Heisenberg Picture unclarities In our lecture about this topic, the two following statements were made:

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*An Operator $O_S$ (Schroedinger Picture) can be time dependent. But only when the operator $O_S$ is time independent, the Operator $O_H$ (Heisenberg Picture) in general depends on time.


*If $H=H_S$ is time independent and $[O_S,H_S]=0$ then it is holds that $O_H=O_S$, in particular $H_S=H_H$$
I have some questions, for both these statements, because they contain some unclarity, at least to me:
Regarding question 1:
What if $O_S=O_S(t)$ ? Isn't it possible to make a transformation, so that we gain an operator in the Heisenberg picture? Is it always the case that the operator in the Schroedinger picture must be time independent so that we have have the same operator in the Heisenberg one? If that's the case, then in the following formula:
$$\frac{dO_H(t)}{dt}=\frac{i}{\hbar}[H_H,O_H]+(\frac{\partial O_S}{\partial t})_H$$
the last term should always be zero, because if it is not, that implies time dependency of the operator in the Schroedinger picture, and, as stated above, we can only speak about the operator in the Heisenber picture, if this operator is time independent in the Schroedinger picture. So, where am I mistaken?
Regarding the 2nd statement:
When we say $O_H=O_S$, should this be understood as: the time independent operator $O_S$ in the schroedinger picture has a corresponding time dependent counterpart $O_H$ in the Heisenberg picture?
What if $H_S\neq H_S(t)$ and $[O_S,H_S]\neq 0$ then can we say $O_S \neq O_H$? What does this inequality mean? Perhaps, that the op. $O_S$ cannot be represented in the Heisenberg picture? Or something else?
 A: TL;DR

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*The operator $\hat{O}_H(t)$ can be time-dependent or independent regardless of whether $\hat{O}_S(t)$ depends on time.


