Ch 7. A Quantum Mechanical Model for the Vibration and Rotation of Molecules MS310 Quantum Physical Chemistry - Schrödinger eq. for the Q.M. harmonic oscillator

Ch 7. A Quantum Mechanical Model Ch 7. A Quantum Mechanical Model for the Vibration and Rotation of for the Vibration and Rotation of

MoleculesMolecules

MS310 Quantum Physical Chemistry

- Schrödinger eq. for the Q.M. harmonic oscillator Schrödinger eq. for the Q.M. harmonic oscillator

- Described by energy spectrum and energy eigen-Described by energy spectrum and energy eigenfunctions of molecules having translational, functions of molecules having translational, vibrational, and rotational degrees of freedomvibrational, and rotational degrees of freedom

- Schrödinger eq. for rotation in 2-D and 3-DSchrödinger eq. for rotation in 2-D and 3-D

- Angular monentum to consider orbitals…Angular monentum to consider orbitals…


7.1 Solving the Schrödinger equation for the Q.M Harmonic Oscillator

Example of vibration in Q.M : chemical bondBonding electron in the simple potential, and equilibrium distance is determined by bond length.

Real potential : anharmonic oscillator(not ideal)

At 300K, 1 or 2 state of vibration occupied → can approximate V(x) as a harmonic oscillator

2

2

1)( kxxV


Schrödinger equation is given by

)()(2

)(

2

2

2

22

xExkx

dx

xdnnn

n

Find the form of solution

)()2

()(

),()(2

)(

2 22

22

22

2

22

xExk

dx

xdxEx

kx

dx

xd

E term can be ignored because of x2 >> E when y2 → ∞Multiply the 2dψ/dx both side and use the product rule


Text p.104


222

2222

2222

22

2

2

22

2

2

222

22

22

2

)dx

d(x)

dx

d(x2

))dx

d(x2x4()

dx

d(x4x4

)2)(dx

]x[d(])

dx

d(x2[

dx

d)

dx

d2(x

])dx

d[(

dx

d)

dx

d2)(

dx

d(),

dx

d2)(

dx

d(])

dx

d(2[

dx

d)

dx

d2)(

dx

d(

)dx

d2(x

k)

dx

d2)(

dx

d(,x

k

dx

d


22

222

2222

2 x2k

)x(dx

dk])

dx

d[(

dx

d,0)

dx

d(x

k])

dx

d[(

dx

d

Assume right term is much smaller than left term

)k

(Ce:0)x(

Ce or Ce,c2

xkln,xdx

kd,xdx

kd

xk

dx

d),x(

k)

dx

d(,0)x(

dx

dk])

dx

d[(

dx

d

22

x

x

2

x

2

x2

222

222

2222

22

2

2

22

lim

Solution : ‘Gaussian’ formTherefore, we can assume the solution

2

2)()(x

exhx


Schrödinger equation is rewritten by

equation Hermite : 0)()2

()(

)2()(

22

2

xhEdx

xdhx

dx

xhd

Hermite equation is already solved : hermite polynomialsSolution is given by

22

22

2

x234

13

3

x224

1

2

x24

13

1

x24

1

0

4

1

nn

x2

nnn

e)x3x2()9()x(,e)1x2()

4()x(

xe)4()x(,e)()x(

)(!n2

1A

,...2,1,0n,e)x(HA)x(

Even state(n:even) : ψ(-x)=ψ(x) : even functionOdd state(n:odd) : ψ(-x)=-ψ(x) : odd function


Eigenvalue is given by ,...2,1,0),2

1()

2

1( nnhn

kEn

There are 2 different phenomenon to classical H.O 1) energy of ground state is not zero : ZPE 2) particle can be found in the classical forbidden region

Probability in the interval ∆x :

