Jacobi Identity

The Poisson bracket possesses the symplectic structure [A,B]=ABBA.[A,B] = AB - BA.

(Note that this algebraic structure is not commutative)

Apply this rule to the brackets:

[A,[B,C]],[A, [B,C]], [B,[C,A]],[B, [C,A]], [C,[A,B]].[C, [A,B]].

Assemble the three brackets to prove that [A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0.[A,[B,C]] + [B, [C,A]] + [C, [A,B]] = 0.


Following this result, we apply the same rules to the following brackets:

[A,[B,C]]=[A,[BCCB]]=A(BCCB)(BCCB)A=ABCACBBCA+CBA\left[ A,\left[ B,C \right] \right] = \left[ A,\left[ BC-CB \right] \right] = A(BC-CB) - (BC-CB)A=ABC-ACB-BCA+CBA

[B,[C,A]]=[B,[CAAC]]=B(CAAC)(CAAC)B=BCABACCAB+ACB\left[ B,\left[ C,A \right] \right] = \left[ B,\left[ CA-AC \right] \right] = B(CA-AC) - (CA-AC)B=BCA-BAC-CAB+ACB

[C,[A,B]]=[C,[ABBA]]=C(ABBA)(ABBA)C=CABCBAABC+BAC\left[ C,\left[ A,B \right] \right] =\left[ C,\left[ AB-BA \right] \right] = C(AB-BA) - (AB-BA)C=CAB-CBA-ABC+BAC

We note that the order of A,B,CA,B,C is significant for the bracket operator. Adding the three brackets will yield 0.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
7 years ago

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