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.


2 Answers 2


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.

  • $\begingroup$ That's what I thought for the Schrödinger case. Is it correct though to say that in the Schrödinger picture if the state vector of a system is initial in an eigenstate (or is a linear the combination of them), then it is the coefficients that evolve in time and the basis kets remain constant? I think what confuses me is that the author uses the eigenstates $\lvert q,t\rangle$ as a basis to expand the state vector onto in the Heisenberg picture, but aren't state vectors time independent in this picture? How can the basis eigenstates be time dependent? ... $\endgroup$
    – Will
    Commented Dec 19, 2015 at 10:54
  • $\begingroup$ ...Is it simply because the corresponding operator (to which they are eigenstates of) evolves in time, and so they must also evolve in time to ensure that they remain eigenstates of the operator? $\endgroup$
    – Will
    Commented Dec 19, 2015 at 10:55
  • 1
    $\begingroup$ @Will: You've confused me with your question now, too ;) You must distinguish between mere vectors in the Hilbert space, and between states that evolve according to their pictures. There are three different things here: A state that evolves according to the picture, an eigenstate of a time-independent observable that's just a vector and doesn't evolve ("eigenstate in the Schrödinger picture") and a time-dependent Heisenberg observable to which we associate the eigenvectors at time $t$ ("eigenstates in the Heisenberg picture"). The latter look like the time-evolved Schr. eigenstates. $\endgroup$
    – ACuriousMind
    Commented Dec 20, 2015 at 15:52
  • $\begingroup$ When we are talk about eigenstates they are always get evolution on time. In Schrodinger picture this is all accounted by the evolution of the initial state. In Heisenberg picture this is accounted by the evolution of the operator that describes the observable. $\endgroup$
    – Nogueira
    Commented Dec 20, 2015 at 18:08
  • $\begingroup$ @ACuriousMind So in the Schrödinger picture the eigenstates remain constant in time, and the state evolves (does this mean that, if the initial state is in an eigenstate, or linear combination, of some operator, then the coefficients evolve in time? $\endgroup$
    – Will
    Commented Dec 20, 2015 at 20:38

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 $a$.


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