6 Introduction to Quantum Mechanics

Introductory Quantum Mechanics

Serway/Jewett chapters 40.8; 41

What are matter waves?• First consider EM radiation• Intensity of a wave is the energy delivered per unit area per unit

time. This may be expressed as:

Intensity ~ (amplitude of the wave)2

• The intensity of EM waves has a natural interpretation in terms of particles, since these beams may be described as a beam of particles – photons. Intensity in the photon model is:

• If N is the [average] density of photons, in the beam (number of photons per unit volume). Consider the number of photons passing through and area A per second. The relevant volume of photons is cΔtA

))((

)(photons) (#

tA

hI

hNc

tA

htANc

tA

hI

tANcVN

))(()(photons) (#

photons) (#

• For light:

– Speed of light, c, is a constant– For a given EM wave, photon energy, hν, is a constant– The only variable reflecting change in the light

intensity is the average density of photons, N, c and ν are constants

• Therefore,

What are matter waves?

2 wave) theof amplitude( hNcI

2 wave)the of amplitude(N

What are matter waves?• N must also be considered from a statistical

point of view. Just as we saw for the cases of partially reflected light or the energy delivered to a screen. Need to ask the question:– What is the probability that a photon is reflected

or of finding a given number of photons in a given volume of the beam?

• The answer is:– The probability of finding a number X of photons

in a volume V is given by ratio X/V

• The average photon density in this case is calculated in the same way: N = X/V !

Since, N is also proportional to the square of the amplitude of the

corresponding wave, we conclude that

The probability of finding photon within a given volume of the beam is

proportional to the square of the amplitude of the

wave associated with this beam

What are the matter waves?• Thus, a wave (or particularly when it what

is waving is abstract, we say wave function) provides the basis for describing the probability density of particles in a beam

• But not only EM radiation has a dual nature. So does matter.

• Max Born extended this interpretation to the matter waves proposed by De Broglie, by assigning a mathematical function, Ψ(r,t), called the wavefunction to every “material” particle

Ψ(r,t) is what is “waving”

Definition of Ψ(r,t)• The probability P(r,t)dV to find a particle

associated with the wavefunction Ψ(r,t) within a small volume dV around a point in space with coordinate r at some instant t is

– P(r,t) is the probability density

• For one-dimensional case

dVtdVtP 2),(),( rr

dxtxdVtxP 2),(),(

Here |Ψ(r,t)|2 = Ψ*(r,t)Ψ(r,t)

Definition of Ψ(r,t)• The probability of

funding a particle somewhere in a volume V of space is– Since the probability to find

particle anywhere in space is 1, we have condition of normalization

• For one-dimensional case, the probability of funding the particle in the arbitrary interval a ≤ x ≤ b is

VV

V dVtdVtPP 2),(),( rr

1),(

2 spaceall

dVtr

dxtxP

dxtxP

everywhere

b

a

ab

2

2

),(

),(

Quantum Mechanics

• The methods of Quantum Mechanics consist in finding the wavefunction associated with a

particle or a system

• Once we know this wavefunction we know “everything” about the

system!

The Uncertainty Principle

• Since we deal with probabilities we have to ask ourselves: “How precise is our knowledge?”

• Specifically, we want to know Coordinate and Momentum of a particle at time t = 0– If we know the forces acting upon the

particle than, according to classical physics, we know everything about a particle at any moment in the future

• But it is impossible to give the precise position of a wave

• A wave is naturally spread out

• Consider the case of diffraction

• Most of the energy arriving at a distant screen falls within the first maximum



• Can we know Coordinate and Momentum (velocity) at some exactly, if we deal here with probabilities?

• The answer in Quantum Mechanics is different from that in Classical Physics, and is encapsulated in the Heisenberg’s Uncertainty Principle

Classical Uncertainty

d

•Consider classical diffraction•Most of light falls within first maximum•The angular limit of the first maximum is at the first zero of intensity which occurs at an angle set by the condition, d sin = , so we can say that the angle of light is between + and -Consider the following:

4~,2~2

have,So

2~and,Now

2~sin

2~sin2

,1~sin

minimumfirst at ,~sin

yy

yy

kyk

y

kkyd

kd

dord

d

As the uncertainty in y increases the uncertainty in the y-component of the k-vector decreases

Classical Uncertainty

~4~

4~)(

by4~

y

y

y

py

ky

kyMultiply

The classical uncertainty relation


An experiment cannot simultaneously determine a

component of the momentum of a particle (e.g., px) and the

exact value of the corresponding coordinate, x.

