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Suppose that I have a system described by a quantum field theory with Hamiltonian $H$ and $U(1)$ charge $Q$. The thermal density matrix is then $$ \rho = \frac{1}{Z} e^{-(H-\mu Q)/T},\quad Z=Tr(e^{-(H-\mu Q)/T}). $$ In the absence of a chemical potential ($\mu=0$), this density matrix is clearly normalizable (assuming we regularize the system e.g. put it in a finite volume) so long as $H$ is bounded from below. But $$ Q = \text{(number of particles)} - \text{(number of anti-particles)}, $$ meaning that $Q$ is not bounded. Now it is much less obvious that $H - \mu Q$ is bounded, yet we typically just assume it is. Note that if it is not bounded, thermal equilibrium does not exist.

Question: Why is it legitimate to assume $H-\mu Q$ is bounded? Are there certain reasonable conditions that $H$ and $Q$ are expected to satisfy that make this bounded?

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  • $\begingroup$ Why not $Z = \text{Tr}~ e^{-\frac{H - \mu Q}{kT}} = e^{\frac{\mu Q}{kT}} ~\text{Tr}~e^{-\frac{H}{kT}}$ ? $\endgroup$ Commented Mar 13, 2023 at 17:51
  • $\begingroup$ $Q$ is a quantum operator, so it can't be taken out of the trace $\endgroup$
    – user105620
    Commented Mar 14, 2023 at 14:47

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