I'm currently studying time-independent perturbation theory and I ran into a couple of doubts that for the life of me I can't seem to solve. A "good quantum number" is the eigenvalue of an eigenvector of an operator O that remains an eigenvector of O with the same eigenvalue as time evolves, according to Wikipedia: https://en.wikipedia.org/wiki/Good_quantum_number. There's a proof in the link above that shows that this is equivalent to having your operator O commute with the hamiltonian H of the problem.

However, I don't understand how these good quantum numbers are related to degenerate perturbation theory. When there is degeneracy, you don't know which states to use in the equations derived in the nondegenerate case, because any linear combination of the degenerate eigenfunctions is also an eigenfunction with the same eigenvalue. Plus, using the nondegenerate equations when there is degeneracy would involve division by zero. How are these two problems solved by the usage of good quantum numbers?

And what exactly all of this have to do with diagonalizing the perturbation matrix?

If possible, please keep your answers simple, I'm just a lowly undergrad working my way through quantum mechanics...

Thanks in advance for any help!


1 Answer 1


I think you should share or point out the source material that is confusing you because we are talking about time independent pertuberation so time is not a factor here.

when there is a degeneracy, this means that we have several eigenvectors $v_1, ... v_k$ with same eigenvalue for the original hamiltonian (i.e. energy), let $\mathbf{V}$ be the subspace spanned by them, it can be proven that the pertuperation $\delta H$ has to be diagonal in this subspace, for that we need to find any set of basis that spans all of $\mathbf{V}$ such that $\delta H$ is diagnonal, this doesn't change the physics because the new basis are still eigenvectors of the original hamiltonian with the same eigenvalue, after that several cases can happen which I won't go in detail into instead I refer you to MIT's 8.06 notes starting from page 15 which goes into detail of first and second order pertuperation theory.

  • $\begingroup$ Thank you for the MIT reference. I was following Griffiths' Introduction to Quantum Mechanics. $\endgroup$ Commented Dec 10, 2021 at 1:21
  • $\begingroup$ I think the reference to time here is simply in the sense that some operator commutes with $H$. $\endgroup$ Commented Dec 10, 2021 at 2:28
  • $\begingroup$ So you can show that the basis you need to choose is the one which diagonalizes the perturbation matrix in the degenerate subspace. I still have a number of questions about this: how does this relate to good quantum numbers? And how does this help solve the problem I mentioned in my question regarding dividing by zero? $\endgroup$ Commented Dec 10, 2021 at 2:57
  • $\begingroup$ I might have missed something but all references I could find in Griffiths to the word "good" are in reference to good basis and how to determine them, division by zero is avoided all together by going into first, second or even higher order pertuberation theory and using these good basis, that's why I mentioned "several cases" which determine how far in the theory (i.e. recursion) you need to go to avoid division by zero and the MIT notes go into detail on these cases for first and second order. $\endgroup$ Commented Dec 10, 2021 at 8:24
  • $\begingroup$ You can find the term "good quantum number" in page 270 of Griffiths' Introduction to Quantum Mechanics (Second Edition), and he uses it again when talking about the Zeeman effect.But it's true, generally he talks about "good quantum states". $\endgroup$ Commented Dec 10, 2021 at 21:27

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