Second order perturbation theory

The result for the non-degenerate second order perturbation theory is $$E_n'= \sum_{m\neq n}\frac{\left|\langle m | H' | n\rangle\right|^2}{E_n-E_m}$$ But does this mean that is doesn't matter if the purturbation is positive or negative? Sorry for not being able to write in clear form.

• – Qmechanic Jul 29 '17 at 18:31

Second-order perturbation theory results depend on $|\lambda|^2$, where $\lambda$ is the coupling to the perturbation ($H = H_0 + \lambda V$). You do get a sign, coming from the denominator. As you can see, the perturbation acts as a "repulsion" between the energy levels, because when $E_m$ is above $E_n$, the shift $E_n^{(2)}<0$ and the levels get further apart in energy. The converse happens when $E_m < E_n$. One says that the levels anti-cross.
For this question, the 1st order correction being 0, we must switch up to calculating 2nd order correction. The order and sign of the correction is wanted in this question, so if I say that the second order correction is proportional to the square of the perturbation term, and from your discussion, it is in principle negative, that the correction is actually a decrease in $$\epsilon^2$$. But from the mathematical definition it shows that for a constant perturbation, it goes out of the integration and $$(<\psi_n|\psi_m>)^2=0$$ since they are orthogonal. Isn't it?