The best one can do is

2))((

xpx


1. The limitations imposed by the uncertainty principle have nothing to do with quality of the experimental equipment

2. The uncertainty principle does imply that one cannot determine the position or the momentum with arbitrary accuracy– It refers to the impossibility of precise

knowledge about both: e.g. if Δx = 0, then Δ

px is infinity, and vice versa

3. The uncertainty principle is confirmed by experiment, and is a direct consequence of the de Broglie’s hypothesis

HOWEVER

• Since the wavefunction, Ψ(x,t), describes a particle, its evolution in time under the action of the wave equation describes the future history of the particle– Ψ(x,t) is determined by Ψ(x, t = 0)

• Thus, instead of the coordinate and velocity at t = 0 we want to know the wavefunction at t = 0– Thus uncertainty is built in from the

beginning

Examples: Bullet

• p = mv = 0.1 kg × 1000 m/s = 100 kg·m/s– If Δp = 0.01% p = 0.01 kg·m/s

– Which is much more smaller than size of the atoms the bullet made of!

– So for practical purposes we can know the position of the bullet precisely

m 1005.1m/skg 0.01

s J1005.1 3234

px

Examples:

• Electron (m = 9.11×10-31 kg) with energy 4.9 eV

• Assume Δp = 0.01% p

– Which is much larger than the size of the atom!

– So uncertainty plays a key role on atomic scale

A10m10

102.11005.1

kg·m/s 101.2 0.01%

s/mkg 102.1J106.19.4kg101.922

4628

34

28-

241931

-px

pp

mEp

Physical Origin of the Uncertainty PrincipleHeisenberg (Bohr) Microscope

• The measurement itself introduces the uncertainty

• When we “look” at an object we see it via the photons that are detected by the microscope– These are the photons that are

scattered within an angle 2θ and collected by a lens of diameter D

– Momentum of electron is changed– Consider single photon, this will

introduce the minimum uncertainty

Heisenberg (Bohr) Microscope

sin2maxphph pp

sin2 phphelectron ppp

hp

hpp

hp

electron

phelectron

ph

2

sin , small for

sin2

sin2

Trying to locate electron we introduce the uncertainty of the

momentum

As a consequence of momentum conservation


• θ~(D/2)/L, L ~ D/2 θ is distance to lens

• Uncertainty in electron position for small θ is

• To reduce uncertainty in the momentum, we can either increase the wavelength or reduce the angle

• But this leads to increased uncertainty in the position, since

2

22.1

22.12

electron

elelctron

x

D

Dxh

pelectron2

DDLxelectron )2)(2/()2(


hxp

xh

p

DxD

hp

electron

electron

electron

22.1))((

22.1222.1

2


• Between energy and time

2))((

tE

EM Waves in Empty Space

• From Maxwell’s equations in differential form one may derive the following partial derivatives for the fields in empty space:

• The simplest solution for the wave equation is a sinusoidal wave: E = Emax sin (kx – ωt)

B = Bmax sin (kx – ωt)– The wave number is k = 2π/ λ (λ is the

wavelength)– The angular frequency is ω = 2πƒ (ƒ =1/T is the

wave frequency)

2 2 2 2

2 2 2 2o o o o

E E B Bμ ε and μ ε

x t x t

1

o o

v cμ ε

= 2.99792 x 108 m/s

Classical Wave Operators

• Consider the wave equation for light

• For a traveling wave solutions are the eigenvectors, sin(kx-t), and eigenvalues of the operator on the left hand side are -k2

• For a standing wave between two reflecting mirros separated by a distance a, the eigenvalues are, -kn

2 = n22/a2

2

2

22

2 ),(1),(

t

txE

vx

txE

• Between perfectly reflecting walls have solutions for the wave equations:

• Solutions satisfying the boundary conditions, E(x=0,a) = 0 require, sin(ka) = 0. This gives, kna = n or kn = n/a and n = ckn = cn/a, n = 1,2,3…

• We may say then that the wave, E(x,t) above is a solution of the wave equation and that the eigenfunctions of the wave equation are

• The eigenvalue of the operator is -kn2 = n22/a2

• The eigenvalue of the operator is -n2/v2

EM Waves between Reflecting Walls

tkxAtkxAtkxAtxE cossin2)sin()sin(),(

txkAtxE nnn cossin2),(

2

2

x

2

2

22

2 ),(1),(

t

txE

vx

txE

2

2

2

1

tv } -kn

2 =-n2/v2

• If particles have a wave description then they should obey a wave equation.