2/

2/

2 )(x

x

dxx


Time-dependent solution

)()()()(),(),(

),(),(

*** xxxxeetxtx

kxetx

nnnntiti

nn

nti

n

: standing wave

Probability density of 12th state of H.O

7.2 Solving the Schrödinger equation for rotation in 2-dimensions


Neglect the coupling, hamiltonian operator is sum of individual operators for the degrees of freedom for the molecule

),(ˆ)(ˆ)(ˆˆCMCMrotinternalvibCMtranstotal HHrHH

And, total energy also can divide to each energy

rotvibtranstotal EEEE

Finally, total wavefunction is product of eigenfunctions of each operator

),()()( CMCMrotinternalvibCMtranstotal r


Set V=0 : no vibration(make easier problem)Rotation : internal motion → motion of reduced mass

),()),(),(

(2 02

2

2

22

yxEy

yx

x

yxrr

Laplacian in 2-dimension 2

2

22 1

)(1

rr

rrr

Fixed r : radial term canceledSolution is given by angular term

ll imim eAeA

Ed

d

r

)(,)(

)()(

2 2

2

20

2

It means clockwise and counterclockwise rotation


Ex) 7.4Normalize the rotational wavefunctions in 2-dimension

Sol)

2

1,1)()(

2

1,1)()(

1)()(

)(,)(

2

0

22

0

2

2

0

22

0

2

2

0

*

AdAdeeA

AdAdeeA

d

eAeA

ll

ll

ll

ll

imim

imim

mm

imim


Boundary condition : ‘quantization’ of angular momentum

Angular momentum must be periodic function because of φ+2π= φ always satisfies. → indistinguishable values φ and φ+2nπ

1,),()2( 2]2[ lll imimim eee

Use Euler’s relation 12sin2cos ll mim

condition of ml : 0, ±1, ±2, ±3, … : quantization of angular momentum


Energy of the rotation

,...3,2,1,0,22

22

20

22

lll

m mI

m

r

mE

l

State of +ml and –ml : same energy and orthogonal each other → 2-fold degenerate with ml ≠ 0 level

22

20

2

2

1

2

||

2

||

II

l

r

lE

l : angular momentum vector, : angular momentum operatorzl̂

)(22

)(ˆ

)(22

)(ˆ

ˆ

liml

im

z

liml

im

z

z

mem

d

deil

mem

d

deil

il

l

l

l

l


Φ+(φ) , Φ-(φ) : eigenfunction of both of hamiltonian and momentum operatorEigenvalue of momentum operator : + mℏ l and - mℏ l

Then, we can obtain the similar form as the C.M

I

l

I

mE l

ml 2

||

2

222

Probability of angular motion

2)

2

1()()()( 2* d

deeddP ll imim

: same for all region


7.3 Solving the Schrödinger equation for rotation in 3-dimension

3-dimensional rigid rotor : similar than 2-dimensional problem

Laplacian in spherical coordinate is given by

Rigid rotor : ‘fixed r’ → r term canceledLike the 2-dimension problem, we can write the Schrödinger equation the

2

2

2222

22

sin

1)(sin

sin

1)(

1

rrr

rrr

),(]),(

sin

1)

),((sin

sin

1[

2 2

2

220

2

EYYY

r


Define the β=2μr02E/ℏ2

Equation is changed by

2

22 ),(

),(sin)),(

(sinsin

Y

YY

Use the separation of variable : Y(θ,φ) = Θ(θ)Φ(φ)

2

22

2

22

2

22

d

)(d

)(

1sin)

d

)(d(sin

d

dsin

)(

1

d

)(d)()()(sin)

d

)(d(sin

d

dsin)(

)()()()(sin)

)()((sinsin

Solve it by the left part and right part is ‘constant’


cd

d

d

d

d

d

2

22 )(

)(

1sin)

)((sinsin

)(

1

Right part : similar to 2-dimensional problem : set c = ml

2

Equation can be change to two ODEs.

