Do anticommutators of operators has simple relations like commutators.

For example: $$[AB,C]=A[B,C]-[C,A]B.$$

But I don't find any properties on anticommutators. Do same kind of relations exists for anticommutators?


closed as off-topic by JamalS, Yashas, Jon Custer, David Hammen, John Rennie Mar 11 '17 at 11:47

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ I'm voting to close this question as off-topic because it shows insufficient prior research with the answer plainly available on Wikipedia and does not ask about any concept or show any effort to derive a relation. $\endgroup$ – JamalS Mar 10 '17 at 21:55
  • 1
    $\begingroup$ very good question $\endgroup$ – Ka-Wa Yip Mar 12 '17 at 0:52

As you can see from the relation between commutators and anticommutators $$ [A,B] := AB-BA = AB - BA -BA + BA = AB + BA - 2BA = \{A,B\} - 2 BA $$ it is easy to translate any commutator identity you like into the respective anticommutator identity. Unfortunately, you won't be able to get rid of the "ugly" additional term.

This is probably the reason why the identities for the anticommutator aren't listed anywhere - they simply aren't that nice.

  • 1
    $\begingroup$ That is all I wanted to know. We always have a "bad" extra term with anti commutators. Thanks ! $\endgroup$ – StarBucK Mar 10 '17 at 21:57
  • $\begingroup$ @user3183950 You can skip the bad term if you are okay to include commutators in the anti-commutator relations. $\endgroup$ – JamalS Mar 10 '17 at 22:00

First, let's prove the result you gave

$$ [AB,C] = ABC-CAB = ABC-ACB+ACB-CAB = A[B,C] + [A,C]B. $$

Notice that $ACB-ACB = 0$, which is why we were allowed to insert this after the second equals sign.

Now let's consider the equivalent anti-commutator $\lbrace AB , C\rbrace$; using the same trick as before we find

$$ \lbrace AB,C \rbrace = ABC+CAB = ABC-ACB+ACB+CAB = A[B,C] + \lbrace A,C\rbrace B $$

Here are a few more identities from Wikipedia involving the anti-commutator that are just as simple to prove: $$ {\displaystyle \{A,BC\}=\{A,B\}C-B[A,C]} $$ $$ {\displaystyle \{AB,C\}=A\{B,C\}-[A,C]B} $$ $$ {\displaystyle [AB,C]=A\{B,C\}-\{A,C\}B} $$ https://en.wikipedia.org/wiki/Commutator#Identities_.28ring_theory.29

Hope this helps.

  • $\begingroup$ I think there's a minus sign wrong in this answer. The correct relationship is $ [AB, C] = A [ B, C ] + [ A, C ] B $. The mistake is in the last equals sign (on the first line) -- $ ACB - CAB = [ A, C ] B $, not $ - [A, C] B $. I think that the rest is correct. $\endgroup$ – QuantumFool Nov 26 '17 at 10:04

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