• A wave equation has a general form, where and are each a function of r or t, say is a function of r and is a function of t

• A wave in 1D of a particular k-vector and frequency is an eigenvector of the wave equation. It will have the form Ψk,(x,t)=Asin(kx-t). Then– Ψk,(x,t) = f(k)Asin(kx-t)– Ψk,(x,t) = g()Asin(kx-t)– Here f() and g() are eigenvectors of the respective

operators

• If particles have a wave description then they should obey a wave equation

• A wave equation has a general form, where and are each a function of r or t, say is a function of r and is a function of t

• A wave in 1D of a particular k-vector and frequency is an eigenvector of the wave equation. It will have the form Ψk,(x,t)=Asin(kx-t).

Quantum Mechanical Operators

),(ˆ),(ˆ21 trOtrO

1O 2O 1O

2O

1O

2O

2O

• For specific boundary conditions only specific values of

k and , kn and n, satisfy the boundary conditions

Physical Quantity Operators

symbol actual operation

Momentum

Total Energy

Coordinate

Potential Energy


)()(ˆ)(

ˆ

ˆ

ˆ

xUxUxU

xxxt

iEE

xipp xx

Guess

Physical Quantity Operators

symbol actual operation

Momentum

Total Energy

Coordinate

Potential Energy


)()(ˆ)(

ˆ

ˆ

ˆ

xUxUxU

xxxt

iEE

xipp xx

Motivation?

Wave Function of Free Particle

• Since the de Broglie expression is true for any particle, we assume that free particles can be described by a traveling wave, i.e. the wavefunction of a free particle is a traveling wave

• For classical waves:]sin[],cos[ tkxAtkxA

Wave Function of Free Particle• However, these functions are not eigenfunctions of the momentum operator,

with them we do not find,

• But let’s try operating on the following wavefunction with ,

• Get same result of course if operate on )exp(]sin[]cos[),( tikxAtkxitkxAtx

),(),(ˆ )()()( txkkAeikAei

Aexi

txp tkxitkxitkxix

xx pphh

kpxi

p

2

2ˆ

)exp(]sin[]cos[),( tikxAtkxitkxAtx

xp

Wave Function of Free Particle

• Similarly can operate on

with ,

txEAeAeiiAet

itxE tkxitkxitkxi ,)(,ˆ )()()(

This wave function is an eigenfunction of bothmomentum and energy

)exp(),( tikxAtx

tiE

ˆ

Expectation Values• Only average values of physical quantities can be

determined (can’t determine value of a quantity at a point)

• These average values are called Expectation Values– These are values of physical quantities that quantum

mechanics predicts and which, from experimental point of view, are averages of multiple measurements

• Example, [expected] position of the particle

1)( with,)(

dxxPdxxxPx

Expectation Values• Since P(r,t)dV=|Ψ(r,t)|2dV, we have a way to

calculate expectation values if the wavefunction for the system (or particle) is known

• In General for a Physical Quantity W– Below Ŵ is an operator (discussed later)

acting on wavefunction Ψ(r,t)

dxtxxtxx

txtxtxdxtxxdxtxxPx

),(),(

),(),(),( since ,),(),(

*

*22

dxtxWtxW ),(ˆ),(*

Expectation Value for Momentum of a Free Particle

• Generally

• Free Particle

dxx

xxip

dxxx

ixdxxpxp

)()(

)()()(ˆ)(

*

**

pkdxAeAekdxAeiki

Aep

dxAexi

Aep

A

dxAeAedxxAex

ikxikxikxikx

ikxikx

ikxikxikx

**

*

*2

n integratio of limits as 0 where

,1)( with )(

Properties of the Wavefunction and its First Derivative

1. must be finite for all x2. must be single-valued for all x3. must be continuous for all x

dxxx

ixpx )()(*

dxxxUxxU )()()()( *

dxtxt

itxE ),(),(*

Schrödinger Equation• Schrödinger developed the wave equation which can be

solved to find the wavefunction by translating the equation for energy of classical physics into the language of waves