22

2

22

)(

)(

1

sin))(

(sinsin)(

1

l

l

md

d

md

d

d

d

Second equation : same as the 2-dimensional problem

...3,2,1,0,)(,)( l

imim meAeA ll

φ part of Y(θ,φ) : depends on ml


)(sin

)()cot( 2

2

2

2

lm

d

d

d

d

Solve the first equation : Legendre’s equation

Set z = cos θ and use it, equation change to

2

22

2

2

sincos)sin(,sindz

d

dz

d

dz

d

d

d

d

d

dz

d

dz

d

d

dz

d

d

0)(]1

2)1[( 2

2

2

22

z

z

m

dz

dz

dz

dz l

Case of ml = 0 : Legendre’s equationUse the power series, write the solution P(z) instead of Θ(z)

0

)(n

nnzazP


Recurrence relation is given by nn ann

nna

)2)(1(

)1(2

If β ≠ integer, this series will not terminate. However, it cannot be solution because it diverge at z=1.Why?

By the ratio test, series diverges at z=1 and it cannot the solution of wavefunction!

If β = integer, well-behaved wavefunction exists and eigenvalue of equation is given by β = l(l+1) (set n=l)

P(z) is called the Legendre polynomials.

12lim

n

n

n a

a

)33035(8

1)(),35(

2

1)(

)13(2

1)(,)(,1)(

244

33

2210

zzzPzzzP

zzPzzPzP


Case of ml ≠ 0 : Associated Legendre polynomialWrite the solution as the Pl

ml(z)

0]1

)1([])1[( 2

22

l

lm

ll

ml P

z

mll

dz

dPz

dz

d

set Plml(z) = (1 - z2)m/2F(z) and equation is given by

)z(F)z1)](1l(lz1

m[

)z(F)z1](1z)1m(m[dz

)z(dF)z1)(1m(z2

dz

)z(Fd)z1(

)z(F)z1](1z)1m(m[

dz

)z(dF)z1)(1m(z2

dz

)z(Fd)z1(]

dz

dP)z1[(

dz

d

dz

)z(dF)z1()z(F)z1(zm

dz

dP

2

m2

2

2l

12

m22

ll2

m2

l2

21

2

m2

12

m22

ll

2

m2

l2

21

2

m2

ml2

212

m2

l

ml

l

lll

l

lll

ll


Finally, equation is changed to

0)()1)(()(

)1(2)(

)1( 2

22 zFmlml

dz

zdFmz

dz

zFdz l

Solution is given by

)(cos)()(

])()1()(0, [consider )()()1()(

)()()(

||22

l

llll

l

l

l

ml

ml

mmlll

mm

ml

lm

PzP

zPzPmzPdz

dzzP

zPdz

dzF

ml must be | ml | ≤ l : if more than l times of differentiation, wavefunction becomes zero and it is not allowed state.Therefore, quantum number is given by β = l(l+1), l = 0,1,2,3… ml = -l, -(l-1), … , -1, 0, 1, …, l-1, l

Wavefunction Ylml(θ,φ) is given by

)()(cos)()(),( l

llm

ml

ml PY


Energy of angular momentum

...3,2,1,0),1(2

),1(2 2

2

20 lll

IEll

Erl

And this notation satisfies, too.

...3,2,1,0),,()1(2

),(ˆ2

lYllI

YH ll ml

mltotal

Case of 2-dimensional rotation : 2-fold degenerecy

In this problem(3-dimentional rotation) : 2l+1 degenerecyThere are 2l+1 ml values per one l value, and these states have same energy!

7.4 The quantization of angular momentum


Energy of angular motion is given by

Difference between |l2| and E : divide by 2ITherefore, hamiltonian and operator also satisfy same relationship.

