## Solution to 1992 Problem 99

 The ground state of hydrogen is nondegenerate, so the first-order correction to the ground state energy is given by\begin{align}\langle \psi_{100}| e E z \psi_{100} \rangle = \int \left|\psi_{100} \right|^2 e E z \label{eqn:1}\end{align}where we have taken the electric field to be in the positive $z$ direction for convenience. Recall the function form of the ground state of hydrogen\begin{align*}\psi_{100} = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}\end{align*}This function is symmetric about the x-y plane (i.e. $\psi_{100}(x,y,-z) = \psi_{100}(x,y,z)$). Therefore, the integral in equation (1) must be zero, because $z$ is antisymmetric with respect to the x-y plane, so the $z$ integral will be an integral of an odd function over the symmetric interval $(-\infty,\infty)$. Therefore, answer (A) is correct.