Dynamics in the heizenberg picture

[dudy]

Hello,
The time-derivative of an operator A(t) = U^\dagger a(t)U in the heizenberg picture is given by:

\frac{dA(t)}{dt} = \frac{i}{\hbar} [H,A(t)] + U^\dagger(\frac{da(t)}{dt})U

Now, I know that under some conditions, we can write:

\frac{dA(t)}{dt} = \frac{i}{\hbar} U^\dagger[H,a(t)]U + U^\dagger(\frac{da(t)}{dt})U

My question is- what are those conditions?

thanks

 

[bapowell]

Under the condition that the Hamiltonian is not dependent on time.

 

[dudy]

is there a more specific rule?
take for example:

H = a^\dagger e^{-i\omega t} + a e^{i\omega t}

(here a is the annihilation operator).

This Hamiltonian is of course time-dependant,
but, non-the-less, it is also true that:

\frac{dA(t)}{dt} = \frac{i}{\hbar} U^\dagger[H,a]U

(where A(t) is the annihilation operator in heizenberg's picture)