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The ordinary trace of an even fermionic operator requires $-\psi$ in the bra of its fermionic coherent state kernel. In one mode the diagonal matrix elements give $⟨-\psi |A|\psi ⟩=A_{00}-\bar{\psi }\psi A_{11}$, whose Gaussian Berezin integral is $A_{00}+A_{11}$. An untwisted bra gives $A_{00}-A_{11}$, the supertrace. This distinction produces the antiperiodic coherent-state thermal boundary conditions.

Let $K=H-\mu N$, take $M$ slices of width $ϵ=\beta /M$, and insert the coherent-state resolution of identity between factors of $e^{-ϵK}$. If $H$ is in normal ordering, each short-time matrix element isCombining the overlap with the Gaussian weight leaves ${\sum }_{j,n}{\bar{\psi }}_{n,j}({\psi }_{n,j}-{\psi }_{n,j-1})$ in the action. The coherent-state time slicing prescription fixes which adjacent labels appear in $H$, rather than allowing an arbitrary ordering change after taking the continuum limit.

The twist $\zeta$ in the thermal trace closes the path with the coherent-state thermal boundary conditions: periodic for bosons, antiperiodic for fermions. Taking the regulated continuum limit givesThe action is dimensionless because $\tau$ has inverse-energy units. For fermions, $\bar{\psi }$ and $\psi$ remain independent Grassmann fields. If the original Hamiltonian is not normally ordered, first express it in normal order, retaining all constants.