Grassmann Functional Integrals for Free Fermions

At a finite regulator, a free fermion functional integral is an exact Berezin integral over finitely many independent odd generators. Once the order of those generators, their measure, the source terms, and the boundary problem are fixed, a paired Gaussian gives a determinant and its normalized source functional contains the inverse regulated Dirac kernel. Ordered odd source derivatives then reproduce the fermion two-point function, including the exchange sign that distinguishes it from a bosonic Gaussian.

The local differential expression is not enough. Vacuum, thermal, and graded traces impose different temporal data and therefore different finite matrices, determinants, and inverses. This page develops the physical free Dirac application entirely at finite regulator, checks it against canonical quantization, and stops before anomalous Jacobians, interacting loops, or an absolute continuum determinant.

Required background. Canonical Quantization of the Free Dirac Field supplies the normalized operator field, CAR, Minkowski vacuum, and particle/antiparticle contractions. Gaussian Fields and Sources supplies finite regulation, normalized generating functionals, inverse kernels, source differentiation, and the role of prescribed boundary data.

Helpful background. Exterior and Graded Algebra, Grassmann Variables, and Berezin Integration fixes the graded sign rule, left derivatives, ordered Berezin measures, determinant and Pfaffian identities, Jacobians, and the algebraic zero-mode cautions used below.

A finite Grassmann Gaussian produces a determinant

Work on a finite regulated index set $I=1,\ldots,N$. An index may combine a time slice, spatial site or mode, spinor component, and any finite internal label. Introduce independent odd generators

$$\psi_I,\quad \bar\psi_I,\quad \eta_I,\quad \bar\eta_I.$$

The bar on $\bar\psi$ and $\bar\eta$ is a label for the paired variables; it does not impose complex conjugation inside the Berezin algebra. Operator adjunction and physical reality enter through the construction that produced the finite kernel and its boundary states.

Inherit the metric, gamma matrices, Dirac adjoint, and Fourier phase from the chapter’s shared declarations. Locally, use left Grassmann derivatives and the paired measure

$$\mathcal D_R(\bar\psi,\psi) = d\bar\psi_N\,d\psi_N\cdots d\bar\psi_1\,d\psi_1,$$

normalized by

$$\int\mathcal D_R(\bar\psi,\psi)\, \psi_1\bar\psi_1\cdots\psi_N\bar\psi_N =1.$$

The subscript $R$ records the regulator and the ordered finite space. It is not shorthand for a formal continuum product measure.

Let $A$ be an invertible $N\times N$ matrix with commuting entries. With the source order shown explicitly, translation invariance and completion of the square give

$$\boxed{ \begin{aligned} \mathcal Z_E[\bar\eta,\eta;A] &= \int\mathcal D_R(\bar\psi,\psi) \exp\!\left( -\bar\psi A\psi +\bar\eta\psi +\bar\psi\eta \right) \\ &= \det A\, \exp\!\left( \bar\eta A^{-1}\eta \right). \end{aligned} }$$

Here and below repeated finite indices are summed. The algebraic step is

$$\begin{aligned} &-\bar\psi A\psi +\bar\eta\psi +\bar\psi\eta \\ &\quad= -(\bar\psi-\bar\eta A^{-1}) A (\psi-A^{-1}\eta) +\bar\eta A^{-1}\eta. \end{aligned}$$

Every exponential is a finite polynomial, so no convergence theorem is being used. Zinn-Justin develops the determinant, source completion, and ordered source-derivative signs in Zinn-Justin 2021, § 1.7, pp. 13–15, eqs. (1.68)–(1.85). Schwartz develops the finite Grassmann algebra and then applies it to the Dirac source integral in Schwartz 2014, § 14.6, pp. 270–272, eqs. (14.87)–(14.105); his implicit source derivative convention is not imported into the explicitly ordered package used here.

