Question: (a) Show that d d [e A B e A ] = e A [ A , B ] e A for all A and
(a) Show that d d [e A B e A ] = e A [ A , B ] e A for all A and B. (b) Derive the expansion e A B e A = B + [ A , B ] + 1 2! [ A , [ A , B ]] + 1 3! [ A ,[ A , [ A , B ]]] + . . . Suggestion: expand F () = e A B e A in Taylor series about = 0, then set = 1
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