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 $$\hat{x}(t) = e^{iHt/\hbar}\hat{x}e^{-itH/\hbar},$$ then $|x,t\rangle$ denotes the eigenstate of $\hat{x}(t)$ at time $t$, that is: $$\hat{x}(t)|x,t\rangle = x|x,t\rangle$$$$\hat{x}(t)|x,t\rangle = x|x,t\rangle.$$
In the Schrödinger equation, states evolve in time, so we can set $|x(t)\rangle = e^{-itH/\hbar}|x\rangle$,$$|x(t)\rangle = e^{-itH/\hbar}|x\rangle,$$ 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 $x(t)$?
