Brownian Motion and Path Integrals

[Anaxandridas]

I am reading "Quantum Mechanics and Path Integrals" (by Feynman and Hibbs) and working out some of the problems... as a hobby of sorts.

I have run into a problem in section 12-6 Brownian Motion. On page 339 (of the emended edition), the authors demonstrate, by example, a method for calculating a probability distribution using a path integral.

The text states that "the integral

P(D,[itex]\theta[/itex])=[itex]\int exp \left\{ -\frac{1}{2R} \int^{T}_{0} \ddot{x}^{2}\left(t\right) dt \right\} D x\left(t\right) [/itex]

is gaussian and becomes an extremum for the path

[itex]\frac{d^{4}\bar{x}}{dt^{4}}[/itex] = 0."

I am having difficulty recognizing from where this path condition arises. Please let me know if you some insight regarding this problem.

 

[eljose79]

Thank you for bringing up this interesting problem from section 12-6 of "Quantum Mechanics and Path Integrals" by Feynman and Hibbs. After looking into the text and the example given, I can offer some insight into the path condition you are having difficulty recognizing.

Firstly, it is important to note that the path integral in question is a functional integral, meaning it involves integrating over all possible paths that the particle could take. In this case, the particle is undergoing Brownian motion, which is a random motion caused by collisions with other particles in the medium it is moving through.

Now, in order to find the probability distribution for this Brownian motion, the authors use a technique called the saddle-point method, which involves finding the extremum (maximum or minimum) of the integrand. In this case, the integrand is a Gaussian function, which is known to have a maximum at its mean value. Therefore, the path that will give the maximum value for the integral is the one that minimizes the exponent in the integrand, which is the same as minimizing the action (the integral in the exponent).

The authors then use the Euler-Lagrange equation to find the path that minimizes the action, which leads to the condition \frac{d^4\bar{x}}{dt^4} = 0. This means that the path that gives the maximum value to the path integral is the one that has no curvature, or in other words, the path that is the straightest.

I hope this explanation helps clarify the path condition that arises in this example. If you have any further questions or concerns, please don't hesitate to reach out. Happy problem-solving!