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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Qmechanic
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Solidification
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You have correctly identified what $\lvert x_i,t_i\rangle$ means - what exactly are you confused about? – ACuriousMind May 03 '21 at 15:35
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Heisenberg picture state $|\psi\rangle_H$ is identified with Schrodinger picture state $|\psi(t)\rangle$ at $t=t_0$.(i.e. $|\psi_H\rangle=|\psi(t_0)\rangle$. https://en.wikipedia.org/wiki/Heisenberg_picture#Summary_comparison_of_evolution_in_all_pictures 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. – Solidification May 03 '21 at 15:43
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1Duplicate. – Cosmas Zachos May 03 '21 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.
Anton Quelle
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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 $t$.
knzhou
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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 $0$). – Solidification May 22 '21 at 17:59
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@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 $x$. – knzhou May 22 '21 at 18:22
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@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 $\mathfrak{Q}$. – knzhou May 22 '21 at 18:23
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But wouldn't you agree that $|x,t_i\rangle$ and $|x,t_f\rangle$ are two different HP states? – Solidification May 22 '21 at 18:24
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@mithusengupta123 Yes, $|x, \mathfrak{Q}_i \rangle$ and $|x, \mathfrak{Q}_f \rangle$ refer to two different states. One is the fixed state at time $t = 0$ that, in Schrodinger picture, evolves into $|x \rangle$ at time $\mathfrak{Q}_i$. The other is the fixed state at time $t = 0$ that, in Schrodinger picture, evolves into $|x \rangle$ at time $\mathfrak{Q}_f$. These are different states, but neither depends on $t$, as both are fixed at $t = 0$. – knzhou May 22 '21 at 18:28
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@knzhou is this a standard notation? That the state $|x, t_i \rangle$ is a time dependent state that yields the value $x$ at time $t_i$, and is "evaluated" or "looked uppon" at time $t=0$ (you use the term "frozen", but for time dependence I think "evaluation" works as well). So, is it standard terminology? Because I think this would clear the confusion I have with Schwartz QFT Book, but he never mentions the meaning that explicitly. – Quantumwhisp Jun 30 '21 at 22:40
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@Quantumwhisp It’s completely standard, textbooks just don’t say it explicitly. – knzhou Jun 30 '21 at 23:56
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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 $\langle x_i, t_i\vert x_j, t_j\rangle$.
ACuriousMind
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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? – Solidification May 03 '21 at 15:50
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2@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. – ACuriousMind May 03 '21 at 15:52