How is the action defined for non-differentiable paths in path integrals?

[Inquisitive_Mind]

In path integrals, how does one deal with non-differentiable paths? Obviously non-differentiable paths are allowed, but with Feymann's formulation, one has to calculate the action for a path, and then sum over all possible paths. How is the action defined (if it is defined at all) for a non-differentiable path?

Also, is it possible to construct path integral vigorously by constructing a measure on the space of possible paths?

 

[selfAdjoint]

Since you are integrating, differentiability of the path doesn't matter. You can compute the length of the perimeter of a square, although the path becomes nondifferentiable at the corners.

 

[Haelfix]

The nondifferentiabality of the paths is not so much a problem (in fact in general one expects that). However the measure is a problem in QFT. No one knows really what exactly *that* is, and is more or less poorly defined mathematically (but curiously, you *can* do a few things with it).

People seem to have more or less given up on that problem for the general case, as its exceedingly hard.

 

[Inquisitive_Mind]

Thanks for your replies!

To selfAdjoint,
Maybe you mistakably thought that I was talking about the ordinary path integral? Or did I miss something?

To Haelfix,
Path integrals are new to me. May you explain why the non-differentiability of paths does not matter? I can imagine that there are many continuous yet everywhere non-differentiable paths (Brownian motion's deterministic counterparts?) that are allowed. I am not sure that they are of measure zero (if the measure is defined at all)?

 

[jtolliver]

Path integrals are new to me. May you explain why the non-differentiability of paths does not matter? I can imagine that there are many continuous yet everywhere non-differentiable paths (Brownian motion's deterministic counterparts?) that are allowed. I am not sure that they are of measure zero (if the measure is defined at all)?

They don't have measure zero. Their contributions cancel out, because the lagrangian involves derivatives, and the action diverges. Its easiest to show that in the case of a Euclidean path integral, where the integrand is $e^{-\infty}=0$ for a non differentiable path.