which lasers are built: basically, when you shine light on an excited molecule, you get more photons out than

you put in.

In order to make much more progress with spectroscopy, we have to

consider some specific choices of the molecular Hamiltonian, H0 , which we

do in the next several lectures. Depending on the system at hand the energy conservation and selection rules give different spectral signatures that ultimately allow us to interpret the spectra of real molecules and to

characterize their physical properties.




VIBRATIONAL SPECTROSCOPY

R

R0 A + B separated atoms

V(R)

Harmonic Approximation

As we’ve emphasized many times in this course, within the Born

Oppenheimer approximation, the nuclei move on a potential energy surface (PES)

determined by the electrons. For example, the potential felt by the nuclei in a

diatomic molecule is shown in cartoon form at right. At low energies, the molecule will sit

near the bottom of this potential energy surface. In

equilibrium bond length this case, no matter what the

detailed structure of the potential is, locally the nuclei will “feel” a nearly harmonic potential. Generally, the motion of the nuclei along the PES is called vibrational motion, and clearly at low energies a good model for the

nuclear motion is a Harmonic oscillator.

Simple Example: Vibrational Spectroscopy of a Diatomic If we just have a diatomic molecule, there is only one degree of freedom (the bond length), and so it is reasonable to model diatomic vibrations using a

1D harmonic oscillator: 2 2P 1 ˆ 2 P 1 2 ˆ 2H = + k R = + mω R

2µ 2 o 2µ 2 o

where ko is a force constant that measures how stiff the bond is and can be approximately related to the second derivative to the true (anharmonic) PES

near equilibrium:

∂2V

k ≈ o 2∂R

R0

Applying Fermi’s Golden Rule, we find that when we irradiate the molecule, the probability of a transition between the ith and fth Harmonic oscialltor

states is: 2

V W

fi ∝

4� fi

2 ⎣⎡δ (E

i − E

f − �ω) + δ (E

i − E

f + �ω)⎦⎤




where ω is the frequency of the light (not to be confused with the frequency of the oscillator, ωo). Because the vibrational eigenstates involve spatial degrees of freedom and not spin, we immediately recognize that it is

the electric field (and not magnetic) that is important here. Thus, we can write the transition matrix element as:

2 22 V = ∫φ f

* µµµµ iE0φ

idτ = E

0 i∫φ f

* µµµµφidτ = E

0 i∫φ f

* eRφidτ

2

fi

Now, we define the component of the electric field, ER, that is along the

bond axis which gives 2 22V = E

R ∫φ f

* eRφidτ

2

= e E ∫φ f

* Rφidτ

2

fi R

So the rate of transitions is proportional to the square of the strength of the electric field (first two terms) as well as the square of the transition

dipole matrix element (third term). Now, because of what we know about the Harmonic oscillator eigenfunctions, we can simplify this. First, we rewrite the position operator, R, in terms of raising and lowering operators:

( ) ( )

( )( )

2 2 22 2 22 * *

22 2

, 1 , 1

ˆ ˆ ˆ ˆ 2 2

1 2

fi R f i R f i

fi R f i f i

e V e E a a d E a a d

e V E i i

φ φ τ φ φ τ µω µω

δ δ µω

+ − + −

+ −

= + = +

⇒ = + +

∫ ∫� �

� 1

1 i

i φ ++ 1i

iφ −

where above it should be clarified that in this expression “i" never refers to √1 – it always refers to the initial quantum number of the system. Thus, we

immediately see that a transition will be forbidden unless the initial and final states differ by one quantum of excitation. Further, we see that the transitions become more probable for more highly excited states. That

is, Vfi gets bigger as i gets bigger.

Combining the explicit expression for the transition matrix element with

Fermi’s Golden Rule again gives: 2

e 2 W ∝ E ((i + 1)δ

f i, +1 + i δ f i, −1 ) ⎣⎡δ (E

i − E

f − �ω) + δ (E

i − E

f + �ω)⎦⎤ fi R

8�µω 2

⇒ W ∝ E {(i + 1)δ f i, +1 ⎡⎣δ (E

i − E

i+1 − �ω) + δ (Ei − E

i+1 + �ω)⎤⎦fi R

+ i δ f i, −1 ⎡⎣δ (E

i − E

i−1 − �ω) + δ (Ei − E

i−1 + �ω)⎤⎦}




0 2

⇒ W ∝ E {(i + 1)δ f i , +1 ⎡⎣δ ( ω−�

o − �ω) + δ (−�ω

o + �ω)⎤⎦fi R

+ i δ f i , −1 ⎡⎣δ (�ω

o − �ω) +

2 ⇒ W ∝ E {(i + 1)δ

f i , +1 + i δ f i , −1}δ (�ω

o − �ω)fi R

Thus, we see that a harmonic oscillator will only absorb or emit photons of frequency �ω

o , where ωo is the frequency of the oscillator. Thus, if

we look at the absorption spectrum, for example, we will see absorption at only one frequency:

ωo Light Frequency (ω)

( ) }o δ ω ω ⎤+ ⎦� �

0

Absorption Intensity

Molecular force constants are typically on the order of an eV per Å, which leads to vibrational frequencies that are typically between 5003500 cm1 and places these absorption

features in the infrared. As a result, this form of spectroscopy is traditionally called IR spectroscopy. We associate the spectrum above as arising from all the

n→n+1 transitions in the Harmonic oscillator (see left). Of course, most of the time the molecule will start in its

ground state, so that the major contribution comes from the 0→1 transition.

