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I'm trying to find the condition necessary for this commutator relationship equality:

$[A,B^2]=2B[A,B]$

So far I've done this: \begin{align*} [A,B^2] & = B[A,B] + [A,B]B \\ & = BAB - BBA + ABB - BAB. \end{align*} Now, its apparent that $B[A,B]$ must equal $[A,B]B$ which is true if $B$ commutes with $[A,B]$. If that's true, what can I say about $[A,B]$? I've played around with identity $BB^{-1}$ and adding things like $AAB - AAB$ but I really can't make that step forward to massage my equation to equal $B[A,B] + B[A,B]$ or $BBB^{-1}[A,B] + BBB^{-1}[A,B]$ Can anyone help me make this more clear or is the answer simply just $B$ must commute with $[A,B]$ so $[A,B]B = B[A,B]$? Thanks.

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    These are supposed to be quantum mechanics operators. Well, I was hoping to show algebraically that [A,B] must necessarily be something like a constant.2012-10-07

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The answer is that $B$ must commute with $[A,B]$. You've shown correctly that $[A,B^2]=B[A,B]+[A,B]B$; for this to be equal to $2B[A,B]$, we need $B[A,B]=[A,B]B$.

This is not true in general for arbitrary Hermitian operators. Take $A= \sigma_1$ and $B=\sigma_2$, where $\sigma_i$ are the Pauli matrices. Then $[A,B]=2 i \sigma_3$ which does not commute with $B=\sigma_2$. Since $B^2=I$, we have $[A,B^2]=0$, but $2B[A,B]=4 i \sigma_2 \sigma_3 \ne 0$.

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    Alright, thanks for the help Logan. I just thought it would require some more rigor. Thanks!2012-10-07