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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?

nightmarish
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1 Answers1

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see here.

When using the definition of the functional determinant you arrive at

$ln(\frac{det(H+m)}{det(H)})=-\frac{d}{da} (\zeta_1 - \zeta_2)|_{a=0}$

with $\zeta_1(s)=tr(H+m)^{-a}$ and $\zeta_2(s)=trH^{-a}$. Now you make a taylor series expansion around $m=0$ and you get your series when you set $\zeta_H(n)=trH^{-n}$.

A generating function of a function $f$ is a function $F$ where the series expansion leads to coefficient that represents another function $f$; here the zeta functions are the coefficients of your series expansion.

kryomaxim
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  • Thank you! Don't you mean $ln(\frac{det(H+m^2)}{det(H)})$ in the first line, i.e. $m^2$ in place of $m$? Does that mean you should taylor expand about $m^{2}=0$ instead of $m=0$? – nightmarish Feb 10 '15 at 12:18
  • yes, sorry I meant $H+m^2$ instead of $H+m$. – kryomaxim Feb 10 '15 at 12:20
  • Thank you so much! I have finally understood what generating functions actually mean! Now, I think I need to Taylor expand. But, I am not sure what to taylor expand or how to taylor expand. If you could just provide a couple of steps, that would be very helpful. – nightmarish Feb 10 '15 at 12:29