Discussion
For a one-dimensional harmonic oscillator, the creation operator \((a^{\dagger})\) acting on the \(n^{th}\) state \(|\psi_{n}\rangle\), where \(n = 0, 1, 2, \ldots\), gives \(a^{\dagger}|\psi_{n}\rangle = \sqrt{n + 1}|\psi_{n+1}\rangle\). The matrix representation of the position operator \(x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^{\dagger})\) for the first three rows and columns is
Answer: C
The matrix is \(\sqrt{\frac{\hbar}{2m\omega}}\) times the matrix with 1 and \(\sqrt{2}\) next to the diagonal:
\[ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{pmatrix} \]
Using \(a^{\dagger}|\psi_{n}\rangle = \sqrt{n+1}|\psi_{n+1}\rangle\) and \(a|\psi_{n}\rangle = \sqrt{n}|\psi_{n-1}\rangle\), the only non-zero elements of \(a + a^{\dagger}\) are
\[ \begin{aligned} \langle n+1|(a + a^{\dagger})|n\rangle &= \sqrt{n+1} \\ \langle n|(a + a^{\dagger})|n+1\rangle &= \sqrt{n+1} \end{aligned} \]
So the elements next to the diagonal are 1 (between \(n = 0\) and 1) and \(\sqrt{2}\) (between 1 and 2), and the diagonal is zero, because \(x\) changes \(n\) by one. The matrix is symmetric, as a Hermitian operator with real elements must be.
The diagonal matrix with 1, \(\sqrt{2}\), \(\sqrt{3}\) is not how \(x\) acts; the option with 1, 1 misses the \(\sqrt{2}\); and the last option couples states 0 and 2, which \(x\) cannot do.
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