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?

  • 2
    $\begingroup$ 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, 2015 at 20:46
  • 4
    $\begingroup$ 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, 2015 at 20:49

1 Answer 1


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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