# Find the eigenkets of degenerate perturbation theory

I guess Im missing something BIG, because it is not explained in any book.

When I study the correction of a perturbed degenerated Hamiltonian I must find first the energy correction. This is well explained: I find the eigenvalues of the "reduced" matrix (here Im calling the "reduced" matrix as the matrix which represents the perturbation in the degenerate subspace).

But how do I find the eigenket correction? One would think that the eigenket corrections are the eigenkets of the reduced matrix, but that matrix has not the dimension of the full Hilbert space, so the eigenket correction would have dimension $n$, being $n$ the degeneracy.

What am I missing?

• Commented Jul 24, 2018 at 20:52
• Commented Jul 24, 2018 at 20:57

I assume you mean time independent perturbation theory. This book by David McIntyre is a great introduction to quantum mechanics (I used it as a supplement to Griffiths'). I especially like how McIntyre uses spin-1/2 as a working example to define the basic concepts of quantum mechanics, and his chapters on perturbation theory are clear and concise.

Below, I describe how to find the first order and second order corrections to energies and energy eigenstates/eigenkets. When $$H$$ is degenerate, one has to locate the degenerate subspace of $$H$$ and work with that directly.

For non-degenerate perturbation theory:

Our original Hamiltonian, $$H$$, and our perturbed Hamiltonian, $$H'$$, then we have the eigenvalue equation,

$$H |n^{(0)}\rangle = E^{(0)}_{n} |n^{(0)}\rangle$$

where the superscript denotes the order of the correction, i.e. (0) denotes the unperturbed system.

So we may expand the energies in terms of the energy corrections,

$$E_{n} = E^{(1)}_{n} + E^{(2)}_{n} + ...$$

and the energy eigenkets,

$$|n\rangle = |n^{(0)}\rangle + |n^{(1)}\rangle + ...$$

So, the elements of the perturbed Hamiltonian are given by

$$H'_{n,n} = \langle n^{(0)}|H'|n^{(0)}\rangle = E^{(1)}_{n} .$$

It can be shown, by taking advantage of the properties of the inner product, that the other energies and the eigenket corrections are

$$E_{n}^{(2)} = \sum\lim_{m \neq n} \frac{|\langle m^{(0)} | H' | n^{(0)}\rangle|^{2}}{E_{n}^{(0)} - E_{m}^{(0)}},$$

$$|n^{(1)}\rangle = \sum\lim_{m \neq n} \frac{\langle m^{(0)} | H' | n^{(0)}\rangle}{E_{n}^{(0)} - E_{m}^{(0)}} |m^{(0)}\rangle.$$

For degenerate perturbation theory:

So, of course, the degenerate subspace of $$H$$ in your question is of smaller dimensionality than $$H$$. This means that not every eigenstate needs a correction! Since you've located the degenerate subspace, now you need to diagonalize the perturbation Hamiltonian $$H'$$ in the degenerate subspace to obtain the corrections to the original eigenkets. Page 339 of McIntyre's book goes over several examples.

If McIntyre's book is not available to you, this website is helpful.