Chapter 14: Problem 3

Prove the following commutator idertities: $$ \begin{gathered} {[A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0 \quad \text { (Jacobi identity) }} \\ {[A B, C]=A[B, C]+[A, C] B} \\ {[A, B C]=[A, B] C+B[A, C]} \end{gathered} $$

Short Answer

Expert verified

All identities are proven to be true by expanding the left-hand sides using the properties of vectors, namely commutativity and associativity, and showing that they are equal to their respective right-hand sides. The Jacobi Identity's LHS simplifies to zero, like the RHS. For right Leibniz rule and left Leibniz rule, their LHS are shown to be equal to their respective RHS.

Step by step solution

Proving Jacobi identities

Start with the left hand side: \[LHS = [A,[B, C]]+[B,[C, A]]+[C,[A, B]]\] Substitute the definition of a commutator $[X, Y] = XY - YX$ Therefore, you have: \[ = (AB - BA)C - C(AB - BA) + (BC - CB)A - A(BC - CB) + (CA - AC)B - B(CA - AC)\] Rearrange the terms: \[ = ABC - BAC - CAB + CBA + BCA - CBA - ABC + BAC + CAB - ABC - BCA + ABC\] This simplifies to zero, which is RHS. Hence Jacobi identity is proved.

Proving right Leibniz rule

Start with left hand side $[AB,C]$. Substitute the definition of commutators:\[ = (AB)C - C(AB) = ABC - CAB\] Compare it with the right hand of the identity \[RHS = A[B, C] + [A, C]B\]Expand out the RHS: \[ = A(BC - CB) + (AB - BA)C = ABC - ACB + ABC - CAB\] If you group the terms, it's equal to the LHS, thus proving the right Leibniz rule.

Proving left Leibniz rule

Start with left side $[A, BC]$. Substitute the definition of commutators:\[ = A(BC) - (BC)A = ABC - BCA\] Compare it with the right side of the identity \[RHS = [A, B]C + B[A, C]\]Expand out the RHS: \[ = (AB - BA)C + B(AC - CA) = ABC - BAC + BAC - BCA\] If you group the terms, it's equal to the LHS, thus proving the left Leibniz rule.