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Path integrals as usually presented - what does it tell us?

  1. Apr 6, 2014 #1
    Typical introductions to path integrals start with asking for the value of [itex]\langle x_1,t_1 | x_2,t_2 \rangle[/itex]. This is usually interpreted as the probability amplitude of observing a particle at x_2 at time time t_2 given that it is located at x_1 at t_1.

    But is this so? I am having trouble relating this to the theory in terms of physically realizable states. Exact position states are not normalizable. So suppose we have a wave function [itex]\Psi (x,t)[/itex] which is a solution to our Schrodinger equation and satisfies the desired initial conditions. It can be considered as the expansion of an abstract state [itex]|\Psi(t)\rangle[/itex] in the basis [itex]|x\rangle[/itex], which are eigenstates of the (time-independent) position operator, so that [itex]\Psi (x,t)=\langle x|\Psi(t)\rangle[/itex]. The time-dependence can be shifted to the basis states by going to the Heisenberg picture so that we have basis states [itex]|x,t\rangle[/itex] that are eigenstates of the time-dependent Heisenberg picture position operator. [itex]\Psi (x,t)=\langle x,t|\Psi\rangle[/itex]. Both cases can be expressed as [itex]\langle x|e^{-iHt}|\Psi\rangle[/itex].

    But when we ask the question of quantum observation, we dont get something which looks like [itex]\langle x_1,t_1 | x_2,t_2 \rangle[/itex]. Which is: given a system in a state corresponding to wave function [itex]\Psi (x,t_1)[/itex] at time t_1, what is the probability amplitude of observing the system in state [itex]\Phi(x,t_2)[/itex] at time t_2?

    The answer is

    [tex]\int \Phi(x,t_2)^* \Psi(x,t_2)d^3 x[/tex]

    But t_1 doesn't enter into this calculation, at least not directly. It only enters as the initial condition for our solution Psi of the Schrodinger equation. Once we have that solution, we know the state time-evolved to t_2 and then calculate the inner product at that time t_2. (and if we know the time-dependence of [itex]\Phi[/itex] we can calculate this integral at any time. It is time-independent.)

    So what sort of question about physically realizable states is [itex]\langle x_1,t_1 | x_2,t_2 \rangle[/itex] the answer to? or what question does [itex]\langle x_1,t_1 | x_2,t_2 \rangle[/itex] help us solve?

    Or, even, what is a quantum question which depends on a time difference t_2 - t_1 ?
     
    Last edited: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2
    If ##|\psi\rangle## and ##|\phi\rangle## are Heisenberg picture states, this amplitude can be written as
    [tex]\langle \phi | \psi \rangle[/tex]
    Inserting complete sets of states,
    [tex]\langle \phi | \psi \rangle = \int dx_2 \int dx_1 \langle \phi | x_2, t_2 \rangle \langle x_2, t_2 | x_1, t_2 \rangle \langle x_1, t_1 | \psi \rangle[/tex]
    This is why the propagator ##\langle x_2, t_2 | x_1, t_2 \rangle## is useful.
     
  4. Apr 6, 2014 #3
    why would we want to use

    [tex]\langle \phi | \psi \rangle = \int dx_2 \int dx_1 \langle \phi | x_2, t_2 \rangle \langle x_2, t_2 | x_1, t_2 \rangle \langle x_1, t_1 | \psi \rangle[/tex]

    ? Why not use

    [tex]\langle \phi | \psi \rangle = \int dx_1 \langle \phi | x_1, t \rangle \langle x_1, t | \psi \rangle[/tex]
     
  5. Apr 6, 2014 #4
    Sure, that is also a valid way of rewriting this amplitude. Computing it in this form requires you to evolve one or both states to the common time ##t## using the Schrodinger equation.

    Solving for the propagator ##\langle x_2, t_2 | x_1, t_2 \rangle## is like solving the Schrodinger equation once and for all. The propagator captures all information about time evolution in the system of interest. Once you have it you can plug it into the equation I gave above and straightforwardly calculate ##\langle \phi | \psi \rangle## for any ##|\psi\rangle## and ##|\phi\rangle##, without solving the Schrodinger equation again.
     
  6. Apr 6, 2014 #5
    Ok. I'll buy that. Thanks.
     
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