• For fixed energy, we obtain the time-independent Schrödinger equation, which describes stationary states

• the energy of such states does not change with time– ψn(x) is an eigenfunction or eigenstate– U is a potential function representing the particle interaction

with the environment

xExxUxx

m

)(

2 2

22

xt

ixxUx

x

m

)(

2 2

22

ExUm

p )(

2

2

Particle in a box with “Infinite Barriers”

• A particle is confined to a one-dimensional region of space between two impenetrable walls separated by distance L– This is a one- dimensional “box”

• The particle is bouncing elastically back and forth between the walls– As long as the particle is inside the box, the

potential energy does not depend on its location. We can choose this energy value to be zero

• U(x) = 0, 0 < x < L, U(x) , x ≤ 0 and x ≥ L

• Since walls are impenetrable, we say that this models a box (potential well) has infinite barriers

Particle in a box with “Infinite Barriers”

• Since the walls are impenetrable, there is zero probability of finding the particle outside the box. Zero probability means that ψ(x) = 0, for x < 0 and x > L

• The wave function must also be 0 at the walls (x = 0 and x = L), since the wavefunction must be continuous– Mathematically, ψ(0) = 0 and

ψ(L) = 0

Schrödinger Equation Applied to a Particle in a Box

• In the region 0 < x < L, where U(x) = 0, the Schrödinger equation can be expressed in the form

• We can re-write it as

xExxUxx

m

)(

2 2

22

xEx

xm

2

22

2

22

22

2

22

2

2

2

mEk

xkx

x

xmE

x

x


• The most general solution to this differential equation is

ψ(x) = A sin kx + B cos kx– A and B are constants determined by the

properties of the wavefunction as well as boundary and normalization conditions

xkx

x 22

2


1. Sin(x) and Cos(x) are finite and single-valued functions2. Continuity: ψ(0) = ψ(L) = 0

• ψ(0) = A sin(k0) + B cos(k0) = 0 B = 0 ψ(x) = A sin(kx)

• ψ(L) = A sin(kL) = 0 sin(kL) = 0 kL = nπ, n = ±1, ±2…

22

22

2

2222

22

822

)(

2

nmL

hn

mLm

nLE

m

kEn

Lk

n

nnn

• The allowed wave functions are given by

• The wave function for one electron is obtianed by normalizing the wave function


x

Ln

A(x) ψn

sin

x

Ln

L (x) ψn

sin

2

Particle in the Well with Infinite Barriers

01

2

22

02

02

2

22

, )1( 2

,2

EEenergyn stategroundmL

EwithnEnmL

En

http://www.falstad.com/qm1d/ http://www.falstad.com/mathphysics.html

http://www.falstad.com/qm1d/

http://www.falstad.com/mathphysics.html

http://www.falstad.com/mathphysics.html

Electron in the 10nm Wide Well with Infinite Barriers

• Calculate E0 for L = 10nm = 10×10-9m

• Assume that a photon is absorbed, and the electron is transferred from the ground state (n = 1) to the second excited state (n = 3)

• What was the wavelengths of the photon?

2

22

012

02

where,mL

EEnEEn

eV 10 meV 1

meV 753eV 00375.0J 106

)1010(101.92

)1005.1(14.3

3-

220

2931

2342

0

.E

E

Electron in a 10 nm Wide Potential Well with Infinite Barriers

• E0 = 3.75 meV

eV 00375.001 EEEground

eV 0.0338eV 00375.093

is state excited Third2

03

3

EE

E

μm 41nm 41333030

1240

eV 03000037500.0338)( 13

.λ

..EEh

Particle in the Infinite Potential Well

22

22

th

2

sin

2

state For the

nmL

E

nL

x

L

n

n

n

Probability to Find particle in the Right Half of the Well

2

1]

2

1

2[

2)(sin

2]sin

2[|)(|

2/

2

2/

2

2/

2 L

Ldxkx

Ldxkx

Ldxx

L

L

L

L

L

L

Average Momentum of Particle in a Box (Infinite Potential Well)