Total energy quantized → |l|2 quantizedWe can write for operator

I

lEtotal 2

|| 2

2l̂

2l̂

),()1(),(ˆ 22 ll ml

ml YllYl

Therefore, value of |l| is given by )1(|| lll

commute hamiltonian, notThen, has 3 component : lx,ly,lz, obtained by the l = r x p

2l̂ l̂l̂


Can calculate this formula

)(ˆ),(ˆ),(ˆx

yy

xilz

xx

zily

zz

yil zyx

Angular momentum operator in spherical coordinate is

)(ˆ),sincot(cosˆ),coscot(sinˆ

ililil zyx

Commutator relation is given by

]ˆ,ˆ[]ˆ,ˆ[

ˆ]ˆ,ˆ[,ˆ]ˆ,ˆ[,ˆ]ˆ,ˆ[

xyyx

yxzxzyzyx

llll

lilllilllill

lx, ly, lz are not commute.


How can obtain the component of angular momentum? → see lz : simplest form(only depends on φ)

lmmeiYl llimm

lzll

,...,3,2,1,0),()()]2

1()[()),((ˆ

Ylml(θ,φ) is eigenfunction of lz

→ Ylml(θ,φ) is eigenfunction of both and lz

Therefore, we can choose and lz can solve the problem easily. Also, we can know the length of angular momentum l and value of z-component lz, but we cannot know the value of x and y component.

Why z component is special? → no special! We can choose another direction and it also commute to . It means z component is simple only in the spherical coordinate and ‘only 1’ component of angular momentum is commute with .

2l̂

2l̂

2l̂

2l̂

7.5 The spherical harmonic functions


We see the spherical harmonic functions

i

i

i

eY

eY

Y

eY

Y

Y

222

12

2

2

11

2

22

10

2

2

11

1

2

10

1

2

10

0

sin)32

15(),(

cossin)8

15(),(

)1cos3()16

5(),(

sin)8

3(),(

cos)4

3(),(

)4

1(),(


Spherical harmonic functions : ‘complex’Make the function ‘real’ by the linear combinationReal wavefunctions(we called it ‘orbital’) are orthonormal, too.

2sinsin16

15)(

2

1

2cossin16

15)(

2

1

sincossin4

15)(

2

1

coscossin4

15)(

2

1

)1cos3(16

5

cos4

3

sinsin4

3)(

2

1

cossin4

3)(

2

1

222

22

222

22

12

12

12

12

202

10

11

11

11

11

22

2

YYi

d

YYd

YYi

d

YYd

Yd

Yp

YYi

p

YYp

xy

yx

yz

xz

z

z

y

x


Shape of p and d orbitals. We can see the each orbital is perpendicular.


Superposition of p and d orbital

Magnitude of pz and py orbital


7.6 The classical harmonic oscillator

Example of oscillator : two masses connect by spring

Introduce the center-of-mass coordinate and relative position

1221

2211 , xxxmm

xmxmxCM


Differentiate by time, we can obtain the center-of-mass velocity and relative velocity

1212

221

21

21

12

21

21

21

1

vvdt

dx

dt

dx

dt

dxv

vmm

mv

mm

m

dt

dx

mm

m

dt

dx

mm

m

dt

dxv CM

CM

Total energy of 2-mass system 222

211 2

1

2

1vmvmE

Change the total energy by the vCM and v.

vmm

mvv)vv

mm

m(

m

mv

m

mmv

vvmm

mvv

m

mmv

m

mmv,v

m

mv

m

mmv

vvv,vmm

mv

mm

mv

21

2CM1CM

21

1

1

2CM

1

211

CM21

12CM

1

212

1

212

1

2CM

1

211

12221

21

21

1CM


222

21

21221

22

21

12

21

12

22

22

21

21

21

21

21

2

21

12

2

21

21

222

211

2

1

2

1

2

1)(

2

1

)(2

1

2

1

)(2

1

2

1

)(2

1)(

2

1

2

1

2

1

vMvvmm

mmvmm

vmm

mmvv

mm

mmvm

vmm

mmvv

mm

mmvm

vvmm

mmv

mm

mvmvmvmE

CMCM

CMCM

CMCM

CMCM

vCM : velocity of ‘whole system’ : independent to motion of internal systemv : relative velocity : dependent to motion of internal systemRestoring force act to ‘internal system’ → restoring force act to reduced mass


