I want to understand the Schrödinger, Heisenberg and interaction picture and have a few questions about them:
So in general you have a time-dependent Hamiltonian $H$, as for example the potential may depend on time. This should not depend on what picture you use, so this is somewhat counterintuitive, as people always say: Schrödinger picture does not include time-dependent operators.
I am not so sure about the Heisenberg picture actually. Clearly, it deals with operators that are time-dependent, but I am not so sure if this includes the Hamiltonian itself. Especially, since the Hamiltonian is determined by the system, I am not so sure whether this operator should satisfy $$\hat A_{\rm H}(t)=\hat U^{\dagger}(t)\,\hat A_{\rm S}(t)\,\hat U(t).$$
So, do we have $$\hat H_{\rm H}(t)=\hat U^{\dagger}(t)\,\hat H_{\rm S}(t)\,\hat U(t)$$ for the Hamiltonian too in the Heisenberg picture?
In the interaction picture with a Hamiltonian $H = H_{0,S} + H_{1,S}$ a state is given by $$| \psi_{I}(t) \rangle = e^{i H_{0, S} t / \hbar} | \psi_{S}(t) \rangle $$
and the operator by $$A_{I}(t) = e^{i H_{0,S} t / \hbar} A_{S}(t) e^{-i H_{0,S} t / \hbar}.
$$
This somehow seems as if $H_{1,S}$ would never appear in the time-evolution of the system which would be counterintuitive. So where does $H_{1,S}$ affect the system and are there any restrictions on $H_{0,S}$ and $H_{1,S}$?
Best Answer
You basically have three questions:
They mean operators like momentum, position, etc not depend on time, not the Hamiltonian.
Yes, they satisfy the relation, so does the Hamiltonian. However because the explicit time dependence of Hamiltonian, the time evolution operator should write as a time ordered operators just like the form in this thread:Evolution operator for time-dependent Hamiltonian
You have to substitute the wave function to the Shordinger equation to see the explicit dependence of $H_{1,S}$, because what you have written is just a unitary transformation, no physics included, the physics comes from the Shordinger equation. Actually you will get:$$i\hbar\frac{\partial}{\partial t}\psi_I(t)=H_{1,S}\psi_I(t)$$