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The thermal trace closes bosonic coherent-state paths periodically and fermionic coherent-state paths antiperiodically: $\psi (\beta )=\zeta \psi (0)$ and $\bar{\psi }(\beta )=\zeta \bar{\psi }(0)$, with $\zeta =1$ for bosons and $-1$ for fermions. The minus sign follows from the fermionic coherent-state trace, not from choosing a different Hamiltonian. These boundary conditions determine the allowed Matsubara frequencies.

The bosonic coherent-state resolution of identity gives the ordinary trace by insertion into a number-state basis. For fermions, the fermionic coherent-state trace requires a sign in the bra. In one mode, for an even operator $A$,Without that sign the result would be the supertrace $A_{00}-A_{11}$. Applying this identity mode by mode to the even operator $A=e^{-\beta (H-\mu N)}$ gives the grand canonical partition function