So when the Hamiltonian is time-independent, we can define the Heisenberg state vectors by evolving the Schrödinger state vectors back in time:
$$ | \psi \rangle_H = \hat{U}^\dagger (t)|\psi(t) \rangle_S=e^{i\hat{H}t} |\psi(t)\rangle_S $$
and we define operators
$$ \hat{A}_H(t) = \hat{U}^\dagger (t) \hat{A}_S \hat{U}(t)$$
which gives us the Heisenberg equation: $$ \frac{d\hat{A}_H(t)}{dt} = -i[\hat{A}_H(t),\hat{H}]. $$
If, in the Schrödinger picture, we have a time-dependent Hamiltonian, the time evolution operator is given by
$$ \hat{U}(t) = T[e^{-i \int_0^t \hat{H}(t')dt'}] $$
If I define the Heisenberg operators in the same way with the time evolution operators and calculate $ dA_H(t)/dt $ I find
$$ \frac{d}{dt} \hat{A}_H(t)= \frac{d\hat{U}^\dagger(t)}{dt} \hat{A}_S\hat{U}(t) + \hat{U}^\dagger(t) \hat{A}_S \frac{d\hat{U}(t)}{dt} \\ = i \hat{U}^\dagger (t) \hat{H(t)} \hat{A}_S \hat{U}(t) - i\hat{U}^\dagger (t)\hat{A}_S\hat{H}(t)\hat{U}(t). $$
At this point, I am not sure how to proceed. I can't commute $\hat{H}(t)$ through $\hat{U}(t) $ because $[\hat{H}(t),\hat{H}(t')] \neq 0$. How do I show derive Heisenberg's equation for a time-dependent Hamiltonian?
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