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$$ \DeclareMathOperator{\dif}{d \!} \newcommand{\ramuno}{\mathrm{i}} \newcommand{\exponent}{\mathrm{e}} \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\braket}[2]{\langle{#1}|{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}|{#2}|{#3}\rangle} \newcommand{\linop}[1]{\hat{#1}} \newcommand{\dpd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dod}[2]{\frac{\dif{#1}}{\dif{#2}}} $$

Using the Schrödinger equation and the definition of the expectation value it can be shown that the time dependence of the expectation value of an observable $A$ for a system in arbitrary state $\ket{\Psi(t)}$ is given by $$ \dod{\langle A \rangle}{t} = \frac{\ramuno}{\hbar} \langle [\linop{H}, \linop{A}] \rangle + \big\langle \dpd{\linop{A}}{t} \big\rangle \, , \tag{1} $$ and this equation shows that, in general, if an operator $\linop{A}$ commutes with the Hamiltonian operator $\linop{H}$ and does not have an explicit time dependence, then the expectation value of the corresponding observable $A$ is time independent.

For stationary states $\ket{\Psi(t)} = \exponent^{-\ramuno E_{k} t / \hbar} \ket{E_{k}}$ the first term in the expression for the time dependence of the expectation value of an observable vanishes $$ \langle [\linop{H}, \linop{A}] \rangle = \bracket{ \Psi(t) }{ \linop{H} \linop{A} }{ \Psi(t) } - \bracket{ \Psi(t) }{ \linop{A} \linop{H} }{ \Psi(t) } = \bracket{ E_{k} }{ \linop{H} \linop{A} }{ E_{k} } - \bracket{ E_{k} }{ \linop{A} \linop{H} }{ E_{k} } = E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } - E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } = 0 \, , $$ and so the time dependence of the expectation value is simply given by $$ \dod{\langle A \rangle}{t} = \big\langle \dpd{\linop{A}}{t} \big\rangle \, . \tag{2} $$

Nevertheless, the statement like the following one

A stationary state is called stationary because the system remains in the same state as time elapses, in every observable way. Wikipedia

is found in many books.

The thing that troubled me is the word every, since from (2) it appears that if an operator $\linop{A}$ carries some explicit time dependence, then the expectation value of the corresponding observable $A$ changes in time. So stationary states are, in fact, not so stationary.

I have the feeling that I am missing something. And our discussion with Bubble helped clarify what's bothering me.

As far as I know operators in the Schrödinger picture usually do not carry an explicit time dependence. Again, usually, but not always. In many books (see, for instance, Griffiths, D.J., Introduction to quantum mechanics, 2nd ed.) one can find that

Operators that depend explicitly on $t$ are quite rare, so almost always $\dpd{Q}{t} = 0$.

And, yet the author claims that

Every expectation value is constant in time.

I feel like there is a gap between operators being almost always explicitly independent of time and every expectation value being constant.

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  • $\begingroup$ You can define linear operators any way you like as long as they act on the appropriate Hilbert space. There are no restrictions. The question is more how to measure these operators. They obviously have to be observables. For instance, you can define $\hat{p}_t(t)=t \hat{p}$, where $\hat{p}$ is momentum. Then you can measure it by measuring momentum at different times and changing your gauge appropriately. $\endgroup$ – Bubble Jul 20 '13 at 21:51
  • $\begingroup$ @Bubble, sorry, but my background in physics is quite limited (I'm chemist), so the last sentence is a dark forest for me. $\endgroup$ – Wildcat Jul 21 '13 at 12:50
  • $\begingroup$ I have a feeling, that when the system is not under the action of some time-varying external potential, then any operator which represents an observable (and not only Hamiltonian) does not carry an explicit time dependence. It can resolve the contradiction. $\endgroup$ – Wildcat Jul 21 '13 at 16:09
  • $\begingroup$ please see my answer. I'm not sure whether I understood the question correctly. Please comment and I will revise the answer. $\endgroup$ – Bubble Jul 21 '13 at 18:16
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    $\begingroup$ As you have correctly pointed out, stationary states of a time independent hamiltonian, can have non zero expectation values for explicitly time dependent operators. I dont quite understand your question however. $\endgroup$ – Prathyush Jul 23 '13 at 9:51
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I think it's expected that you have a bit of common sense about this.

Let's take the operator $O(t)$ which is the position operator when $t$ is between 9:00 and 10:00 in the morning, and the momentum operator the rest of the day.

Now take a system in a stationary state, and ask "What is the expectation value of $O(t)$ at each time $t$"? Whoa, the expectation value changes dramatically each day at 9:00, then changes again at 10:00!

Does that mean the state is not "stationary" at 9:00 or 10:00? No, of course it doesn't mean that!

When an operator has explicit time dependence, as $O(t)$ does, it means that you have a different -- possibly totally unrelated -- operator at each time $t$. Wikipedia says "the system remains in the same state as time elapses, in every observable way". That's correct. I don't think a reasonable person reading that sentence would infer that if you calculate the expected position at 9:30, and then you calculate the expected momentum at 10:30, the two calculations should have the same answer for a stationary state.

