Question

Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C, A, B) = 0

Answer 1

Answer:

Three linear operators A,B, and C will satisfy the condition $[[A, B],C] + [[B,C), A] + [[C, A], B] = 0$.

Explanation:

According to the question we have to prove.

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

Now taking Left hand side of the equation and solve.

$[[A, B],C] + [[B,C), A] + [[C, A], B]$

Now use commutator property on it as,

$=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0$

Therefore, it is proved that $[[A, B],C] + [[B,C), A] + [[C, A], B] = 0$.