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The time evolution of states under a time-dependent Hamiltonian $H_S(t)$ in the Schrödinger picture is determined by $$ \label{TDS} i\hbar \frac{d |{\psi(t)}\rangle}{dt} = H_{\mathrm{S}}(t) |\psi(t)\rangle, $$ and the canonical coordinates $\mathbf{q}$, $\mathbf{p}$ are time-independent operators in this picture. Now consider a general time-dependent unitary transformation $R(t)$ (with $R^{\dagger}R=\mathbb{I}$) applied to the vectors in the Hilbert space, in particular $$ |\widetilde{\psi}(t)\rangle = R(t) |\psi(t)\rangle $$ Substituting the inverse relation into \eqref{TDS}, one finds that $|\widetilde{\psi}(t)\rangle$ satisfies the equation $$ i\hbar\frac{d |\widetilde{\psi}(t)\rangle}{dt} = \left(R H_{\mathrm{S}} R^{\dagger} + i \hbar \frac{d R}{dt} R^{\dagger}\right) |\widetilde{\psi}(t)\rangle = \widetilde{H_1}(t) |\widetilde{\psi}(t)\rangle $$ Now let $A_{\mathrm{S}}(t)$ be an explicitly time-dependent operator in Schrödinger picture. The relevant matrix elements can be found as usual via $$ \langle i |A_{\mathrm{S}}(t)| j \rangle = \langle \tilde{i} |\widetilde{A}(t)| \tilde{j} \rangle $$ where the transformed operator is defined by $\widetilde{A}(t) = R A_{\mathrm{S}}(t) R^{\dagger}$. The time-dependence of this transformed operator is given by $$ \label{TDTO} \frac{d \widetilde{A}(t)}{dt} = \frac{d R}{dt} A_{\mathrm{S}} R^{\dagger} + R \frac{\partial A_{\mathrm{S}}}{\partial t} R^{\dagger} + R A_{\mathrm{S}} \frac{d R^{\dagger}}{dt} $$ since for an operator in Schrödinger picture $\frac{d A_{\mathrm{S}}}{dt} = \frac{\partial A_{\mathrm{S}}}{\partial t}$.

Now as far as I am aware, every unitary operator can be written as some (possibly time-ordered) complex exponential of a Hermitian operator, so I think that for any unitary operator it should hold that $$ i \hbar \frac{d R(t)}{dt} = G(t) R(t) $$ for some Hermitian operator $G(t)$ (can someone confirm this with a reference?). Then one finds that \eqref{TDTO} can be written $$ \frac{d \widetilde{A}(t)}{dt} = -\frac{i}{\hbar} \left[\widetilde{A}(t), \widetilde{G}(t)\right] + R \frac{\partial A_{\mathrm{S}}}{\partial t} R^{\dagger} = -\frac{i}{\hbar} \left[\widetilde{A}(t), \widetilde{H_2}(t)\right] + \widetilde{\left(\frac{\partial A}{\partial t}\right)} $$ The Heisenberg picture is of course a special case of this, for which one chooses $R(t) = U^{\dagger}(t,t_0)$, where $U(t,t_0)$ is the time-evolution operator that solves \eqref{TDS}. In this case, one gets $\widetilde{H_1}(t) = 0$ and $\widetilde{H_2}(t) = H_\mathrm{H}(t) = U^{\dagger}(t,t_0) H_{\mathrm{S}}(t) U(t,t_0)$. So for the Schrödinger and Heisenberg pictures, there seems to be a clear assignment of the Hamiltonian operator possible.

Now to my question: in the general case, where $R(t)$ is an arbitrary unitary time-dependent transformation (which is sometimes called the transformation to a rotating frame), does it make sense to talk about $\mathit{the}$ Hamiltonian operator? It seems to me that there are two operators $\widetilde{H_1}(t)$ and $\widetilde{H_2}(t)$ that have features similar to a Hamiltonian operator in that they determine the dynamics of the state vectors and operators in the rotating frame.

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Yes what you have written makes sense. In fact, what you are thinking about is actually the realization of canonical transformations in quantum mechanics.

An exercise which you might find informative if you've never worked it out: start with a Lagrangian theory. We know that if $L$ is the Lagrangian, then $L^\prime=L+\frac{d f}{dt}$ where $f=f(q,t)$ generates the same classical dynamics. But then you might wonder: if $L$ and $L^\prime$ generate the same Euler-Lagrange equations, do they generate the same Hamiltonian via the Legendre transform? Do they even have the same momenta?

You will find that the answer to both questions is no, yet both the conjugate momenta and Hamiltonian transform in such a way that the new set of equations remains equivalent to the old ones if we change variables appropriately.

The transformation you find in the Hamiltonian should look exceptionally familiar.

I will also mention that there is a bit of a loophole in the quantum case which doesn't carry over from the classical calculation. However, I admit that it's been a while since I did this and don't quite remember what it is. I do remember thinking at the time that it probably has something to do with the appearance of anomalies in quantum theories though.

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