However, the other transitions occur at the same frequency and also contribute to the absorption.

This is the classic paradigm for IR vibrational spectroscopy: each diatomic molecule absorbs radiation only at one frequency that is characteristic of

the curvature of the PES near its minimum. Thus, in a collection of different molecules one expects to be able differentiate one from the other by looking for the frequency appropriate to each one. In particular there is




a nice correlation between the “strength” of the bond and the frequency at which it will absorb.

Of course, we are not always or even usually interested in diatomics, and even diatomics are not perfect Harmonic oscillators. Thus, there are a number of reasons why IR absorption spectra do not really look like the

classic Harmonic oscillator spectrum shown above, but more like:

Heterogeneity The primary reason the real spectrum above looks different than the model

is because the real spectrum was taken in solution. The model is correct for a single diatomic, or for many, many copies of identical diatomic molecules. However, in solution, ever molecule is just slightly different, because every

molecule has a slightly different arrangement of solvent molecules around it. These solvent molecules subtly change the PES, slightly shifting the vibrational frequency of each molecule and also modifying the transition

dipole a bit. Thus, while a single hydrogen fluoride molecule might have a spectrum like the model above, a solution with many HF molecules would look something like:





Going over to the situation where there are 1023 HF molecules and

recognizing that our spectra will tend to add the intensity of lines that are closer together than our spectrometer can differentiate, we anticipate that

for a diatomic molecule in solution, the vibrational spectrum should look something like:




The resulting feature in the spectrum is usually called a lineshape. It

primarily reflects the distribution of different environments surrounding your oscillators. Thus, by analyzing the lineshape of a wellknown type of vibration (such as a C=O stretch) one can get an idea about the environments

those CO groups live in: How polar are the surroundings? Are they near electron withdrawing groups? What conformations give rise to the spectrum? Finally, we should note that vibrational spectra recorded in the

gas phase have very narrow linewidths, qualitatively resembling our model above.

Anharmonicity




Another reason real spectra differ from our model is that assuming the PES is harmonic is only a model. If we want high accuracy, we need to account for anharmonic terms in the potential:

V (R) = 1 2 2 1 3 1 4

2 mω

o (R − R0 ) +

6 α (R − R

0 ) + 24

β (R − R0 ) + ...

One can investigate the quantitative effects of the anharmonic terms on the spectrum by performing variational calculations. However, at a basic level

there are two ways that anharmonic terms impact vibrational spectra: 1) The energy differences between adjacent states are no longer

constant. Clearly, the eigenvalues of an anharmonic Hamiltonian

will not be equally spaced – this was a special feature of the Harmonic system. Thus, for a real system we should expect the 0→1 transition to have a slightly different frequency than 1→2,

which in turn will be different that 2→3 …. Generally, the higher transitions have lower (i.e. redshifted) energies because of the

shape of the molecular PES – rather than tend toward infinity at large distances as the harmonic potential does, a molecular PES tends toward a constant dissociation limit. Thus, the higher

eigenstates are lower in energy than they would be for the corresponding harmonic potential. Taking this information, we would then expect a single anharmonic oscillator to have a

spectrum something like:

Absorption

Intensity

Light Frequency (ω)ωo

0→1

1→2

2→3

where we note that while the rate of, say, 1→2 is about twice that of 0→1 (because the transition dipole is twice as big) the intensity of 0→1 is greater because the intensity is (Probability of i being occupied)x(Rate of i→f) and at room temperature the system




spends most of its time in the ground vibrational state. The 1→2, 2→3 … lines in the spectrum are called hot bands.

2) Anharmonicity relaxes the Δn=±1 selection rule. Note that the

rules we arrived at were based on the fact that a+φ

i ∝ φ

i+1 . This is

only true for the Harmonic oscillator states. For anharmonic eigenstates a

+φ ∝ φ + ε φ + ..... . Thus transitions with Δn=±2,±3… i i+1 1 i+2

will no longer be forbidden for anharmonic oscillators. Rather, in

the presence of a bit of anharmonicity, they will be weakly allowed. Combining this observation with point 1) above results in a more complete picture for what the IR spectrum of an anharmonic

oscillator should look like:


Light Frequency (ω)ωo

0→1

1→2

2→3

2ωo

0→2

1→3

2→4

The peaks at around 2ωo are called overtones. Meanwhile, those at around ωo are called fundamentals.

Polyatomic Molecules

The final difference between the model above and a general IR spectrum is that in chemistry, we are not always dealing with diatomic molecules. For a polyatomic molecule, we can still think of the potential as a Harmonic

potential, but it has to be manydimensional – it has to depend on several variables R1, R2, R3, …. The most general Harmonic potential we can come up with is then of the form:

k R + k R R + k R R + .... + k R R + k R + k R R + .... V (R1, R

2, R

3,... ) = 1

2 11 1

2 1

2 12 1 2 1

2 13 1 3 1

2 21 2 1 1

2 22 2

2 1

2 23 2 3

+ 1 k R R + 1 k R R + 1 2k R + ...