• Can evaluate the integral and show it is zero • Can note that the right hand side is either 0 or

imaginary, but momentum cannot be imaginary so it must be zero

0)cos()sin(2

sin2

]sin2

[)()(

0

00

**

L

LL

dxkxkxkiL

dxx

kxL

ikx

Ldxx

xixp

Finite Potential Well• The potential energy is

zero (U(x) = 0) when the particle is 0 < x < L (Region II)

• The energy has a finite value (U(x) = U) outside this region, i.e. for x < 0 and x > L (Regions I and III)

• We also assume that energy of the particle, E, is less than the “height” of the barrier, i.e. E < U

Finite Potential WellSchrödinger Equation

I. x < 0; U(x) = U

II. 0 < x < L; U(x) = 0

III. x > L; U(x) = 0

IIII E

dx

dm

2

22

2

III EU

dx

d

m

2

22

2

xExxUx

xm

)(2 2

22

IIIIIIIII EU

dx

d

m

2

22

2

Finite Potential Well: Region II

• U(x) = 0– This is the same situation as

previously for infinite potential well

– The allowed wave functions are sinusoidal

• The general solution of the Schrödinger equation is

ψII(x) = F sin kx + G cos kx

– where F and G are constants

• The boundary conditions , however, no longer require that ψ(x) be zero at the sides of the well

Finite Potential Well: Regions I and III• The Schrödinger equation for these regions is

• It can be re-written as

• The general solution of this equation is

ψ(x) = AeCx + Be-Cx

– A and B are constants– Note (E-U) is the negative of kinetic energy, -Ek – In region II, C is imaginary and so have sinusoidal solutions we found– In both regions I and III,

and ψ(x) is exponential

EU

dx

d

m

2

22

2

222

22

2 )(2 where,

)(2

EUm

CCEUm

dx

d

mU

C2

Finite Potential Well – Regions I and III

• Requires that wavefunction, ψ(x) = AeCx + Be-Cx not diverge as x ∞

• So in region I, B = 0, and ψI(x) = AeCx

– to avoid an infinite value for ψ(x) for large negative values of x

• In region III, A = 0, and ψIII(x) = Be-Cx

– to avoid an infinite value for ψ(x) for large positive values of x

Finite Potential Well

• Thus, we equate the two expressions for the wavefunction and its derivative at x = 0, L – This, together with the

normalization condition, determines the amplitudes of the wavefunction and the constants in the exponential term.

– This determines the allowed energies of the particle

• The wavefunction and its derivative must be single-valued for all x– There are two points at which wavefunction is given by

two different functions: x = 0 and x = L

Ldx

dL

dxd

LLdx

ddx

d

IIIII

IIIII

III

III

)()(

00

)0()0(

Finite Potential WellGraphical Results for ψ (x)

• Outside the potential well, classical physics forbids the presence of the particle

• Quantum mechanics shows the wave function decays exponentially to zero

Finite Potential WellGraphical Results for Probability

Density, | ψ (x) |2

• The probability densities for the lowest three states are shown

• The functions are smooth at the boundaries

• Outside the box, the probability of finding the particle decreases exponentially, but it is not zero!

http://phys.educ.ksu.edu/vqm/html/probillustrator.html




Tunneling• The potential energy has a

constant value U in the region of width L and zero in all other regions

• This a called a barrier• U is the called the barrier

height. Classically, the particle is reflected by the barrier– Regions II and III would be forbidden

• According to quantum mechanics, all regions are accessible to the particle– The probability of the particle being in a classically

forbidden region is low, but not zero– Amplitude of the wave is reduced in the barrier – A fraction of the beam penetrates the barrier– http://phys.educ.ksu.edu/vqm/html/qtunneling.html– http://phet.colorado.edu/web-pages/simulations-base.html

http://phys.educ.ksu.edu/vqm/html/qtunneling.html

http://phet.colorado.edu/web-pages/simulations-base.html

More Applications of TunnelingResonant Tunneling Device

• Electrons travel in the gallium arsenide• They strike the barrier of the quantum dot

from the left• The electrons can tunnel through the barrier

and produce a current in the device

More Applications of Tunneling Scanning Tunneling Microscope

• An electrically conducting probe with a very sharp edge is brought near the surface to be studied

• The empty space between the tip and the surface represents the “barrier”

• The tip and the surface are two walls of the “potential well”

