# Why are the eigenkets of perturbed hamiltonian the eigenkets of the perturbation matrix?

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.

• 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? Commented Mar 29, 2022 at 14:22
• So what are the eigenvectors of $I+\lambda \sigma_1$? Commented Mar 29, 2022 at 14:30
• Commented Mar 29, 2022 at 14:32
• Commented Mar 29, 2022 at 19:55