2 31 3 1 2 32 3 2 2 33 3

where it is important to notice the cross terms involving, say R1 and R2, which couple the different vibrations. At first sight, it seems like we




can’t solve this Hamiltonian; the only manydimensional Harmonic potential we would know how to solve would be one that is separable:

V� ( R1, R

2, R

3,... ) = 1

2 k11 R1

2 + 1

2 k22 R2

2 + 1

2 k33 R3

2 + ...

If the Harmonic potential were of this form, we would be able to write down

the eigenstates as products of the 1D eigenstates and get the energies as sums of the 1D eigenenergies. As it turns out, by changing coordinates we

can turn a quadratic system with offdiagonal cross terms (like the first potential) into one with no cross terms (like the second). These new coordinates, in terms of which the Hamiltonian separates, are called normal

modes and they allow us to reduce a polyatomic molecule to a collection of independent 1D oscillators.

First, we note that V can be rewritten concisely in matrix notation [Note: it may be useful to consult McQuarrie’s supplement on Matrix Eigenvalue problems if the following seems unfamiliar.]:

V ( R1, R

2, R

3,... ) = 1

2 RT iK Ri

⎛ R1 ⎞ ⎛ k11 k12 k

13 ... ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ R2 ⎟ ⎜ k21 k

22 k23 ... ⎟R ≡ K ≡

⎜ R3 ⎟ ⎜ k

31 k23 k

33 ... ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ � ⎠ ⎝ � � � �⎠

Now, the Hamiltonian is of the form:

TH = ∑

P i

2

+ 1

R iK Ri ˆ i 2µ

i 2

It is convenient to first transform to mass-weighted coordinates:

P ij pi ≡ i xi ≡ µi R

i k� ij ≡ k

µ µ µ i i j

in terms of which we can write the Hamiltonian:

H = ∑ p

i

2

+ 1

ˆT i � i ˆx K x

i 2 2

As is clear from the kinetic energy above, in these coordinates, every degree of freedom has the same reduced mass.

Now we perform the normal mode transformation. We want to write:




⎛ k11

' 0 0 ... ⎞

⎜ ' ⎟ x

T i � i ˆ = y T

iK ′ ⎜ 0 k22 0 ... ⎟K x ˆ iy K ′ =

'⎜ 0 0 k33 ... ⎟

⎜ ⎟⎜ ⎟ ... ... ... ... ⎝ ⎠ Further, we will assume there is a matrix U that transforms from x to y

y = U x i ⇒ y T = x T iUT

where in the second equality, we recall the general rule that the transpose of a product is the product of the transposes, but in the opposite order. Combining these two equations:

iK x � ˆ x ′ i ˆxT i ˆ = y T

iK ′iy = ˆT iU

T iK iU x

⇒ K� = UT iK ′iU

The last equation is a common problem encountered in linear algebra: the quest to take a given matrix ( K� ) and place it in diagonal form (the right hand side). For a symmetric matrix like K� the solution to this problem is

well known: the diagonal entries of K ′ are the eigenvalues of K� and the columns of the transformation matrix U are the eigenvectors of K� . The transformed variables y are called the normal modes. These modes are

linear combinations of the local degrees of freedom R1, R2, R3, … that we

started out with. Thus, while the initial motions might correspond clearly to local stretching of one bond or bending of an angle, the normal modes will generally be complicated mixtures of different molecular motions. We can

visualize this in the simple case of two degrees of freedom. The local modes R1, R2 can be thought of as the two orthogonal axes in a plane. Meanwhile, the normal modes y1, y2 are also orthogonal axes, but rotated from the

original set:

R1

R2

R1

R2

y2

y1

Local modes Normal modes

The local modes have interactions between each other: local stretches are

coupled to local bends, etc. As a result, the Hamiltonian is not separable in terms of the local modes R1, R2. However, by design the Hamiltonian is separable when written in terms of the normal modes:




H = ∑ p

i

2

+ 1

kii ′ y

i

2

i 2 2 ’ The eigenvalues, kii , tell us the “stiffness” of the PES in the particular

direction yi. Note that the Hamiltonian above is exactly the same Hamiltonian as the one we started with. The coupling terms have simply been rotated away by changing the coordinates. As discussed before, we can

immediately interpret the spectra of this Hamiltonian in terms of a sum of many independent oscillators. Thus, for a polyatomic molecule within the harmonic approximation we expect to see lines at each of the normal mode

frequencies:


Light Frequency (ω)ω3 ω5ω1 ω2 ω4

Where we have noted that the different oscillators will typically also have different transition dipoles. (For obvious reasons, in vibrational

spectroscopy the square of the transition dipole is often called the oscillator strength) We can, of course, combine this polyatomic picture with the anharmonicity effects above to get a more general picture that looks

like:

2ω1 2ω2 ω5-ω1 ω2+ω3

ω1 ω2 ω3 ω4 ω5 Light Frequency (ω)





where we predict the existence of various hotbands and overtones for each of the normal mode oscillators in the molecule. Note that while the overtones always involve multiple quanta, the quanta need not come from the

same normal mode – hence we expect not only overtones at 2ω1 , but also a

combination bands at ω2 + ω

3 and ω5 − ω

1 . The picture above is qualitatively

correct for the IR spectrum of a single molecule. In solution, heterogeneity leads to a smearing out and broadening of the peaks, leading to the complex

IR fingerprints we are used to.

