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)
Best Answer
I think there is a semantic problem between you and the author reagarding:
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.