where A H ( t ) ≡ ≡ A ( t ) = e i ℏ ℏ t H A e − − i ℏ ℏ t H in this particular case. The equation is solved 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 ] + 1 2 ! ( i t ℏ ℏ ) 2 [ H , [ H , A ] ] + 1 3 ! ( i t ℏ ℏ ) 3 [ H , [ H , [ H , A ] ] ] + ⋯ ⋯