Zeta regularisation voodoo
August 13, 2019. A technical post on path integrals, zeta function regularisation of determinants, and an application to hot oscillators.
Motivation
Prerequisites: quantum field theory, path integrals.
In physics, mathematics, and life in general, we often need to evaluate Gaussian integrals. The simplest integral is onedimensional:
\[\int_{\infty}^\infty \frac{dx}{\sqrt{2\pi}} \, e^{\tfrac{1}{2}\alpha x^2} = \frac{1}{\sqrt{\alpha}}\]We have written an extra factor of $1/\sqrt{2\pi}$ in the measure $dx$ for reasons that will become clear in a moment. In $n$ dimensions, we can replace $\alpha x^2$ with any bilinear $x_i A_{ij} x_j$, where $A_{ij}$ is a diagonalisable matrix. By going to the eigenbasis of $A$, the onedimensional result immediately generalises to
\[\int_{\infty}^\infty \frac{dx}{(2\pi)^{n/2}} \,e^{x_i A_{ij} x_j} = \frac{1}{\sqrt{\det A}} = \prod_i \lambda_i^{1/2},\]where the $\lambda_i$ are eigenvalues of $A$. In quantum field theory, we take things even further, and evaluate the integral for an infinitedimensional vector space. In this case, we replace the discretely indexed vector $x_i$ with a continously indexed function $x(\tau)$, for say $\tau \in [0, L]$, and the integral over all vectors $x_i$ becomes an integral over all paths $x(\tau)$. This is called a path integral.
Let’s define the integral measure for the path as the limit of the discretisations of the path $x(\tau)$ into chunks evaluated at $\tau_n := nL/N$, or
\[\mathcal{D}x(\tau) := \lim_{N\to\infty}(2\pi)^{N/2}\prod_{n=1}^N dx(\tau_n).\]We might cross our fingers and hope that the infinitedimensional Gaussian path integral, for a bilinear operator $A(\tau, \tau’)$, has exactly the same expression as above:
\[\int \mathcal{D}x \, \exp\left[\int_0^L d\tau\, d\tau'\, x(\tau)A(\tau,\tau') x(\tau')\right] = \frac{1}{\sqrt{\det A}} = \prod_i \lambda_i^{1/2}.\]We now see why we shifted the $\sqrt{2\pi}$ onto the left: in the infinitedimensional limit, it is a nuisance factor which blows up unless absorbed into the path integral. But we can still get infinities! In general, the infinite product of eigenvalue $\lambda_i$ will not converge.
But quantum physicists are not cowed by infinities. The usual trick is to regulate the divergence in some way, essentially, finding some way to smoothly ignore the behaviour of very large eigenvalues which are typically nonphysical (i.e. they occur where we expect the model, or the experiment, to break down). Here, we will consider a class of infinite products,
\[\det \mathcal{X}^a_{\xi} = \prod_{k=a} \xi (k+a),\]and show how to regulate them using a generalisation of the Riemann zeta function. We finish with some simple applications to quantum statistical mechanics.
