1. 2. Solving Schrödinger’s Equation
Superposition
Given a few solutions of Schrödinger’s equation, we can make more of them
Let 1 and 2 be two solutions of Schrödinger’s equation
Let c1 and c2 be any pair of complex numbers
Then the following is also a solution of Schrödinger’s equation
We can generalize this to an arbitrary number of solutions:
We can even generalize to a continuum of solutions:

2. 2A. The Free Schrödinger’s Equation We know a lot of solutions
Consider the free Schrödinger equation:
We know a lot of solutions of this equation:
By combining these, we can make a lot more solutions*
The function c(k) can be (almost) any function of k
This actually includes all possible solutions
*The factor of (2)3/2 is arbitrary and inserted here for convenience. In 1D, it would be (2)1/2

3. We can (in principle) always solve this problem
Goal: Given (r,t = 0) = (r), find (r,t) when no potential is present
Set t = 0:
This just says that c(k) is the Fourier transform of (r)
Substitute it in the formula above for (r,t) and we are done
Actually doing the integrals may be difficult

4. Sample Problem
A particle in one dimension with no potential has wave function at t = 0 given by
What is the wave function at arbitrary time?
Need 1D versions of these formulas:
From Appendix:
Find c(k):

5. Sample Problem (2)
A particle in one dimension with no potential has wave function at t = 0 given by
What is the wave function at arbitrary time?
Now find (x,t):

6. 2B. The Time Independent Schrödinger Eqn. Separation of Variables
Suppose that the potential is independent of time:
Conjecture solutions of the form:
Substitute it in
Divide by (r)(t)
Left side is independent of r
Right side is independent of t
Both sides are independent of both! Must be a constant. Call it E

7. Solving the time equation
The first equation is easy to solve
Integrate both sides
By comparison with e-it, we see that E =  is the energy
Substitute it back in:

8. The Time Independent Schrödinger Equation
Multiply the other equation by (r) again:
The Strategy for solving
Given a potential V(r) independent of time, what is most general solution of Schrödinger’s time-dependent equation?
First, solve Schrödinger’s time-independent equation
You should find many solutions n(r) with different energies En
Now just multiply by the phase factor
Then take linear combinations
Later we’ll learn how to find cn

9. Why is time-independent better?
Time-independent is one less variable – significantly easier
It is a real equation (in this case), which is less hassle to solve
If in one dimension, it reduces to an ordinary differential equation
These are much easier to solve, especially numerically

10. Probability Conservation
2C. Probability current
Probability Conservation
Recall the probability density is:
This can change as the wave function changes
Where does the probability “go” as it changes?
Does it “flow” just like electric charge does?
Want to show that the probability moves around
Ideally, show that it “flows” from place to place
A formula from E and M – can we make it like this?
To make things easier, let’s make all functions of space and time implicit, not write them

11. The derivation (1) Start with Schrödinger’s equation
Multiply on the left by *:
Take complex conjugate of this equation:
Subtract:
Rewrite first term as a total derivative
Cancel a factor of i
Left side is probability density

12. The derivation (2) Consider the following expression:
Use product rule on the divergence
Substitute this in above
Define the probability current j:
Then we have:

13. Why is it called probability current?
Integrate it over any volume V with surface S
Left side is P(r  V)
Use Gauss’s law on right side
Change in probability is due to current flowing out j V
If the wave function falls off at infinity (as it must) and the volume V becomes all of space, we have

14. Calculating probability current
This expression is best when doing proofs
Note that you have a real number minus its complex conjugate
A quicker formula for calculation is:
Let’s find  and j for a plane wave:

15. Sample Problem
A particle in the 1D infinite square well has wave function
For the region 0 < x < a. Find  and j.

16. Sample Problem (2)
A particle in the 1D infinite square well has wave function
For the region 0 < x < a. Find  and j.
In 1D:

17. Sample Problem (3)
A particle in the 1D infinite square well has wave function
For the region 0 < x < a. Find  and j.
After some work …

18. 2D. Reflection from a Step Boundary
The Case E > V0: Solutions in Each Region
A particle with energy E impacts a step-function barrier from the left: I
incident
transmitted
reflected II
Solve the equation in each of the regions
Assume E > V0
Region I
Region II
Most general solution:
A is incident wave
B is reflected wave
C is transmitted wave
D is incoming wave from the right: set D = 0

19. Step with E > V0: The solution
incident
transmitted
reflected
Schrödinger’s equation: second derivative finite
(x) and ’(x) must be continuous at x = 0 II
We can’t normalize wave functions
Use probability currents!

