Discussion
A particle of mass m is in a potential \(V(x) = \frac{1}{2}m\omega^{2}x^{2}\) for \(x \gt 0\) and \(V(x) = \infty\) for \(x \le 0\), where \(\omega\) is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
Answer: A
The ratio is \(\frac{3}{7}\).
The infinite wall at \(x = 0\) forces the wavefunction to vanish there. Of the full oscillator's eigenfunctions, only the odd ones (\(n = 1, 3, 5, \ldots\)) vanish at the origin, and on \(x \gt 0\) they still satisfy the Schrodinger equation. So the allowed energies are
\[ \begin{aligned} E &= \left(n + \tfrac{1}{2}\right)\hbar\omega,\ n = 1, 3, \ldots \\ \frac{E_{1}}{E_{3}} &= \frac{\frac{3}{2}\hbar\omega}{\frac{7}{2}\hbar\omega} = \frac{3}{7} \end{aligned} \]
\(\frac{1}{3}\) is the ratio for the full oscillator (\(\frac{1}{2}\) to \(\frac{3}{2}\)), and \(\frac{3}{5}\) uses consecutive levels \(n = 1, 2\), ignoring that even states are forbidden.
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