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I'm dealing with a quantum gas, thought as a system of N non-interacting particles.

I would be tempted to say that the total energy of the system equals the sum of the energies of the single particles, as they do not interact, but then I thought that some kind of field may be stepping in, and that there may be some more additional energy stored in the field.

Is this correct? That is, is my assumption valid just where there aren't magnetic or electric field messing things up?

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  • $\begingroup$ Classically it might be true, but quantum mechanically, it is not so. There are quantum interactions even in an Ideal quantum gas. Refer to some Quantum statistics, you might understand it. $\endgroup$ – user35952 Mar 23 '14 at 15:15
  • $\begingroup$ Prof. David Tong's lectures notes on statistical physics (damtp.cam.ac.uk/user/tong/statphys/sp.pdf) offer a good introduction to quantum gases (see section 3). $\endgroup$ – JamalS Mar 23 '14 at 15:25
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If you are really dealing with N non-interacting particles, the total energy is just the sum of the individuals'. You can see this by writing the Hamiltonian of the system as

\begin{equation} H=\sum_{i=1}^N\epsilon_i c_i^\dagger c_i \end{equation} where $\epsilon_i$ is the energy of state $i$ and $c_i$ is the annihilation operator of state $i$.

For both bosons and fermions, $[c_i^\dagger c_i,c_j^\dagger c_j]=0$ for $i\neq j$, then you can diagonalize the Hamiltonian by diagonalizing each term in the summation separately, which simply means the total energy is just the sum of the individuals'.

However, if you are dealing with interacting quantum gases, the problem immediately becomes complicated.

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if you are dealing with N non-interacting particles, the total energy is just the sum of the individuals. However there is a catch because all these particles are identical you can exchange 2 of them and still end up with the same physical state. Therefore in statistical physics you need factor that in. $$E =\frac{1}{N!} \sum^N_{i=1} E_i$$ see also Mixing paradox

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