Simple Harmonic Oscillator

• To explain blackbody radiation Planck postulated that the energy of a simple harmonic oscillator is quantized– In his model vibrating charges act as

simple harmonic oscillators and emit EM radiation

• The quantization of energy of harmonic oscillators is predicted by QM

Simple Harmonic Oscillator (SHO)• Let’s write down the Schrödinger Equation for SHO• For SHO the potential energy is

• Time independent Schrödinger Equation for SHO in 1D

mk

xmkxxU

22

)(222

xExxm

x

xm

22

22

2

22

Simple Harmonic Oscillator• Solutions of time-independent Schrödinger

equation for 1D harmonic oscillator

xExxm

x

x

m

22

22

2

22

Simple Harmonic Oscillator• Planck’s expression for

energy of SHO

• Energy of SHO obtained from the solution of the Schrödinger equation– Thus, the Planck

formula arises from the Schrödinger equation naturally

– n = 0 is the ground state with energy ½hν

2 ;2

,...3,2,1,021

21

h

n

hnnE

nhE

Term ½hν tells us that quantum SHO always oscillates. These are

called zero point vibrations

Simple Harmonic Oscillator• Energy of SHO from the Schrödinger equation

• The zero point energy ½hν is required by the Heisenberg uncertainty relationship

• The term of ½hν is important for understanding of some physical phenomena

• For example, this qualitative explains why helium does not become solid under normal conditions – the “zero point vibration” energy is higher than the

“melting energy” of helium

• Force between two metal plates

hnhE21

Quantum Model of the Hydrogen Atom

• Potential Energy

• Time-independent Schrödinger Equation

• Schrödinger Equation in so-called spherical or polar coordinates

2222

here ,)( zyxrre

krU e

Ere

krrr

rrrm e

2

2

2

2222

2

2

sin

1

sin

112

E

zyx

ek

zyxm e

222

2

2

2

2

2

2

22

2

• The wavefunction is the function of three variables now and can be written as

• We had one variable (quantum box or simple harmonic oscillator) one “quantum number”

• Here we can assume that both wavefunction and energy should in general depend on three “quantum numbers”, corresponding to each coordinate

Wavefunction of the Hydrogen Atom

)()()(,, gfrRr

Wavefunction of the Hydrogen Atom

• The three “quantum numbers”, corresponding to each coordinate are

– n is the “principal quantum number”; it corresponds to coordinate r

– l is the “orbital quantum number”; associated with the coordinate θ, and determines the magnitude of the electron’s angular momentum, L.

– ml is the “magnetic quantum number”; associated with the coordinate φ, and it determines the orientation of L in the magnetic field

The Hydrogen Atom• Thus we can write in general:

• Quantum numbers determine the “quantum state”– Often we say that the electron is in the state (n, l, ml)

• The energy of the particle in a quantum state depends on all quantum numbers

),,(),,( ,, rrlmln

lmlnEE ,,

The Hydrogen Atom• An electron in a hydrogen atom only has

physically reasonable solutions when E, l and ml have the values given indicated below:

lm

nlnl

n

ae

kh

meE

n

EE

rr

l

emln

mln

l

l

..., ,2 ,1 ,0

) i.e.( 1 ..., ,2 ,1 ,0

... ,3 ,2 ,1

28 ,

),,(),,(

0

2

20

4

020

,,

,,

The Hydrogen Atom

• The electron energy in the hydrogen atom depends only on the principal quantum number, n

• Same result as in Bohr’s model.

20

,, n

EE

lmln

Hydrogen Atom: From Bohr to Schrödinger

• The electron orbital (angular) momentum is

– L is thus quantized – This is more general than the Bohr’s Postulate 3,

since it allows for orbits with angular momentum of zero (!!)

• For large values of l (l >> 1):

)1( llL

lllLl )1(

Hydrogen Wavefunction

The Hydrogen Atom: From Bohr to Schrödinger

1. Remember: Electrons “simply exists” within the atom, and we can only know the probability of finding the electron at a certain coordinate

2. Bohr’s orbits correspond to the coordinate, where the probability of finding the electron is the largest1. Moreover, Bohr’s theory was

limited to the states with highest angular momentum, l = n – 1

• The wavefunction gives physical meaning of the “orbits”:


• There are no orbits!!!– We describe the electron

through the quantum states – The energy of the electron is

constant when electron in a given quantum state

– Thus, no energy can be “taken” from electron by radiation

• Why doesn’t the accelerating electron emit EMW when in states corresponding to Bohr’s Orbits?


