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Insert the coherent-state resolution of identity between short-time evolution operators. For a normally ordered Hamiltonian, the kernel uses the next bra label and current ket label: $⟨{\psi }_{j+1}|e^{-ϵH}|{\psi }_j⟩=\exp ({\bar{\psi }}_{j+1}{\psi }_j-ϵH({\bar{\psi }}_{j+1},{\psi }_j)+O({ϵ}^2))$. Together with the Gaussian measure this yields the first-order derivative term $\bar{\psi }{\partial }_{\tau }\psi$. The adjacent-label prescription retains operator ordering information that a formal continuum expression alone can hide.

Use natural units $\hslash =k_B=1$ for the action conventions of this question. Insert the coherent-state resolution of identity between imaginary-time steps and take the thermal trace. Coherent-state time slicing gives the normally ordered interaction and first-order time term:The fields obey the coherent-state thermal boundary conditions $\psi (\beta ,\mathbf{r})=\psi (0,\mathbf{r})$ and $\bar{\psi }(\beta ,\mathbf{r})=\bar{\psi }(0,\mathbf{r})$. Bosonic Matsubara frequencies are therefore $2\pi n/\beta$. A periodic spatial box can be chosen for the homogeneous bulk calculation; temporal periodicity follows from the trace independently of that spatial choice. The barred and unbarred labels belong to the coherent-state integral prescription, with the usual conjugate contour for bosonic Gaussian integral.

Write $O=\int d[\bar{\psi },\psi ]e^{-\sum \bar{\psi }\psi }|\psi ⟩⟨\psi |$. For bosons the measure is ${\prod }_{n}d\mathrm{Re}{\psi }_{n}d\mathrm{Im}{\psi }_n/\pi$ over $\mathbb{C}$ in each mode, with ${\bar{\psi }}_n={\psi }_n^{\ast }$. For fermions it is an ordered Berezin integral over independent ${\bar{\psi }}_n,{\psi }_n$; choose $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$, so $\int d\bar{\psi }d\psi e^{-\bar{\psi }\psi }=1$.

For one bosonic mode, $a|\psi ⟩=\psi |\psi ⟩$ and $⟨\psi |a={\partial }_{\bar{\psi }}⟨\psi |$. Integration by parts in the Gaussian measure gives $aO=Oa$; the conjugate argument gives $a^{†}O=O{a}^{†}$. Boundary terms vanish because of the Gaussian weight.

For one fermionic mode, put $n=a^{†}a$ and move Grassmann coefficients to the left. The weighted projector isIts ordinary commutators are $[a,W]=-a+\psi I$ and $[a^{†},W]=a^{†}-\bar{\psi }I$. Their Berezin integrals vanish, so again $[a,O]=[a^{†},O]=0$. Equivalently the sole surviving coefficient in $\int W$ is $-\bar{\psi }\psi I$, giving $I$ directly.

The modes factorize. In the irreducible Fock space representation, commuting with every creation and annihilation operator makes $O$ a scalar multiple of the identity. Its vacuum matrix element is the normalized Gaussian integral, equal to one. Thus the coherent-state resolution of identity isFor infinitely many modes, this argument first uses a finite-mode regulator.

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

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.

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.

An unnormalized bosonic coherent state is $|\psi ⟩=e^{\psi a^{†}}|0⟩$ for a complex label $\psi$. The canonical commutation relation gives $a|\psi ⟩=\psi |\psi ⟩$ and overlap $⟨\psi |{\psi }^{\prime }⟩=e^{\bar{\psi }{\psi }^{\prime }}$. Multiplication by $e^{-|\psi {|}^2/2}$ produces a normalized coherent state. The unnormalized form makes the coherent-state resolution of identity and thermal time slicing particularly simple.