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

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

3. Wave Equation for Photons: Electric Field E
2nd space derivative
2nd time derivative
Propose Solution:
Calculate Derivatives:
After Substitution: 

4. Schrödinger Eqn. for Electrons+: Wave Function Y
2nd space derivative
1st time derivative
Propose Simple Solution for constant V:
Calculate Derivatives:
After Substitution: 

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

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 Y coefficients so that normalization occurs.

7. Infinite Square Well Potential: Visual Solutions
Wave and Probability Solutions
Energy Solutions
Yn(x)
Yn2(x)
n = 3
n = 2
n = 1

8. Infinite Square Well: Solve general Y from S.Eqn.
Inside Well:
(V = 0)
where
Oscillatory
Outside Well:
(V = )
y cannot penetrate barriers!

9. Infinite Square Well: Satisfy B.C. and Normalization
Satisfy boundary conditions
Quantized Energy Solutions
Satisfy normalization
using identity
and
Wave Solutions

10. Finite Square Well Potential: Visual Solutions
Wave and Probability Solutions
Energy Solutions
“leaks”
outside barrier
High energy
particles
“escape” E Vo E3 E2 E1
Yn(x)
Yn2(x)
n = 3
n = 2
n = 1
Energy vs. width:
Energy vs. height:

11. Finite Square Well: Solve general Y from S.Eqn.
Inside Well:
(V = 0)
where
Oscillatory
Outside Well:
(V = Vo)
where
y can penetrate barriers!
Decaying

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).

13. Finite Square Well: Example Problem
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.

14. Simple Harmonic Well Potential: Visual Solutions
Wave and Probability Solutions
Yn(x)
Yn2(x)
(different well widths)
Energy Solutions
n = 2
n = 1
n = 0

15. Simple Harmonic Well: Solve Y from S.Eqn.
NEW!
Inside Well:
where
Y(x) is not a simple trigonometric function.
Outside Well:
where
Y(x) is not a simple decaying exponential.

16. Step Potential: Y(x) outside step
V(x) = 0
where
Y(x) is oscillatory
Case 1
Case 2
Energy
Y(x)

17. Step Potential: Y(x) inside step
V(x) = Vo
where
Y(x) is oscillatory for E > Vo
Y(x) is decaying for E < Vo
Case 1
Case 2
Energy
E > Vo
E < Vo
Y(x)
Scattering at Step Up:
Scattering at Well - wide:
Scattering at Well - various:

18. Step Potential: Reflection and Transmission
At a step, a particle wave undergoes reflection and transmission (like electromagnetic radiation!) with probability rates R and T, respectively.
R(reflection) + T(transmission) = 1
Reflection occurs at a barrier (R  0), regardless if it is step-down or 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).

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.

20. Barrier Potential Y(x) is oscillatory Y(x) is decaying
where
Outside Barrier:
V(x) = 0
Y(x) is oscillatory
Inside Barrier:
V(x) = Vo
where
Y(x) is decaying
Energy
Transmission is Non-Zero!
Y(x)
Single Barrier:

21. Barrier Potential: Example Problem
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:
Resonant Barrier:
Double Barrier + :

22. Scanning Tunneling Microscopy: Schematic
Tip
VDC
Bias voltage e-
Constant current
contour
Distance s
Sample e e e e
Tunneling current  e -2ks
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.

23. STM: Ultra-High Vacuum Instrument
Coarse Motion
Sample
Scanner
Tip
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.

24. STM: Data of Si(111)7×7 Surface
empty
STM
7×7 Unit
18 nm
7 nm
= adatom
STM topograph shows rearrangement of atoms on a Si(111) surface.
Adatoms appear as bright “dots” when electrons travel from sample to tip.

25. Expectation Values and Operators
By definition, the “expectation value” of a function is:
“Operate” on Y(x) to find expectation value (or average expected value) of an “observable.”
Observable Symbol Operator
Position
Momentum
Kinetic energy
Hamiltonian
Total Energy x p K H E x

26. Expectation Values: Example Problem
Find <p>, <p2> for ground state Y1(x) of infinite well (n = 1) L
-L/2
+L/2
<p> =
<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.

27. Expectation Values: Example Problem, cont.
= 1 by normalization
<p2> =

28. Complex Number Tutorial: Definitions
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.
Cartesian Form: z = x + iy
Polar Form: z = r(cosq + i sinq)
where r = (x2 + y2)1/2 and tanq = y/x
Exponential Form: z = rei q
Conjugate: z* = x – iy = rcosq – i rsinq = re– i q
where (z*)(z) = (x – iy)(x + iy) = x2 + y2 (real!)

29. Complex Number Tutorial: Taylor Series
Proof of equivalence for polar and exponential forms:

30. Schrödinger Eqn.: Derivation of Space & Time Dependence
Schrödinger Equation is 2nd Order Partial Differential Equation
Assume y is separable [i.e. V(x) only]
Substitution of y
Partial derivatives are now ordinary derivatives
Divide by y(x)0(t)
Space dependence ONLY
Time dependence ONLY

31. Schrödinger Eqn.: Derivation of Space & Time Dependence
Left and right sides have only space (x) and time (t) dependence now
Space:
Set each side of equation equal to a constant C
Time:
Space Equation
 need V(x) to solve!
Time Solution:
Check by substitution!