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For a Hamiltonian like,

$$\hat{H}=\sum_{k}\hbar\omega_{k}b_{k}^{\dagger}b_{k}$$

What does it mean to say that the frequencies $\omega_{k}$ must be positive if $b_{k}$, $b_{k}^{\dagger}$ are boson operators for $\hat{H}$ to have a minimum? What would be different if these were fermion operators?

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The point is that the operator $b^\dagger_k b_k$ has only positive eigenvalues, so if one of the $\omega_k$ were negative, you could find an eigenstate which has eigenvalues of $b^\dagger_k b_k$ as large as you want and the corresponding term in the Hamiltonian would become lower and lower. This implies that there is no ground state for such a theory.

If the operators were fermion operators, then $b^\dagger_k b_k$ could only take the eigenvalues $1$ or $0$, so that would not be a problem anymore.

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  • $\begingroup$ Photons are bosons and their frequencies do not have to be positive. The hamiltonian should contain $\left| \omega_k \right|$. $\endgroup$
    – my2cts
    Jun 16, 2018 at 11:43

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