Integrated kernel and spectral zeta function

[grilo]

I was looking at a paper about strong-coupling expansion (N. F. Svaiter, Physica (Amsterdam) 345A, 517 (2005) ) and it claims that
[tex] -\int d^d x \int d^d y (-\Delta + m^2)\delta^d(x-y) = \textbf{Tr} I + \left.\frac{d}{ds}\zeta(s)\right|_{s=0} [/tex]
where [tex]\zeta(s)[/tex] is the spectral zeta function, and [tex]I[/tex] is the identity matrix.

It is clear to me that the derivative of the zeta function is related to the logarithm of the determinant of the operator in the left-hand side. What is *not* clear is how that double integral is related to that.

 

[eljose79]

Thank you for bringing up this interesting paper on strong-coupling expansion. After carefully reviewing the equation you mentioned, I can provide some insights to help clarify the relationship between the double integral and the logarithm of the determinant.

Firstly, the double integral on the left-hand side represents the Green's function of the operator (-\Delta + m^2) in d-dimensional space, where \delta^d(x-y) is the Dirac delta function. This Green's function is commonly used to solve differential equations in physics and is related to the spectral zeta function \zeta(s) through the formula:

-\int d^d x \int d^d y (-\Delta + m^2)\delta^d(x-y) = \int_0^\infty dt\, \mathrm{e}^{-tm^2} \mathrm{Tr}\left[ \mathrm{e}^{-t(-\Delta)}\right]

This formula can be found in the paper by N. F. Svaiter (Eq. 3.5) and is derived using the heat kernel method. The right-hand side of the equation is the trace of the heat kernel, which is related to the spectral zeta function through the Mellin transform.

Now, to understand the relationship between the double integral and the logarithm of the determinant, we can use the fact that the logarithm of the determinant of an operator can be expressed as the integral of its trace over the complex plane. This is known as the Witten index and is given by:

\log \det (-\Delta + m^2) = \int_0^\infty \frac{dt}{t} \mathrm{Tr}\left[\mathrm{e}^{-t(-\Delta + m^2)}\right]

Comparing this equation with the previous one, we can see that the double integral on the left-hand side is equivalent to the logarithm of the determinant of the operator (-\Delta + m^2).

In summary, the double integral in the equation you mentioned is related to the Green's function of the operator (-\Delta + m^2), which is in turn related to the logarithm of the determinant through the Witten index. I hope this helps clarify the relationship between these quantities.