# A question about the commutator

I'm self-studying quantum mechanics and have a question regarding the commutator.

Since the commutator of two operators is defined as

$$[A,B]$$ = $$AB$$ - $$BA$$

Assuming that these operators do not commute, does that mean that the value of, say $$AB$$ = (commutator) $$+BA$$

For example, the commutator between position $$x$$ and momentum $$p$$ is $$[x,p] = xp-px$$ = $$i\hbar$$

Does it follow that $$xp = i\hbar + px$$

• $\uparrow$ Yes. – Qmechanic Dec 23 '18 at 7:56
• Just to be careful, note that these are operators: $\hat x \hat p = i\hbar \mathbf 1 + \hat p \hat x$ – GodotMisogi Dec 23 '18 at 8:07

## 1 Answer

Yes, that is correct. Just as $$x=y-z$$ implies that $$y=x+z,$$ the identity $$[A,B]=AB-BA$$ implies $$AB=[A,B]+BA.$$

• Thank you. I know it's an elementary arithmetic operation but I wasn't sure if this was still true with operators. – Steven Dec 23 '18 at 8:14
• @Steven. Yes, it's easy to get confused and uncertain about new things even if they are quite trivial. – md2perpe Dec 23 '18 at 10:07
• @Steven. It's actually quite common to use $AB = [A,B] + BA,$ e.g. like $$A^2 B = A(AB) = A([A,B]+BA) = A [A,B] + ABA = A [A,B] + (AB) A \\ = A[A,B] + ([A,B]+BA)A = A [A,B] + [A,B] A + BA^2$$ which shows that $$[A^2,B] = A[A,B]+[A,B]A.$$ – md2perpe Dec 23 '18 at 10:10
• @Steven There's no need to be so careful. Almost everything you know about arithmetic still works. The only change is that you can't swap the order anymore, $AB \neq BA$. – knzhou Dec 23 '18 at 11:33