Evaluation of a path integral

Question

I am stuck weeks in the following calculation from a paper:

Path integral $I$ is written as

$$ I = \int_{\mathbf{r}_0 = \mathbf{0}}^{\mathbf{r}_L=\mathbf{R}}\exp\left[-A\int_0^L\dot{\mathbf{r}}_t^2\,dt+\frac{i\eta}{2L^2}\int_0^L\int_0^L (\mathbf{r}_t-\mathbf{r}_s)^2\,ds\,dt\right]{\cal D}\mathbf{r}.$$

Since the contribution of the path $\mathbf{r}_t$ to $I$ becomes most dominant when $\mathbf{r}_t$ makes the exponent stationary, we have $\mathbf{r}_t$ as the solution of the following Euler-Lagrange equation subject to the boundary conditions $\mathbf{r}_0 = \mathbf{0}$ and $\mathbf{r}_L = \mathbf{R}$:

$$\ddot{\mathbf{r}}_t + \alpha^2\mathbf{r}_t - \beta\int_0^L\mathbf{r}_s\,ds=0,$$ where $\alpha^2 = \dfrac{i\eta}{AL}$ and $\beta = \dfrac{i\eta}{AL^2}$. With this solution, $I$ is expressed as $$I = \exp(-A\mathbf{r}_L\dot{\mathbf{r}}_L).$$

I don't have any problem with the Euler-Lagrange equation but cannot obtain the last expression. Anyone help me out of this?

Answer 1

One thing we can do is solve the E-L DE. If we differentiate it w.r.t. $t$, and substitute $\mathbf{x}=\dot{\mathbf{r}}_t$, we get simply $\ddot{\mathbf{x}}+\alpha^2\mathbf{x}=0$, the solution of which is $\mathbf{x}=B\cos(\alpha t)+C\sin(\alpha t)$. Integrating to get $\mathbf{r}_t$ yields $$\mathbf{r}_t=\frac{B}{\alpha}\,\sin(\alpha t)-\frac{C}{\alpha}\,\cos(\alpha t)+D.$$ Now we impose three conditions on $\mathbf{r}_t$ to get the three constants: $\mathbf{r}_0=0,\;\mathbf{r}_L=\mathbf{R},$ and the E-L DE itself. This imposes \begin{align*} D-\frac{C}{\alpha}&=0 \\ \frac{B \sin (\alpha L)-C \cos (\alpha L)+\alpha D}{\alpha}&=\mathbf{R} \\ \frac{\alpha^4 D-\beta \left(\alpha^2 D L+B\right)+B\,\beta \cos (\alpha L)+C \beta \sin (\alpha L)}{\alpha^2}&=0, \end{align*} the solution of which is \begin{align*} B&= -\frac{\alpha \mathbf{R} \csc ^2\left(\frac{\alpha L}{2}\right) \left(\alpha^3-\alpha L \beta+\beta \sin (\alpha L)\right)}{-2 \alpha \cot \left(\frac{\alpha L}{2}\right) \left(\alpha^2-L \beta\right)-4 \beta},\\ C&= \frac{\alpha \mathbf{R} \beta (\cos (\alpha L)-1)}{\alpha \sin (\alpha L) \left(L \beta-\alpha^2\right)+2 \beta (\cos (\alpha L)-1)},\\ D&= \frac{\mathbf{R} \beta (\cos (\alpha L)-1)}{\alpha \sin (\alpha L) \left(L \beta-\alpha^2\right)+2 \beta (\cos (\alpha L)-1)}. \end{align*} You can check and see that this solves the two boundary conditions plus the E-L DE.

But here's the problem: the way you've defined $I$, there is an indefinite integral w.r.t. $s$ buried in there, whereas $I=\exp(-A\mathbf{r}_L\dot{\mathbf{r}}_L)$ is a constant. Are you sure that's correct? Hopefully, my solution of the DE will push you along a little bit.

Did you mean $$\int_0^L\left[\int(\mathbf{r}_t-\mathbf{r}_s)^2\,ds\right]\Bigg|_{s=t}\,dt$$ inside the exponential?

Answer 2

I assume the bounds of the integral with respect to $s$ are also $0$ and $L$.

First, we integrate the first term inside the exponential by parts. We get

$$-A\int_0^L \dot{\mathbf{r}}_t^2 dt = \left[ -A\mathbf{r}_t \dot{\mathbf{r}}_t \right]_0^L +A\int_0^L \mathbf{r}_t \ddot{\mathbf{r}}_t dt = -A\mathbf{r}_L \dot{\mathbf{r}}_L +A\int_0^L \mathbf{r}_t \ddot{\mathbf{r}}_t dt,$$

since $\mathbf{r}_0 = 0$.

Next, we expand the square inside the second term. We get

$$\frac{i\eta}{2L^2} \int_0^L \int_0^L \left( \mathbf{r}_t^2 + \mathbf{r}_s^2 - 2\mathbf{r}_t\mathbf{r}_s \right) ds\: dt.$$

The $\mathbf{r}_t^2$ term doesn't depend on $s$, so we can integrate it with respect to $s$ getting a factor of $L$. As for the $\mathbf{r}_s^2$ term, we can switch the dummy variables $s$ and $t$ so it becomes identical to the $\mathbf{r}_t^2$ term. With the definitions of $\alpha$ and $\beta$, we arrive at

$$\frac{i\eta}{2L^2} \int_0^L \left( L \mathbf{r}_t^2 + L \mathbf{r}_t^2 - 2\mathbf{r}_t \int_0^L \mathbf{r}_s ds \right) dt = \int_0^L \mathbf{r}_t \left(A\alpha^2 \mathbf{r}_t -A\beta \int_0^L \mathbf{r}_s ds \right) dt.$$

Finally, putting all together we find that the whole expression inside the exponent is

$$-A\mathbf{r}_L \dot{\mathbf{r}}_L + A \int_0^L \mathbf{r}_t\left(\ddot{\mathbf{r}}_t + \alpha^2 \mathbf{r}_t -\beta \int_0^L \mathbf{r}_s ds \right) dt.$$

The expression inside the parentheses clearly vanishes for the "classical" path satisfying the Euler-Lagrange equation. Since we are only considering this path's contribution to the path integral, we arrive at the desired solution

$$I=\exp(-A\mathbf{r}_L \dot{\mathbf{r}}_L).$$