Evaluating Functional Determinants in a Thermal Field Theory using the Heat-Kernel

Introduction

In this post, we will discuss how to compute fluctuation determinants is a finite temperature field theory. The general form of an operator that we will investigate is:

$$ \begin{align} \mathcal{O} &= -\nabla^2 + \omega_{n}^2 + m^2 + V(x) \end{align} $$

where $m$ is a zero-temperature mass, $\omega_{n} = 2\pi n T$ for bosons and $2\pi(n+1/2)T$ for fermions and $V(x)$ is some space-dependent function. Often, $V(x)$ will take the form of a background- or low-mass-field-dependent mass for the fluctuation fields of the problem. For example, consider a $\phi^4$ theory at finite temperature with the following action:

$$ \begin{align} S[\phi] &= \int_{0}^{\beta}d\tau\int d^{3}x \dfrac{1}{2}(\partial_{\mu}\phi)^2 -\dfrac{1}{2}m^2\phi^2 +\dfrac{\lambda}{8}\phi^4 \end{align} $$

where $\phi = \phi(\tau, x)$. If one expands the field $\phi$ as a Fourier series in the imaginary time coordinate $\tau$,

$$ \begin{align} \phi(\tau, x) &= \sum_{n=-\infty}^{\infty}\phi_{n}e^{i\omega_{n}\tau} \end{align} $$

and wishes to integrate over the heavy Matsubara modes ($n\neq0$ modes), then they will need to compute a thermal fluctuation determinant of an operator of the form:

$$ \begin{align} \mathcal{O} &= -\nabla^2 + \omega_{n}^2 + m^2 + \dfrac{3}{2}\lambda\phi_{0}^2 \end{align} $$

Where $\phi_{0}$ is the zero-mode ($n=0$ Matsubara mode.) In this case, $V(x) = (3\lambda/2)\phi_{0}^2(x)$. The question we would like to answer is: how does one evaluate the following fluctuation determinant

$$ \begin{align} \mathrm{tr}\log\left(-\nabla^2 + \omega_{n}^2 + m^2 + V(x)\right) \end{align} $$

We will answer this question in the limit as $T\to\infty$.

Heat Kernal Expansion

The method we will employ to evaluate the fluctuation determinant of the operator $\mathcal{O} = -\nabla^2 + \omega_{n}^2 + m^2 + V(x)$ is the so called heat-kernel expansion. The idea is to use the following identity:

$$ \begin{align} \log(\lambda) = -\int_{0}^{\infty}\dfrac{ds}{s}e^{-\lambda s} + \mathrm{constant} \end{align} $$

where the constant we left implicit is some infinite constant which will play no role in our analysis. To derive this expression, one can consider the following:

$$ \begin{align} \int_{0}^{\infty}\dfrac{ds}{s}e^{-\lambda s} &= \int_{0}^{\infty}ds\int_{\lambda}^{\infty}d\lambda’e^{-\lambda’ s} \end{align} $$

Switching the order of integration and integrating over $s$, we find:

$$ \begin{align} \int_{0}^{\infty}\dfrac{ds}{s}e^{-\lambda s} &= \int_{\lambda}^{\infty}d\lambda’\dfrac{1}{\lambda’} = \lim_{\lambda’\to\infty}\log(\lambda’) - \log(\lambda) \end{align} $$

Thus, up to an infinite constant, the identity holds. Note that if we take the trace of a log of a determinant, we are equivalently summing over the log of the eigenvalues of the operator. Therefore, we can say that:

$$ \begin{align} \mathrm{tr}\log(\mathcal{O}) = \sum_{\lambda}\log(\lambda) = -\int_{0}^{\infty}\dfrac{ds}{s}\sum_{\lambda}e^{-\lambda s} = \int_{0}^{\infty}\dfrac{ds}{s}\mathrm{tr}e^{-\mathcal{O} s} \end{align} $$

We will define the factor of $e^{-\mathcal{O}s}$ as the heat kernel:

$$\begin{align} K(s, x, y) \equiv \langle x|e^{-\mathcal{O}_{x} s}|y\rangle \end{align}$$

where the subscript $x$ denotes that the derivatives and function evaluation of $\mathcal{O}$ to be evaluated with $x$. Then, the trace of $e^{-\mathcal{O} s}$ is just the integral over $x$ of $K(s,x,x)$:

$$\begin{align} \mathrm{tr}e^{-\mathcal{O} s} = \int d^{d}x K(s,x,x) \end{align}$$

