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 $xp = i\hbar + px$

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

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

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  • $\begingroup$ Thank you. I know it's an elementary arithmetic operation but I wasn't sure if this was still true with operators. $\endgroup$ – Steven Dec 23 '18 at 8:14
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    $\begingroup$ @Steven. Yes, it's easy to get confused and uncertain about new things even if they are quite trivial. $\endgroup$ – md2perpe Dec 23 '18 at 10:07
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    $\begingroup$ @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 $$[A^2,B] = A[A,B]+[A,B]A.$$ $\endgroup$ – md2perpe Dec 23 '18 at 10:10
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    $\begingroup$ @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, $AB \neq BA$. $\endgroup$ – knzhou Dec 23 '18 at 11:33

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