What is the meaning of the ket states in the notation $\langle x_f,t_f|x_i,t_i\rangle$?

Question

Path-integral amplitudes are denoted by the inner product $\langle x_f,t_f|x_i,t_i\rangle$ where $|x_i,t_i\rangle$ is a time-independent position eigenstate of the time-dependent Heisenberg picture operator $\hat{x}(t)$ at time $t_i$ with eigenvalue $x_i$. In short, $$\hat{x}(t_i)|x_i,t_i\rangle=x_i|x_i,t_i\rangle.$$ I have some trouble with this notation. Usually, we identify the the Schrodinger picture state $|\psi(t_0)\rangle_S$ at time $t_0$ to coincide with the Heisenberg picture state $|\psi\rangle_H$. In this notation, are we using two different reference times $t_i,t_f$ instead of a single $t_0$? I am confused.

Answer 1

Note that in the Heisenberg picture the operator $x(t):= e^{itH}x(0)e^{-itH}$ is time dependent, so in general the operators $x(t)$ and $x(t')$ are different, and therefore will have different eigenvectors. Labelling a state $|x,t\rangle$ does not imply that the state has a time-dependence, it merely labels which of the position operators it is an eigenvector of, since we now have infinitely many.

Since the Heisenberg and Schroedinger pictures are equivalent, we can, of course move back and forth. Choosing $t=0$ such that $\hat x(0)=\hat x$ where $\hat x$ is the schroedinger picture operator, then we have $|x,0\rangle=|x\rangle(0)$, where on the right hand it is a Schroedinger picture state, and $|x\rangle(t)=e^{-itH}|x\rangle(0)$. Then we can compute $\hat x(t)|x\rangle(-t)=e^{itH}\hat x(0)e^{-itH}e^{itH}|x\rangle(0)=e^{itH}\hat x(0)|x,0\rangle=xe^{itH}|x,0\rangle=x e^{itH}|x\rangle (0)=x|x\rangle (-t)$. This just says that $|x,t\rangle=|x\rangle(-t)$. This just confirms that the states $|x,t\rangle$ really are Heisenberg states, they are not equal to the Schroedinger picture states, though they contain the same information if you know them for all t.

Answer 2

The state $|x, t \rangle$ is a Heisenberg picture state. Concretely, it means the state at $t = 0$ which will evolve into the state $|x \rangle$ at time $t$.

Answer 3

An amplitude $\langle x_i, t_i\vert x_j, t_j\rangle$ is the probability amplitude that answers the question "How likely is it I will measure a state at position $x_j$ at time $t_j$ when I measured it to be at $x_i$ at time $t_i$?".

The measurement at $t_i$ results in my initial state being $\lvert x_i,t_i\rangle$ - the corresponding eigenstate of the position operator at the time of measurement. When I measure position at $t_j$, the Born rule tells us the probability amplitude to measure $x_j$ is $\langle x_i, t_i\vert x_j, t_j\rangle$.