I am attempting to integrate this path integral in Euclidean variable $\tau $ (but this need not be the same as the $X^0$ field):

$$Z=\int _{X(0)=x}^{X(i)=x'}DX\exp \left(-\int _0^i d\tau \left[\frac{1}{2L}\partial _\tau X^\mu \partial _{\tau} X_{\mu} +\frac{L}{2}m^2\right ]\right).$$

The indices are being contracted in the Minkowski metric of dimension $D$. What I did so far was I moved the $X^\mu $ field to the left using integration by parts (dropping boundary terms) and pulled out the constant mass term:

$$Z=\exp \left(-\frac{L}{2}m^2\right)\int _{X(0)=x}^{X(i)=x'}DX \exp \left(\int _0^i d\tau \frac{1}{2L}X^\mu \partial _\tau ^2 X_\mu \right).$$

The answer might be related to:

$$Z=\frac{\exp \left(-\frac{L}{2}m^2\right)(D-1)\sqrt{2L}}{\sqrt{\det \partial _\tau ^2}}.$$

This is around where I got stuck. How does one go about finding the eigenvalues of $\partial _\tau ^2$? Should I have Fourier transformed earlier in the process? How does one incorporate the boundary fields $x$ and $x'$ into the answer?

Ideally, I would like to utilize zeta-function regularization. Supposing I can find and order the eigenvalues of $\partial _\tau ^2$ as $0<\lambda _1\leq \lambda _2\leq \ldots $ and $\lambda _n\rightarrow \infty $. Then, the zeta-function is defined as: $$\zeta _{\partial _\tau ^2}(s)\equiv \sum _{n=1}^\infty \frac{1}{\lambda _n^s}.$$ (Is there a typo on the Wikipedia for the index?)

Then, differentiating term-by-term yields: $$\zeta '(s)=\frac{d\zeta }{ds}=\sum _{n=1}^\infty \frac{-\ln \lambda _n}{\lambda _n^s}.$$

Then, the determinant of the operator is: $\det \partial _\tau ^2=\exp \left[-\zeta _{\partial _\tau ^2}'(0)\right]$. This equation made sense as basically the determinant is the product of the eigenvalues however it seems really unclear to me how to use this process to regularize the path integral, as well as the issue of solving for the actual eigenvalues of the operator themselves.


1 Answer 1


Since $X^\mu$ does not vanish on the boundaries and is not periodic, you cannot just ignore boundary terms. You need to be more careful and you can do so by moving Fourier space.

We consider the path integral \begin{equation} \begin{split} K ( x , x' ; T ) = \int\limits_{X(0) = x}^{X(T)=x'} [DX] \exp \left( - \int_0^T d \tau \left[ \frac{1}{2L} \partial_\tau X^\mu \partial_\tau X_\mu + \frac{L}{2} m^2 \right] \right) . \end{split} \end{equation} Given the boundary conditions in the path integral, we can expand $X^\mu(\tau)$ in a Fourier series expansion as $$ X^\mu(\tau) = x^\mu + \tau \left( \frac{x'^\mu - x^\mu}{T} \right) + L \sum_{n\neq0} X_n^\mu \exp \left( \frac{2\pi i n \tau}{T} \right) , \tag{1} $$ with $$ \sum_{n\neq0} X_n^\mu = 0 . \tag{2} $$ Note that the sum is over all integers not equal to zero since the zero mode has been included in (1) separately. This zero mode is entirely fixed by the boundary conditions and we do not path integrate over this mode. Finally, a factor of $L$ has been included in (1) to make $X_n$ dimensionless.

The path integral measure is \begin{equation} \begin{split} \int\limits_{X(0) = x}^{X(T)=x'} [DX] &= \int \frac{d^D p}{(2\pi)^D} \int \left( \prod_{n>0} (4\pi)^D d^D a_n d^D b_n e^{ i p \cdot a_n } \right) . \end{split} \end{equation} Since the $X_n$'s are complex, we have broken up the integral into its real part $a_n$ and its imaginary part $b_n$. Further since $(X^\mu_n)^* = X^\mu_{-n}$ (due to reality of $X^\mu(\tau)$), we restrict only to $n > 0$. The integral over $p$ imposes the constraint (2).

Plugging in all of these explicit formulas into the path integral and evaluating, we find \begin{equation} \begin{split} K ( x , x' ; T ) &= \exp \left( - \frac{1}{2} \left[ \frac{( x - x' )^2}{TL} + TL m^2 \right] \right) \prod_{n>0} \left( \frac{ 4 \pi T }{ n^2 L } \right)^{D/2} . \end{split} \end{equation} The infinite product can be written as \begin{equation} \begin{split} \prod_{n>0} \left( \frac{ 4 \pi T }{ n^2 L } \right)^{D/2} &= \exp \left[ \frac{D}{2} \sum_{n>0} \ln \frac{ 4 \pi T }{ n^2 L } \right] = \exp \left[ \frac{D}{2} \ln \frac{ 4 \pi T }{L} \sum_{n>0} 1 - D \sum_{n>0} \ln n \right] \end{split} \end{equation} Each of these infinite sums can be regulated using the zeta function. In the usual physics way, we write \begin{equation} \begin{split} \sum_{n>0} 1 \quad &\to \quad \left( \sum_{n>0} n^{-s} \right) \bigg|_{s=0} = \zeta(0) = - \frac{1}{2} , \\ \sum_{n>0} \ln n \quad &\to \quad - \frac{d}{ds} \left( \sum_{n>0} n^{-s} \right) \bigg|_{s=0} = - \zeta'(0) = \frac{1}{2} \ln (2\pi) . \end{split} \end{equation} Thus, \begin{equation} \begin{split} \prod_{n>0} \left( \frac{ 4 \pi T }{ n^2 L } \right)^{D/2} \quad \to \quad \left( \frac{ 16 \pi^3 T }{L} \right)^{-D/4} . \end{split} \end{equation} The full result is therefore \begin{equation} \begin{split} \boxed{ K ( x , x' ; T ) = \left( \frac{ 16 \pi^3 T }{L} \right)^{-D/4} \exp \left( - \frac{1}{2} \left[ \frac{( x - x' )^2}{TL} + TL m^2 \right] \right). } \end{split} \end{equation}

WARNING: I have not double checked this calculation. I could have easily gotten numerical factors incorrect. Please verify all of this yourself!

  • $\begingroup$ Thanks so much. $\endgroup$ Sep 22, 2023 at 23:27

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