1. 12. Approx. Methods for Time Ind. Systems
12A. The Variational Principle
A Great Way to Find the Ground State
Given an arbitrary time-independent Hamiltonian H, we want to approximate
The eigenstates |i
The eigen-energies Ei
Idea behind the variational principle:
The ground state is the state with the lowest energy
The first excited state is the lowest energy state orthogonal to the ground state
Etc.
The method:
Choose a large number of state vectors
Measure their energy
Pick the one with the lowest energy

2. The Ground State Has the Lowest Energy
Imagine we knew the eigenstates and energies of the Hamiltonian
These states are assumed to be complete and orthonormal
Assume the energies are ordered:
For an arbitrary state |, we use completeness to write:
Let’s find the expectation value of the Hamiltonian for |:
It follows that
Consider the inner product:
We therefore have:

3. The Variational Principle
Select a wide range of trial vectors |
Calculate the expectation values at right
Choose the one with the lowest ratio at right
Use this ratio as an estimate of the ground state energy
Use the corresponding state vector | (normalized) as an estimate of ground state vector |1
The variational part:
We want as many state vectors as possible, ideally infinitely many
Best way to do this is to make | have one or more variational parameters:
Calculate the energy as a function of these parameters
Find the value of min that minimizes E()
Then the estimate of the energy and wave function is:

4. Theory vs. Practice For homework problems:
Pick a set of trial vectors described by a small number ( 2) of parameters
Find E() analytically
Find the minimum using derivatives
Substitute back in min
For research problems:
Pick a set of vectors described by a large number (100’s) of parameters
Find E() numerically
Find the minimum using multi-dimensional search algorithms (simplex method, for example)

5. Good Trial Wave Functions
How do you pick a good trial wave function?
Discontinuous functions will have infinite derivative
Can show this yields infinite P2 and hence infinite H
Don’t use discontinuous functions
Non-smooth functions are okay, but can be tricky to evaluate
Discontinuous first derivatives have infinite second derivative at a point
This will contribute non-trivially to
Unless  vanishes at the non-smooth point
When in doubt, you can avoid this problem with:
No!
Danger!
Fine!

6. Sample Problem (1)
A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the ground-state energy.
The potential rises suddenly
Try a function that disappears suddenly
No need to normalize in this formalism
We now need to calculate
Work out the pieces, one at a time

7. Sample Problem (2)
A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the ground-state energy.
Put the pieces together:
Minimize with respect to a
Substitute back in to get E(a), an estimate of the energy:

8. Some comments on how we did
We picked a terrible trial wave function
We still did pretty well (20% error)
The error in the energy is caused by the square of the amount of bad wave functions in the wave function
Small things squared are very small
The state vector has first order errors
Not as reliable
We got an overestimate of the energy
This will always happen
With more parameters, you can do much better
This method is powerful; most realistic problems are solved with (advanced) variational approaches

9. Sample Problem – Hydrogen Estimate (1)
A particle of mass m in 3D lies in potential V = - kee 2/r. Estimate the ground-state energy.
Do this in class
Trial wave functions:
First function tried:
This is not normalizable, | = 
Second function tried:
This looks very promising
Things we need to find:

10. Sample Problem – Hydrogen Estimate (2)
A particle of mass m in 3D lies in potential V = - kee 2/r. Estimate the ground-state energy.
We now calculate the energy function:
Find the minimum:
Substitute in to get the minimum energy:
We got it exactly right!
The wave function is
Also exactly right

11. Can We Get Beyond the Ground State?
Is there a way to get states beyond the ground state?
Yes, if we pick states orthogonal to the ground state
Removing the approximate ground state:
Assume you have an estimate of the true normalized ground state found by variational method
Create a set of states that you estimate are close to the next excited state
Remove the portion of these states in the approximate ground state
This state is, by construction, orthogonal to |1
Calculate the energy for this state:
Minimize this energy and find min
Then we approximate the first excited state energy as

