My question is about the calculation of a functional integral (which looks like a partition function).

If we have the operator $A$ having discrete spectrum, and eigenvectors $\phi_{i}$ and eigenvalues $\lambda_i$, and the eigenvalues have density $\rho(\lambda)=\sum_i \delta(\lambda_i-\lambda)$. Then the functional integral $\int D\phi \exp\{-(\phi,A\phi)\}$ is written as integral over the "fourier" coefficients $a_i = (\phi_i,\phi)$, where we use $\phi = \sum_{i} a_i \phi_i$:

$\int D\phi \exp\{-(\phi,A\phi)\} = \int_{-\infty}^{\infty}\prod_i da_i \exp\{-\lambda a_i^2\}$

This is product of gaussian integrals, so if we put "UV" cutoff on eigenvalue, $\Lambda$, then I think the result of this integral is the product of the inverse square root of eigenvalues:

$ \int^{\Lambda}\prod_i da_i \exp\{-\lambda a_i^2\}= \prod_i^{\Lambda}\lambda_i^{-1/2}$

However, my TA wrote without deriving that the correct result is this:

$N(\Lambda) \ e^{-1/2 \sum_\lambda^{\Lambda}\ln \lambda}$

where $N(\Lambda) = \int \rho(\lambda) \ d\lambda$ is the number of eigenvalues below $\Lambda$.

I don't understand this result, especially this appearance of this counting number $N(\Lambda)$. Why does the functional integral not give the product of eigenvalues, and where is this $N(\Lambda)$ coming from? Is this related to the density of states that is sometimes written in a partition function? I would appreciate any help in understanding this, and any pointer to books that explain as well.


1 Answer 1


I think you TA is either wrong, or you misunderstood what he wrote. You are correct up the factor of $(\sqrt \pi)^n$ where $n=\int_0^\Lambda \rho(\lambda) d\lambda$ from the $n$ Gaussian integrals. I expect that the TA just meant that there is some normalization factor $N(\Lambda)$ that depends on $\Lambda$, but that factor is the one I gave above, and not $\int_0^\Lambda \rho(\lambda) d\lambda$

  • $\begingroup$ Thank you for your reply. I think it is likely that yes the factor $N(\Lambda)$ is not the integral of the density of eigenvalue. Otherwise it doesn't make sense. In general however, when a sum is transformed into integral over eigenvalues using $\rho$, then the measure will be $\int \rho(\lambda) d\lambda (...)$, right? This means that the integral depends on $N(\Lambda)$ correct? Thank you again. $\endgroup$
    – TJNY699
    Mar 3, 2019 at 18:33
  • $\begingroup$ @TJNY699 The integral depends on the normalization, yes. But the correlators etc. do not, as it cancels out when you divide by the partition function. $\endgroup$
    – mike stone
    Mar 3, 2019 at 21:03

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