Non-differentiable path in path integral?

kof9595995

It seems obvious in path integral, the paths include some non-differentiable path (some even discontinuous, I think), wouldn't it cause any serious problem? For example, the classical lagrangian as the phase factor, is defined on differentiable paths, isn't it?

HallsofIvy

if there are only a finite number of points on which the path is not smooth, then you can integrate on each segment and sum. If there are a countably infinite number of points where the path is not smooth, then that sum will be an infinite sum which may not converge. If there are an uncountable number of points where the path is not smooth, then you have a real problem!

kof9595995

Then would "uncountable number of points where the path is not smooth" happen? It seems quite possible to me, since there's not at all any restriction on how you choose |q><q| to insert.

lugita15

In fact, I think *almost all* of the paths would have uncountably many discontinuities. But for each discontinuous path, there will be many smooth paths which are arbitrarily close to the discontinuous one. So I don't think there's any reason to worry.

kof9595995

Why not? In principle we need to evaluate all possible paths, don't we?

DarMM

The path integral for quantum mechanics is only supported on nondifferentiable functions. It assigns measure zero to the set of differentiable functions. However it is also supported only on continuous functions.

The support is on even more singular objects in quantum field theory, where the measure is only supported on distributions, not functions.

kof9595995

@DarMM: so you mean only non-differentiable paths gives non-zero contributions to the integral? But it doesn't seem to answer my question, what happened if the integrals for some paths do not exist?

The_Duck

I know very little about this, but in the elementary introductions I have read to the path integral, I believe it is equal to an integral over all continuous paths composed of straight line segments, in the limit where the number of line segments composing the path goes to infinity. You calculate the action for each line segment, add up all the line segments to get the action for the entire path, and then integrate the exponential of the action over all such paths. You ignore the "kinks" joining line segments. In the limit of many line segments, this integral looks like an expression you can write down in QM for the amplitude for the particle to propagate from the start point to the end point.

DarMM

Yes, only non-differentiable give a non-zero contribution. However this doesn't mean the path integral doesn't exist. The measure the phase factor (in Euclidean time) is [tex]e^{-S[q]}[/tex] which doesn't make any sense on those paths, since the Lagrangian contains derivatives as you said. However the phase factor and the measure [tex]\mathcal{D}[q][/tex] together make sense on those paths. So the path integral exists in quantum mechanics.

kof9595995

Could you brief me how is this so? Well, but you should know I know very little about measure theory(mathematical analysis is in my reach), please explain it in a layman style if possible, many thanks.