# Eigenstates of $|x(t)\rangle$

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 $$x(t)$$?

• Which operator is appearing in your second equation $\hat{x}|x(t)\rangle =x(t)|x(t)\rangle$. It is not necessarily (or even typically) true that $|x(t)\rangle$ will be an eigenvalue of $\hat{x}$ in the schrodinger picture Nov 4, 2021 at 15:11

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!