Page 1

Topic 5: Schrödinger Equation

• Wave equation for Photon vs. Schrödinger equation for Electron+

• Solution to Schrödinger Equation gives wave function 2 gives probability of finding particle in a certain region

• Square Well Potentials: Infinite and Finite walls oscillates inside well and is zero or decaying outside well, E n2

• Simple Harmonic Oscillator Potential (or parabolic) is more complex, E n

Example Infinite Well Solution

Page 2

Schrödinger Equation

• Step Potential of Height V0

is always affected by a step, even if E > V0

– For E > V0, oscillates with different k values outside/inside step.

– For E < V0, oscillates outside step and decays inside step.

• Barrier Potential of Height V0

oscillates outside and decays inside barrier.

• Expectation Values and Operators

• Appendix: Complex Number Tutorial

Page 3

2nd time derivative2nd space derivative

Wave Equation for Photons: Electric Field E

Propose Solution:

Calculate Derivatives:

After Substitution:

where and E p k

2 2

2 2 2

1E E

x c t

0, cosE x t E kx t

2

2 202

cos ,E

E kx t E x tt

22

2,

Ek E x t

x

22

2k

c

kc E pc

Page 4

1st time derivative2nd space derivative

Schrödinger Eqn. for Electrons+: Wave Function

Propose Simple Solution for constant V:

Calculate Derivatives:

After Substitution:

TOTok EVE

22

2

, ,, ,

2

x t x tV x t x t i

m x t

, cos sini kx tx t A A ke x t i kx t

2

2 22

i kx t i kx ti Ae i ik Ae kt x

2

20

2 2

0

2

2

k V i im

kV

m

where and E p k

Page 5

Schrödinger Equation: Applications

• Now, find the eigenfunctions and eigenvalues E of the Schrödinger Equation for a particle interacting with different potential energy shapes. (assume no time dependence)

22

2

( )( ) (

2)( )V x x x

m x

xE

• Possible potential energies V(x) include:

• Infinite and Finite square wells (bound particle).

• Simple Harmonic or parabolic well (bound particle).

• Step edge (free particle).

• Barrier (free particle).

Page 6

Schrödinger Equation: Definitions

• Wave function has NO PHYSICAL MEANING!

• BUT, the probability to find a particle in width dx is given by:

• Normalization of – Probability to find particle in all space must equal 1.

– Solve for coefficients so that normalization occurs.

1x x dx

2( , ) , , ,P x t dx x t x t dx x t dx

Page 7

n(x) n2(x)

Infinite Square Well Potential: Visual Solutions

2 2 2 2

2

2

2 2n

k

m mLE n

L

xn

Lxn

sin2

Wave and Probability Solutions Energy Solutions

n = 2

n = 1

n = 3

Page 8

xkx 2" where

2

22 2

pmE

k

kxAkxAx cossin 21

xEdx

xd

m

2

22

2

cannot penetrate barriers! 0x

Infinite Square Well: Solve general from S.Eqn.

Outside Well:(V = )

Inside Well:(V = 0)

Oscillatory

Page 9

00

000cos 22 AA L

nknkLkLA n

0sin1

2

222

22

22 mLn

m

kEn

1sin0

22

dxL

xnAdx

L

n

2sin

0

2

dmL

An2

using identity andL

x

Infinite Square Well: Satisfy B.C. and Normalization

• Satisfy boundary conditions

L

xn

Lxn

sin2

• Satisfy normalization

kxAkxAx cossin 21

0L

Wave Solutions

Quantized Energy Solutions

Page 10

n(x) n2(x)

Finite Square Well Potential: Visual Solutions

Wave and Probability Solutions Energy Solutions

2 2 2

2

2

22

n

m L

nE

k

E

Vo

E3

E2

E1

“leaks” outside barrier

High energyparticles“escape”

n = 2

n = 1

n = 3

Energy vs. width: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07a.htmlEnergy vs. height: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_07b.html

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12" x k x where 1

22

2mEk

1 11 2sin cosx A k x A k x

xEdx

xd

m

2

22

2

can penetrate barriers!

Finite Square Well: Solve general from S.Eqn.

Inside Well:(V = 0)

22" x k x where 2

22

20o

mk V E

2 21 2

k x k xx B e B e

ax ax

Outside Well:(V = Vo)

Oscillatory

Decaying

Page 12

Finite Square Well: Example Problem

(a) Sketch the wave function (x) for the n = 4 state for the finite square well potential.

(b) Sketch the probability distribution 2(x).