The heat-kernel turns out to satisfy the heat equation (hence the name). This is easy to see:

$$ \begin{align} \dfrac{\partial}{\partial s}e^{-\mathcal{O}_{x} s} = -{\mathcal{O}_{x}} K(s,x,y) \quad \implies \quad \left(\dfrac{\partial}{\partial s} + {\mathcal{O}_{x}}\right)K(s,x,y) = 0 \end{align} $$

The right-most equality is simply the heat equation with a source term of $\omega_{n}^2 + m^2 + V(x)$. The boundary condition of the heat equation can be found by setting $s=0$:

$$ \begin{align} K(s=0,x,y) = \langle x|y\rangle = \delta^d(x-y) \end{align} $$

The heat-kernel can be determined exactly if we set $V(x) = 0$. Let’s call:

$$ \begin{align} \mathcal{M}^2 = \omega_{n}^2 + m^2 \end{align} $$

for ease of notation. The heat-kernel for the operator:

$$ \begin{align} \mathcal{O}_{0} = -\nabla^2 + \mathcal{M}^2 \end{align} $$

is given by the following:

$$ \begin{align} K_{0}(s,x,y) = \dfrac{1}{(4\pi s)^{d/2}}\exp\left(-\dfrac{|x-y|^2}{4s}-s\mathcal{M}^2\right) \end{align} $$

This can be verified by differentiation. Additionally, one can check that this expression satisfies the boundary condition (to do this, integrate some function $f(x)$ by Taylor expanding the function. The results are a set of gaussian integrals which can easily be evaluate to obtain $f(y)$.) Further more, $K_{0}(s, x, x)$ is:

$$ \begin{align} K_{0}(s,x,x) = \dfrac{1}{(4\pi s)^{d/2}}e^{-s\mathcal{M}^2} \end{align} $$

We can use this result to find the general result. Recall the Zassenhaus formula (a special case of the Baker-Campbell-Hausdorff formula):

$$ \begin{align} e^{-t(X + Y)} &= e^{-t X}e^{-t Y}e^{-\frac{t^2}{2}[X,Y]} e^{-\frac{t^3}{3!}(2[Y,[X,Y]]+ [X,[X,Y]])}\cdots \end{align} $$

where the $\cdots$ represent terms of order $t^{4}$ and higher. If we break up our general operator into:

$$ \begin{align} \mathcal{O} = \mathcal{O}_{0} + V(x), \end{align} $$

then we find that the general heat-kernel is given by:

$$ \begin{align} e^{-s\mathcal{O}} &= e^{-s\mathcal{O}_{0}}e^{-sV}e^{-\frac{s^2}{2}[\mathcal{O}_{0},V]} e^{-\frac{s^3}{3!}(2[V,[\mathcal{O}_{0},V]]+ [\mathcal{O}_{0},[\mathcal{O}_{0},V]])}\cdots \end{align} $$

We can easily evaluate the various commutators by considering their action on some test function $f(x)$. The results for the commutators shown are:

$$ \begin{align} [\mathcal{O}_{0},V] &= -\nabla^2 V\\ 2[V,[\mathcal{O}_{0},V]]+ [\mathcal{O}_{0},[\mathcal{O}_{0},V]] &=-4V\nabla^2V+\nabla^4V \end{align} $$

The higher order terms can easily be evaluated as well. Thus, we have that:

$$ \begin{align} K(s,x,x) &= e^{-s(\mathcal{M}^2 + V)}e^{\frac{s^2}{2}\nabla^2V} e^{-\frac{s^3}{3!}\left(\nabla^4V-4V\nabla^2V\right)}\cdots \end{align} $$

In order to make progress on this expression, we need to have control over which terms in the expression are important. Let’s re-introduce our original definitions:

$$ \begin{align} \mathcal{M}^2=\omega_{n}^2+m^2 = \tilde{\omega}_{n}^2T^2 + m^2 \end{align} $$

where we defined:

$$ \begin{align} \tilde{\omega}_{n}^2 = \omega_{n}^2/T^2 \end{align} $$

Next, we will define a scaleless variable $z=sT^2$. In terms of $z$, our expression is:

$$ \begin{align} K(s,x,x) &= e^{-z\tilde{\omega}_{n}^2}e^{-\frac{z}{T^2}(m^2 + V)}e^{\frac{z^2}{2T^4}\nabla^2V} e^{-\frac{z^3}{3!T^6}\left(\nabla^4V-4V\nabla^2V\right)}e^{\mathcal{O}(T^{-6})} \end{align} $$

