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In M. Nakahara's book Geometry, Topology and Physics on page 19, the probability amplitude for a particle to move from $x_i$ at time $t_i$ to $x_f$ at time $t_f$ is given as

$$ \tag{1} \langle x_f, t_f | x_i, t_i \rangle $$

where the Heisenberg picture vectors are defined

$$\tag{2} \hat{x}(t_i)|x_i, t_i\rangle = x_i|x_i, t_i\rangle $$

(and similarly for $x_f$) with $\hat{x}(t)$ is the position operator.

I have never seen $(1)$ before. Is that the formula for calculating probability amplitudes in the Heisenberg picture? Why are the states dependent on time?

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    $\begingroup$ I don't understand your question - (1) is simply the inner product of the initial with the final state, whose square gives the transition probability as always between two states. $\endgroup$ – ACuriousMind Sep 13 '15 at 19:45
  • $\begingroup$ @ACuriousMind: But why are the states dependent on time in the Heisenberg picture? Why is it enough to just take the inner product? I expect a time-dependent operator in between. $\endgroup$ – Bass Sep 13 '15 at 20:00
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    $\begingroup$ This tripped me up when I first saw it; $|x, t\rangle$ means the Heisenberg state that has a particle at $x$ at time $t$. That means it's the Schrodinger state that, if it existed at time $t = 0$, would evolve into a particle at $x$ at time $t$. $\endgroup$ – knzhou Sep 13 '15 at 20:46
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    $\begingroup$ To make contact with the definition you know, rewrite these Heisenberg states as Schrodinger states evaluated at $t = 0$. Then you can show that the amplitude is equal to $\langle x_f|U_s(t_f, t_i)|x_i\rangle$ where all the states here are Schrodinger states, which should be familiar. $\endgroup$ – knzhou Sep 13 '15 at 20:49
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I) OP's question (v1) seems to be spurred by a common confusion: The Heisenberg instantaneous position eigenstate $|x,t_0\rangle_H $ does not evolve in time $t$ but does depend on a time parameter $t_0$. In detail,

$$\tag{1} | x, t_f \rangle_{H}~=~ e^{i\hat{H}\Delta t/\hbar} | x, t_i \rangle_{H}, \qquad\Delta t~:=~t_f-t_i, $$

where we for simplicity have assumed that the Hamiltonian $\hat{H}$ has no explicit time dependence.

II) So

$$\tag{2} K(x_f, t_f ; x_i, t_i) ~=~{}_{H}\langle x_f, t_f | x_i, t_i \rangle_{H} ~=~{}_{H}\langle x_f, t_0 |e^{-i\hat{H}\Delta t/\hbar} | x_i, t_0 \rangle_{H} $$

is the amplitude for a particle going from the initial position $x_i$ at initial time $t_i$ to final position $x_f$ at final time $t_f$.

We usually identify the Schrödinger and Heisenberg picture at some fiducial time $t_0$, cf. above comment by Kevin Zhou. The amplitude (2) does not depend on the fiducial time parameter $t_0$.

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  • $\begingroup$ Okay, I think I understand the notation. But why is the probability amplitude equal to $\langle x_f, t_f | x_i, t_i \rangle$? I know that the inner product between a state and an eigenstate of the observable gives the probability that this observable is measured. But why can the inner product "predict" the time evolution? Shouldn't this amplitude contain the Hamiltonian? $\endgroup$ – Bass Sep 15 '15 at 6:59
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Sep 15 '15 at 11:23

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