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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.

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  • $\begingroup$ You have correctly identified what $\lvert x_i,t_i\rangle$ means - what exactly are you confused about? $\endgroup$
    – ACuriousMind
    May 3, 2021 at 15:35
  • $\begingroup$ Heisenberg picture state $|\psi\rangle_H$ is identified with Schrodinger picture state $|\psi(t)\rangle$ at $t=t_0$.(i.e. $|\psi_H\rangle=|\psi(t_0)\rangle$. en.wikipedia.org/wiki/… What are we doing here? Are we using two different reference times here $t_i$ and $t_f$ where we match the S. P. state with the H. P. state? I thought H P states are defined once and for all at only one time. $\endgroup$ May 3, 2021 at 15:43
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    $\begingroup$ Duplicate. $\endgroup$ May 3, 2021 at 16:03

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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.

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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 $\langle x_i, t_i\vert x_j, t_j\rangle$.

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  • $\begingroup$ This is fine. I am confused about the states. $|x,t\rangle$. It seems to be a HP state. But this means we are matching the SP state at two different fiducial instants? $\endgroup$ May 3, 2021 at 15:50
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    $\begingroup$ @mithusengupta123 Neither of these states is being evolved in time. The Heisenberg and Schrödinger picture differ in whether we apply the time evolution to states or to operators, but they do not differ about what constitutes a state or an operator at any instant. The distinction between "HP states" and "SP states" you seem to imagine doesn't exist in that form. $\endgroup$
    – ACuriousMind
    May 3, 2021 at 15:52
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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 $t$.

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  • $\begingroup$ Please see my comment below ACuriousMind's answer (and also the question). I am troubled by the fact that usually, we define the HP state to be a state frozen at one and only one fiducial instant of time $t_0$ (often taken to be $0$). $\endgroup$ May 22, 2021 at 17:59
  • $\begingroup$ @mithusengupta123 It is frozen, as I said, at time $t = 0$. $\endgroup$
    – knzhou
    May 22, 2021 at 18:20
  • $\begingroup$ @mithusengupta123 The name of the state contains the letter $t$, but that's not the same thing as the state itself depending on time. For example, you are used to labeling states as $|x \rangle$, but that doesn't mean the state of a system depends on the position $x$. Instead, here the letter $x$ labels a particular state which is peaked at the position $x$. $\endgroup$
    – knzhou
    May 22, 2021 at 18:22
  • $\begingroup$ @mithusengupta123 If it helps, we can change the letter. For example, we could just as easily define $|x, \mathfrak{Q} \rangle$ to be the fixed, Heisenberg state at time $t = 0$, which would evolve in Schrodinger picture into $|x \rangle$ at a time equal to $\mathfrak{Q}$. $\endgroup$
    – knzhou
    May 22, 2021 at 18:23
  • $\begingroup$ But wouldn't you agree that $|x,t_i\rangle$ and $|x,t_f\rangle$ are two different HP states? $\endgroup$ May 22, 2021 at 18:24

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