An independent convention round trip is useful. Srednicki places $d\chi$ before $d\bar\chi$ within each pair and writes a plus quadratic form. In Srednicki 2007, § 44, pp. 276–281, eqs. (44.3)–(44.40), set $M=-A$ and identify $(\chi,\bar\chi)$ with $(\psi,\bar\psi)$. The determinant contributes $\det(-A)=(-1)^N\det A$, while reversing the $N$ within-pair differentials contributes the second $(-1)^N$. They cancel and reproduce the site formula.

For $N=1$, the whole sign structure is visible:

$$\begin{aligned} \int d\bar\psi\,d\psi\, e^{-a\bar\psi\psi} &= \int d\bar\psi\,d\psi\, (1+a\psi\bar\psi) \\ &=a. \end{aligned}$$

This one-pair calculation is the quickest check after translating any external convention.

Ordered sources generate the inverse kernel

Divide by the zero-source value without changing the kernel, measure, or boundary problem:

$$Z_E[\bar\eta,\eta;A] \equiv \frac{\mathcal Z_E[\bar\eta,\eta;A]} {\mathcal Z_E[0,0;A]} = e^{\bar\eta A^{-1}\eta}.$$

With left derivatives, the two elementary insertion rules are

$$\begin{aligned} \frac{\partial^L\mathcal Z_E} {\partial\bar\eta_I} &= \int\mathcal D_R\, \psi_I e^{(\cdots)}, \\ \frac{\partial^L\mathcal Z_E} {\partial\eta_J} &= -\int\mathcal D_R\, \bar\psi_J e^{(\cdots)}. \end{aligned}$$

The second minus sign follows from $\partial^L_{\eta_J}(\bar\psi_I\eta_I)=-\bar\psi_J$. Keeping the order of the desired insertion gives

$$\begin{aligned} \left\langle\psi_I\bar\psi_J\right\rangle_E &= \left. \frac{\partial^L}{\partial\eta_J} \frac{\partial^L}{\partial\bar\eta_I} Z_E \right|_{\eta=\bar\eta=0} \\ &=(A^{-1})_{IJ}. \end{aligned}$$

The rightmost derivative acts first. Reversing the two odd insertions or the two odd derivatives changes the sign:

$$\left\langle\bar\psi_J\psi_I\right\rangle_E = -(A^{-1})_{IJ}.$$

For one pair this becomes

$$\frac{ \int d\bar\psi\,d\psi\, \psi\bar\psi\,e^{-a\bar\psi\psi} }{ \int d\bar\psi\,d\psi\, e^{-a\bar\psi\psi} } = \frac1a,$$

while the reversed insertion is $-1/a$. This is a stronger sign check than remembering an unlabelled functional-derivative formula.

The determinant belongs to the unnormalized integral. It cancels from $Z_E=\mathcal Z_E/\mathcal Z_E[0]$ only because numerator and denominator refer to the same finite problem. It remains in ratios between different kernels, boundary conditions, regulators, or retained mode spaces. At a convergent Euclidean finite regulator, the comparison is

$$\begin{array}{c|c} \text{quadratic variables} & \text{kernel dependence}\\ \hline \text{real commuting} & (\det A)^{-1/2}\\ \text{complex commuting} & (\det A)^{-1}\\ \text{paired Grassmann} & \det A \end{array}$$

up to the declared reference-measure constants and, for commuting variables, the required convergence cycle. The opposite determinant power is an exact finite result, not yet a statement about a determinant of a differential operator Srednicki 2007, § 44, pp. 279–281, eqs. (44.27), (44.37)–(44.38).

The regulated Dirac integral reproduces the Feynman propagator

Let $K_{F,R}$ be an invertible finite matrix representing the free Dirac quadratic problem after the regulator and the Feynman vacuum boundary data have been chosen. The composite finite index now includes its spinor label. Write

$$S_{0,R}=\bar\psi K_{F,R}\psi,$$

and use the site’s Lorentzian weight and source convention:

$$\begin{aligned} \mathcal Z_{L,R}[\bar\eta,\eta] &= \int\mathcal D_R(\bar\psi,\psi) \\ &\quad\times \exp\!\left[ i\bar\psi K_{F,R}\psi +i\bar\eta\psi +i\bar\psi\eta \right]. \end{aligned}$$

