Discussion
If \(H\) is the Hamiltonian for a free particle with mass \(m\), the commutator \([x, [x, H]]\) is
Answer: B
The commutator is \(-\hbar^{2}/m\).
For a free particle \(H = p^{2}/2m\). Using \([x, p] = i\hbar\):
\[ \begin{aligned} [x, H] &= \frac{1}{2m}[x, p^{2}] = \frac{1}{2m}(2i\hbar p) = \frac{i\hbar p}{m} \\ [x, [x, H]] &= \frac{i\hbar}{m}[x, p] = \frac{i\hbar}{m}(i\hbar) = -\frac{\hbar^{2}}{m} \end{aligned} \]
The positive results miss the sign from \(i^{2} = -1\), and the results with \(2m\) forget that \([x, p^{2}] = 2i\hbar p\), whose factor 2 cancels the 2 in \(2m\).
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