Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer 1

up vote 1 down vote accepted

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 $2B[A,B]=4 i \sigma_2 \sigma_3 \ne 0$.

share|improve this answer
    
Alright, thanks for the help Logan. I just thought it would require some more rigor. Thanks! –  tquarton Oct 7 '12 at 22:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.