I've just started studying perturbation theory, and of course have now encountered the case where degeneracy arises. I understand why we have to diagonalize the perturbation matrix, and the concept of finding a "good" basis, but I'm not quite sure why the good bases of the unperturbed Hamiltonian that diagonalise the perturbation are now the new eigenkets of the new total Hamiltonian $(H_{0} + \hat{V})$ beyond the degeneracy subspace?

Shouldn't we do some extra work, and use $$ |n^{(1)}\rangle = \sum\lim_{E_m \neq E_n} \frac{\langle m^{(0)} | H' | n^{(0)}\rangle}{E_{n}^{(0)} - E_{m}^{(0)}} |m^{(0)}\rangle.$$ to find the corrections from all the other eigenkets outside the degeneracy subspace?

I'll try and make this clearer with the following example:

Let's consider a 2D simple harmonic oscillator and its Hamiltonian H : $ \frac{1}{2m}(p_{x}^2 +p_{y}^2) + K(x^2 +y^2)$.

Completely analogous to the 1D SHO, the energies are simply $E_{np} = \hbar\omega(n+p+1)$. One can clearly see that there is a two-fold degeneracy for the 1st excited state, i.e. $E_{10} = E_{01}$. Therefore, when we apply the perturbation $\hat{V} = xy$, we need to apply the degenerate perturbation theory machinery. Elements of the perturbation matrix can be found in terms of 1st excited states (See below).

$$ H'=K'\begin{pmatrix} \langle 10|xy|10\rangle & \langle 10|xy|01\rangle \\ \langle 01|xy|10\rangle & \langle 01|xy|01\rangle \end{pmatrix}= \mathbb E\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\\mathbb E = \frac{K'}{2\beta^2}, \beta^2=\frac{m\omega_0}{\hbar} $$

Now, the eigenvalues of this perturbation matrix, are the first order energy corrections. Makes sense. But now, why are the eigenkets of this matrix, the new eigenkets of the new Hamiltonian?

$$ E'=+\mathbb E\leadsto\varphi_1=\frac{1}{\sqrt2}(\psi_{10}+\psi_{01}) \\E'=-\mathbb E\leadsto\varphi_2=\frac{1}{\sqrt2}(\psi_{10}-\psi_{01}), $$ where are the corrections due the other, non-degenerate eigenkets, as described in my summation at the top?

Any clarification would be much appreciated.

  • $\begingroup$ No, my question is why are the eigenvectors of V, now the new perturbed eigenkets? Why are we not considering the contributions from all the other, non-degenerate eigenkets? $\endgroup$
    – jambajuice
    Commented Mar 29, 2022 at 14:22
  • $\begingroup$ So what are the eigenvectors of $I+\lambda \sigma_1$? $\endgroup$ Commented Mar 29, 2022 at 14:30
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – jambajuice
    Commented Mar 29, 2022 at 14:32
  • $\begingroup$ Near duplicate. $\endgroup$ Commented Mar 29, 2022 at 19:55

1 Answer 1


I asked my professor the same exact question like a couple weeks ago, and she responded with yes; technically you should use the good basis as your zeroth order wf and apply non-degenerate perturbation theory to get an approximate wf for the perturbed system (perturbation plus original potential). However, she also added that usually we were going to focus on finding the energy corrections to see how degeneracies become “lifted” for different cases of perturbations and so for now we were just going to stop with the “good basis” since that’s all you need to find in order to use non-degenerate perturbation theory to find the energy corrections (aka for finding the fine, hyperfine structure and Zeeman effect energy corrections you can just work with a good basis for each circumstance and use non-degenerate perturbation theory successively to get first order E correction). However, unless I’m understanding something wrong (and if so, let me know please), you are correct for the wf correction (in general) but maybe your class just isn’t going that far at the moment so they don’t really focus on that?

  • 1
    $\begingroup$ Yeah I think this is correct, also Cosmas attached a link which pretty much clears up my confusion. Thanks! $\endgroup$
    – jambajuice
    Commented Mar 29, 2022 at 21:23

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