# Dynamics in the heizenberg picture

1. Feb 17, 2012

### 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

Last edited: Feb 17, 2012
2. Feb 17, 2012

### bapowell

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

3. Feb 19, 2012

### 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)