I have for quite some time now tried to find a satisfactory answer to this, but I haven't yet. In perturbation theory, with small parameter $\lambda$, we expand the eigenstate as

$$| E \rangle=| E^{(0)} \rangle + \lambda | b \rangle + ...$$

Where $| E^{(0)} \rangle$ is an eigenstate of the unperturbed Hamiltonian. The problem is that when there is degeneracy, there is a choice of these eigenstates. I know the answer is that they are chosen to be eigenvectors of the perturbing Hamiltonian, but my question is why.

  • $\begingroup$ DanielSank's answer is just fine. I'd just like to add that you do not get these kind of issues when you do Brillouin-Wigner (BW) perturbation theory rather than Rayleigh-Schrödinger (i.e. expansion in $\lambda$). BW perturbation theory is more easily generalized to situations with (quasi-)degeneracy and rarely treated in books. A basic introduction can be found here: phys.ufl.edu/~kevin/teaching/6646/04spring/bw.pdf $\endgroup$ Jun 18, 2014 at 23:11
  • $\begingroup$ Yeah, I had read that. One thing isn't clear though, is the zeroth order state in degenerate BWPT arbitrary then? As in, it'll give the right answer whatever state I choose in the degenerate subspace? $\endgroup$
    – guillefix
    Jun 18, 2014 at 23:39
  • $\begingroup$ Nope, BW and RS will give you the same results. I just mentioned it b/c it's more automatic and you don't have to worry about these degeneracy problems. In particular, I wasn't aware of this until today as I never learned anything other than BW perturbation theory. $\endgroup$ Jun 19, 2014 at 6:36
  • $\begingroup$ I wasn't say it will give different results. I was just saying that the only way that I see it can circumnavigate the problem of which to choose (in the degenerate subspace for the zeroth order) is that the choice doesn't matter. $\endgroup$
    – guillefix
    Jun 19, 2014 at 11:36
  • $\begingroup$ For the record, I have found an explanation of the BW perturbation theory that has made me understand me. It comes from a set of lecture notes on quantum mechanics which is the clearest I have found: bohr.physics.berkeley.edu/classes/221/1112/notes/pertth.pdf (All notes in: bohr.physics.berkeley.edu/classes/221/1112/221.html) $\endgroup$
    – guillefix
    Mar 20, 2015 at 20:48

2 Answers 2


This is really a great question.

Take a look at the figure attached here. The circle indicates a 2-dimensional degenerate subspace at $\lambda=0$.

In red we indicate two possible basis states for the subspace. In blue we show another possible choice of basis states.

Now look at the curvy green lines, these are the states as they evolve for $\lambda > 0$. The green curves do not connect with the red states at $\lambda=0$. The perturbation series is a Taylor series, which is a continuous function (it's a polynomial), so there is no way to make the series go from the red states at $\lambda=0$ to the green states for $\lambda > 0$.

Clearly, we have to start the perturbation series at $\lambda=0$ with the blue states, because those connect up with the green ones when $\lambda > 0$. The green states are, by definition, eigenstates of the perturbed Hamiltonian, so the blue states at $\lambda=0$ must also be eigenstates of the perturbed Hamiltonian.

Degenerate perturbation Degenerate perturbation theory states. We show the states as functions of the perturbation parameter $\lambda$ (green), and various choices of the unperturbed states (blue, red).

  • $\begingroup$ Thanks! I like this way of thinking about limits! One more thing. We have to go from "they have to be eigenstates of the perturbed Hamiltonian" as $\lambda$ goes to $0$ to the standard calculation of the eigenstates restricting the perturbing Hamiltonian to the degenerate subspace. Is the argument for that, that the action of the perturbing Hamiltonian approaches that of the restricted one as $\lambda$ approaches 0? $\endgroup$
    – guillefix
    Jun 18, 2014 at 23:32
  • $\begingroup$ @guillefix: I don't understand what you're asking. Your sentence is really long and too grammatically complex for my little brain. $\endgroup$
    – DanielSank
    Jun 18, 2014 at 23:48
  • $\begingroup$ I was just trying to justify the way we compute the eigenstates. You say they have to be eigenstates of the perturbed Hamiltonian. However, when we compute it, we actually consider only its restriction to the degenerate subspace. And I just said that I think that's fine because the perturbed Hamiltonian approaches its restriction as $\lambda$ goes to $0$. And sorry for my long sentences, maybe because my first language is Spanish :P $\endgroup$
    – guillefix
    Jun 18, 2014 at 23:54
  • $\begingroup$ @guillefix: "The eigenstates" are eigenstates of the perturbed Hamiltonian because that's what we mean by "eigenstates". When you compute the perturbed states, we not restrict them to the unperturbed degenerate subspace. $\endgroup$
    – DanielSank
    Jun 19, 2014 at 0:26
  • $\begingroup$ I thought we diagonalized $V_{i j}= \langle E_{i} | V | E_{j} \rangle$ where the $\{ | E_{i} \rangle \}$ are the degenerate states right? That is the restriction of V to the degenerate subspace. $\endgroup$
    – guillefix
    Jun 19, 2014 at 15:16

As I cannot comment (reputation>50 needed), I put my comment here, as this might be considered an answer, too.

The answer and picture of DanielSank is very nice, so just to add what I think is a satisfactory answer as well, just in words.

On page 139 of my favourite textbook:-), the explanation is right in the beginning of the "Secular equation" paragraph.

"We denote by $\psi_n^{(0)}$, $\psi_{n'}^{(0)},\dots$ the eigenfunctions belonging to the same eigenvalue $E_n^{(0)}$ of the energy. The choice of these functions is, as we know, not unique; instead of them we can choose any $s$ (where $s$ is the degree of degeneracy of the level $E_n^{(0)}$) independent linear combinations of these functions. The choice ceases to be arbitrary, however, if we subject the wave functions to the requirement that the change in them under the action of the small applied perturbation should be small. ... The coefficients in these combinations are determined, together with the corrections in the first approximation to the eigenvalues, as follows."

Then it turns out that the coefficients are obtained as a solution of the characteristic (secular) equation and that such eigenvectors are the eigenvectors of the perturbed Hamiltonian.

  • $\begingroup$ Can you tell us what you "favorite textbook" is? The link is broke. $\endgroup$
    – dturn805
    Jan 20 at 23:39
  • $\begingroup$ @dturn805 Quantum Mechanics: A Shorter Course of Theoretical Physics by L D Landau, E. M. Lifshitz. Paragraph 33, The secular equation. I have updated the link. $\endgroup$
    – wondering
    Jan 21 at 11:54

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