Discussion
Given \([x_{i}, p_{j}] = i\hbar\delta_{ij}\), i, j = 1, 2, 3, \([x_{1}, p_{1}^{3}]\) is :
Answer: C
The commutator is \(3i\hbar p_{1}^{2}\).
Use the product rule \([A, BC] = [A, B]C + B[A, C]\) with \([x_{1}, p_{1}] = i\hbar\), a number that commutes with everything:
\[ \begin{aligned} [x_{1}, p_{1}^{3}] &= [x_{1}, p_{1}]p_{1}^{2} + p_{1}[x_{1}, p_{1}]p_{1} \\ &\quad + p_{1}^{2}[x_{1}, p_{1}] \\ &= 3i\hbar\, p_{1}^{2} \end{aligned} \]
In general \([x, p^{n}] = n i\hbar p^{n-1}\), which is \(i\hbar\, \partial(p^{n})/\partial p\). The option \((i\hbar)^{3}p_{1}\) multiplies the commutator three times instead of adding three terms, \(i\hbar^{2}p_{1}\) has the wrong power of p, and zero would apply only to different components such as \([x_{1}, p_{2}^{3}]\).
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