Solution to 1992 Problem 89

The wave function corresponding to energy level n is
\begin{align*}\psi_n = \left(\text{const} \right) H_n \left(\xi \right) e^{-\xi^2/2}\end{align*}
where \xi = \left(\text{const} \right) x and H_n is the nth Hermite polynomial. These wavefunctions represent all possible solutions to the harmonic oscillator potential that go to 0 when \left|\xi\right| or equivalent \left|x\right| goes to infinity. Therefore, these wavefunctions represent all possuble solutions to the harmonic oscillator potential on a domain that is infinite in at least one direction (like in this problem). Now, Hermite polynomials are even functions if n is even and odd functions if n is odd. Therefore, only those \psi_n where n is odd satisfy the boundary condition \psi_n\left(0 \right) = 0. Therefore, only these \psi_n are solutions for the potential in this problem. Therefore, answer (E) is correct.

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