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.

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    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
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    I scanned through that paper and also couldn't find anything relevant. You might want to ask this on http://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 Answers1


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

Tom Copeland
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    Could you be more specific? I don't see where that answers the question. – joriki Nov 24 '11 at 05:59
  • 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 07:03
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    I think the emphasis there is on "sketch heuristically". I don't see anything that justifies this connection in any more rigour than what anon had already stated. Could you point to something more specific plesase? – joriki Nov 24 '11 at 07:19
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    Maybe you can state clearly what the question is that Anon asked. I gleaned from what he says that he wants some intuition on the realtionship, which I thought might be provided by Cartier's paper in which Cartier clearly states in the last few sentences at the end that there is a lack of rigor and the relationship is under active investigation. I certainly don't pretend that I can answer the question myself, so I suggested the paper and will leave it up to Anon to judge whether it is helpful to him. – Tom Copeland Nov 24 '11 at 08:24
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    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 08:36
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    Perhaps http://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'm not sure the OP is notified when you write a comment under an answer without pinging her or him -- you might want to write this as a comment to the question. – joriki Dec 11 '11 at 11:16
  • I couldn't find a comment button for the question--one reason I posted an answer rather than a comment initially. – Tom Copeland Dec 11 '11 at 11:48
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    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