The Feynman path integral (for a one dimensional space such as the $x$-axis) is an infinite dimensional integral where the number of dimensions is countably infinite (it is an $n$-dimensional integral in the limit where $n$ approaches infinity). Hence, we are sampling each path into a time-ordered set of positions where the size of the set is countably infinite, not uncountably infinite. So, we are not including every real-valued point, of any path, in the calculation, as that would require sampling each path into a time-ordered set of positions where the size of the set is uncountably infinite.

Does the above imply that the object, which we are analyzing via the path integral, can only move in paths that are continuous in time (including backtracking as part of the path), except for a finite number of discontinuities?

That is, a "crazy" path such as the object starts out at $x_0$ at $t_0$, ends at $x_1$ at $t_1$ (all four values are rational numbers), and follows a path that is a continuous polynomial in $t$, for values of $t$ that are rational (ending up at $x_1$ at $t_1$), but follows a different function of $t$ for values of $t$ that are irrational (still ending up at $x_1$ at $t_1$) is not allowed? The reason I ask is that I learned about the path integral as incorporating all imaginable paths, but I think some are clearly out of the question, if they contain more than a countably infinite size of information.

  • $\begingroup$ From what I remember every path except Hölder-2 continuous paths are of measure $0$ $\endgroup$
    – Slereah
    Apr 12, 2018 at 21:15
  • 1
    $\begingroup$ The functional integral, when it makes sense (euclidean time), uses Wiener processes, i.e. Brownian motion. Therefore the considered paths are continuous (but essentially nowhere differentiable). $\endgroup$
    – yuggib
    Apr 13, 2018 at 13:20
  • $\begingroup$ @yuggib Since the path is continuous and bounded, then I think that a countably infinite number of points can adequately represent it, even if it is differentiable nowhere. Does that make sense to you? $\endgroup$
    – David
    Apr 14, 2018 at 5:47
  • $\begingroup$ @David The continuous image of a separable space is separable, so yes, you can approximate a continuous path from $\mathbb{R}$ to $\mathbb{R}^d$ with a countable set of points. $\endgroup$
    – yuggib
    Apr 14, 2018 at 6:50
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/185445/2451 , physics.stackexchange.com/q/444133/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 30, 2018 at 14:55


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