As should be clear from the above discussion, IR spectra contain a wealth of

information about the molecule: the stiffness of each normal mode, the degree of anhormonic effects, the character of the local environment felt

by the oscillators …. Of course, in order to extract this information, one must be able to assign the spectrum – i.e. one must be able to distinguish hotbands from overtones and associate the various normal modes (at least

qualitatively) with physical motions of the molecule. This task can be extremely challenging – and computation must be used as a guide in many cases – but when it is accomplished, one typically has a very sensitive

fingerprint of molecule under consideration.



1 5.61 Physical Chemistry Lecture #36 Page

NUCLEAR MAGNETIC RESONANCE Just as IR spectroscopy is the simplest example of transitions being induced by light’s

oscillating electric field, so NMR is the simplest example of transitions induced by the oscillating magnetic field. Because the strength of mattermagnetic field interactions are typically two orders of magnitude smaller than the corresponding electric field interactions, NMR is a much more delicate probe of molecular structure and properties. The NMR spin Hamiltonians and wavefunctions are particularly simple, and permit us to demonstrate several fundamental principles (about raising and lowering operators, energy levels, transition

probabilities, etc.) with a minimal amount of algebra. The principles and procedures are applicable to other areas of spectroscopy electronic, vibrational, rotational, etc. – but for

these cases the algebra is more extensive.

Nuclear Spins in a Static Magnetic Field

For a single isolated spin in a static magnetic field, the contribution to the energy is:

H = − m B îˆ i = − γ I B 0 0 0

where γ is called the gyromagnetic ratio. If we choose our z axis to point in the direction of E

nerg

y

the magnetic field then:

H = − m B ˆˆ = − γ I B 0 z 0 z 0

If we assume the nuclear spin is ½ (As it is for a proton) then we can easily work out the energy levels of this Hamiltonian:

1 1 �ωE± = ± 2 γ �B

0 ≡ ±

2 0

where ω0 = γB0 is called the nuclear Larmor frequency (rad/sec). Now, nuclei are never isolated in chemistry – they are always surrounded by electrons. As we learned for the hydrogen atom, the electrons near the nucleus shield the outer electrons from the bare

electric field produced by the nucleus. Similarly, the electrons shield the nucleus from the bare electric field we apply in the laboratory. More specifically, the electron circulation produces a field, B’ opposed to B0 and of

magnitude equal to σ B0 where σ is a constant. bare nucleus with

nucleus electrons Thus, the effective field, B, at the nucleus is B

0 B

0(1-σ)

B = (1 −σ )B0

Note that σ is different for each chemically different nuclear spin – this is the famous

hω0 hω

0 (1 − σ )chemical shift – and permits resolution of lines in

NMR spectra corresponding to chemically different sites. The Hamiltonian is modified accordingly

ˆ ˆH = − m B ˆ (1 − σ ) = − γ I B (1 − σ )0 z 0 z 0

Zero Field High Field

Thus, instead of “seeing” a magnetic field of magnitude B0, a proton in a molecule will see a



5.61 Physical Chemistry Lecture #36 Page 2

magnetic field of magnitude (1-σ)B0 and the associated Hamiltonian and spin state energies

will become: 1 1E± = ± 2 γ �B

0 (1− σ ) ≡ ± 2 �ω

0 (1−σ ) This is illustrated in the figure above. Note the sign of the Hamiltonian is chosen so that the

α state (spin parallel to B0) is lower in energy than the β state ( spin antiparallel to B0).

Now, in the simplest NMR experiment, we probe this system with an oscillating magnetic field perpendicular to the static field. By convention, we take this field to be along the x axis:

H1 (t ) = − ˆ i (t ) = − γ î

1 (t ) γ ˆ x cos (ωm B

1 I B = − I B x t )

We use Fermi’s Golden Rule to describe the spectrum of the spin in the oscillating field. The selection rule is:

2 22

V = ∫φ * m B i 1φidτ = γ Bx i∫φ f

* I x φidτfi f

Now, we recall that I x can be written in terms of the raising and lowering operators for

angular momentum: ˆ ˆ Î ∝ ( I + I )x + −

So that: 2

* ˆ ˆV ∝ γ Bx i∫φ f ( I+ + I− )φidτ 2

fi

We immediately see that the integral is nonzero only if the initial and final spin states differ by ±1 quantum of angular momentum (i.e. ΔM = ± 1 ) , because the operator must either

raise or lower the eigenvalue of I z . Thus, there are two possible transitions: α→β and β→α.

