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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.

Differentiate the subtracted grand potential of the reduced BCS model with respect to the squared pairing amplitude. The displayed fermionic Matsubara sum converts the derivative of the trace logarithm into ${\sum }_k\tanh (\beta E_k/2)/(2E_k)$. Setting it equal to $V/\lambda$ gives the uniform nonzero-gap saddle. The limit E→0 is finite at positive temperature, $\beta /4$. The sum relies on antiperiodic temporal boundary conditions from the fermionic coherent-state trace.

Let $D_n={\hat{n}}_{n\uparrow }{\hat{n}}_{n\downarrow }$. On a site, empty and doubly occupied states have spin zero, while either singly occupied state has spin $1/2$. ThusThe on-site spin-square normal-ordering identity rewrites the interaction as $3U{\sum }_n{D}_n$. Its contact coefficient is $3U$ in this parameterization, rather than the $U$ convention often used for the Hubbard model.

Using a fermionic coherent-state trace and time slicing, replace each normal-ordered monomial by its Grassmann field symbol. With ${\xi }_{\mathbf{k}}={ϵ}_{\mathbf{k}}-\mu$,The independent fields have antiperiodic thermal boundary conditions. Normal ordering is crucial: the explicit $3\hat{n}/4$ subtraction removes the spin-square contraction, so there is no extra chemical-potential shift from it.

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

For one bosonic mode, define the unnormalized bosonic coherent state $|z⟩=e^{z{a}^{†}}|0⟩$. The commutator $[a,a^{†}]=1$ gives $a|z⟩=z|z⟩$ and $⟨z^{\prime }|z⟩=e^{{\bar{z}}^{\prime }z}$. The normalized state is $e^{-|z{|}^2/2}|z⟩$; it is not orthogonal to a different coherent state. The coherent-state resolution of identity follows from Gaussian moments,For many modes take products of these states and measures. The complex amplitudes have the natural scale of square root occupation; coherent states turn normal-ordered matrix elements into classical symbols without replacing the quantum trace by a single classical configuration.

For fermions, $\{c_i,c_j^{†}\}={\delta }_{ij}$ and a nilpotent annihilation operator cannot have an arbitrary nonzero ordinary complex eigenvalue. Introduce independent Grassmann variables ${\eta }_i,{\bar{\eta }}_i$, anticommuting among themselves and with the fermionic operators, and define $|\eta ⟩=\exp (-{\sum }_i{\eta }_i{c}_i^{†})|0⟩$. Then $c_i|\eta ⟩={\eta }_i|\eta ⟩$ and $⟨{\bar{\eta }}^{\prime }|\eta ⟩=e^{{\sum }_i{\bar{\eta }}_i^{\prime }{\eta }_i}$. With an ordered Berezin integral measure satisfying $\int d\bar{\eta }d\eta \bar{\eta }\eta =-1$ for one pair,for an even operator $O$; this is the fermionic coherent-state trace. In one mode the trace integrand is $e^{-\bar{\eta }\eta }(O_{00}-\bar{\eta }\eta O_{11})$, whose integral is $O_{00}+O_{11}$. Without the sign reversal it would be the supertrace, not the thermal trace. These formal Grassmann-valued coherent states do not represent ordinary complex-valued fermion amplitudes.

To construct the many-body thermal path integral, write $\mathcal{Z}=\mathrm{Tr}{e}^{-\beta K}$ with $K=H-\mu \hat{N}$ and divide the imaginary-time interval $\mathcal{T}=\beta \hslash$ into slices of width $\delta \tau$. Insert the appropriate coherent-state resolution of identity between factors. For a normal-ordered Hamiltonian, the adjacent-slice kernel iscoherent-state time slicing makes the overlap and Gaussian measure combine into the first-order time derivative. The grand-canonical coherent-state action in the regulated continuum limit isFor bosons the integration variables are complex, with the conjugate labels on the original integration contour; for fermions the barred and unbarred Grassmann variables are independent. The continuum notation retains the time-sliced ordering prescription. For example, $n(n-1)=a^{†2}{a}^2$ has normal symbol ${\bar{z}}^2{z}^2$, whereas $n^2$ has that symbol plus $\bar{z}z$. Dropping that distinction changes a theory. The coherent-state thermal boundary conditions follow because the bosonic trace closes paths periodically and the fermionic sign in the trace closes them antiperiodically:The barred fields obey the same respective boundary conditions. Hence Matsubara frequencies are $2\pi n/(\beta \hslash )$ and $(2n+1)\pi /(\beta \hslash )$. The derivative term encodes the canonical structure; omitting it would remove the quantum dynamics.^[ref]

As a check, a free diagonal mode with $K=\xi a^{†}a$ gives $Z_B=(1-e^{-\beta \xi }{)}^{-1}$ for $\xi >0$, whereas $K=\xi c^{†}c$ gives $Z_F=1+e^{-\beta \xi }$. Gaussian integration produces an inverse determinant for bosons and a determinant for independent complex fermions, with their respective temporal boundary conditions. Differentiating $\log Z$ recovers Bose-Einstein and Fermi-Dirac occupations. Interactions become higher powers in the normal-ordered symbols, and sources generate time-ordered correlations. The coherent-state construction preserves statistics, ordering and boundary conditions together; a formal continuum expression without its measure and temporal prescription is incomplete.