How can one prove that $$ (\log\det\cal A=) \operatorname{Tr} \log \cal{A} = \int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s \mathcal{A}},$$ for a sufficiently well-behaved operator $\cal{A}?$ How mathematically rigorous is the expression?

I'm looking at the $d=2$ Euclidean case, as discussed for $\cal{A}=-\Delta + m^2$ in paragraph 32.2.1 of the book Mirror Symmetry by Vafa et al.

  • $\begingroup$ Looks an awful lot like a Laplace transform. $\endgroup$
    – John
    Feb 11, 2014 at 17:41

5 Answers 5


First of all, I think there should be a minus sign in front of $\Delta$ : $A= (- \Delta + m^2)$ so that it is positive.

Assume that $A$ has pure point spectrum as it happens for the operator you wrote in compact manifolds with dimension $n$ (there is a unique self-adjoint extension of the said $A$ when defining the initial domain to be the space of $C^\infty$ functions on that manifold). It is possible to prove that the eigenvalues diverge as fast as pictured by a known asymptotic formula due to Weyl: $$\lim_{j \to +\infty}\lambda_j^{n/2}/j = C_n >0\qquad (1)$$ where eigenvalues are counted taking their (finite) multiplicity into account: $$0 \leq ... \leq \lambda_j \leq \lambda_{j+1} \leq ...\to +\infty$$ So $e^{-tA}$ is trace class because has non-negative eigenvalues $e^{-t\lambda_j}$ with finite multiplicity such that $\sum_j e^{-t\lambda_j} < +\infty$ as you can prove form (1). $$\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s A} = \int_{\epsilon}^\infty \frac{{d}s}{s}\sum_j e^{-s\lambda_j} = \sum_j \int_{\epsilon}^\infty \frac{{d}s}{s} e^{-s\lambda_j} $$ I have swapped the symbol of integral and that of sum, because, once-again (1) easily implies that the function $(j,s) \mapsto e^{-s\lambda_j}/s$ is (absolutely) integrable in the product measure, so I could exploit Fubini-Tonelli theorem. Since: $$\ln \lambda = \lim_{\epsilon \to 0^+}\left(\int_\epsilon^{+\infty} \frac{e^{-\lambda t}}{t} dt + (\gamma -\ln \epsilon)\right)$$ we can write: $$\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s A} = \sum_j \ln \lambda_j + \sum_j O(\ln \epsilon)\:.$$ Up to a divergent part one has to renormalize, the found result can be re-written as: $$\int_{0^+}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s A} = tr \ln A$$ Actually, the point is that $tr\ln A$ is not defined,for $A= (-\Delta +m^2)$ and needs to be regularized.

  • $\begingroup$ I had to search a while to find the relation between $\ln\lambda$ and the integral - it's not even on the lengthy Wikipedia page on the incomplete Gamma function. Do you use the Exponential integral $\mathrm{Ei}$ often? $\endgroup$
    – Nikolaj-K
    Feb 11, 2014 at 20:41
  • $\begingroup$ Actually in the past I made research in heat kernel theory, zeta function regularization and all that and that integral relation I used was quite standard, I just searched through some old paper of mine containing these formulas. Yes, I remember that there is some relation with $Ei$... $\endgroup$
    – V. Moretti
    Feb 11, 2014 at 20:55
  • $\begingroup$ You're right, there is a small typo in the book: the covariant derivative should be $D_\mu=\partial_\mu -i e A_\mu$ such that $A_\mu$ is real, then the operator we get is $\phi^* \cal{A} \phi = - \phi^* D_\mu D_\mu \phi,$ which as you correctly pointed out is the negative of the laplacian, and has thus positive eigenvalues. Thanks for the nice answer $\endgroup$
    – jj_p
    Feb 12, 2014 at 11:12

Here is an absolutely non-rigorous start of an explanation.

Assume that all the eigenvalues $a_i$ of $\mathcal A$ are positive (which is the case of the inverse propagator you give), then $$\int_\epsilon^\infty \frac{ds}{s}\operatorname{Tr} e^{-s\mathcal A}=\sum_i\int_\epsilon^\infty \frac{ds}{s}e^{-s a_i}=-\sum_i Ei(-\epsilon a_i),$$ where $Ei(x)$ is the Exponential integral.

Now, in the limit $\epsilon \to 0$ (that is assumed here ?), we get $$-\sum_i Ei(-\epsilon a_i)=div.-\sum_i \ln a_i, $$ where $div.=-\sum_i (\gamma+\ln \epsilon)$ is a divergent contribution in the limit $\epsilon\to0$ ($\gamma$ is the Euler constant). So, up to a sign and an infinite constant, we're good...

  • $\begingroup$ You could clean this up by noting that $e^{-s a}/s$ is $\int_{0}^a e^{-s a} - 1$, then flipping the order of integration. This gets rid of the Exponential integral, which I always like to do, and it involves talking an integral with respect to a constant, which I can't resist. Also should note that you don't need eigenvalues, just the diagonal elements $a_{nn} = \langle n | \mathcal{A} |n \rangle$, so you probably don't even need $\mathcal{A}$ to have eigenvalues/be Hermitian/whatever. $\endgroup$
    – webb
    Feb 11, 2014 at 18:44
  • $\begingroup$ @webb: I don't really see the point of doing that, how does it help ? And I'm pretty sure that the eigenvalues are needed, or you have to tell me how you define the log of a matrix. $\endgroup$
    – Adam
    Feb 11, 2014 at 19:03

Hint: Try a Laplace Transformation.


Usually one starts these things by using $\operatorname{tr} \ln A = \ln\, \operatorname{det} A$.


@adam: "you have to tell me how you define the log of a matrix"

you define log of a matrix the same way you define, for instance, its exponential - via Taylor expansion, when it exists. Now, the nice thing about that is you get terms that are powers of the matrix (remember, trace of a sum of terms is the sum of traces). For these, if the matrix can be written as $A = UVU^{-1}$ you end up with $Tr(A^n) = Tr(UVU^{-1}\ldots UVU^{-1}) = Tr(UV^nU^{-1}) = Tr(V^n)$ where $V$ is diagonal (hopefully positive definite for the log to exist). This will clearly mean that $Tr(\log(A)) = Tr(\log(V))$ and since $V = diag\{V_i|i\in \overline{1,n}\}$ is diagonal it gives $Tr(\log(V)) = Tr(diag\{\log(V_i)\}) = \Sigma_i \log(V_i) = \log(\Pi_i V_i) = \log(\det(V)) = \log(\det(A))$


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