Futher, the energy conservation rule tells us that these transitions will only occur when the photon energy matches the energy gap between the two states. As a result, we can

This is perhaps not all that shocking: there is only one transition here, and so we might have guessed that the spectrum would involve the frequency of that transition. However, we note

two generalizations of this result. First, we note that if we had chosen to apply the oscillating field parallel to the static field, we would not have generated any transitions; we

immediately draw the spectrum of a single shielded spin:

Intensity

Frequency (ω)(1−σ)ωo




only changed the spin state because we could decompose the xoscillating field into raising

and lowering operators. If the field was zoscillating, then we would have had 22

V ∝ γ Bx i∫φ f

* I z φidτfi

which is only nonzero for the trivial α→α and β→β transitions. Second, we note that if the

spin was bigger than ½ (e.g. a spin3/2 nucleus) then our selection rule above would be precisely the same. Thus, we would have allowed transitions −

2

3 ↔ − 12 , − 1

2 ↔ + 1

2 and

+ 12

↔ + 3

2 and all of these transitions would occur at the same frequency. Thus, spin3/2

transitions like − 2

3 ↔ + 12 or −

2

3 ↔ + 2

3 are strictly forbidden.

Now, as noted above, depending on their environment, different protons will be shielded differently, resulting in a spectrum that will look qualitatively like:

Intensity

Frequency (ω)(1−σ3)ωo (1−σ4)ωo(1−σ2)ωo(1−σ1)ωo

We note that the transition moment above is independent of the chemical environment: it does not depend on shielding or any other property of the molecule. Thus, the area under an

NMR peak is strictly proportional to the number of spins that have transitions at that frequency. This stands in contrast to IR spectroscopy, where the intensity of each oscillator depended on the character of the oscillator, the initial state ….

Two Spins – J Couplings Now, we are not usually interested in two isolated spins. For two uncoupled spins with

different chemical shifts ( σ1≠ σ2 ) in an external field we obtain a Hamiltonian of the form:

H0

= − γ I1z B0 (1 − σ

1 ) − γ I2 z B0 (1 − σ

2 )

Because this Hamiltonian is separable, we can immediately work out the energies:

E ↓↓

= E 1↓

+ E 2↓

= + ωο [(1 − σ

1) + (1 − σ

2)]

2

E ↓↑

= E 1↓

+ E 2↑

= + ωο [(1 − σ

1) − (1 − σ

2)]

2

E ↑↓

= E 1↑

+ E 2↓

= − ωο [(1 − σ

1) − (1 − σ

2)]

2

E ↑↑

= E 1↑

+ E 2↑

= − ωο [(1 − σ

1) + (1 − σ

2)]

2




where we have assumed for simplicity that σ1< σ2 so that E ↓↑

> E ↑↓ . Now, the selection rule is

the of the same form as for a single spin, but Ix decomposes into a sum of Ix for spin 1 and an

Ix for spin 2: 22

* ˆ ˆV = γ Bx i∫φ f

* I x φ

idτ = γ B

x i∫φ

f ( I1x + I

2 x )φidτ

2

fi

* ˆ ˆ ˆ ˆ∝ ∫φ f ( I1+ + I1− + I

2+ + I2− )φidτ

2

The remaining integral is only nonzero if ΔM1

= ±1 or ΔM2

= ±1 , because the operators must

raise or lower the spin state of either spin 1 or spin 2 (but not both). If we wanted to change

both spins, we would need an operator like I1+ I2− , which would allow us to raise 1 while also

lowering 2. Since we do not have any of these cross terms, we conclude only one or the other

spin can flip in an allowed transition – any twospin transitions are forbidden.

Combining these results for two uncoupled spins, we obtain the picture at left. We note that

the ↑↑↔↓↓ and ↓↑↔↑↓ transitions are forbidden, since they require flipping both spins simultaneously. For the allowed transitions, we can easily work out

the energies:

E ↓↓

− E ↑↓

= + ωο [(1 − σ

1) + (1 − σ

2)] +

ωο [(1 − σ1) − (1 − σ

2)] = ωο (1 − σ

1)

2 2

E ↓↑

− E ↑↑

= + ωο [(1 − σ

1) − (1 − σ

2)] +

ωο [(1 − σ1) + (1 − σ

2)] = ωο (1 − σ

1)

2 2

E ↓↓

− E ↓↑

= + ωο [(1 − σ

1) + (1 − σ

2)] −

ωο [(1 − σ1) − (1 − σ

2)] = ωο (1 − σ

2)

2 2

E ↑↓

− E ↑↑

= − ωο [(1 − σ

1) − (1 − σ

2)] +

ωο [(1 − σ1) + (1 − σ

2)] = ωο (1 − σ

2)

2 2

So we have only two transition energies, corresponding to each of the isolated transitions,

ΔΔΔΔM2=±1

ΔΔΔΔM2=±1ΔΔΔΔM1=±1

ΔΔΔΔM1=±1

E↓↑

E↑↓

E↓↓

E↑↑

just as predicted above:

Intensity

Frequency (ω)(1−σ1)ωo(1−σ2)ωo

Where we note that there are actually two degenerate transitions contributing to each line.