Now it is clear how to proceed. We expand this expression in inverse powers of $T$. If we expand in inverse powers of $T$ and integrate over $s$, we find that, for bosons:

$$ \begin{align} &\dfrac{T}{2}\sum_{n=-\infty}^{\infty} \mathrm{tr}\log(-{\nabla^2}+(2\pi n T)^2 + m^2 +V({x}))\\\ &\hspace{1cm}= -\dfrac{T}{2}\sum_{n=\infty}^{\infty} \int d^{d}{x}\left(\dfrac{T}{4\pi}\right)^{d/2}\int_{0}^{\infty}dz\dfrac{K(z,{x},{x})}{z^{d/2+1}}\notag\\ &\hspace{1cm}\underset{d\to3-2\epsilon}{=} \int d^{3}{x}\bigg{[} -\frac{\pi^2 T^4}{90}+\frac{T^2(m^2 + V)}{24} -\frac{\left(m^2 + V\right)^2 \left(\frac{1}{\epsilon}+\log(\frac{\mu^2}{4\pi T^2e^{-\gamma_{E}}})\right)}{64\pi^2}+ \mathrm{O}({T^{-2}}) \notag\\ &\hspace{3cm} + \left(-\dfrac{T(m+V)^{3/2}}{12\pi} + \dfrac{T({\nabla}^2V)}{32\pi\sqrt{m^2+V}} + \dfrac{T(-4V{\nabla}^2V + {\nabla}^2V)}{192\pi(m^2+V)^{3/2}} + \cdots\right) \bigg{]}\notag \end{align} $$

and for fermions:

$$ \begin{align} &-\dfrac{T}{2}\sum_{n=-\infty}^{\infty} \mathrm{tr}\log(-{\nabla^2}+(2\pi (n+1/2) T)^2 + m^2 +V({x}))\\\ &\hspace{1cm}=\dfrac{T}{2}\sum_{n=\infty}^{\infty} \int d^{d}{x}\left(\dfrac{T}{4\pi}\right)^{d/2}\int_{0}^{\infty}dz\dfrac{K(z,{x},{x})}{z^{d/2+1}}\notag\\ &\hspace{1cm}\underset{d\to3-2\epsilon}{=} \int d^{3}{x}\bigg{[} -\frac{7\pi^2 T^4}{720} + \frac{T^2(m^2 + V)}{48} + \frac{\left(m^2 + V\right)^2 \left(\frac{1}{\epsilon}+\log(\frac{4e^{-\gamma_{E}}\mu^2}{\pi T^2})\right)}{64\pi^2}+\mathrm{O}({T^{-2}})\bigg{]}\notag \end{align} $$

It is useful to note that, for bosons, the term proportional to $(m^2+V)^2$ can be written as:

$$ \begin{align} &-\frac{\left(m^2 + V\right)^2 \left(\frac{1}{\epsilon}+\log(\frac{\mu^2}{4\pi T^2e^{-\gamma_{E}}})\right)}{64\pi^2}\\ &\qquad= \frac{\left(m^2 + V\right)^2}{64\pi^2}\left[-\left(\frac{1}{\epsilon}+\log(4\pi e^{-\gamma})\right)+2\left(\dfrac{3}{4}+\log(4\pi)-\gamma_{E}\right)+\left(\log\left(\dfrac{T^2}{\mu^2}\right)-\dfrac{3}{2}\right)\right]\notag \end{align} $$

where we’ve isolated the typical subtraction term (containing the $1/\epsilon + \log(4\pi e^{-\gamma_{E}})$) and the Coleman-Weinberg-like potential term (the $\log(T^2/\mu^2) - 3/2$). We can do the same for fermions, obtaining:

$$ \begin{align} &\frac{\left(m^2 + V\right)^2 \left(\frac{1}{\epsilon}+\log(\frac{4\mu^2e^{-\gamma_{E}}}{\pi T^2})\right)}{64\pi^2}\\ &\qquad= \frac{\left(m^2 + V\right)^2}{64\pi^2}\left[ \left(\frac{1}{\epsilon}+\log(4\pi e^{-\gamma})\right)-2\left(\dfrac{3}{4}+\log(\pi)-\gamma_{E}\right)-\left(\log\left(\dfrac{T^2}{\mu^2}\right)-\dfrac{3}{2}\right)\right]\notag \end{align} $$

Even though we stopped our expansion at order $T^{0}$, one can easily see how to include higher order term: simply cary out the expansion of the heat-kernel to higher orders in $1/T$. For expand, the next term in the non-zero mode bosonic expansion is:

$$ \begin{align} \dfrac{\zeta(3)}{768\pi^2 T^2}\left[\nabla^4V+2V\nabla^2V+(m^2+V)^3\right] \end{align} $$

and the next term in the fermionic expansion is the same thing multiplied by 7.