Eigenstates of $|x(t)\rangle$

Question

In the Heisenberg picture, operators depend on time. Let $\hat{x}$ be the position operator and $$\hat{x}(t) = e^{iHt/\hbar}\hat{x}e^{-itH/\hbar},$$ then $|x,t\rangle$ denotes the eigenstate of $\hat{x}(t)$ at time $t$, that is: $$\hat{x}(t)|x,t\rangle = x|x,t\rangle.$$

In the Schrödinger equation, states evolve in time, so we can set $$|x(t)\rangle = e^{-itH/\hbar}|x\rangle,$$ where $|x\rangle$ is the eigenstate of $\hat{x}$. I'd like to know if $|x(t)\rangle$ is itself eigenstate of some operator. I mean, I already found some expressions of the form $\hat{x}|x(t)\rangle = x(t)|x(t)\rangle$, but what exactly is the meaning of this expression and this eigenvalue $x(t)$?

Answer 1

1. Let us for simplicity assume that the Hamiltonian $\hat{H}$ has no explicit time dependence. Then the time-evolution operator is simply $$ \hat{U}(t)~=~\exp\left(-\frac{i}{\hbar}\hat{H}t\right).\tag{1}$$

2. The position operator in the Heisenberg picture is $$\hat{x}_H(t)~=~\hat{U}(t)^{-1} \hat{x}_S\hat{U}(t).\tag{2}$$

3. The Heisenberg instantaneous eigenstate $|x,t\rangle_H$ satisfies $$\hat{x}_H(t)|x,t\rangle_H ~=~x|x,t\rangle_H,\tag{3}$$ where $$|x,t\rangle_H~=~\hat{U}(t)^{-1}|x\rangle~=~\hat{U}(-t)|x\rangle.\tag{4}$$

4. The time-evolution of the state $|x\rangle$ in the Schrödinger picture is $$ |x (t)\rangle_S~=~\hat{U}(t)|x\rangle~=~|x,-t\rangle_H \tag{5}$$ with an opposite $t$-dependence!

References:

1. J.J. Sakurai, Modern Quantum Mechanics, 1994; Chapter 2.