L

n = 4(x)

L/2

xL

n = 4

L/2 x

2(x)

Page 13

Finite Square Well: Example Problem

E

(x)

x1

x

V1

V2

x2

Region 1 Region 2

Sketch the wave function (x) corresponding to a particle with energy E in the potential well shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 2.

(x) oscillates inside the potential well because E > V(x), and decays exponentially outside the well because E < V(x).

• The frequency of (x) is higher in Region 1 vs. Region 2 because the kinetic energy is higher [Ek = E - V(x)].

• The amplitude of (x) is lower in Region 1 because its higher Ek gives a higher velocity, and the particle therefore spends less time in that region.

Page 14

n(x) n2(x)

(different well widths)

Simple Harmonic Well Potential: Visual Solutions

Wave and Probability Solutions

Energy Solutions

n = 1

n = 0

n = 2

1

2nE n

Page 15

Simple Harmonic Well: Solve from S.Eqn.

Inside Well:

xx 2" where 0])([2

22 ExV

m

Outside Well:

xkx 2" where 2

2

2mk E V x

NEW!

(x) is not a simple decaying exponential.

(x) is not a simple trigonometric function.

12 22

1 1exp , exp , etc.2

( ) (Hermite Polynomial) (Gaussian Functi

2

on)

o o

Kx K KxA A x

x

Page 16

Step Potential:(x) outside step

Outside Step:V(x) = 0

22

2S. Eqn: ( )

2

d xV x x E x

m dx

xkx 21

" where 2

21

2mk E

(x) is oscillatory

(x)

Energy

Case 1 Case 2

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Step Potential:(x) inside step

Inside Step:V(x) = Vo

22" x k x where 2 2

2 2o

mk E V

(x) is oscillatory for E > Vo

Case 1 Case 2

(x)

(x) is decaying for E < Vo

Energy

E < VoE > Vo

Scattering at Step Up: http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_06b.htmlScattering at Well - wide: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05d.htmlScattering at Well - various: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_05b.html

Page 18

2

1 2

1 2

Rk k

k k

1 22

1 2

4T

k k

k k

Step Potential: Reflection and Transmission

R(reflection) + T(transmission) = 1

• Reflection occurs at a barrier (R 0), regardless if it is step-downor step-up.

– R depends on the wave vector difference (k1 - k2) (or energy difference), but not on which is larger.

– Classically, R = 0 for energy E larger than potential barrier (Vo).

• At a step, a particle wave undergoes reflection and transmission (like electromagnetic radiation!) with probability rates R and T, respectively.

Page 19

Step Potential: Example Problem

A free particle of mass m, wave number k1 , and energy E = 2Vo is traveling

to the right. At x = 0, the potential jumps from zero to –Vo and remains at

this value for positive x. Find the wavenumber k2 in the region x > 0 in

terms of k1 and Vo. In addition, find the reflection and transmission

coefficients R and T.

22 2

1

1

31 121 2

31 2 1 12

2

2 2 42

2 2 2 2 3 3

2

0.225(1% reflected)

2.225

1 1 0.0102 0.99 (99% transmitted

6

0.0102

)

o o

o o o o

m V mVmEk and

m V E m V V m Vor k

k kk k

k k k k

T R

mVk

R

Page 20

Barrier Potential

Outside Barrier:V(x) = 0 (x) is oscillatory

Energy

(x)

(x) is decaying

12" x k x where 1

22

2mk E

22" x k x where 2

22

20o

mk V E

Inside Barrier:V(x) = Vo

22 aT ke

Transmission is Non-Zero!

http://www.sgi.com/fun/java/john/wave-sim.html

Single Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/samples/107_12c.html

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Barrier Potential: Example Problem

E

(x)

x

Region 1 Region 3

Region 2

Vo

Sketch the wave function (x) corresponding to a particle with energy E in the potential shown below. Explain how and why the wavelengths and amplitudes of (x) are different in regions 1 and 3.

(x) oscillates in regions 1 and 3 because E > V(x), and decays exponentially in region 2 because E < V(x).

• Frequency of (x) is higher in Region 1 vs. 3 because kinetic energy is higher there.

• Amplitude of (x) in Regions 1 and 3 depends on the initial location of the wave packet. If we assume a bound particle in Region 1, then the amplitude is higher there and decays into Region 3 (case shown above).

Non-resonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-5.htmlResonant Barrier: http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-6.htmlDouble Barrier + : http://www.kfunigraz.ac.at/imawww/vqm/pages/supplementary/107S_resonance-0.html

Page 22

Scanning Tunneling Microscopy: Schematic

• STM is based upon quantum mechanical tunneling of electrons across the vacuum barrier between a conducting tip and sample.