12. Comments on Excited State(s)
Is the resulting energy guaranteed to be an overestimate?
In general no:
The state is guaranteed to contain none of the estimated ground state |1 
But it could contain a small mixture of the actual ground state |1 
Does this work as well for excited states as it did for the ground state?
Error from previous step gets compounded with this step
Over many steps, errors accumulate
An exception where it does work
Suppose the problem has some symmetry
Then all eigenstates can be classified by their eigenvalues under this symmetry
Choose trial state vectors that have this symmetry eigenvalue
The energies you find will be true overestimates of the energy of the lowest state with each symmetry eigenvalues

13. Sample Problem (1)
A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the first excited state energy. Is it an overestimate?
The potential is symmetric under parity
The ground state, previously discussed, has even parity
Let’s find the lowest energy odd parity state
Try an odd wave function
Work out the pieces we need, one at a time

14. Sample Problem (2)
A particle of mass m in 1D lies in potential V = m2x2/2. Estimate the first excited state energy. Is it an overestimate?
Minimize the energy:
Substitute it back in:
Since our state is odd, the lowest odd energy state must be lower than this value
Actual coefficient is 1.5 (15% off)

15. 12B. The WKB Approximation A Method For High-Energy States
The variational method is good for ground state and other low energy states
The WKB method is good for highly excited states
Idea behind the WKB approximation
If the energy is large, the wave function will be quickly oscillating
The potential will then look like it’s slowly varying
Start from Schrödinger’s equation in 1D:
Define:
Then Schrödinger’s equation becomes:
If we think of k(x) as a constant, this solution would be like eikx
This suggests breaking  into a magnitude and a phase
A(x) and (x) are real functions

16. Breaking it into two equations
Substitute in (denote derivatives with primes):
Match real and imaginary parts:
Multiply second equation by A
Anything with a zero derivative is a constant
Rename  as W, then

17. 0th Order WKB Approximation
Expand out that second derivative
Substitute it back in
If energy is large, so is k(x)
To zeroth order, assume that k2 dominates the other two terms

18. Bound Problems with Steep Boundaries
Suppose we want to consider bound states
For now, assume the potential goes to infinity suddenly at x = a and x = b
For bound states, we generally prefer real wave functions:
Sines, cosines, or some linear combination
We have two additional constraints:
Wave function must vanish at x = a
Wave function must vanish at x = b
Vanishing at x = a requires that
Then vanishing at x = b requires that
Where n = 0, 1, 2, …
Recall
So we have

19. Bound Problems with Steep Boundaries
Suppose we want to consider bound states
For now, assume the potential goes to infinity suddenly at x = a and x = b
For bound states, we generally prefer real wave functions:
Sines, cosines, or some linear combination
We have two additional constraints:
Wave function must vanish at x = a
Wave function must vanish at x = b
Vanishing at x = a requires that
Then vanishing at x = b requires that
Where n = 0, 1, 2, …
Recall
So we have

20. Near a Soft Boundary: Airy Functions
Consider now the case where the potential rises smoothly across the boundary
This is the typical case
We assumed k2 is large, but this is not true near x = a, b
By definition, k2 vanishes there
Close to x = a, Taylor expand k2(x)
We are trying to solve
Compare to the Airy equation
General solution to Airy equation is
Ai and Bi are called Airy functions
The general solution for  will be Ai Bi

21. Phase Shift Near One Soft BoundaryAi Bi
Bi diverges as x  
We want only Ai, so
The Ai function as x  – behaves like
This implies as x goes above a, we would have
Compare this with the WKB wave function in the allowed region
We therefore conclude:

22. Quantization Condition For Soft Boundaries
When we had steep boundaries, we demanded wave function vanishes at x = a and x = b
This implied
The softening of the boundary at x = a, changed the boundary condition there to
Similarly, the softening of the boundary at x = b changes the boundary condition there to
Recall the definition of k(x):
We therefore have:

23. Three Different Cases
We can get quantization conditions for three types of problems:
Hard boundaries:
Soft boundaries:
One hard, one soft:

24. How to use these formulas
Given a potential V(x), can we estimate the energy eigenstates?
Pick the energy E
Find the classical turning points a and b, the points where the particle would stop classically
Write the relevant integral, usually
Do the integral
Solve for En as a function of n

25. Sample Problem
Estimate the eigenenergies of a particle of mass m in the 1D Harmonic oscillator with V(x) = m2x2/2
We pick an energy E
We solve for the turning points:
Substitute into the integral:
Make a trigonometric substitution:
By coincidence, we got it exactly right