• Since the energy of electron in a hydrogen does not depend on different l and ml (in general for other atoms it does), It is naturally to assume that photons are only emitted or absorbed with energies corresponding to the difference between various energy sates

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The Hydrogen Atom• Since, the energy depends only on the

principal quantum number, the energy of an electron in different quantum states - with the same n, but different l and/or ml is the same

• The different sates having the same energy are called Degenerate States

• The number of such states (having the same energy or energy level) is referred to as “degeneracy of the energy level”

• In general, the degeneracy arises from the symmetry in a system

Degeneracy: The Hydrogen Atom

• Degree of degeneracy for the hydrogen depends on value of the principal quantum number, n

• Indeed, the larger n, the more values can l and, thus, ml take on

Degeneracy: The Hydrogen Atom

• Example: n = 3 there are 9 states (described by different wavefunctions) that have the same energy

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Degeneracy• Degeneracy can be “lifted” if

additional interactions (forces) applied to a system– Consider a hydrogen atom in the magnetic

field, B, applied along z-direction. The energy of electron is then

– This energy now depends on ml : we say degeneracy over ml was lifted

– How many energy levels with different degeneracy are there for n = 3?

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More Features of the Atomic Wavefunctions

• Electrons in the different quantum states are unlikely to be found in the same spatial regions

• Electrons in the lower angular momentum states are more likely to be found closer to the nucleus than those in states with higher angular momentum

• In multi-electron atoms the degeneracy over angular momentum is lifted and the sates with the same n but different l have slightly different energies

Space Quantization• l determines the magnitude of the angular

momentum, L:

• ml determines orientation of L in space when atom is in a magnetic field– Taking the magnetic field along z-direction, we

can show that

– Thus, not only is the angular momentum quantized, but also its component in some direction, usually taken to be the z-direction

)1( llLl

lz mL

Space Quantization

• Therefore, L, in an atom, cannot have any arbitrary orientation (in magnetic field) with respect to z-direction, but rather have only discrete orientations: Space Quantization

• However, as long as we do not have a “preferred” direction (e.g. defined by magnetic field) space quantization is meaningless

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Space Quantization: Example, l = 2

Space Quantization: Zeeman Effect

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Space Quantization: Stern-Gerlach Experiment

• Silver atoms studied in the non-uniform magnetic field with the gradient along z-direction (dB/dz)

• Although the atoms are neutral, they possess a magnetic dipole, and the inhomogeneous magnetic field acts on this dipole with force

Stern-Gerlach Experiment• The atoms are deflected as they travel through the

magnet• The stronger the force, the greater the deflection, and

thus the father away from the center of screen the atoms would land as detected on the screen

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Stern-Gerlach Experiment• If Lz is quantized (Lz = mlħ), as it follows

from the Schrödinger equation, the force Fz is quantized

• Thus, one would observe discrete positions on the screen where atoms land, and the number of positions is 2l+1 (The expected number of the positions is the same as the number of different ml numbers)

• In classical physics case we would expect smooth distribution of the positions on the screen

Stern-Gerlach Experiment

• Only two lines were observed on the screen on which the atoms land: one above and one below of the center !! – Moreover, there were no central line and the

distance from the center to each positions was the same

– Stern-Gerlach Experiment

http://mutuslab.cs.uwindsor.ca/schurko/nmrcourse/animations/stern-gerlach/sgpeng.htm



Stern-Gerlach Experiment

• Stern-Gerlach experiment indeed showed discrete nature of the momentum, confirming space quantization

• However, the result contradict quantitative prediction of the Schrödinger

• The experiment by Phipps and Taylor with hydrogen atoms, where ml = 0 (and thus Lz = 0) showed the two emission lines in the spectrum, suggesting that the effect is NOT due to Orbital Momentum

• Thus, there may be a magnetic dipole other than the one associated with orbital momentum!

