# 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.

• Show your work so far, I don't know where you are stuck. Commented Jun 25, 2013 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 $q>b,q'>b$ Commented Jun 26, 2013 at 9:13

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 ;)