Apply the finite identity with

$$A=-iK_{F,R}, \qquad \bar\xi=i\bar\eta, \qquad \xi=i\eta.$$

No formal manipulation of a continuum measure is needed. The exact result is

$$\begin{aligned} \mathcal Z_{L,R}[\bar\eta,\eta] &= \det(-iK_{F,R}) \\ &\quad\times \exp\!\left[ -i\bar\eta K_{F,R}^{-1}\eta \right]. \end{aligned}$$

Define the delta-normalized inverse and the raw ordered correlator by

$$G_{D,R}=K_{F,R}^{-1}, \qquad S_{F,R}=iG_{D,R}.$$

Then the normalized source functional is especially compact:

$$Z_{L,R}[\bar\eta,\eta] = \frac{\mathcal Z_{L,R}[\bar\eta,\eta]} {\mathcal Z_{L,R}[0,0]} = e^{-\bar\eta S_{F,R}\eta}.$$

The all-left insertion rules are

$$\psi_I \longleftrightarrow -i\frac{\partial^L}{\partial\bar\eta_I}, \qquad \bar\psi_J \longleftrightarrow +i\frac{\partial^L}{\partial\eta_J}.$$

Thus

$$\boxed{ \begin{aligned} \left\langle \mathrm T\,\psi_I\bar\psi_J \right\rangle_R &= \left. \frac{\partial^L}{\partial\bar\eta_I} \frac{\partial^L}{\partial\eta_J} Z_{L,R} \right|_{0} \\ &=(S_{F,R})_{IJ}. \end{aligned} }$$

Again the rightmost derivative acts first. Reversing the derivative order returns $-(S_{F,R})_{IJ}$. The one-pair Lorentzian check is

$$\int d\bar\psi\,d\psi\, e^{ik\bar\psi\psi} =-ik, \qquad \left\langle\psi\bar\psi\right\rangle =\frac{i}{k}.$$

This fixes both the phase of the unnormalized determinant and the factor of $i$ in the raw correlator.

Suppose the regulated bulk inverse approaches the free Minkowski-vacuum boundary value. In the continuum notation inherited by this chapter,

$$\begin{aligned} G_D(x-y) &= \int\frac{d^4p}{(2\pi)^4} \frac{p\!\!\!/+m} {p^2-m^2+i0} e^{-ip\cdot(x-y)}, \\ S_F(x-y) &=iG_D(x-y). \end{aligned}$$

Therefore

$$S_F(x-y) = \int\frac{d^4p}{(2\pi)^4} \frac{i(p\!\!\!/+m)} {p^2-m^2+i0} e^{-ip\cdot(x-y)},$$

and the decisive contact check is

$$\left( i\gamma^\mu\partial_\mu-m \right)S_F(x-y) = i\delta^{(4)}(x-y)\mathbf1_4.$$

Schwartz writes the same field/source integrand and derives the momentum kernel in Schwartz 2014, § 14.6, p. 272, eqs. (14.100)–(14.105), treating $i\epsilon$ as vacuum boundary data. His grouped measure and implicit derivative package are not copied here: the source exponent and derivative order above follow from the declared finite identity and the one-pair checks. Independently, The Fermion Propagator derives the same $S_F$ from the mode expansion, CAR, spin sums, and graded time ordering. Agreement of the contact term, pole prescription, and negative-time exchange sign checks the functional construction by a second route.

Boundary data are part of the kernel

The first-order Dirac expression does not select an inverse by itself. For the unsymmetrized action, varying $\psi$ produces the temporal boundary form contained in

$$i\int_{\partial M}d\Sigma_\mu\, \bar\psi\gamma^\mu\delta\psi.$$

A coherent-state transition kernel naturally fixes a ket label $\psi$ at the initial end and a bra label $\bar\psi$ at the final end; coherent-state overlaps and endpoint factors supply the remaining data. Fixing both members of each independent pair at both ends would overstate the first-order boundary problem.

