# Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions.

The first definition is based on zeta function regularization. If an operator $S$ has the spectrum of eigenvalues $\{\lambda_i\}$ then the associated zeta function is formally the operator trace

$$\zeta_S(s)=\mathrm{tr}(S^{-s})=\sum_{i=1}^\infty \lambda_i^{-s}.$$

The sum converges only when the $\mathrm{Re}(s)$ is sufficiently large, so $\zeta_S$ is defined by analytic continuation elsewhere on $\mathbb{C}$. Formally, this means that (in symbolic appearance at least)

$$\det S =e^{-\zeta_S'(0)}=\prod_{i=1}^\infty\lambda_i.$$

Though this isn't literally convergent, it does establish an intuitive basis for why the quantity may be called a determinant through analogy with the case in finite dimensions.

On the other hand, the following path integral quantity is a second possible avenue to defining the determinant for suitable operators:

$$\frac{1}{\sqrt{\det S}} \propto\int e^{-\pi\langle \phi,S\phi\rangle}\, \mathcal D\phi.$$

In finite-dimensional Euclidean space, the proportion is an actual equality, which can be seen by writing the inner product as $\langle x, Sx\rangle=\lambda_1x_1^2+\cdots+\lambda_nx_n^2$, separating the integral and then observing that each factor is either $\int dx_i=\infty$ when $\lambda_i=0$ or a rescaled Gaussian integral otherwise. However, in the infinite-dimensional case we can only compare determinants of operators in relative proportion to each other, so that divergent constants cancel appropriately.

It is stated that the results of these two definitions agree with each other, and Wikipedia cites the paper Extremals of Determinants of Laplacians as having established this fact. However, of what little in the paper that I can genuinely follow, I don't see any demonstration that the zeta regularization and path integral formulation agree with each other, so either I'm so out of my depth I can't even recognize the proof let alone understand it, or the Wikipedia article is misguided.

The former is very much a possibility - I understand what manifolds are and can do some basic tensor manipulations to, say, derive the geodesic equation, but other than this I'm not educated in differential geometry, and I am likewise ignorant to all but the basic construction of a path integral. I'd be appreciative if someone could shine a light on the underlying theory in play here at a level I can understand, if possible.

• I retagged from (path-integrals) to (functional-integration) as the former can be misinterpreted to mean line-integrals in the context of multivariable calculus and complex analysis. – Willie Wong Jul 30 '11 at 13:58
• I scanned through that paper and also couldn't find anything relevant. You might want to ask this on theoreticalphysics.stackexchange.com, where people will probably be more familiar with functional determinants. – joriki Nov 23 '11 at 22:38
• Tom left a comment about an arXiv paper underneath his answer. – joriki Dec 11 '11 at 12:11

## 1 Answer

Try Pierre Cartier's Mathemagics (A Tribute to L. Euler and R. Feynman).

• Could you be more specific? I don't see where that answers the question. – joriki Nov 24 '11 at 5:59
• Anon states "I don't see any demonstration that the zeta regularization and path integral formulation agree with each other ....I'd be appreciative if someone could shine a light on the underlying theory in play here at a level I can understand, if possible." The paper I've suggested leads up to page 69 and 70 which sketch heuristically the connection between Feynman' functional integration and regularized determinants he presents on pages 67-59 related to zeta function regularization I believe. – Tom Copeland Nov 24 '11 at 7:03
• I see. Answers are usually used to answer the question completely, or at least in large parts, or as far as it is likely to be answered. If you merely wanted to point out a relevant paper without claiming to answer the question, a comment would have been the more usual avenue. – joriki Nov 24 '11 at 8:36
• Perhaps arxiv.org/abs/quant-ph/0011059 has the answers you are looking for: J. Casahorran, "Quantum-mechanical tunneling: differential operators, zeta-functions and determinants" – Tom Copeland Dec 11 '11 at 10:47
• I see, sorry, I keep forgetting there's a reputation threshold (of 50) for comments -- it doesn't apply to your own posts, that's why you can comment here but not under the question. I wrote a short comment under the question pointing to your comment. – joriki Dec 11 '11 at 12:12