The Spin

• The idea of an intrinsic magnetic dipole was first introduced by “hand” by Uhlenbeck and Goudsmit as “internal orbital momentum” called “Spin”

• The existence of spin and the laws determining its behavior arise naturally from the relativistic version of Quantum Mechanics, developed by Paul Dirac

The Spin• By analogy with the orbital momentum and to

explain the experiments, Uhlenbeck and Goudsmit postulated that there is a magnetic dipole moment associated with the spin: the spin angular momentum

• This is analogous to the orbital angular momentum

• The spin can have only two projections on the z-axis, thus, explaining the Stern-Gerlach and Phipps-Taylor experiments– The existence of the spin also explained earlier observations

such as the Fine Structure of Spectral Lines

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with,)1( ssz mmSsssS

...;L );1(...)1( with ,)1( mmnnL z

The Spin• The Spin is a [quantum] property of the

electron. • Spin quantum number “is always there”• Other quantum numbers can change

depending on the specifics of the potential the electron is in.

• For instance:– In hydrogen, in addition to the spin, an electron is

characterized by three quantum numbers n, l, ml

– In a 1D Infinite potential well, in addition to the spin, an electron is characterized by only one quantum number n

• The spin is intrinsic to the electron!

The Hydrogen Atom Wavefunction: Revisited

• Now we need to add another quantum number, associated with the electron spin

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The Ground State of the Hydrogen Atom

• For the hydrogen ground state we have

• Reminder: The hydrogen atom is unusual since states with the same l have the same energy.

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The Ground State of Atoms other than Hydrogen

• As soon as an atom contains more than one electron, the states with different l no longer have the same energy, and degeneracy over orbital quantum number is lifted

• What is the ground sate of a multi-electron system?– Perhaps in the lowest energy state all electrons

in the atom have the same four quantum numbers

• NO!!!

Pauli’s Exclusion Principle• NO TWO ELECTRONS IN A SYSTEM CAN BE

IN THE SAME QUANTUMS STATE• In other words, no two electrons can have

the same values of the quantum numbers

• For atoms these are:

• This allows us to understand the Periodic Table

• The electron configuration of any atom must satisfy the Pauli’s exclusion principle

sl mmln ; ; ;

Pauli’s Exclusion Principle: Atoms• Definitions:

– a shell is the set of states with the same n (thus different l, ml and ms)

– a sub-shell the set of states with the same n and l (thus different ml and ms)

• Using the rules for the atomic quantum numbers, we determine that the number of electrons in

– a sub-shell is 2(2l+1) – a shell is 2n2

Additional Rules for Determining the Ground State-Configuration

1. The total energy of the electron increases with increasing n (energy less negative)

2. Within a given shell (a given n), the l = 0 states always have the lowest energy

3. The energy of the sub-shells generally increases with l :

El=0 < El=1 < El=2< El=3

since electron with lowest value of l can be closer to nucleus and does not feel shielding of (+) nuclear charge by (–) electron cloud

Electronic Configuration of the Atomic Ground State: Examples

• We shall use letters instead of numbers to denote the orbital quantum number:

l = 0 s; l = 1 p; l = 2 d; l = 3 f

• We shall use superscript above the letter to indicate the number of electrons in a given sub-shell

• Hydrogen (1): 1s1

• Lithium (3): 1s2 2s1

• Nitrogen (7): 1s2 2s2p3

• Argon (18): 1s2 2s2p6 3s2p6


• Potassium (19): 1s2 2s2p6 3s2p6 4s1

• Calcium (20): 1s2 2s2p6 3s2p6 4s2

• Scandium (21): 1s2 2s2p6 3s2p6 d1 4s2

– Note that 3d sub-shell starts to fill in after 4s sub-shell is filled up

– The same is true for 4d vs 5s sub-shells


• Nickel (28): 1s2 2s2p6 3s2p6d8 4s2 • Copper (29): 1s2 2s2p6 3s2p6d10 4s1 • Silver (47): 1s2 2s2p6 3s2p6d8

4s2p6d10 5s1 – Note that there is no nd9 state for either

Cu, Ag, or Au (they “loose” one s-electron from the next shell)

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Bees• This should be a learning experience. So let’s discuss your

project before the due date• After ~ 600 seconds some bees have reached the boundary

and escaped • Up to this time would expect that the variance of the average

that variance begins to approach the diffusion relation derived by Einstein,

D is the diffusion coefficient, is the mean free path, and is the mean free time

• Could you describe one characteristic of the bees’ motion that you choose. For example, you could consider the time dependence of the number of bees escaping per unit time or the spatial distribution of bees at various times.

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6 Introduction to Quantum Mechanics