Temporal data select different fermionic inverses

Problem	Temporal data	Output
Vacuum in–out	Vacuum endpoint states, Euclidean caps, or an equivalent Feynman tilt	Time-ordered Feynman inverse
Thermal trace	Antiperiodic Euclidean-time identification	Thermal fermion kernel and ordinary partition function
Graded trace	Periodic Euclidean-time identification	Trace with a fermion-parity insertion
Response or nonequilibrium evolution	Retarded or closed-time-path contour data	A different inverse or matrix of contour correlators

Even though a finite Grassmann exponential always terminates, the Feynman $i0$ cannot be erased: it selects the vacuum state and inverse, and at finite regulator it also changes the corresponding determinant. An absolute continuum determinant phase requires the additional regulator and normalization data discussed below. A periodic time lattice computes a different object.

Exact boundary check: one fermionic level

Let

$$H=\omega a^\dagger a, \qquad \Delta\tau=\frac{\beta}{M}, \qquad r=e^{-\Delta\tau\omega}.$$

Use unnormalized coherent states

$$|\xi\rangle=e^{-\xi a^\dagger}|0\rangle, \qquad \langle\bar\xi| = \langle0|e^{-a\bar\xi},$$

with

$$\langle\bar\xi'|\xi\rangle =e^{\bar\xi'\xi}, \qquad \mathbf1 = \int d\bar\xi\,d\xi\, e^{-\bar\xi\xi} |\xi\rangle\langle\bar\xi|.$$

The exact one-step kernel is

$$\langle\bar\xi'| e^{-\Delta\tau H} |\xi\rangle = e^{r\bar\xi'\xi}.$$

The ordinary trace contains an endpoint minus sign,

$$\operatorname{Tr}A = \int d\bar\xi\,d\xi\, e^{-\bar\xi\xi} \langle-\bar\xi|A|\xi\rangle.$$

After inserting $M-1$ resolutions of the identity, use

$$\mathcal D_M = d\bar\xi_M\,d\xi_M\cdots d\bar\xi_1\,d\xi_1.$$

The exact finite action is

$$\begin{aligned} S_{\mathrm{AP}} &= \sum_{j=1}^{M}\bar\xi_j\xi_j -r\sum_{j=2}^{M}\bar\xi_j\xi_{j-1} \\ &\quad+ r\bar\xi_1\xi_M. \end{aligned}$$

Equivalently,

$$S_{\mathrm{AP}} = \sum_{j=1}^{M} \bar\xi_j(\xi_j-r\xi_{j-1}), \qquad \xi_0=-\xi_M.$$

For $M\ge2$, the matrix entries are

$$(K_{\mathrm{AP}})_{jk} = \begin{cases} 1,&j=k,\\ -r,&j=k+1,\\ +r,&j=1,\ k=M,\\ 0,&\text{otherwise}. \end{cases}$$

Only the identity permutation and the full temporal cycle contribute to its determinant. The wrap $+r$, the $M-1$ ordinary links $-r$, and the cycle sign $(-1)^{M-1}$ give

$$\det K_{\mathrm{AP}} = 1+r^M = 1+e^{-\beta\omega}.$$

This equals the operator answer $\operatorname{Tr}e^{-\beta\omega a^\dagger a}$ for the empty and occupied states. With a periodic wrap the corner entry is $-r$ and

$$\det K_{\mathrm P} = 1-r^M = 1-e^{-\beta\omega}.$$

That is $\operatorname{Tr}[(-1)^F e^{-\beta H}]$, not the ordinary thermal trace. At $\beta=0$, the two determinants are $2$ and $0$, matching $\operatorname{Tr}\mathbf1$ and $\operatorname{Tr}(-1)^F$. At $\omega=0$, the periodic kernel has a zero mode while the antiperiodic thermal kernel does not.

