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# Jacobi Identity

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

(Note that this algebraic structure is not commutative)

Apply this rule to the brackets:

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

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

Solution

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

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

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

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

We note that the order of $$A,B,C$$ 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
3 years, 8 months ago

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