# Generating function for the zeta function of the Hamiltonian

1. Feb 10, 2015

### spaghetti3451

Given a Hamiltonian $H$, with a spectrum of eigenvalues $\lambda$, you can define
its zeta function as $\zeta_H(s) = tr \frac{1}{H^s} = \sum_{\lambda}^{} \frac{1}{\lambda^s}$.

Subsequently, the log determinant of $H$ with a spectral parameter $m^2$ acts as a generating function for the zeta functions:

$ln(\frac{det(H+m^2)}{det(H)}) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}m^{2n} \zeta_{H}(n)$.

I understand that the zeta function for the Hamiltonian is defined in analogy to the Riemann zeta function. However, I do not understand how the log determinant can be used as a generating function for the zeta functions.

What exactly is a generating function? Can somebody prove the second relation, please?

2. Feb 10, 2015

### DrDu

I would try to use the identity ln det=Tr ln.

3. Feb 10, 2015

### spaghetti3451

So far, I have gone through the following steps in proving the second relation:

$ln(\frac{det(H+m^2)}{det(H)})\\ = ln(det(H+m^2)) - ln(det(H))\\ = [-\zeta_{H+m^2}^{'}(0)] - [-\zeta_{H}^{'}(0)]\\ = -\zeta_{H+m^2}^{'}(0) +\zeta_{H}^{'}(0)\\ = -\frac{d}{ds}\zeta_{H+m^2}(s)|_{s=0}+\frac{d}{ds}\zeta_{H}(s)|_{s=0}\\ =-\frac{d}{ds}(tr\frac{1}{(H+m^2)^s})|_{s=0}+\frac{d}{ds}(tr\frac{1}{(H)^s})|_{s=0} \\= -\frac{d}{ds}(tr\frac{1}{(H+m^2)^s}-tr\frac{1}{(H)^s})|_{s=0}\\= -\frac{d}{ds}(tr(H+m^2)^{-s}-tr(H)^{-s})|_{s=0}\\=??\\=??\\=??\\=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}m^{2n} \zeta_{H}(n)$

I think I am supposed to Taylor-expand the expression around $m^2 = 0$ to obtain the RHS, but I am not sure how to proceed. Could someone please write the next line or two to show me the way?

Last edited: Feb 10, 2015
4. Feb 10, 2015

### DrDu

I think that's easier: $\ln \det(H+m^2)-\ln \det H=\mathrm{Tr}\ln ((H+m^2)/H)=\mathrm{Tr} \ln (1+m^2/H)$
Taylor expansion of your logarithm should give you what you want.

5. Feb 10, 2015

### spaghetti3451

Alright! Let me try to reproduce the solution according to your specification:

$ln(\frac{det(H+m^2)}{det(H)})\\=ln(det(H+m^2))-ln(det(H))\\=tr(ln(H+m^2))-tr(ln(H))\\=tr(ln(\frac{H+m^2}{H}))\\=tr(ln(1+\frac{m^2}{H}))\\=tr(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(\frac{m^2}{H})^{n})\\=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}m^{2n}tr(\frac{1}{H^n})\\=\\=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}m^{2n}\zeta_{H}(n)$

It's working fine, DrDu. Thank you! I hope there aren't any mistakes in my derivation.

6. Feb 10, 2015

### spaghetti3451

I have one other question.

The first and second zeta functions correspond to the leading two coefficients in the small $m^2$-expansion in the above relation (that gives the generating function for the zeta functions).

Therefore, they are simply determined by taking one and two derivatives of the functional determinant with respect to $m^2$ as

$\zeta_{H}^{L}(1)=+(\frac{\partial}{\partial m^2})^{1}ln(det(H+m^2))|_{m^{2}=0},\\ \zeta_{H}^{L}(2)=-(\frac{\partial}{\partial m^2})^{2}ln(det(H+m^2))|_{m^{2}=0}.$

I understand how the two equations come about when you differentiate both sides of the generating function relation with respect to $m^2$.
However, I do not understand why you need to take the resulting derivative to zero. Could you please help me out with it?

7. Feb 10, 2015

### DrDu

Come on, that's easy! Write down the derivative of the Taylor series explicitly.

8. Feb 10, 2015

### spaghetti3451

Indeed, it's easy! As you mentioned, if I write down the terms of the Taylor series (in the RHS of the generating function relation) explicitly, and then differentiate this series with respect to $m^2$, then all terms except the first term depend on $m$. To drop these other terms, we should take $m^{2}=0$.

I get it! :) Thank you!