The one-level system, its time slicing, antiperiodic trace, source kernel, and periodic graded-trace alternative are developed in Zinn-Justin 2021, §§ 4.5.3–4.6, pp. 83–85, eqs. (4.86)–(4.106). That source expands a short step as $1-\Delta\tau\,\omega$; using the exact transfer factor $r$ above makes the finite-$M$ determinant an exact benchmark with the same continuum limit. Normalized coherent states or a Hamiltonian shift redistribute endpoint and overall factors, so those conventions must be translated as a package.

Zero modes change the integration problem

If $\det A=0$ or $\det K_{F,R}=0$, stop before writing an inverse or dividing by the zero-source integral. A Grassmann zero mode makes the unsaturated integral vanish rather than diverge. For one zero-mode pair,

$$\int d\bar\zeta\,d\zeta\,1=0, \qquad \int d\bar\zeta\,d\zeta\, \zeta\bar\zeta=1.$$

Insertions or sources can saturate the zero-mode generators, but then the formula containing $A^{-1}$ is not the applicable identity. A valid treatment must identify the left and right zero modes, fix the orientation of their measure, saturate or project them explicitly, and define any $\det' A$ and pseudoinverse on the stated complement. Adding a mass, changing temporal boundary conditions, or taking a volume limit defines a different problem; none is an algebraic repair performed after inversion.

For paired Dirac variables the nonsingular finite Gaussian gives a determinant. A single set of variables with an antisymmetric quadratic form gives a Pfaffian instead. Relating that algebra to a physical Majorana field also requires the reality structure developed on Majorana Fields and Reality Conditions; the determinant or Pfaffian phase in an interacting or continuum problem lies beyond this page.

A reproducible finite-regulator workflow

Use the method in this order:

1. Specify the finite problem. Name the retained modes or sites, temporal slices, spinor and internal labels, reference measure, state, and boundary conditions.
2. Fix every odd order. Declare the generator list, paired measure, quadratic form, source order, and left or right derivatives before doing algebra.
3. Test invertibility. Compute or bound the smallest singular value on the declared finite space. If it vanishes, switch to an explicit zero-mode treatment rather than writing $K^{-1}$.
4. Evaluate and normalize. Apply the finite Gaussian identity, retaining $\det(-iK)$ until the question justifies division by the identical zero-source problem.
5. Differentiate in the written order. Check the result first on one pair, then verify $KK^{-1}=I$ on the full regulated space.
6. Run a physical cross-check. Compare the contact equation, exchange sign, and boundary prescription with canonical quantization or an exact transfer-matrix calculation.
7. Only then discuss limits. State which volume, spacing, cutoff, mass, and $i0$ limits are taken and which claim establishes their existence.

For a dense $N\times N$ kernel, an LU-based determinant and factorization cost order $N^3$ operations and order $N^2$ storage. If only selected propagator columns are needed, solve $Kx=e_J$ rather than materializing the full inverse. Sparse structure can reduce the cost, but near-zero singular values make both determinants and solves ill-conditioned; that numerical warning is often the first practical zero-mode diagnostic.

The continuum symbol $\det(i\gamma\cdot\partial-m)$ does not acquire a value merely because every finite determinant exists. Its regulator, reference normalization, boundary conditions, zero modes, phase, and renormalization must be specified; a cutoff that controls propagator integrals need not by itself regulate the determinant Zinn-Justin 2021, § 12.7.2, p. 275, after eq. (12.56). Chiral anomalies require a separate regulator-aware treatment—the obstruction to continuing all four-dimensional $\gamma_5$ identities is one warning sign Zinn-Justin 2021, § 12.8, p. 276, eq. (12.57)—and belong to the later Symmetry treatment, not to this free finite identity.

Common pitfalls

Treating $\bar\psi$ as the value of $\psi^\dagger$ during integration. The paired generators are independent. Their physical adjoint relation is encoded by the coherent-state or field-theory construction, not imposed as a restriction of the finite Berezin algebra.

Quoting a Gaussian without its order. Reversing two differentials, sources, insertions, or odd derivatives changes a sign. State the measure and derivative convention, then pass the one-pair benchmark.

