Discussion
Let \(|m\rangle\) and \(|n\rangle\) denote the energy eigenstates of a one-dimensional simple harmonic oscillator. The position and momentum operators are \(\hat{X}\) and \(\hat{P}\), respectively. The matrix element \(\langle m|\hat{P}\hat{X}|n\rangle\) is non-zero when
Answer: B
The matrix element is non-zero when \(m = n\) or \(m = n \pm 2\).
Write \(\hat{X} \propto (a + a^{\dagger})\) and \(\hat{P} \propto i(a^{\dagger} - a)\). Then
\[ \begin{aligned} \hat{P}\hat{X} &\propto (a^{\dagger} - a)(a + a^{\dagger}) \\ &= a^{\dagger 2} - a^{2} + a^{\dagger}a - aa^{\dagger} \\ &= a^{\dagger 2} - a^{2} - 1 \end{aligned} \]
\(a^{\dagger 2}\) raises \(n\) by 2, \(a^{2}\) lowers it by 2, and the constant \(-1\) (from \([a, a^{\dagger}] = 1\)) gives a diagonal element. Indeed \(\langle n|\hat{P}\hat{X}|n\rangle = -\frac{i\hbar}{2}\), not zero.
'\(m = n \pm 2\) only' misses that diagonal term; \(n \pm 1\) would come from a single operator, and \(n \pm 3\) from three.
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