The equation is solved by the A (t) defined above, as evident by use of the standard operator identity, e B A e − − B = A + [ B , A ] + 1 2 ! [ B , [ B , A ] ] + 1 3 ! [ B , [ B , [ B , A ] ] ] + ⋯ ⋯ . which implies A ( t ) = A + i t ℏ ℏ [ H , A ] − − t 2 2 ! ℏ ℏ 2 [ H , [ H , A ] ] − − i t 3 3 ! ℏ ℏ 3 [ H , [ H , [ H , A ] ] ] + … …