# Why do particles travel along smooth paths in quantum mechanics?

LukeD
I haven't been studying quantum mechanics for very long, and I've only just started reading about the path integral formulation, so I don't know many of the details yet, but I noticed something peculiar.

The path integral formulation is possible because of the similarities of the Schroedinger Equation with the Diffusion Equation. The probability density in the Diffusion Equation can be arrived at by considering a path integral over Brownian motion. Almost every path that contributes to the integral is a nowhere differential path (and the smooth paths actually contribute 0 to the integral)

An elementary property of Brownian Motion is that $$\int_0^T (x(t+dt) - x(t))^2 dt \propto T$$ for almost every path (the probability of this not being true for a path is 0). On the other hand, for any differentiable path, this integral gives 0 due to the derivative being bounded.

The Feynman Path Integral is built using the same math. Now we arrive at the wave function by considering a path integral over Brownian motion. The function that we integrate however is a complex function and the wave function we compute is not a probability distribution but a square root of a probability distribution.

Now keeping that in mind, I compute $$<\psi|(U^\dagger(\Delta t) x U(\Delta t) - x)^2|\psi> \rightarrow (\Delta t)^2$$. This is actually a pretty easy calculation, it follows from the unitarity of U and time independence of the Hamiltonian.

However, if I integrate this quantity over some finite time, I get 0. If I had some contribution to this average from the nowhere smooth paths in Brownian Motion, I would get something proportional to the time of travel.

What happened to my nowhere smooth paths? How did they all cancel out? (Also, am I correct in my conclusion that they do cancel out?)

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LukeD
If the above argument is valid, their actions are obviously distributed evenly over both S in [0, hbar*Pi) and S+Pi (or perhaps as my professor Bob Suter suggested, completely randomly), but I can't see this...

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peteratcam
An elementary property of Brownian Motion is that $$\int_0^T (x(t+dt) - x(t))^2 dt \propto T$$ for almost every path (the probability of this not being true for a path is 0). On the other hand, for any differentiable path, this integral gives 0 due to the derivative being bounded.
Maybe my stochastic calculus is not good enough, but that result looks bizarre. I don't understand why there are two 'dt's on the LHS, nor why the integral of an infintesimal quantity over a finite region is finite.

LukeD
Sorry, I was being a little loose with my notation (though I think you'd probably read that integral the way I intended it. Now that I think about it, that second dt shouldn't have been there)

The way that integral is written in Steven Shreve's books is

$$\lim_{\| \Pi \| \rightarrow 0} \sum_{j=0}^{n-1} [f(x_{j+1}) - f(x_j)]^2$$

It is a peculiar property of nowhere smooth paths, that this integral is generally non-zero. (Though for smooth paths, you are correct that (f(x_{j+1}) - f(x_j))^2 is infinitesimal, and the integral gives 0)