26. Probability Density in WKB and Classical
Consider the wave function in the classically allowed region in the WKB approximation:
The probability density is:
WKB works for high energy, k(x) large
Rapidly oscillating sine function:
So we have
k is like momentum, which is like velocity
Classically, fraction of time spent in region of size dx is given by

27. 12C. Time Independent Perturbation Theory
A Series Expansion
Consider a situation where the Hamiltonian can be split into a large, easily solved piece, and a small piece:
Replace W by W, where  = 1
The large part is assumed to have known eigenvalues and complete, orthonormal eigenvectors:
Let |n be the exact eigenstates of H with energies En
In the limit   0, the |n’s will be |n’s and the En’s will go to n’s
It makes sense, therefore, to imagine a series expansion in terms of the parameter  for |n and |n

28. The Idea Behind The Method
We write out Schrödinger’s equation as follows:
Expand to some order:
Since this must be true for all , the coefficients of every power of  must match

29. A Small Ambiguity Problem
Schrödinger’s time-independent equation does not completely determine the eigenstates |n
We can always multiply by an arbitrary complex number c:
Therefore these states are slightly ambiguous
In the limit W = 0, we know that
One way to resolve the ambiguity is to simply demand
The problem: this means final states |n are not normalized
Can always be normalized later
Irrelevant for energy computation
Only relevant in 2nd or higher order state vector

30. The Procedure For each order (p): Act on the left with n|
Let H0 act on n|:
Act on the left with m| for m  n:
Reconstruct |n(p) using completeness
Iterate order by order
When done, reconstruct |n and En using  = 1:

31. First and Second Order:
First order:
Second order:

32. Third Order Energy and Summary:
Put it all together:

33. Sample Problem
Find the ground state energy to second order and eigenstate to first order for a particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small.
Split Hamiltonian into H0 and perturbation W:
Find eigenstates and energies of H0:
We now need to find

34. Sample Problem (2)
Find the ground state energy to second order and eigenstate to first order for a particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small.
We therefore have:

35. Validity of Perturbation Theory
Perturbation theory can only be trusted if subsequent terms get smaller
Let’s compare first and second order energy contribution:
Let  be smallest gap between n and m:
Then we have
Compare to 'n
Roughly, perturbation theory works if
So we need   W

36. 12D. Degenerate Perturbation Theory The Problem and Its Solution
Look at the second order expression for energy or first order for state vector:
If two states have the same unperturbed energy, we will get in trouble
The problem disappears if we get “lucky” by having
If we have such states, then the unperturbed eigenstates are in fact ambiguous
We can use this ambiguity to change basis for our unperturbed states
Thereby ensuring the problem goes away
Then proceed with perturbation theory as normal

37. Procedure for Degenerate Perturbation Theory
Suppose we have a set of degenerate states
Define the reduced W matrix:
Find g normalized eigenvectors and eigenvalues for this matrix:
Change basis to:
Then in the new basis, we have:
And:
So we can use these as our new basis states!

38. Working in the New Basis States
In the new basis, the matrix elements of W vanish for any pair of distinct states with the same energy
This gets rid of problem terms
Note that the basis states we start with are determined partly by the perturbation
Even to leading (0th) order, we need to include perturbation theory
The perturbation breaks the degeneracy and determines our states
Note that the first correction to the energies are just
First order energy easy to find just from eigenvalues of

39. Sample Problem
A particle of mass m in two dimensions has Hamiltonian as given at right, where b is small
(a) What are the eigenstates and energies in the limit b = 0?
(b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b.
H0 is two harmonic oscillators
The x-direction has frequency 
The y-direction has frequency 2
The unperturbed states and energies are:
The first two states are nondegenerate
The second excited states are degenerate
Need to use degenerate perturbation theory!

40. Sample Problem (2)
(b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b.
We place our degenerate states in some order:
We then calculate the perturbation matrix

41. Sample Problem (3)
(b) For the first pair of degenerate states, determine the eigenstates to leading order and energies to first order in b.
We now find eigenstates and eigenvalues of this matrix:
The energies and eigenstates are therefore:
Note that to find the energies, all we needed was the eigenvalues