# Commutator relationship proof $[A,B^2] = 2B[A,B]$

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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I think you meant $B[A,B] + [A,B]B=BAB-BBA+ABB-BAB$? – draks ... Oct 7 '12 at 21:58
Yes I did, sorry. – tquarton Oct 7 '12 at 21:59
These are elements of what, exactly? The answer you got is pretty good: what sort of alternative condition did you envision? – Mariano Suárez-Alvarez Oct 7 '12 at 22:02
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. – tquarton Oct 7 '12 at 22:05

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! – tquarton Oct 7 '12 at 22:12