The phase space version of Feynman's path integral expression for the free particle propagator involves a (formal) sum over paths in phase space with fixed $q$ endpoints and (as far as I'm aware) arbitrary $p$ behaviour. Does this mean that the paths summed over do not necessarily represent paths of a particle with a fixed total time? (i.e. the time you would obtain by integrating $1/p$ over $q$)?


Comments to the question (v2):

  1. In (non-relativistic) quantum mechanics, time is a parameter (as opposed to a selfadjoint operator), cf. e.g. this Phys.SE post and links therein.

  2. In the phase space path integral $$ K(q_f,t_f;q_i,t_i) ~\equiv~\langle q_f,t_f \mid q_i,t_i\rangle ~=~\int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~ {\cal D}p~ \exp\left(\frac{i}{h}S[q,p]\right) \tag{1} $$ the total time $\Delta t=t_f-t_i$ is fixed.

  3. It follows from the derivation of the phase space path integral (1) from the operator formulation, that one should not impose boundary conditions on the momentum. Intuitively, this agrees with Heisenberg's uncertainty principle: Since the initial and final positions $q(t_i)=q_i$ and $q(t_f)=q_f$ are known, the initial and final momenta $p(t_i)$ and $p(t_f)$ are unknown.

  4. For a classical solution$^1$ $(q_{\rm cl},p_{\rm cl})$ to Euler-Lagrange (EL) eqs. satisfying the pertinent Dirichlet boundary conditions, it is of course simple to extract the classical initial and final momenta $p_{\rm cl}(t_i)$ and $p_{\rm cl}(t_f)$. However, quantum mechanically, this is not the whole story. Feynman instructed us to sum over all histories, not just the classical paths, but also virtual paths. And the initial and final momenta $p(t_i)$ and $p(t_f)$ remain unknown. Phrased differently: Quantum mechanically, it does not make sense to claim that the particle followed a specific path, cf. e.g. the double-slit experiment.


$^1$ In principle, there can be instanton, i.e. more than one classical solution satisfying the pertinent Dirichlet boundary conditions.

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