Negative energies and a partition function

Question

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:

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

Answer 1

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.

These two issues aside, the resolution to your problem comes from noting that even though certain mathematical solutions exist to equations you write down, your job as a physicist is to rule out those that are unreasonable on physical grounds. Having a negative energy state of possibly arbitrarily low negative energy is not valid. So we only keep the positive energy states.