Solution to 1986 Problem 96

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.

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