�

�


We now permit the two spins to be coupled to one another in a simple way. We include a J

coupling of the spins: J

2 I1z I2z , where the factors of � are included so that J has units of

energy. Thus, the Hamiltonian assumes the form

H0

= − γ I1z

B0 (1 − σ

1 ) − γ I2 z

B0 (1 − σ

2 ) + J

2 I1z I2z

Now, we can work out the eigenvalues of this new Hamiltonian quite easily because we know

the eigenvalues of I1z

and I2z . For example, for the ↓↓ state:

H0φ

↓↓ = ⎛⎜ −γ I

1z B0 (1 − σ1 ) − γ I

2 z B0 (1 − σ2 ) +

J 2

I1z I2z

⎞⎟φ

↓↓⎝ � ⎠

⎛ J ⎛ −� ⎞ ⎛ −� ⎞⎞ = ⎜ E

1↓ + E

2↓+

2 ⎜ ⎟ ⎜ ⎟⎟φ↓↓

⎝ � ⎝ 2 ⎠ ⎝ 2 ⎠⎠

= ⎜⎛

E 1↓

+E 2↓

+ J ⎟⎞φ

↓↓⎝ 4 ⎠

Similar algebra for the other states gives:

⎛ J ⎛ � ⎞ ⎛ −� ⎞⎞ ⎛ J ⎞ H

0φ

↑↓ = ⎜ E

1↑ + E

2↓+

2 ⎜ ⎟ ⎜ ⎟⎟φ↑↓

= ⎜ E 1↑

+E 1↓

− ⎟φ↑↓

⎝ � ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 4 ⎠

⎛ J ⎛ −� ⎞ ⎛ � ⎞⎞ ⎛ J ⎞ H

0φ

↓↑ = ⎜ E

1↓ + E

2↑+

2 ⎜ ⎟ ⎜ ⎟⎟φ↓↑

= ⎜ E 1↓

+E 2↑

− ⎟φ↓↑

⎝ � ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 4 ⎠

⎛ J ⎛ � ⎞ ⎛ � ⎞⎞ ⎛ J ⎞ H

0φ

↑↑ = ⎜ E

1↑ + E

2↑+

2 ⎜ ⎟ ⎜ ⎟⎟φ↑↑

= ⎜ E 1↑

+E 1↓

+ ⎟φ↑↑

⎝ � ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 4 ⎠

Thus, in the presence of the coupling, our energy diagram changes:

ΔΔΔΔM2=±1


ΔΔΔΔM1=±1

E↓↑

E↑↓

E↓↓

E↑↑

ΔΔΔΔM2=±1


ΔΔΔΔM1=±1

E↓↑

E↑↓

E↓↓

E↑↑

Uncoupled J Coupling

Where we have noted that states where the spins are parallel shift upward in energy and

those where the spins are antiparallel shift down, and we have exaggerated the magnitude of the shift for visual effect. Note that the selection rules do not change, because the states have not changed – only the energies are different with the coupling on. The energies of the

allowed transitions are:

− οE ↓↓

− E ↑↓

= + ωο

⎢⎡ (1 − σ

1) + (1 − σ

2) +

J ⎥⎤ ω

⎢⎡−(1 − σ

1) + (1 − σ

2) −

J ⎥⎤

= ωο (1 − σ1) +

J

2 ⎣ 4 ⎦ 2 ⎣ 4 ⎦ 2




E ↓↑

− E ↑↑

= + ωο

⎢⎡ (1 − σ

1) − (1 − σ

2) −

J ⎥⎤

− ωο

⎢⎡−(1 − σ

1) − (1 − σ

2) +

J ⎥⎤

= ωο (1 − σ1) −

J

2 ⎣ 4 ⎦ 2 ⎣ 4 ⎦ 2

E ↓↓

− E ↓↑

= + ωο

⎢⎡ (1 − σ

1) + (1 − σ

2) +

J ⎥⎤

− ωο

⎢⎡ (1 − σ

1) − (1 − σ

2) −

J ⎥⎤

= ωο (1 − σ2) +

J

2 ⎣ 4 ⎦ 2 ⎣ 4 ⎦ 2

ω ⎡ J ⎤ −

ωο ⎡ J ⎤ J E

↑↓ − E

↑↑ = ο

⎢−(1 − σ1) + (1 − σ

2) − ⎥ ⎢−(1 − σ

1) − (1 − σ

2) + ⎥ = ωο (1 − σ

2) −

2 ⎣ 4 ⎦ 2 ⎣ 4 ⎦ 2

Thus, whereas we had two doubly degenerate transitions in the absence of coupling, in the

presence of coupling we have four distinct transitions:

Intensity

Frequency (ω)(1−σ1)ωo(1−σ2)ωo

J J

where here we have noted the physical fact that J is typically much smaller than the

difference in chemical shielding σ between distinct protons. Thus, we see that the splitting of NMR peaks is mediated by the coupling between the nuclear spins. This coupling is typically mediated via the electrons – nucleus 1 pushes on the electrons, which are delocalized

and in turn push on nucleus 2. While one can routinely compute these couplings via DFT or HF, it is much more common to use empirical rules to determine which protons will be coupled and how large we expect the coupling to be. We should note that the magnitude of the Jsplitting is independent of the magnetic field strength. Meanwhile, the Larmor frequency increases with increasing B0. Thus, in a strong enough magnet, the peaks with shielding near σ1

will be very far from those with shielding σ2.

