# Potentials in Feynman path integral

I am trying to understand the Feynman path integral by reading the book from Leon Takhtajan.

In one of the examples, there is a full explanation of the calculation of the propagator

$$K(\mathbf{q'},t';\mathbf{q},t) = \frac{1}{(2\pi\hbar)^n} \int_{\mathbb{R}^n} e^{\frac{i}{\hbar}(\mathbf{p}(\mathbf{q'}-\mathbf{q})-\frac{\mathbf{p}^2}{2m}T)} d^n\mathbf{p},\quad T=t'-t.$$

in the case of a free quantum particle with Hamiltonian operator

$$H_0 = \frac{\mathbf{P}^2}{2m},$$

and the solution is given by

$$K(\mathbf{q'},t';\mathbf{q},t) = \left(\frac{m}{2\pi i \hbar T}\right)^{\frac{n}{2}} e^{\frac{im}{2\hbar T}(\mathbf{q}-\mathbf{q'})^2}.$$

Could you please help me to understand how to perform the calculation in the case where the Hamiltonian is given by

$$H_1 = \frac{\mathbf{P}^2}{2m} + V(\mathbf{Q})$$

where $V(\mathbf{Q})$ is the potential defined by

$$V(\mathbf{Q})=\left\{ \begin{array}{cc} \infty, & \mathbf{Q} \leq b \\ 0, & \mathbf{Q}>b. \\ \end{array} \right.$$

Update :

I've read the article provided by Trimok, and another one found in the references, but I am still annoyed with the way the propagator is computed. I may be mistaken, but it seems that in that kind of articles, they always start the computation from the scratch, without using what they already know about path integral.

I am actually trying to write something about the use of path integrals in option pricing. From Takhtajan's book, I know that for a general Hamiltonian $H=H_0 + V(q)$ where $H_0 = \frac{P^2}{2m}$, the path integral in the configuration space (or more precisely the propagator) is given by

$$\begin{array}{c} \displaystyle K(q',t';q,t) = \lim_{n\to\infty}\left(\frac{m}{2\pi\hbar i \Delta t}\right)^{\frac{n}{2}} \\ \displaystyle \times \underset{\mathbb{R}^{n-1}}{\int \cdots\int} \exp\left\{\frac{i}{\hbar}\sum_{k=0}^{n-1}\left(\frac{m}{2}\left(\frac{q_{k+1} - q_k}{\Delta t}\right)^2 - V(q_k)\right)\Delta t\right\} \prod_{k=1}^{n-1} dq_k.\\ \end{array}$$ I would like to start my computation from this result, and avoid repeating once again the time slicing procedure. So du to the particular form of the potential, I think I can rewrite the previous equation as $$\begin{array}{c} \displaystyle K(q',t';q,t) = \lim_{n\to\infty}\left(\frac{m}{2\pi\hbar i \Delta t}\right)^{\frac{n}{2}} \\ \displaystyle \times \int_0^{+\infty} \cdots\int_0^{+\infty} \exp\left\{\frac{i}{\hbar}\frac{m}{2}\sum_{k=0}^{n-1}\frac{(q_{k+1} - q_k)^2}{\Delta t}\right\} \prod_{k=1}^{n-1} dq_k.\\ \end{array}$$ Then I need a trick to go back to full integrals over $\mathbb{R}$ and use what I already know on the free particle propagator. However, since the integrals are coupled, I don't find the right way to end the calculation and find the result provided by Trimok.

Could you please tell me if I am right or wrong ? Thanks.

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Show your work so far, I don't know where you are stuck. –  Chris Gerig Jun 25 '13 at 19:44
You have an example in this reference - page 2 -Chapter "The propagator for a free particle in a restricted domain". The idea is in fact the use of the "image method". Your result should be $K_1(q',t',q,t) = K(q',t',q,t) - K(q',t', 2b - q,t)$ for $q>b,q'>b$ –  Trimok Jun 26 '13 at 9:13
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## 1 Answer

The "easiest" way is to relate the path integral problem to a PDE using Feynman-Kac formula, then you're actually solving diffusion in half space, which is usually solved by odd expansion of the solution to the whole space (image or reflection method)

The other way to comprehend where these canceling paths in $2B-x$ comes from is from stochastic process consideration, using the reflection principle of Brownian motion. Description of this concept is available everywhere ;)

Careful derivation for path integrals was made by Schulman as far as I can remember.

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