# In QM perturbation theory, is the system generally in an eigenstate of the perturbing Hamiltonian, $\hat H_1$?

In my notes the derivation of the second order energy correction we don't do the following:

$$\sum_k a_{nk}\langle\phi_n|\hat H_1|\phi_k\rangle=\sum_ka_{nk}E_k^{(1)}\langle\phi_n|\phi_k\rangle$$

where $$\phi_n$$ and $$\phi_k$$ are eigenstates of the unperturbed Hamiltonian and $$\hat H_1$$ is the perturbing Hamiltonian. Is this because the eigenstates of the unperturbed system are not necessarily eigenstates of the perturbation? I'm not sure why else this wouldn't work.

• Yes, generally we choose the state as eigenstate of the known hamiltonian, the perturbation is done evaluating the correction on that states. The strength of perturbation theory is that we can, if the perturbation is small, always consider the states of the total hamiltonian as being given by the state of the known hamiltonian plus the correction given by the perturbative part. Mar 25 '20 at 14:15
• Also, if you want, take a look at the lectures of the course of Quantum physics III given by Zwiebach at MIT, the first part on perturbation theory is very clear, it helped me a lot at the time. Mar 25 '20 at 14:21
• I will definitely take a look at that, I've seen some of Zwiebach's lectures he's really excellent, thanks. Mar 25 '20 at 14:23

Indeed, the eigenstates of the unperturbed Hamiltonian ($$\hat{H}_0$$) are not the eigenstates of the perturbation ($$\hat{H}_1$$) - otherwise one wouldn't need the perturbation theory, since the perturbing Hamiltonian is diagonal: it does not mix the states and merely adds a correction to the energy levels of $$\hat{H}_0$$.
Let me further note that frequently when presenting the perturbation theory one assumes that the diagonal part of $$\hat{H}_1$$ is already incorporated into $$\hat{H}_0$$, i.e. $$\langle n |\hat{H}_1 |n\rangle = 0$$. In some treatments it may appear as the first order correction $$E_n^{(1)} = \langle n |\hat{H}_1 |n\rangle$$ - note that the states on the right and the left here are the same, unlike in the equation that you wrote, which is probably incorrect or based on a different definition of $$E_n^{(1)}$$.