*Whether $\hat{H}_S$ is time-dependent or not, $\hat{H}_S(t)=\hat{H}_H(t)$ but to have $\hat{O}_H(t) = \hat{O}_S$ you need a time-independent $\hat{O}_S$.
Explanation:
I think to understand this it's better to go back to the definitions and derive the Heisenberg picture from the more familiar Schrödinger picture. The time-evolution in the Schrödinger picture is governed by the equation \begin{equation} 
i\hbar\frac{\mathrm{d}|\Psi(t)\rangle}{\mathrm{d}t} = \hat{H}_S(t)|\Psi(t)\rangle
\end{equation} which I have written using the state vector $|\Psi(t)\rangle$ of the system and $\hat{H}_S(t)$ is the (possibly time-dependent) Hamiltonian. Formally, the wavefunction of the system is $\Psi(\mathbf{x},t)=\langle \mathbf{x} | \Psi(t) \rangle$, but it doesn't matter for now. You can get the expectation value of an observable which is a Hermitian operator $\hat{O}_S(t)$, that itself may or may not be a time-dependent according to \begin{equation} \langle\hat{O}_S(t)\rangle = \langle \Psi(t) |\hat{O}_S(t)| \Psi(t) \rangle, \end{equation} if only you could compute $| \Psi(t) \rangle.$
The simpler case is when the Hamiltonian is time-independent, i.e. $\hat{H}_S(t)=\hat{H}_S$ for all times $t$. Then the solution of the Schrödinger equation is formally given as \begin{equation} |\Psi(t)\rangle = \exp(-i\hat{H}_St/\hbar) |\Psi(0)\rangle, \end{equation} where the exponentiated Hamiltonian is defined via the Taylor expansion \begin{equation}\exp(-i\hat{H}_St/\hbar) = \sum_{k=0}^{\infty} (-i\hat{H}_St/\hbar)^k/k!.\end{equation} With this in hand, you can now write down the equation for the time-evolution of the expectation value, \begin{equation} \frac{\mathrm{d}\langle\hat{O}_S(t)\rangle}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \langle \Psi(t) | \hat{O}_S(t) | \Psi(t) \rangle = \frac{\mathrm{d}}{\mathrm{d}t} \langle \Psi(0) | e^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar} | \Psi(0) \rangle, \end{equation} which by the product rule for derivatives evaluates to (here you can just differentiate the exponential of an operator as if it were a function of an ordinary variable - you can prove this by differentiating the Taylor expansion) \begin{equation} \frac{\mathrm{d}\langle\hat{O}_S(t)\rangle}{\mathrm{d}t} = \langle \Psi(0) | \frac{i\hat{H}_S}{\hbar} e^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar} + e^{i\hat{H}_St/\hbar} \frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t} e^{-i\hat{H}_St/\hbar} + e^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar} \frac{-i\hat{H}_S}{\hbar} | \Psi(0) \rangle. \end{equation} Collecting terms and noting that a function of an operator commutes with the operator itself (to prove, use the Taylor expansion), you end up with \begin{equation} \frac{\mathrm{d}\langle\hat{O}_S(t)\rangle}{\mathrm{d}t} = \frac{i}{\hbar} \langle \Psi(0) | e^{i\hat{H}_St/\hbar} [\hat{H}_S,O_S(t)] | e^{-i\hat{H}_St/\hbar} \Psi(0) \rangle + \frac{i}{\hbar} \langle \Psi(0) | e^{i\hat{H}_St/\hbar} \frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t} e^{-i\hat{H}_St/\hbar} |\Psi(0) \rangle. \end{equation} Now, this gives you an idea to define transformed operators \begin{equation} \hat{O}_H(t) = e^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar}, \end{equation} because taking a time derivative, you get \begin{equation} \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} = \frac{i}{\hbar}\hat{H}_Se^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar} + e^{i\hat{H}_St/\hbar} \frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t} e^{-i\hat{H}_St/\hbar} + e^{i\hat{H}_St/\hbar} \hat{O}_S(t) e^{-i\hat{H}_St/\hbar}\frac{-i}{\hbar}\hat{H}_S,\end{equation} that is \begin{equation} \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} = \frac{i}{\hbar}[\hat{H}_H(t),\hat{O}_H(t)] + \left(\frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t}\right)_H.\end{equation} Here I used that $\hat{H}_S\hat{O}_S(t) = \hat{H}_Se^{i\hat{H}_St/\hbar}e^{-i\hat{H}_St/\hbar}\hat{O}_S(t)$ and that by the previous notation \begin{equation} \left(\frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t}\right)_H = e^{i\hat{H}_St/\hbar}\frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t} e^{-i\hat{H}_St/\hbar}. \end{equation} Going back to the equation for the expectation value, you can now rewrite it as \begin{equation} \frac{\mathrm{d}\langle\hat{O}_S(t)\rangle}{\mathrm{d}t} = \langle \Psi(0) | [\hat{H}_H(t),\hat{O}_H(t)] + \left(\frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t}\right)_H| \Psi(0) \rangle = \langle \Psi(0) | \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} | \Psi(0) \rangle \end{equation} or even more tellingly \begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} \langle \Psi(t) | O_S(t) | \Psi(t) \rangle = \langle \Psi(0) | \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} | \Psi(0) \rangle. \end{equation}
So to answer question 1, the time-dependence of operators in the Scrödinger and Heisenberg pictures don't have much to do with each other, as is clear from the derivation. The operator $\hat{O}_H(t)$ can be time-dependent or independent regardless of whether $\hat{O}_S(t)$ depends on time.
To answer question 2, note that \begin{equation} [\hat{H}_H(t),\hat{O}_H(t)] = ([\hat{H}_S,\hat{O}_S(t)])_H \end{equation} and if $[\hat{H}_S,\hat{O}_S(t)]=0$ the Heisenberg equation of motion reduces to \begin{equation} \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} = \left( \frac{\mathrm{d}\hat{O}_S(t)}{\mathrm{d}t} \right)_H.\end{equation} Now if $\mathrm{d}\hat{O}_S(t)/\mathrm{d}t=0$ then indeed \begin{equation} \frac{\mathrm{d}\hat{O}_H(t)}{\mathrm{d}t} = 0 \end{equation} and the solution of this is just $\hat{O}_H(t) \equiv \hat{O}_H(0) = \hat{O}_S$ for all $t$.
Hope this helps. For Hamiltonians time-dependent even in the Schrödinger picture the situation is slightly but not much more trickier, but case of the time-independent $\hat{H}_S$ should illustrate the point.
