It's actually an interesting problem you raise, so let me try to expand a bit on my comment.
In general, the program for “showing that [first-order] perturbation theory (PT) works” is the following:
- Solve the perturbed problem exactly, yielding solutions $H\psi_n = E_n\psi_n$.
- Differentiate the results by the perturbation parameter, i.e. compute $dE_n/dv$ and $d\psi_n(x)/dv$.
- Compare this to the PT results $E_n^{(1)}$ and $\psi_n^{(1)}(x)$.
In the present case, this will not be possible analytically. Write an Ansatz for the full solution, split in two regions, (L) $V(x)=v$, and (R) $V(x)=0$. As in the regular infinite potential well, for each region you can eliminate one of the linearly independent solutions by the boundary conditions $\psi_L(0) = \psi_R(a) = 0$.
You are left with three parameters (an “amplitude” for each region, and the energy) and three relations to fix them ($\psi_R(a/2) = \psi_L(a/2)$, $\psi'_R(a/2) = \psi'_L(a/2)$, $\int\!dx\,\psi(x) = 1$). Simple enough in principle, but not analytically solvable in this case because the energy is given by an implicit transcendental equation.
The reason I say your problem is interesting is that, as you increase $v$, it will change the character of the lower-lying states. In the unperturbed case, all states are “oscillatory”, albeit cut off at the edges. In the perturbed case, this will still be true on the right side, but once $v$ is big enough, you will get states that decay exponentially on the left. A sensible guess would be that PT breaks down for these states.
As for $V_{nm} = \langle\phi_n|V|\phi_m\rangle$ and orthogonality: The $\phi$ are even/odd functions with respect to the well center. Split the integral $\langle\phi_n|\phi_m\rangle = \int_0^a\!dx\phi_n\phi_m = L + R$ into two parts corresponding to left and right side. As you say, $L + R = 0$; and $V_{nm} = vL$. If $n$ and $m$ have the same parity, $L=R$ and so $V_{nm} = 0$; but if the parity is different, $L=-R$ and $V_{nm}$ is non-zero.
1) Your perturbation operator does not conserve the particle number of the phonons, so only even powers of it will contribute to equilibrium expectation values.
Since you are interested only in the ground state, which doesn't have any phonons excited, this means that you have to create a phonon first. After that either another phonon can be created or the former can be destroyed. Only processes that return to the ground state contribute to the expectation value and for this you obviously need to apply the interaction operator an even amount of times.
2) In a path integral the phonons can be integrated out, leaving an effective electron-electron interaction. That too is proportional to the square of the perturbations matrix element.
Best Answer
Using the resolution of identity: $$1=\sum_m|m\rangle\langle m|,$$ we can write $$ \langle n|H'(E_n^0-H^0)^{-1}H'|n\rangle = \sum_{m,m'}\langle n|H'|m\rangle\langle m|(E_n^0-H^0)^{-1}|m'\rangle\langle m'|H'|n\rangle = \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle m'|H'|n\rangle}{\langle m|E_n^0-H^0|m'\rangle}=\\ \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle m'|H'|n\rangle}{(E_n^0-E_m^0)}\delta_{m,m'}= \sum_{m}\frac{\langle n|H'|m\rangle\langle m|H'|n\rangle}{(E_n^0-E_m^0)} $$ Thus, the expression in the book is essentially correct: the one thing missing is the projection operator that assures that $m\neq n$, but it might be that this is mentioned somewhere in the text or superfluous due to the nature of the problem.