Are the eigenstates of an operator time independent?

Question

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.

Answer 1

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.

Answer 2

You already know that exist two standar ways of encodes time evolution in the operator language of quantum mechanics. You may try to combine the two description in something called interaction picture. All this ways to encodes time evolution is based upon this following equation: $$ \langle A(t)\rangle=\langle\psi|U^{\dagger}(t)\hat{A}\,U(t)|\psi\rangle $$ Where $\hat{A}$ is the operator associated to the observable $A$ at time $t=0$ and $|\psi\rangle$ is the state of the system in time $t=0$ (preparation). $U(t)=e^{\frac{H\,t}{i\hbar}}$ with $H$ being the hamiltonian of the system.

So, first, is obviously that the ray determined by the eigenstate of $U(t)$ is independent of time, after all, is just a phase contribution. The eigenstate get some sort of evolution like $e^{\frac{E\,t}{i\hbar}}|E\rangle$, but the eigenspaces of any eigenvalue is preserved through time.

Now, in the Schrodinger picture, all the unitary operator $U(t)$ are absorbed by the definition of the state at different times. $$ \langle\psi|U^{\dagger}(t)\hat{A}\,U(t)|\psi\rangle=\langle\psi(t)|\hat{A}|\psi(t)\rangle $$ with this choice, the operator $A$ is independent of time. The spectra and the eigenvectors of $A$ do not depend on time as well, after all the operator is independent on time. Yes, in Schrodinger picture the eigenvectors are independent on time, but the states - and eigenstates - are not!

This is crucial. If you want a state that at time $t$ is the eigenvector of $A$ in the Schrodinger picture then: $$ AU(t)|\psi\rangle=A|\psi(t)\rangle=a|\psi(t)\rangle $$

We can identify the eigenvector of $A$ by $|a\rangle$, where $a$ is the eigenvalue. The eigenstate of $A$ at time $t_0$ is $U(t-t_0)|a\rangle$

In the Heisenberg picture, (my favourite one), the observables absorbs the unitary operators. $$ \langle\psi|U^{\dagger}(t)\hat{A}\,U(t)|\psi\rangle=\langle\psi|\hat{A}(t)|\psi\rangle $$

Now, for each time, we have a different operator that describes the same observable. Is expected that two different operators has differents eigenvalues and eigenvectors. So, the eigenvectors associated to the observables change through time.

If you want an state that is an eigenstate of $A$ at time $t$ you need to take the eigenvector of $A(t)$:

$$ U^{\dagger}(t)AU(t)|\psi\rangle=A(t)|\psi\rangle=a|\psi\rangle $$

In Heisenberg picture, define an eigenvector by $|a\rangle$ is not sufficient because we have a lot of operators associated with one physical observable at different times. We can define $|a,\,t\rangle$, mean that is an eigenvector of $A(t)$ with eigenvalue $a$.