$$[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0$$

This proof is not wanted, since there is an attachment of some variables.

How can you prove the equation?


closed as off-topic by Waffle's Crazy Peanut, Emilio Pisanty, Manishearth Sep 18 '13 at 18:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, Manishearth
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Is $[\cdot,\cdot]$ supposed to be the Poisson bracket as it is in the proof you referenced or do you mean the quantum mechanical commutator? $\endgroup$ – Jonas Greitemann Sep 18 '13 at 12:10
  • $\begingroup$ In general you can't. But if [,] is the commutator, why dont just just expand the Jacobi formula? $\endgroup$ – jinawee Sep 18 '13 at 12:13
  • 1
    $\begingroup$ I'd consider this a math question. $\endgroup$ – Nikolaj-K Sep 18 '13 at 12:25
  • $\begingroup$ Comments to the question (v1): Echoing Jonas's comment, is this a question about Poisson brackets or operator commutators? The proof is very different depending on which. And what is 'generalized' about the Jacobi identity? $\endgroup$ – Qmechanic Sep 18 '13 at 15:58

You can use the definition of the QM commutator $[X,Y]=XY-YX$, then expand all the commutators and simplify.


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