In the path integral formulation of quantum mechanics, the amplitudes of all possible paths from starting point A to end point B are added up.

For a free particle, how can all these paths be physically possible? If the particle is to conserve momentum, it must travel in a straight line. It is not interacting with anything else, so the path cannot be a curve in flat spacetime. Why does the integral include physically impossible paths?

Image from wikipedia

  • 1
    $\begingroup$ "Physically" is not synonymous with "classically". $\endgroup$ Jun 23, 2021 at 2:00
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/571421, physics.stackexchange.com/q/571458 $\endgroup$ Jun 23, 2021 at 5:13
  • $\begingroup$ @ConnorBehan: I'm not aware of any mechanism that allows a free particle to "physically" violate momentum conservation. Please illuminate me. $\endgroup$
    – user37222
    Jun 23, 2021 at 16:41
  • $\begingroup$ There doesn't need to be one. Every time a new theory becomes accepted, it's because of a prediction which the old theory couldn't make. In this case, the virtual paths you drew have no way of being predicted by definite phase space trajectories. But experiments have shown that these don't exist. $\endgroup$ Jun 23, 2021 at 18:55

1 Answer 1

  1. First of all, in the path integral we integrate over all off-shell virtual paths. The virtual paths came from how we derived the path integral from the operator formalism by inserting infinitely many completeness relations, see any decent textbook on the matter.

  2. In the semiclassical limit $\hbar\to 0 $, the path integral is dominated by contributions from on-shell classical paths, i.e. solutions to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post.

  3. In QM with an external potential, the paths don't necessarily conserve momentum, due to external forces. However, in the free case [which OP seems to be asking about] the classical [but not necessarily the virtual] paths conserve momentum.

  4. By the way, in QFT in the Fourier momentum representation of spacetime [which OP doesn't seem to be asking about] the momentum conservation of off-shell virtual paths is typically a consequence translation invariance of spacetime.

  • $\begingroup$ 1. Off-sell paths still conserve momentum. It violates only $E^2 - p ^2 = m^2$. Besides, in the non-relativistic formulation, this is not relevant. 2. The size of the contribution is not the problem here unless paths not conserving momentum have zero contribution. 3. The question is about the simplest case - the free particle. 4. The non-relativistic free particle lagrangian is translational invariant. $\endgroup$
    – user37222
    Jun 23, 2021 at 16:33
  • $\begingroup$ 1. In addition, inserting infinitely many completeness relations has the same problem. The "propagations" don't all conserve momentum. $\endgroup$
    – user37222
    Jun 23, 2021 at 16:46
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 23, 2021 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.