The first-order correction to the nondegenerate state $\psi_0'$ in time-independent nondegenerate perturbation theory is

$\begin{align*}\psi_0' = \sum_{n \neq 0} \frac{\langle \psi_n | H' | \psi_0 \rangle}{ E_0 - E_n} \psi_n\end{align*}$
where $E_n$ is the energy of the unperturbed state $\psi_n$ . In this case, $H' = V'$ , which is an even function. $\psi_0$ is also an even function. Therefore, whenever, $\psi_n$ is odd, the inner product will be zero. Therefore, answer (B) is correct.

Solution to 1986 Problem 96