20. Summary: Step with E > V0
incident
transmitted
reflected II

21. Step with E < V0 I II What if V0 > E? Region I same as before
Region II: we have I II
incident
reflected
evanescent
Most general solution:
A is incident wave
B is reflected wave
C is damped “evanescent” wave
D is growing wave, can’t be normalized
(x) and ’(x) must be continuous at x = 0:
No transmission since evanescent wave is damped

22. Step Potential: All cases summarized
For V0 > E, all is reflected
Note that it penetrates, a little bit into the classically forbidden region, x > 0
This suggests if barrier had finite thickness, some of it would bet through
Reflection probability: R T

23. 2E. Quantum Tunneling Setting Up the Problem V(x)
Barrier of finite height and width: V0
Solve the equation in each of the regions
Particle impacts from left with E < V0
General solution in all three regions: I II
III
- d/2
+ d/2 x
Match  and ’ at x = -d/2 and x = d/2
Why didn’t I include e-ikx in III? Why did I skip letter E?
Solve for F in terms of A

24. Skip this Slide – Solving for F in terms of A
Multiply 1 by ik and add to 2
Multiply 3 by  and add to 4
Multiply 3 by  and subtract 4
Multiply 5 by 2 and substitute from 6 and 7

25. Barrier Penetration Results
We want to know transmission probability
For thick barriers,
Exponential suppression of barrier penetration

26. Unbound and Bound State
For each of the following, we found solutions for any E
No potential
Step potential
Barrier
This is because we are dealing with unbound states, E > V()
Our wave functions were, in each case, not normalizable
Fixable by making superpositions:
We will now consider bounds states
These are when E < V()
There will always only be discrete energy values
And they can be normalized
Usually easier to deal with real wave functions

27. 2F. The Infinite Square Well
Finding the Modes
V(x)
Infinite potential implies wave function must vanish there
In the allowed region, Schrödinger’s equation is just x a
The solution to this is simple:
Because potential is infinite, the derivative is not necessarily continuous
But wave functions must still be continuous:

28. Normalizing Modes and Quantized Energies
We can normalize this wave function:
Note that we only get discrete energies in this case
Note that we can normalize these
Most general solution is then

29. The 3D Infinite Square Well
In allowed region:
Guess solution:
Normalize it:
This is product of 1D functions
Energy is
This is sum of 1D energies c b a

30. 2G. The Double Delta-Function Potential
Finding Bound States
V(x)
-a/2
a/2 I
First, write out Schrödinger’s Equation: II
III x
Bound states have E < V() = 0
Within each region we have:
General solution (deleting the parts that blow up at infinity):

31. Dealing with Delta Functions
V(x)
-a/2
a/2 I II
III
To deal with the delta functions, integrate Schrödinger’s equation over a small region near the delta function:
For example, near x = +a/2
Do first term on right by fundamental theorem of calculus
Do second term on right by using the delta functions
Take the limit   0
Left side small in this limit

32. On the right side of the equation above, is that I, II, or III?
Simplifying at x = ½a
V(x)
-a/2
a/2 I II
III
Since there is a finite discontinuity in ’,  must be continuous at this boundary
On the right side of the equation above, is that I, II, or III?
Write these equation out explicitly:
Substitute first into second:

33. Repeating at x = – ½a Repeat the steps we did, this time at x = –½a
Note these equations are nearly identical:
The only numbers equal to their reciprocal are 1

34. Graphical Solution
Right side is two curves, left side is a straight line
Right side, plus
Right side, minus
Left side
Black line always crosses red curve, sometimes crosses green curve, depending on parameters
Sometimes two solutions, sometimes one
Normalize to finish the problem
Note one solution symmetric, one anti-symmetric