Spin Dynamics and Pulsed NMR One of the extremely appealing aspects of NMR is we can exactly work out virtually any property we’re interested in knowing. In particular, we can get a picture of the dynamics of

the spin in an external magnetic field. This gives us a qualitative picture of what we are doing when we take an NMR spectrum and also serves as the basis for modern pulsed NMR experiments. Consider an arbitrary initial state written as a linear combination of the two

spin states: ψ (t ) = cα (t )ψα + cβ (t )ψ β

where we have noted that the time dependence of the state comes through the time dependence of the coefficients. We can write this in matrix mechanics:




ψ�

( )t = ⎜⎛ cα (t )

⎟⎞

⎜ cβ ( )t ⎟⎝ ⎠ We can also write the timedependent Schrödinger equation in Matrix mechanics:

i ψ t = ˆ t ⎛ c

�

�α (t ) ⎞

= ⎛⎜⎜

Hαα Hαβ ⎞

⎟ ⎛⎜⎜

cα (t ) ⎞ � � ( ) Hψ ( ) ⇒ i�

⎝⎜⎜

β ( )t ⎟⎟ Hβα Hββ

⎟⎝ c ( )

⎟⎟c ⎠ ⎝ ⎠ β t ⎠

Now, for a spin in a static field, we know the Hamiltonian

⎛ −�ω� 0 ⎞ ⎜ 0 ⎟

H = − ω (1 − σ ) I ≡ − ω� I ⇒ H = − ω� I = ⎜ 2 ⎟0 z 0 z 0 z

⎜ �ω� 0 ⎟

⎜ 0 ⎟⎝ 2 ⎠

Thus, the TDSE becomes:

⎛ −�ω� 0 ⎞ ⎛ c�α ( ) ⎜ 0 ⎟ ⎛ c ( )t ⎞ 2 α t ⎞

i� ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ c�β ( )t ⎠ ⎜ −�ω�

0 ⎟ ⎝ cβ ( )t ⎠⎜ 0 ⎟⎝ 2 ⎠

Which reduces to two independent differential equations for the coefficients:

0 0i�c�α ( )t =−�ω�

cα ( )t i�c�β ( )t =+�ω�

cβ ( )t 2 2

These equations can easily be integrated to yield: +iω�0t −iω�0t

c t = e 2 c 0 c t = e 2 c 0α ( ) α ( ) β ( ) β ( ) where we will assume for simplicity that the initial values, cα (0) , cβ (0) are real. Thus, the

magnitude of each coefficient is constant with time; we only acquire a phase factor for each coefficient. However, these coefficients completely describe the time evolution of an

arbitrary spin state in the static magnetic field.

Now that we have solved for the coefficients of the time dependent wavefunction, let’s look

at some interesting properties of the system. First, let’s compute the zcomponent of the spin:

⎛ � ⎞ 2

I ( ) = (c ( ) * c ⎜ ⎟⎟ = ⎢⎣

cα ( )t − c ( )t 2

⎦⎤ = � ⎢⎡

α 02

t t ( )t *) ⎜ 2

0

⎟⎟ ⎜⎜⎛ cα ( )t ⎞ � ⎡

⎥ ⎣ c ( ) − c ( )0

2 ⎤⎦⎥z α β β β

2⎜ � ⎟ ⎝ cβ ( )t ⎠ 2 ⎜ 0 − ⎟⎝ 2 ⎠

Thus, the zcomponent of the spin does not change with time! This is perhaps a bit

surprising. We continue to compute the x and y components:

⎛ � ⎞ ⎜ 0 2 ⎟ ⎛ cα ( )t ⎞ �

I t = c t * c t * = ⎡c t * c t + c t * c t ⎤( ) ( α ( ) β ( ) ) ⎜⎜ � ⎟

⎟ ⎝⎜⎜cβ ( )t ⎠

⎟⎟ 2 ⎣ α ( ) β ( ) β ( ) α ( )⎦x

⎜ 0 ⎟⎝ 2 ⎠

= �

c ( ) cβ ( )0 ⎡⎣e −iω�0t + e

+iω�0t

⎦ = �c 0 cβ ( )0 cosω tα 0 ⎤ α ( ) �

02



�


⎛ −i� ⎞ ⎜ 0 2 ⎟ ⎛ cα ( )⎞ �t −i

I t = c t * cβ t * ⎜ = ⎡c t * c t − c t * cα t ⎤( ) ( α ( ) ( ) ) i�

⎟ ⎜⎜c ( )t ⎟⎟ 2 ⎣ α ( ) β ( ) β ( ) ( )⎦y

⎜ 0 ⎟ ⎝ β ⎠⎜ ⎟⎝ 2 ⎠

= c 0 c 0 ⎡e − e ⎤ = −�c 0 c 0 sin ω� t −i�

α ( ) β ( ) ⎣ −iω�0t +iω�0t

⎦ α ( ) β ( ) 02

Thus, the x and y components oscillate with time at the shielded Larmor frequency ω� 0. It is

convenient to define a magnetization vector that contains these three expectation values: ˆM

