# 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.

## The setup

Like the non-degenerate case, we are covering the time independent perturbation methods from section 16.1 of the text [1].

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

so that

Setting requires

for

We rescale our kets

where

The normalization of the rescaled kets is then

One can then construct a renormalized ket if desired

so that

## The meat.

We continue by renaming terms in 1.10

where

Now we act on this with the Hamiltonian

or

Expanding this, we have

We want to write this as

This is

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

With

or

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.

# References

[1] BR Desai. *Quantum mechanics with basic field theory*. Cambridge University Press, 2009.