This
$$[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0$$
This proof is not wanted, since there is an attachment of some variables.
How can you prove the equation?
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Abhimanyu Pallavi Sudhir
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asked Sep 18, 2013 at 12:05
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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 GreitemannCommented Sep 18, 2013 at 12:10
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$\begingroup$ In general you can't. But if [,] is the commutator, why dont just just expand the Jacobi formula? $\endgroup$– jinaweeCommented Sep 18, 2013 at 12:13
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1$\begingroup$ I'd consider this a math question. $\endgroup$– Nikolaj-KCommented Sep 18, 2013 at 12:25
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$\begingroup$ 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 ♦Commented Sep 18, 2013 at 15:58
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You can use the definition of the QM commutator $[X,Y]=XY-YX$, then expand all the commutators and simplify.