Time independent perturbation theory with degeneracy
Posted by peeterjoot on September 30, 2011
Time independent perturbation with degeneracy.
In class it was claimed that if we repeated the derivation of the first order pertubation with degenerate states, then we’d get into (divide by zero) trouble if the state we were perturbing had degeneracy. Here I alter the previous derivation to show this explicitly.
Like the non-degenerate case, we are covering the time independent perturbation methods from section 16.1 of the text .
We start with a known Hamiltonian , and alter it with the addition of a “small” perturbation
For the original operator, we assume that a complete set of eigenvectors and eigenkets is known
We seek the perturbed eigensolution
and assumed a perturbative series representation for the energy eigenvalues in the new system
Note that we do not assume that the perturbed energy states, if degenerate in the original system, are still degenerate after pertubation.
Given an assumed representation for the new eigenkets in terms of the known basis
and a pertubative series representation for the probability coefficients
We rescale our kets
The normalization of the rescaled kets is then
One can then construct a renormalized ket if desired
We continue by renaming terms in 1.10
Now we act on this with the Hamiltonian
Expanding this, we have
We want to write this as
So we form
and so forth.
Zeroth order in
Since , this first condition on is not much more than a statement that .
First order in
How about ? For this to be zero we require that both of the following are simultaneously zero
This first condition is
From the second condition we have
Utilizing the Hermitian nature of we can act backwards on
We note that . We can also expand the , which is
I found that reducing this sum wasn’t obvious until some actual integers were plugged in. Suppose that , and , then this is
Observe that we can also replace the superscript with in the above manipulation without impacting anything else. That and putting back in the abstract indexes, we have the general result
Utilizing this gives us, for
Here we see our first sign of the trouble hinted at in lecture 5. Just because does not mean that . For example, with and we would have
We’ve got a unless additional restrictions are imposed!
If we return to 1.33, we see that, for the result to be valid, when , and there exists degeneracy for the state, we require
(then 1.33 becomes a equality, and all is still okay)
And summarizing what we learn from our conditions we have
Second order in
Doing the same thing for we form (or assume)
We need to know what the is, and find that it is zero
Utilizing that we have
From 1.37, treating the case carefully, we have
Again, only if for do we have a result we can use. If that is the case, the first sum is killed without a divide by zero, leaving
We can now summarize by forming the first order terms of the perturbed energy and the corresponding kets
Notational discrepency: OOPS. It looks like I used different notation than in class for our matrix elements for the placement of the indexes.
 BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.