The maths
Define the Hurwitz function by
\[\zeta(s, a) = \sum_{k \geq 0}\frac{1}{(k+a)^s}, \quad a > 0, \quad \Re(s) > 1.\]This is a generalisation of the Riemann zeta function $\zeta_\text{R}$, since $\zeta_\text{R}(s) = \zeta(s, 1)$. Although the definition of the Hurwitz function above blows up at $s = 0$, we can analytically continue so that it is defined at $s = 0$. From the DLMF (or otherwise), it obeys
\[\begin{align} \zeta(0, a)= \frac{1}{2}  a, \quad \zeta'(0, a) = \ln \left(\frac{\Gamma(a)}{\sqrt{2\pi}}\right),\label{zeta0} \end{align}\]where $\Gamma$ is the Gamma function. Consider an operator $\mathcal{X}^a_\xi$ with spectrum $\lambda_k = \xi(k+a)$, $k \in \mathbb{Z}_{\geq 0}$. We can define the associated spectral zeta function
\[\begin{align} \zeta_\mathcal{X}(s) & = \sum_{k \geq 0} \lambda_k^{s}\label{zeta1}\\ & = \sum_{k \geq 0}\frac{1}{\xi^s(k+s)^s} = \xi^{s}\zeta(s, a). \label{zeta2} \end{align}\]We can differentiate (\ref{zeta1}), (\ref{zeta2}) with respect to $s$, and equate them to get
\[\begin{align} \zeta'_\mathcal{X}(s) & = \sum_{k \geq 0} \lambda_k^{s}\log\lambda_k \label{zeta3}\\ & = \xi^{s}\zeta'(s,a)  \xi^{s}\log \xi \cdot \zeta(s, a). \label{zeta4} \end{align}\]From (\ref{zeta3}) and the elementary identity
\[\det A = \prod_i a_i = e^{\sum_i \log a_i} = e^{\mathrm{Tr} \log A},\]where $a_i$ are the eigenvalues of $A$, we also have
\[\begin{equation} \zeta'_\mathcal{X}(0) = \sum_{k \geq 0} \log \lambda_k = \log \det \mathcal{X}. \label{zeta5} \end{equation}\]Setting $s = 0$ in (\ref{zeta4}), using (\ref{zeta0}), and equating with (\ref{zeta5}), we find
\[\begin{align} \det \mathcal{X}^a_{\xi} & = \prod_{k=a} \xi (k+a) \notag \\& = \exp \left[\zeta'(0, a)  \log \xi \cdot \zeta(0,a)\right] \notag \\ & = \exp \left[\ln \left(\frac{\Gamma(a)}{\sqrt{2\pi}}\right) +\log \xi\left(\frac{1}{2}  a\right) \right] \notag \\ & = \frac{\sqrt{2\pi}}{\Gamma(a) }\xi^{1/2a}. \label{final} \end{align}\]This is our nice, simple final result!
Hot harmonic oscillators
A simple example is a single boson in a thermal state. The physics is governed by the partition function for the system,
\[Z[\beta] = \mathrm{Tr}[e^{\beta \hat{H}}].\]One can show that this is equivalent to a weighted sum over paths which are periodic in imaginary time,
\[Z[\beta] = \int_{x(0)=x(\beta)} \mathcal{D}x(\tau) \, e^{S_E[x(\tau)]}\]where $S_E$ is the action in Euclidean signature. To specialise, we can consider the simplest case of a bosonic harmonic oscillator, with
\[S_E[x(\tau)] = \int_0^\beta d\tau\, \frac{1}{2}\left(\dot{x}(\tau)^2+\omega^2 x^2\right) = \int_0^\beta d\tau\, \frac{1}{2}x(\tau)\mathcal{J} x(\tau),\]where $\mathcal{J} =\partial_\tau^2 + \omega^2$ after integrating by parts. The partition function is a Gaussian:
\[Z[\beta] = \exp\left[\int_0^L d\lambda\, d\mu\, x(\lambda)\mathcal{J} x(\mu)\right] = \frac{1}{\sqrt{\det \mathcal{J}}}.\]Let’s calculate the regulated determinant of the operator $\mathcal{J}$. To find the eigenvalues, we first determine the eigenfunctions $f$, recalling they are subject to the periodicity constraint $f(0) = f(\beta)$:
\[\begin{align} \mathcal{J} f(\tau) & = \ddot{f} + \omega^2 f = \lambda f, \quad f(0) = f(\beta) \notag \\ \Longrightarrow \quad f_k(\tau) &= e^{2\pi k \tau/\beta}, \quad \lambda_k = \left(\frac{2\pi k}{\beta}\right)^2+\omega^2 \notag \end{align}\]for any integer $k$. But then
