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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:

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

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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)$? –  NiftyKitty95 Feb 13 '12 at 15:40
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1 Answer

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.

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you are assuming that negative energy states decay to states of even lower negative energy. The few examples of negative energy systems that we have to study (casimir effect, unstable cavitation, etc) show us exactly the opposite: states with zero total energy are more stable that systems with negative energy. So you have to be careful what you mean here with physical grounds. –  lurscher Feb 13 '12 at 16:10
    
I was careful: the problem is specifically with "Having a negative energy state of possibly arbitrarily low negative energy." This is absolutely true. –  josh Feb 13 '12 at 16:48
    
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
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