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The commutator $[A^2,B]$ can be written as $A[A,B]+[A,B]A$. So if $[A,B]=0$, $[A^2,B]$ is also zero. But is the converse also true? If $[A^2,B]$ is given to be zero, then is [A,B]=0?

Let $C=[A,B]$. If $[A^2,B]=0$, then $A[A,B]+[A,B]A=0$, so $AC+CA=0$. But then I don't know how to show that $C$ is/is not equal to zero. Can someone please show me the mathematical proof?

thanks

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    $\begingroup$ hint: let $A$ be a square root of the identity matrix (e.g., $A=\sigma_x$ and $B=\sigma_y$) $\endgroup$
    – AccidentalFourierTransform
    Commented May 18, 2016 at 19:39
  • $\begingroup$ Thanks. I can prove that $[A,B]=0$ using your method (so I guess the answer is that $[A^2,B]=0$ does imply $[A,B]=0$ if $A^2$ is nonzero? ). But are there more general ways of proving it? Cause we need to assume A and B to have particular forms. $\endgroup$
    – Physicist
    Commented May 18, 2016 at 19:48

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It's not true. For example take

$$ A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \qquad\qquad\qquad B = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} $$

you have $A^2=0$ so that $[A^2,B]=0$, but

$$ [A,B] = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. $$

You must add some condition, for example if you know also

$$ [A,[A,B]]=0 $$

and $A$ is invertible, then it's ok.

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  • $\begingroup$ What about if I require $A^2$ to be non zero? Under what conditions is it true/untrue? $\endgroup$
    – Physicist
    Commented May 18, 2016 at 19:50
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    $\begingroup$ Also if $A^2$ is not zero, as @AccidentalFourierTransform is telling you. Let $A=\sigma_x, B=\sigma_y$ then $[A,B]=\sigma_z$ but $A^2=1$ and so it does commute with $B$! $\endgroup$
    – MaPo
    Commented May 18, 2016 at 19:53
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No, it's not true and a simple counter example suffices. Let $A = \vec L^2$, the square of the angular momentum operator, and $B = L_z$, the z component of the angular momentum operator.

$[\vec L ^ 2, L_z] = 0$, but $[\vec L , L_z] != 0$ because the components of the angular momentum operator do not commute with each other.

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  • $\begingroup$ In what sense are you claiming that $\widehat{L^2}$ is the square of an "operator" $\left(\widehat {L_x},\widehat {L_y},\widehat {L_z}\right)$? And clearly the rule $[\hat A\circ \hat A,\hat B ]=\hat A[\hat A,\hat B]+[\hat A,\hat B]\hat A$ doesn't hold since the latter is a vector operator and the former is a scalar operator. $\endgroup$
    – Timaeus
    Commented May 19, 2016 at 16:44
  • $\begingroup$ And you even picked a vector "operator" whose components don't commute with each other, so the alleged vector operator isn't even an observable since it lacks a basis of eigenstates. A vector observable isn't an ordered tuple of observables. You can't do a measurement to observe the angular momentum vector. It's not like position or momentum. $\endgroup$
    – Timaeus
    Commented May 19, 2016 at 16:50

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