# Eigenvalue counting number in Functional Integral

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.

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 $$\int_0^\Lambda \rho(\lambda) d\lambda$$
• 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. – TJNY699 Mar 3 '19 at 18:33