# Negative energies and a partition function

I'm writing down the partition function for a system, for which I know the dispersion relation

$$E \left( \mathbf{k} \right) = \sqrt{ \left| \mathbf{k} \right|^2 + m^2 + \cdots }$$

The exact form is not important, what matters is that technically, as the dispersion relation is the solution of a 2nd degree polynomial equation, there should be a $\pm$ sign before the square root.

When I write the partition function I get:

$$Z = \sum_{\mathbf{k}} \exp \left( - \frac{E \left( \mathbf{k} \right)}{\beta} \right)$$

but if I wanted to include the negative energies too, I would get:

$$Z = \sum_{\mathbf{k}} \exp \left( - \frac{E \left( \mathbf{k} \right)}{\beta} + \frac{E \left( \mathbf{k} \right)}{\beta} \right) = \sum_{\mathbf{k}} 1$$

which is clearly absurd: my system has dynamics! ;-) Now my question is: are the negative really unphysical? Wouldn't it be more correct to keep track of the two-branch dispersion relation with something along these lines:

$$Z = \sum_{\mathbf{k}} \exp \left( - 2 \frac{E \left( \mathbf{k} \right)}{\beta} \right)$$

-
If you really demand to also add the negative energy states, why wouldn't the sum be $Z = \sum_{\mathbf{k}} \exp \left( - \frac{E \left( \mathbf{k} \right)}{\beta} \right) +\sum_{\mathbf{k}} \exp \left( \frac{E \left( \mathbf{k} \right)}{\beta} \right)=2\ \sum_{\mathbf{k}}\cosh \left( \frac{E \left( \mathbf{k} \right)}{\beta} \right)$? –  NikolajK Feb 13 '12 at 15:40

There are many problems here. First, one typically takes $\beta = 1/T$ and so you want a partition function like $$Z(\beta) = \sum_n \exp(-\beta E_n)\,.$$ The next technical problem is that $\exp(-\beta E_1) + \exp(-\beta E_2) \neq \exp[-\beta(E_1+E_2)]$ as you claim it does.
that is, assuming that the $\exp( - \beta E_{n} )$ factor is functionally unaltered in the negative energy region. I'm not so sure as you seem to be. –  lurscher Feb 13 '12 at 17:05