2
$\begingroup$

I have encountered a statement which I don't understand: When the energy spectrum is continuous above a certain minimum energy, the partition function $Z$ diverges for all $Re(\beta)>0$, where

$Z=Tr{e^{-\beta H}}$.

For example, this is the main argument used in https://arxiv.org/abs/0712.0155 in calculating the partition function in $AdS_3$ gravity. (In particular see P. 21)

Naively, let's say the density of states $g(E)$ is a power law, $g(E)=E^a$, then

$Z=\int_0^\infty dE g(E)e^{-\beta E}$ is finite.

I know why $g(E)$ (and its integral) has to grow slower than exponential, but I don't see why the energy spectrum must be discrete.

So my first question is: why must the energy spectrum must be discrete for a finite $Z$?

Second question, in the paper mentioned above, the authors claim that the partition function they obtained for AdS3 gravity (P. 21 eq. 2.31) is unphysical because the tail goes like $1/\beta^s$, but I don't understand why. After all a power law $g(E)$ would give me sth like this.

Maybe I am missing sth trivial in my thinking, but I hope someone can enlighten me. Thanks!

$\endgroup$

1 Answer 1

4
$\begingroup$

When Hamiltonian spectrum is continuous in some interval, the weight function $g(E)$ has to be infinite there, as there is infinity of eigenvalues in arbitrarily small interval.

This is a consequence of how partition function is defined in quantum statistical physics - sum of $e^{-\beta E_i}$ for every linearly independent eigenfunction $i$ of the Hamiltonian. Finite weight is given to every single possible function. This works well if the spectrum is discrete, but fails when it is continuous.

In classical statistical physics, partition function is defined differently - as integral of $e^{-\beta E_i}$ over space of possible states $i$. Finite weight is given not to single possible state, but to set of them that has finite measure. That's why partition sum in classical statistical physics can be finite even if calculated over states that have all energies from some interval.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.