\[\begin{align} \det \mathcal{J} &= \prod_{k\in\mathbb{Z}} \lambda_k = \omega^2\prod_{k\geq 1}\left(\frac{2\pi k}{\beta}\right)^4 \cdot \prod_{k\geq 1}\left[1 + \left(\frac{\beta\omega}{2\pi k}\right)^2\right]^2,\notag \end{align}\]where we pick up the $\omega^2$ from the $k = 0$ term. The second product can be evaluated using the result from residue calculus,
\[\prod_{k\geq 1}\left[1 + \left(\frac{z}{n\pi}\right)^2\right] = \frac{\sinh z}{z}.\]The first product diverges, but can be evaluated by voodoo for $\xi = 2\pi/\beta$ and $a = 1$:
\[\prod_{k\geq 1}\left(\frac{2\pi k}{\beta}\right) = \det \mathcal{X}^1_{2\pi/\beta} = \sqrt{2\pi}\cdot \sqrt{\beta/2\pi} = \sqrt{\beta}.\]Combining these results, we find
\[Z[\beta] = \frac{1}{\sqrt{\det \mathcal{J}}} = \frac{1}{\beta\omega}\prod_{k\geq 1} \left[1 + \left(\frac{\beta\omega}{2\pi k}\right)^2\right]^{1} = \frac{1}{2\sinh (\beta\omega/2)}.\]Of course, it is much simpler to go the traditional route! We have
\[Z[\beta] = \sum_{n\geq 0} e^{\beta E_n} = e^{\beta\omega/2}\sum_{n\geq 0} e^{\beta \omega n} = \frac{e^{\beta\omega/2}}{1e^{\beta \omega}} = \frac{1}{2\sinh(\beta\omega/2)}.\]While this is not exactly an advertisment for path integrals, it is a small advertisment for voodoo. Usually, one has to insert factors by hand to get the Hamiltonian and path integral method to agree. Now the two approaches agree automatically!
Exercise 1. Hot fermionic oscillator.
A Fermionic oscillator has the “square root” Lagrangian
\[L_E = \bar{\psi}\mathcal{Q}\psi, \quad \mathcal{Q} = \partial_\tau + \omega,\]where $\bar{\psi}$ and $\psi$ are Grassman (anticommuting) classical coordinates for the fermion.

Starting with the onedimensional Gaussian integral for Grassman numbers $\bar{\theta}, \theta$,
\[\int d\bar{\theta} \, d\theta \, e^{\alpha \bar{\theta}\theta} = \alpha,\]argue that
\[Z[\beta] = \int \mathcal{D}\bar{\psi}(\tau)\, \mathcal{D}\psi(\tau)\, e^{S_E[\bar{\psi},\psi]} = \det\mathcal{Q}.\]  What are the boundary conditions for the fields $\bar{\psi}, \psi$? Hint. Consider two identical fermions on the thermal circle. What happens if they are exchanged?
 Find the eigenvalues of $\mathcal{Q}$, subject to the boundary conditions you found in the previous question, and write a product expression for $Z[\beta]$.

Simplify using voodoo and the product formula
\[\cosh \left(\frac{x}{2}\right) = \prod_{k \geq 0} \left[1+ \frac{x^2}{\pi^2(2n+1)^2}\right].\]  Check your result agrees with the fermionic oscillator Hamiltonian, which has energy eigenvalues $E_\pm = \pm \omega/2$. You should find in both cases that
Exercise 2. (For the author) Anyone for anyons?
Anyons are particles with fractional spin $a$. This means that after going around the thermal loop, they pick up a phase $e^{2\pi i a}$.

Determine the natural boundary conditions for spin $a$, and reproduce the result of BoschiFilho and Farina (1994) for the determinant of $\mathcal{J}$ with these boundary conditions,
\[\det\mathcal{J} = 4 \left[\cosh^2 \left(\frac{\beta\omega}{2}\right)  \sin^2(\pi a)\right]\]using voodoo rather than Green’s functions.

(Open) Determine whether the anyonic path integral can be written
\[Z_a[\beta] = [\det \mathcal{J}]^{p(a)}\]for some power $p(a)$. For bonus points, what is $p(a)$?
References
 Geometry, Topology and Physics (2003), Mikio Nakahara.
 An Introduction to Quantum Field Theory (1995), Michael Peskin and Daniel Schroeder.
 Quantum Field Theory and the Standard Model (2013), Matthew Schwartz.