D.79 Derivation of perturbation theory

This note derives the perturbation theory results for the solution of
the eigenvalue problem

First, small

is not a valid mathematical term. There
are no small numbers in mathematics, just numbers that become zero in
some limit. Therefore, to mathematically analyze the problem, the
perturbation Hamiltonian will be written as

where

The idea is now to start with a good eigenfunction good

is still to be defined), and
correct it so that it becomes an eigenfunction of

If

Before proceeding with the practical application, a disclaimer is
needed. While it is relatively easy to see that the eigenvalues
expand in whole powers of

When the above power series are substituted into the eigenvalue
problem to be solved,

the net coefficient of every power of

These are the equations to be solved in succession to give the various terms in the expansion for the wave function

Note that all it takes is to solve problems of the form

The equations for the unknown functions are in terms of the unperturbed Hamiltonian

For difficult perturbation problems like you find in engineering, the
use of a small parameter

where you are assumed to just imagine that

first order small,

Now consider each of these equations in turn. First, (D.55)
is just the Hamiltonian eigenvalue problem for

where the coefficients

The problem (D.56) becomes

where the left hand side was cleaned up using the fact that the

And that is exactly the first order correction to the energy claimed in {A.38.1};

To get the coefficients

The coefficients

Note however that if the problem is degenerate, there will be
eigenfunctions

The eigenvalues of this submatrix are the correct first order energy changes. So, if all you want is the first order energy changes, you can stop here. Otherwise, you need to replace the unperturbed eigenfunctions that have energy

You will need to rewrite the Hamiltonian perturbation coefficients in terms of these new eigenfunctions. (Since the replacement eigenfunctions are linear combinations of the old ones, no new integrations are needed.) You then need to reselect the eigenfunction

With the coefficients

The entire process repeats for higher order. In particular, to second
order (D.57) gives, writing

To get the second order contribution to the energy, take again an inner product with

This gives the second order change in the energy stated in {A.38.1}, if

In the degenerate case, when taking an inner product with a

will need to be diagonalized, (the rest of the equation needs to be zero). Its eigenvalues give the correct second order energy changes. To proceed to still higher energy, reselect the eigenfunctions following the same general lines as before. Obviously, in the degenerate case the entire process can become very messy. And you may never become sure about the good eigenfunction.

This problem can often be eliminated or greatly reduced if the
eigenfunctions of good

quantum number

To see why, consider two example eigenfunctions

and since

can be replaced by just

The Hamiltonian perturbation coefficient