0, 2

2

2

2

kxdt

xd

dt

xdaF

Therefore, we can divide this motion by two motions.(whole motion : motion of center of mass + motion of reduced mass)

Our focus is motion by restoring force.If oscillator is harmonic oscillator, force is given by F = -kx

Solution is given by

tk

btk

b

tk

itk

ctk

itk

cecectx tkitki

sincos

)sin(cos)sin(cos)(

21

21/

2/

1


If initial condition is x(0)=0,v(0)=v0

tk

vk

tx

vk

bvk

bkk

bdt

dxvb

kbx t

sin)(

,0cos|)0(,00cos)0(

0

02022011

)sin()(or sincos 2sin2cos)(

2,2

11 ,2

2121

tAtxtbtbT

tb

T

tbtx

kk

TkT

Potential & kinetic energy of harmonic oscillator

22

2

1,

2

1, vEkxkxdxFdxEkxEF kineticpotentialpotential

Classical harmonic oscillator : ‘continuous energy spectrum’


7.7 Angular motion and the classical rigid rotor


Rotation of 2-particle : centered at center of mass

Consider the constant r = r1+r2

Kinetic energy of system is given by

22

21221221

21

221

2

12

1

22

1

21

2

121

22

1

222

21

2

121

22

1

21

21

2121

2

21

21

21

222

211

22112121

221121

2

1)(

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

12

1

2

1

,,,

vvvvvvvvv

vvm

mv

m

mv

m

mv

m

mv

vm

mvv

m

mv

vm

mm

mm

mmv

m

mm

mm

mm

vmvmE

vmvmvvdt

dr

dt

dr

dt

drvrmrmrrr


t

v

tt

vva

vvv

tt

012

12

0

12

limlim

Centripetal acceleration r

tva lcentripeta

2|)(|

Angular velocity and angular acceleration is given by

2

2

,dt

d

dt

d

dt

d

r

dt

rdvt

t

r

t

sv

,0,


Direction of angular velocity and angular acceleration : right-hand rule

2222

2000

0

2

1

2

1

2

12

1,

IrvE

ttt

t

kinetic

Case of constant acceleration

I = μr2 : moment of inertia


Angular momentum l is defined by

prl x : cross product

sinsin vrprl

Magnitude or Angular momentum is

φ : angle between p and r

Kinetic energy is given byI

l

r

lpE

222

2

2

22

Classical rigid rotor : continuous energy

7.8 Spatial quantization


See the angular momentum.First, we see the semiclassical description

angular momentum cannot lie on the z-axis. Why?| ml | ≤ l is condition of ml and magnitude of l is given by

Therefore, if the case of ml = l (extreme case) → z-component cannot be same as the magnitude of angular momentum.

Angular momentum lie on the z-axis : x, y component = 0→ know 3 component simultaneouslyBut it cannot be possible because commutator is not zero!

)1( ll

lll mmm )1(


If we know the total angular momentum and z-component, then we cannot know the x and y component and only we know the

→ cone has an open end

Finally, we can see the l=2 case(d orbital, too), vector model of angular momentum

Vector of angular momentum only have certain orientation in space. → spatial quantization

c.f) classical case : possible l values make the surface of sphere, not a cone

222222 ])1([ lyxz mllllll


- Quantum mechanics is used to study the vibration and rotation of a diatomic molecules.

- Vibrational degree of freedom modeled by the harmonic oscillator was considered.

- The harmonic oscillator has a discrete energy spectrum like the particle in the box in Q.M

- The Q.M model for rotational motion providing a basis for understanding the orbital motion of electrons around the nucleus of an atom as well as the rotation of a molecule about its principal axes was formulated.

Summary Summary

Documents

Ch 7. A Quantum Mechanical Model for the Vibration and Rotation of Molecules MS310 Quantum Physical Chemistry - Schrödinger eq. for the Q.M. harmonic oscillator