Copying the Euclidean source exponent into Lorentzian signature. The Lorentzian weight $e^{iS}$ gives $\det(-iK)$ and $e^{-i\bar\eta K^{-1}\eta}$. The raw ordered correlator is $S_F=iK^{-1}$, whereas the delta-normalized inverse is $G_D=K^{-1}=-iS_F$.

Calling a local operator its own inverse. Feynman, thermal, retarded, and closed-time-path data select different kernels. The contact equation checks an inverse only after its domain and boundary data are fixed.

Normalizing away a determinant that is being compared. Division by $\mathcal Z[0]$ removes the determinant only at fixed regulator, kernel, measure, modes, and boundary problem. Changing any of them can make the determinant ratio the quantity of interest.

Using the invertible formula in the presence of zero modes. If the zero-source integral vanishes, neither $K^{-1}$ nor $\mathcal Z[\bar\eta,\eta]/\mathcal Z[0,0]$ exists on the full space. Saturate or project the zero modes explicitly before proceeding.

Check your understanding

Check 1: recover the inverse and its exchange sign

For the normalized Euclidean functional $Z_E=e^{\bar\eta A^{-1}\eta}$, use only left derivatives to obtain $\langle\psi_I\bar\psi_J\rangle_E$ and the reversed insertion.

Solution

The derivative nearest $Z_E$ acts first:

$$\left. \frac{\partial^L}{\partial\eta_J} \frac{\partial^L}{\partial\bar\eta_I} Z_E \right|_0 = (A^{-1})_{IJ}.$$

The two derivative operators are odd. Reversing them gives

$$\left. \frac{\partial^L}{\partial\bar\eta_I} \frac{\partial^L}{\partial\eta_J} Z_E \right|_0 = -(A^{-1})_{IJ},$$

which equals $\langle\bar\psi_J\psi_I\rangle_E$. The sign is the same one seen by exchanging the two field generators directly.

Check 2: locate the Lorentzian factor of i

For one pair with nonzero commuting $k$, evaluate the zero-source integral with exponent $ik\bar\psi\psi$ and its normalized $\psi\bar\psi$ insertion.

Solution

Since $\bar\psi\psi=-\psi\bar\psi$,

$$e^{ik\bar\psi\psi} = 1-ik\psi\bar\psi.$$

Therefore

$$\mathcal Z_L[0,0]=-ik, \qquad \frac{1}{\mathcal Z_L[0,0]} \int d\bar\psi\,d\psi\, \psi\bar\psi e^{ik\bar\psi\psi} = \frac{i}{k}.$$

Thus $G_D=k^{-1}$ is the delta-normalized inverse and $S_F=ik^{-1}$ is the raw ordered correlator.

Check 3: diagnose the periodic zero mode

Set $\omega=0$ in the one-level Euclidean trace. Compare antiperiodic and periodic boundary conditions and identify the operator traces they compute.

Solution

At $\omega=0$, $r=1$. The antiperiodic determinant is

$$\det K_{\mathrm{AP}}=1+r^M=2,$$

which is $\operatorname{Tr}\mathbf1$ on the empty and occupied states. The periodic determinant is

$$\det K_{\mathrm P}=1-r^M=0.$$

It computes $\operatorname{Tr}(-1)^F=1-1$ and has a constant temporal zero mode. The vanishing result is correct boundary physics, not a failed approximation; the inverse formula must stop.

Where the method is used next

• Fermion Signs and Closed Loops develops interacting reordering signs and the extra sign of a closed fermion loop.
• Ghosts, Auxiliary Fields, and Gauge-Parameter Dependence represents Faddeev–Popov determinants with Grassmann fields.
• Fermion Determinants, Pfaffians, and Measure Positivity develops spectral positivity, determinant phases, Pfaffians, and flavor powers.
• Coherent-State Path Integrals for Many-Body Systems develops time slicing, endpoint data, and interacting many-body uses.
• Naive Fermions and Species Doubling studies what a local lattice Dirac kernel gains and duplicates under discretization.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. DOI.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007. DOI.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 5th ed. Oxford University Press, 2021. DOI.