Discussion
If \(\bar{A}\) and \(\bar{B}\) are quantum mechanical operators, then \([[\bar{A}, \bar{B}], [\bar{B}, \bar{A}]]\) is equal to :
Answer: B
The commutator is 0.
Let \(C = [\bar{A}, \bar{B}]\). Then \([\bar{B}, \bar{A}] = -C\), and
\[ [C, -C] = -(CC - CC) = 0 \]
since every operator commutes with itself and with any multiple of itself. This holds for any pair of operators.
The first option is what you get by expanding carelessly: it is not identically zero in general, so it cannot equal a quantity that always vanishes. The third option and \([A, B]^{2}\) are likewise non-zero for general operators, for example position and momentum, where \([x, p]^{2} = -\hbar^{2}\).
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