I'm reading Modern Quantum Mechanics by Jun John Sakurai and in section 2.2 he talks about Base Kets and Transition Amplitudes. He goes to show, that $|a',t\rangle=\mathcal{U}^\dagger|a'\rangle$, (with $\mathcal{U}=e^{-\frac{iHt}{\hbar}}$ being the time evolution operator) and the derivation itself is totally fine, but the result bothered me a bit, because earlier in that derivation when he talked about the Schrödinger picture and said

"In the Schrödinger picture, A does not change, so the base kets, obtained as the solutions to this eigenvalue equation at $t = 0$, for instance, must remain unchanged".

Obviously if you can apply the Time-Evolution Operator to a gerneral state ket, you can also apply it to a base ket, which would then be changing by a complex multiple, which of course doesnt change the eigenvalue Equation. Is that just poor wording on his part to say it doesn't change? and is it true then, that $$\mathcal{U}^\dagger|a',t\rangle_S=|a',t=0\rangle=\mathcal{U}|a',t\rangle_H$$ so the base kets evolve differently in the two pictures?

One last note: I guess the reason this bothered me so much, is because if my thoughts on this are correct, we end up in the Schrödinger picture with an Operator that doesn't change, but base kets that do, but in the Heisenberg picture, both Operator and base kets change with time? (although I know that the phase shift doesn't affect the probabilities)

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    $\begingroup$ As you do not define the operators $\mathcal{U}$ and $U$, your question is not self-contained. You cannot expect other people to guess the meaning or having your book at hand. Assuming that $U$ is a typo and $\mathcal{U} \equiv \mathcal{U}(t)=e^{-iHt/\hbar}$, you can answer your question yourself by combining $A_H(t)=e^{i H t/\hbar} A_H(0) e^{-iHt/\hbar}$, $A_S=A_H(0)$, $A_S |a\rangle =a |a\rangle$ with the definition $|a,t\rangle:= e^{iHt/\hbar}|a\rangle$. $\endgroup$
    – Hyperon
    Commented Mar 12 at 12:33
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    $\begingroup$ @Hyperon Wow, I didn't even notice that, my mind just recognized unitary time development. Good example of how people often interpret rather than actually "see". $\endgroup$ Commented Mar 12 at 12:45

2 Answers 2


I think there is a semantic problem between you and the author reagarding:

"In the Schrödinger picture, A does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged".

As I do understand it, what he is referring to, is that, if an operator $\hat{A}$ does not change with time, then calculating the eigenstates of $\hat{A}$ will give you the same eigenstates independently of what value you assign to your time argument, because the time argument just does not occur in the calculation.

This is not to be confused with the statement, that a system prepared in one of the eigenstates of $\hat{A}$ would not change with time. In fact it would change at least by a time depended phase factor, if $\hat{A}$ commutes with $\hat{H}$ or in more complicated ways, if it does not commute. As a consequence a system in an eigenstate of $\hat{A}$ will generally not remain in a eigenstate of $\hat{A}$ after time evolution. In Schrödinger-picture the eigenstate problem of $\hat{A}$ is time-independent and independent of your System in general, but the Eigenstates of $\hat{A}$ still will develop in time according to the systems time-evolution operator, but they won't remain eigenstates after time-evolution. I do agree that the formulation "The eigenstates of A do not change with time." is ambigous and I hope I could clarify how it is to understand.

Regarding your second question, in Heisenberg picture the kets do not show any time development, which means that $ |a'\rangle_H = |a',t=0\rangle$. The Heisenberg state does have no time development, as the time evolution operator is included in the Heisenberg-operators instead.

Edit: The states describing a system state do show no time evolution in Heisenberg picture as stated in the above paragraph. Now when looking at the eigenvalue problem of an operator $\hat{A}$ in Heisenberg picture, as we did for the Schrödinger picture before, they indeed will show a time development because the operator $\hat{A}_H(t)$ will change with time in a way that $\langle a|_H | \psi \rangle_H = \langle a|_S | \psi \rangle_S = \langle a(t=0)| \hat{U} |\psi(t=0) \rangle $. Taking the hermitian adjoint one gets $| a \rangle_H = \langle a|_H^\dagger = (\langle a| \hat{U}) ^\dagger = \hat{U}^\dagger | a \rangle$. So, as mentioned in the comment of Albertus Maguns, the ket-states created by the eigenvalue problem of an operator in the Heisenberg picture, do indeed show a time development inversely to the one, that the corresponding eigenstate of the same operator in Schrödinger-picture would show, when developed according to the Schrödigner equation of the System. This would suggest that your line $$U^\dagger|a',t\rangle_S=|a',t=0\rangle=U|a',t\rangle_H$$ might be true. However from my understanding using such a time dependend state to desribe the physical state of your system means you are no longer operating in the Heisenberg picture, which shouldn't be a problem physically, so who cares. But nomenclature really seems to get quite convoluted there.

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    $\begingroup$ You are incorrect about time development of Heisenberg base kets. The notation $|a^\prime\rangle$ suggests that the ket in question is an eigenstate as it is labeled by the eigenvalues of $A$ (at least this is the notational convention in Sakurai). In the Heisenberg picture, the base kets do show time development, i.e. they move oppositely to that of the Schrodinger state kets. See the table on page 89 of Sakurai's text for clarification. $\endgroup$ Commented Mar 12 at 15:16
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    $\begingroup$ It is also worth noting that within the context of Sakurai's second chapter, the time development operator is defined as $e^{-i\hat H t/\hbar}$ (cf. Sakurai, pg. 73). Thus, the Hamiltonian is independent of time and commutes with the time development operator so that the eigenstates remain eigenstates even after application of the time development operator, i.e. they are stationary states. Thus, your statement that: "... won't remain eigenstates after time-evolution." though true in general, is false within the context of the book's chapter. $\endgroup$ Commented Mar 12 at 15:47
  • $\begingroup$ This is not your fault as it is Sakurai's, it seems to me that he should have placed this discussion prior to page 73. since it can easily be understood in the context of time dependent Hamiltonians as well. $\endgroup$ Commented Mar 12 at 15:49

When Sakurai talks about the base kets not changing, he is basically talking about the solutions of the eigenvalue problem, of which the time independent Schrodinger equation is a noted special case.

For example, when you are working in the Schrodinger picture, the operators carry no time dependence and the base kets are found from the eigenvalue problem, $$\hat A|a\rangle=a|a\rangle.$$ Where $|a\rangle$ is an eigenstate of $\hat A$ for some time $t-t_0$. In the Schrodinger picture, the state kets move with time, but not the base kets.

Whenever you decide that you are going to take one of the base kets and apply the time development operator to it, as you most certainly can, you have left the Schrodinger picture and have begun to work in the Heisenberg picture. Se for example, equation (2.2.44b) on pg. 88 of Sakurai's text: $$c_{a^\prime}(t)=(\langle a^\prime|\mathcal U)\cdot|\alpha, t_0=0\rangle.$$ Note how the tine development operator is applied to the base bra, instead of to the state ket as it would be in the Schrodinger equation. So essentially, in the two respective pictures: the operators are stationary in the Schrodinger picture and the state kets are moving; in the Heisenberg picture the operators are moving and the state kets are stationary. See table 2.1 on pg. 89 for a convenient summary of what's moving and what's staying still in the Schrodinger and Heisenberg pictures.

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    $\begingroup$ If he was only talking about "the solutions of the time independent Schrödinger equation" why would he call the operator A and not H? I think he is really talking more generally. $\endgroup$
    – Zaph
    Commented Mar 12 at 12:31

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