(b) Show that [1, 1 =0 Hint: The following commutator identities are helpful. [B, A]= -[A, B], [A, A] =0 [A, B + C]=[A, B]+ [A, C] [A + B, C]=[A, C]+[B,C] [A, BC]= [A, B]C + B[A, C] [AB, C]=[A, C]B + A[B, C ] [AB, CD]=[A, C]BD + A[B, C]D + C[A, D]B + AC[B, D]