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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
Page 11
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
Page 17
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
Page 21
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