� ( )t ≡ ( I t( ) I ( )t I ( )t )x y z

It is fairly easy to see that the magnetization is precessing about the magnetic field: the projection onto the magnetic field axis is constant, while the perpendicular motion is tracing out a circular path. This is precisely the behavior one would expect from a classical magnetic

moment in a magnetic field. In this case, the magnetic field would exert a torque on the magnetic moment according to:

effdM �

(t )= M �

( )t ×γ B �

dt

where we note that the magnetic moment feels the shielded magnetic field Beff . This gives us three differential equations for the components of the magnetization, called Bloch Equations:

dM x (t )

γ ⎡⎣My ( ) Beff − Mz ( ) Beff ⎤⎦ = γ M

yt

eff= tz

ty ( ) Bz

dt

dM y (t )

⎡ t t ⎤ ( ) Beff eff eff= − γ ⎣Mx ( ) Bz − M

z ( ) Bx ⎦ = − γ Mx

t z

dt

dM

dt

z (t ) ⎣ x ( ) y

eff

y ( ) x

eff

⎦ = γ ⎡M t B − M t B ⎤ = 0

Where we have noted that only the zcomponent of the magnetic field is nonzero. Further,

it is easy to see by substitution that our quantum mechanical predictions for I x (t ) and

I (t ) satisfy the equations above for Mx (t ) and M

y (t ) , respectively (try it and see). Thus,y

we find that the quantum evolution of the average spin exactly follows the classical

equations of motion! We find comfort in this conclusion, because it is usually much easier to think in terms of classical properties whenever possible, giving us a very nice semiclassical way of interpreting NMR.

This rather surprising result turns out to be true for a single spin evolving in an arbitrary time depenedent magnetic field Beff

(t). To prove this, we have to use Ehrenfest’s theorem, which states that for an arbitrary operator O, the time dependent average value of O

satisfies d

⎡ ˆ , ˆ ⎤ tO ( )t = i ⎣H O⎦ ( )

dt



�

�


We proved a shortened version of this on one of the Problem Sets. Applying Ehrenfests

theorem to the three operators I x , I

y , I

z in an arbitrary magnetic field Beff

(t) gives equations

of motion that are exactly the same as the classical equations for M (t ) . Thus, one can prove

quite rigorously that the classical picture is exactly right for describing spin dynamics in a

magnetic field.

What does this gain us? Well, with this result in hand it is relatively easy to derive the

correct differential equations for our favorite time dependent magnetic field: eff eff

B (t ) = − B − B cos (ωt )z x

This is the magnetic field we apply in an NMR experiment and being able to visualize the dynamics will help us understand how the experiment works. It is relatively straightforward to work out the associated Bloch Equations for this magnetic field. They are:

dM t x ( )

= γ M ( ) t Beff

dt y z

dM y (t )

= − γ ⎡⎣Mx ( ) z

eff + Mz ( ) Bx )⎦t B t cos (ωt ⎤

dt

z dM

dt

(t ) = − γ M

y ( ) t Bx

cos (ωt )

These equations can actually be solved analytically to obtain the magnetization as a function of time. From these equations we obtain the picture below:

ω � ω ω = ω0

ω � ω00

Here, we are plotting the magnetization as a function of time for various choices of the frequency of the oscillating magnetic field component. If our field oscillates too quickly (first case) then the magnetization just sees the average field and noting interesting

happens – we just get precession about the average field. If the oscillating field is too slow, the magnetization oscillates around the instantaneous field and we get a sort of hulahoop motion of the magnetization. However, if we hit the frequency just right (middle) we can get

the magnetization to invert – to go from “up” to “down”. Thus, we see that the absorption condition in NMR is associated with flipping the magnetization of the system.

Now, we note that at resonance, with the field on continuously, the spin will actually flip from “up” to “down” and back to “up” and back to “down”… as a function of time. It is this




oscillation that shows up in our NMR spectrum. However, it is possible to turn the

oscillating field on and off as a function of time. Thus, for example, if we kept the field on for exactly π/ω0 then the system would only have time to flip one time – all the up spins would be converted to down and vice versa. Such a pulsed magnetic field is called an inversion

pulse, for obvious reasons. Meanwhile, if we kept the field on for exactly π/2ω0 we could drive all the magnetization into the xy plane. This is called a π/2 pulse. Further, we note

these pulses only work if we are on resonance with a particular proton’s Larmor frequency; from the above figure it is clear that if we are off resonance, we can’t get the spins to flip. Thus, one can imagine fairly complex sequences of inversion pulses and π/2 pulses applied at

various frequencies being used to isolate different couplings within a complicated molecule (like a protein). Thus, it should not be surprising that cutting edge NMR experiments are all timeresolved in order to extract the maximum information from the molecule.



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