• To form image, tip is raster-scanned across surface and tunneling current is measured.

Constant current contour

Bias voltage

e- Distance s

eSample

eee

Tip

Tunneling current e -2s

VDC

Page 23

STM: Ultra-High Vacuum Instrument

• Well-ordered, clean surfaces for STM studies are prepared in UHV.

• Sample is moved towards tip using coarse mechanism, and the tip is moved using a 3-axis piezoelectric scanner.

Tip

Scanner

Sample

Coarse Motion

Page 24

STM: Data of Si(111)7×7 Surface

• STM topograph shows rearrangement of atoms on a Si(111) surface.

• Adatoms appear as bright “dots” when electrons travel from sample to tip.

7×7 Unit

STM empty

7 nm18 nm

= adatom

Page 25

dxxxfxf(x)

Position

Momentum

Kinetic energy

Hamiltonian

Total Energy

xx

p

K

H

E

xi

2

2

2

2m x

2

2

2

2V x

m x

Expectation Values and Operators

• By definition, the “expectation value” of a function is:

• “Operate” on (x) to find expectation value (or average expected value) of an “observable.”

Observable Symbol Operator

ti

Page 26

1 1

0

L

x i xx

dxL

x

Lxi

L

x

L

L

sin

2sin

2

0

dxLL

x

L

xi

L

L

cossin

2

0

<p> = 0 by symmetry (odd function over symmetric limits)

Note: The average momentum goes to zero because the “sum” of positive and negative momentum values cancel each other out.

<p> =

0 L

-L/2 +L/2

Expectation Values: Example Problem

• Find <p>, <p2> for ground state 1(x) of infinite well (n = 1)

Page 27

21 1 12

0

2 where sin

L xx x dx x

x L L

2 2

21 1 1 12

0

where L

x x dx x xL x L

2

1 1

0

L

x x dxL

<p2> =

<p2> =

2

L

= 1 by normalization

Expectation Values: Example Problem, cont.

Page 28

• Imaginary number i given by: i2 = –1 ( i3 = –i, i4 = 1, i–1 = –i )

• Complex number z is composed of a real and imaginary parts.

Complex Number Tutorial: Definitions

2 1 –1

i

2i

– i

(2, i)

30º or /6

2x + iy 5(cos30º + isin30º)

5 ei/6

Conjugate: z* = x – iy = rcos – i rsinre– i

where (z*)(z) = (x – iy)(x + iy) = x2 + y2 (real!)

Cartesian Form: z = x + iy

Polar Form: z = r(cos + i sin

where r = (x2 + y2)1/2 and tan = y/x

Exponential Form: z = rei

Page 29

• Proof of equivalence for polar and exponential forms:

Complex Number Tutorial: Taylor Series

2 3

2 3

4

4

2 4 3 5

54

2

5

3

2 3

exp( ) 1 .2! 3! 4!

cos( ) 1 . and

sin .3! 5!

sin( ) .2! 4! 3! 5!

cos 1 . 2! 4!

12! 4!

12

.

exp

3! 5

( )

!

!

x xx e

x x xx x et

x xi x i x et

c

t

x x x xx etc x x etc

etc

c

ix ixi

c

x xx

ixix

ixx

42 .

3!

sin

1 

exp( )

4!

cos

etc where i

ix i x

x

x

Page 30

22

2

1 1( )

2

d x d tV x i

m x t dtdx

Schrödinger Equation is 2nd Order Partial Differential Equation

Assume is separable [i.e. V(x) only]

Substitution of

Partial derivatives are now ordinary derivatives

Space dependence ONLY Time dependence ONLY

Divide by (x)0(t)

Schrödinger Eqn.: Derivation of Space & Time Dependence

22

2

, ,( , ) ,

2

x t x tV x t x t i

m tx

22

2( )

2

x t x tV x x t i

m tx

,x t x t

22

2( )

2

d x d tt V x x t i x

m dtdx

Page 31

Left and right sides have only space (x) and time (t) dependence now

Schrödinger Eqn.: Derivation of Space & Time Dependence

Set each side of equation equal to a constant C

Space:

Time:

Space Equation need V(x) to solve!

Time Solution:

22

2

1 1( )

2

d x d tV x i

m x t dtdx

22

2

1( )

2

d xV x C

m x dx

1 d t

i Ct dt

22

2( )

2

d xV x x E x

m dx

Check by substitution! iEtt e