In the Schrodinger picture, are the eigenstates of an operator time independent? Is it their expectation values that evolve in time rather than the actual eigenstates?
For example, say I have an operator $\hat{A}$ with a set of eigenstates $\lbrace\lvert n\rangle\rbrace$, such that $\hat{A}\lvert n\rangle =a_{n}\lvert n\rangle$. Is it the expectation values $a_{n}$ that evolve in time and the eigenstates remain stationary?
The reason I ask is that in an introduction to the path integral formulation of quantum mechanics that I've been reading, the author discusses state vectors in the Heisenberg picture and how they relate to the Schrödinger picture. Now I know that they are related in general (for a time independent Hamiltonian) by $$\lvert\psi\rangle_{H}=e^{i\hat{H}t}\lvert\psi (t)\rangle_{S}$$ but here they consider eigenstates $\lvert q\rangle$ of the position operator $\hat{q}_{S}$ in the Schrödinger picture, such that $\hat{q}_{S}\lvert q\rangle =q\lvert q\rangle$. They then relate this to the operator in the Heisenberg picture, $\hat{q}_{H}(t)= e^{i\hat{H}t}\hat{q}_{S}e^{-i\hat{H}t}$ at an instant in time $t$, and since the eigenvalues should be the same in both pictures, i.e. $$\hat{q}_{H}(t)\lvert q,t\rangle = q\lvert q,t\rangle$$ this suggests that the corresponding eigenstates in the Heisenberg picture are related to those in the Schrödinger picture by $$\lvert q,t\rangle =e^{i\hat{H}t}\lvert q\rangle$$ where the notation $\lvert q,t\rangle$ denotes an eigenstate of $\hat{q}_{H}(t)$ at an instant in time $t$.
To me this suggests that the eigenstates in the Schrödinger picture are time independent.
Sorry if this is a stupid question, I've just got myself confused.
Best Answer
Somewhat confusingly, "eigenstates" in the respective pictures behave exactly in the opposite way as actual states of the system. That is because eigenstates belong to operators:
An "eigenstate" in the Schrödinger picture is time-independent simply because it is not meant to be time-dependent. It would be non-sensical to evolve an eigenstate of an operator $A$ that doesn't commute with $H$ in time - there is no guarantee at all that it will stay an eigenstate of $A$, this would be a useless notion!
Conversely, an "eigenstate" in the Heisenberg picture has to be time-dependent since the observable it belongs to changes in time, and again, if it doesn't commute with $H$, it would not act as $A(t)\lvert a\rangle = a\lvert a \rangle$ on an $\lvert a \rangle$ that isn't time-dependent.