In path integral formulation we say that we are summing over all possible ways for the system get from initial to final state. Now if we just write the amplitude and then insert complete set of states, is it then true that this insertion of sets of states for every step is actually where we sum over all possible states? Of course, to get all possible paths we have to multiply, but is it in this moment that this can be seen as a sum over all paths?

To expand a little. In a book by A. Zee there is a fun derivation of path integral formulation where we start by observing the amplitude for a particle in some inital state I to be found in some final state F after some time T. Now, if these states are just eigenstates of a position vector then we can to this: we can split our time translation operator for T into small time intervals and the between those put for every interval a complete set of states. By doing so, are we explicitly doing integral over all possible paths and can it be seen as a result of combining, which is made, as usual, by multiplication of all possible realizations of position? I am pretty sure that this is it. But not completely sure.

  • $\begingroup$ Can you try to ask this question in a way such that "Yes." is not a potentially complete answer (it's too short to even submit as an answer)? $\endgroup$
    – ACuriousMind
    May 11 '19 at 11:25

As far as I understand, it is along the right track but not complete.

It is not enough to insert the complete set of position-eigenstates at each time-slice. One also needs to insert a complete set of momentum-eigenstates at each time-slice. Assuming that the Hamiltonian is Weyl ordered, this allows you to treat the full propagator as a phase-space path integral over all the (phase-space) paths starting at a specific position with an arbitrary momentum and terminating at a specific position with an arbitrary momentum.

Now, if the Hamiltonian is of the form $\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})$ such that the functional integral over momenta can be carried out separately, the full propagator can now be treated as a configuration-space path integral over all the paths from the initial specific position to the final specific position.

For this final simplified case, the full propagator can be treated as a summation over all the paths that Zee has beautifully illustrated. In the cases where the integration over momenta cannot be carried out separately, the full propagator wouldn't simply be a summation over those illustrated paths (as those are, of course, configuration-space paths).

But yes, in essence, the picture of talking in terms of summation over all paths arise from the act of inserting complete sets of eigenstates at each time-slice.

  • $\begingroup$ Yes tnx. Of course in order to apply hamiltonian you can insert also momentum eigenkets. This I know. Tnx for this confirmation. $\endgroup$ May 11 '19 at 18:37
  • $\begingroup$ @ŽarkoTomičić To be clear, you "have to" not just "can" insert momentum eigenstates. $\endgroup$
    – Dvij D.C.
    May 11 '19 at 22:31
  • 1
    $\begingroup$ Of course. Wrong choice of words. $\endgroup$ May 11 '19 at 22:45

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