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    $\begingroup$ Your example is trivial. But, is there any operator, which is explicitly time dependent and represents the same observable at all times? Not like your $O(t)$ which represents position during some time periods and momentum during others, but rather like the Hamiltonian, which represents the same observable, total energy, all the time even when it carries an explicit time dependence. Now, if there is some operator of such kind, then the expectation value of the corresponding observable evolves in time even if the system is in stationary state. $\endgroup$ – Wildcat Jul 23 '13 at 14:35
  • $\begingroup$ Well, "stationary", in this context, means stationary with respect to operators not having an explicit time dependence. I suggested to Wildcat to focus on the math and not the words during our discussion. An eigenstate of the Hamiltonian is as "stationary as possible". $\endgroup$ – Bubble Jul 23 '13 at 17:52
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    $\begingroup$ @Wildcat - There is no operator like that, not even H. Explicit time dependence always means representing different things at different times. For example, maybe I have an electron in a variable applied electric field, and I treat the electron quantum-mechanically and E-field classically. Then there's a time-dependent Hamiltonian H. At $t=0$ H would represent, say, "the energy of an electron in an environment where there's a uniform 1.3V/m electric field". At $t=1$ H would represent, say, "the energy of an electron in an environment where there's a uniform 2.8V/m electric field". $\endgroup$ – Steve Byrnes Jul 23 '13 at 17:56
  • $\begingroup$ @SteveB, so operators of observables are time-independent, unless an observable carries some "external" time-dependence. By external I mean that this time-dependence is coming not from the system itself, but from outside of it. Doesn't it mean then that $t$ in $|\Psi(t)>$ and $t$ in $\hat{A}(t)$ are two different parameters: the first is related to the system, the second to the environment? $\endgroup$ – Wildcat Jul 23 '13 at 18:48
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    $\begingroup$ @Wildcat - I think your term "external" is about right. Most of the time that somebody writes down an operator with explicit time dependence, it's because something "externally" (to the quantum system) is changing. (Not because it's mathematically necessary, but because that's the situation where such operators are useful, unlike my $O(t)$ example which is useless.) I wouldn't use the term "two different parameters", but yes you shouldn't confuse explicit time dependence with implicit time dependence. BTW the latter is different in the "Schrödinger picture" vs "Heisenberg picture". $\endgroup$ – Steve Byrnes Jul 23 '13 at 22:57
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I'm not really sure anymore what you're asking so I will try to answer all possible variations of your question that I can think of.

  1. If you are asking whether there exist observables which are explicitly time dependent while the Hamiltonian of your system is time independent then the answer is yes. For instance, let's say that you can define momentum,$\hat{p}$, for your system then the operator $\hat{O}=t \hat{p}$, which is linear and Hermitian, is an observable. You can measure it somehow. For instance, you can measure momentum and then multiply it by the time you see on your clock. After measurement the system will be in one of the momentum eigenstates, of course.

  2. If you are asking can a Hamiltonian in the Schrödinger picture evolve an operator so that the operator is time dependent (either explicitly or implicitly) then the answer is no. All of the time dependence is in the state vector by definition.

You can put a field which varies with time, for instance, in the Hamiltonian, but then you can not use $|\psi(t)>=e^{−iE_kt}|E_k⟩$ anymore. Note that the article assumes that you have a time independent Hamiltonian. You have to use Dyson series to solve the equation of motion. And in any case, if you are in the Schrödinger picture the time dependence will again be carried by the state vector, except for the operators you defined to have explicit time dependence.

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  • $\begingroup$ Does not the first statement contradicts with the second one? The first is: yes, we do have observables represented by time-dependent operators. The second: but all time dependence is in the state vector. $\endgroup$ – Wildcat Jul 21 '13 at 18:47
  • $\begingroup$ Aha, I see the problem now. In the first case the observable does not play a part in the time evolution of the system, if it did the Hamiltonian would be time dependent. $\endgroup$ – Bubble Jul 21 '13 at 18:49
  • $\begingroup$ Also, the Hamiltonian does not evolve this observable, we defined it to be like that. The time evolution played no part in the observable's time dependence. $\endgroup$ – Bubble Jul 21 '13 at 19:00
  • $\begingroup$ Hmmm... By saying that "the time evolution played no part in the observable's time dependence" do you mean that the second term from (1) is absent in Schrodinger picture? $\endgroup$ – Wildcat Jul 21 '13 at 19:48
  • $\begingroup$ If the observable has no explicit time dependence then yes! In my first example I meant that $t \hat{p}$ is time dependent because we defined the observable like that. It will be like that in any picture because that is its definition. For this observable the second term in your equation (1) will give you the expectation value of the momentum operator. I contrasted this with the statement that the Hamiltonian took some operator without any explicit time dependence (e.g., $\hat{p}$) and gave it time dependence (in the Schrodinger picture), which is false and can not happen. $\endgroup$ – Bubble Jul 21 '13 at 19:54
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Imagine a ball, laying still on the ground. While you are looking at it, slowly walking around. The state of the system is staying the same, while the things that you see depends on your aim.

State of the system -- only it matters. And that is what your wiki quote clearly utters.

Looking for an answer drawing from credible and/or official sources.

Edit: All this discussion is one big nitpicking. Remark you are making is not really sticking. For "official sources" it won't worth the fuss. If you want clarity, then just use the math.

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  • $\begingroup$ I guess there's an argument you're loosing somewhere. Hoping for a draw by splitting a hair? $\endgroup$ – Kostya Jul 23 '13 at 12:35
  • $\begingroup$ After reading Steve B answer, I have a feeling that, perhaps, I do not properly understand the meaning of an explicit time dependence of operators. Does the fact that some operator is explicitly time dependent means that it represent different observables at different times? Or is it possible for such operators to represent the same observable each and every time? $\endgroup$ – Wildcat Jul 23 '13 at 14:50
  • $\begingroup$ then can you advice me any credible source (book, notes, whatever) where the right language is used? $\endgroup$ – Wildcat Jul 23 '13 at 15:08

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