Grassmann Gaussian integral

A Grassmann Gaussian integral satisfies $\int D\bar{\psi }D\psi e^{-\bar{\psi }A\psi }=\det A$. Adding Grassmann sources and completing the square produces an exponent containing $A^{-1}$.

For an antisymmetric matrix $A$, a Berezin integral extracts the top coefficient of the exponential. Expansion into pairings gives the Pfaffian, with a constant sign depending on the order of the differentials. This is a polynomial in the entries of $A$, not the positive square root of its determinant: in dimension two it is proportional to $A_{12}$ and changes sign with that entry. The identity $\mathrm{Pf}(A{)}^2=\det A$ determines its square but does not discard its orientation.

For an even-dimensional antisymmetric matrix $M$, put $\theta =(u+iv)/\sqrt{2}$ and $\bar{\theta }=(u-iv)/\sqrt{2}$, with independent Grassmann variables $u,v$. Anticommutation cancels the mixed terms, giving ${\bar{\theta }}^{T}M\theta =(u^{T}Mu+v^{T}Mv)/2$. The constant Grassmann change-of-variables formula changes only the normalization. The two real Pfaffians multiply to $\mathrm{Pf}(M{)}^2=\det M$, agreeing with the complex Grassmann Gaussian integral. In odd dimension $\det M=0$ and both top-degree integrals vanish.

Use the Berezin integral orientation $\int {\prod }_{i}d{\bar{\theta }}_{i}d{\theta }_i{\theta }_1{\bar{\theta }}_1\cdots {\theta }_n{\bar{\theta }}_n=1$. For an invertible matrix $B$, the normalized weight $e^{-\bar{\theta }B\theta }$ has $⟨{\theta }_i{\bar{\theta }}_j⟩=(B^{-1}{)}_{ij}$. Differentiate its determinant with respect to $B_{ji}$ to obtain this formula; two derivatives give the displayed four-point formula. The second pairing carries a fermionic sign. These are ordered contractions, so exchanging two odd factors changes their sign.

For $C=B^{-1}$, ordered Grassmann Gaussian contractions give $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$ and $⟨(\bar{\theta }M\theta )(\bar{\theta }N\theta )⟩=\mathrm{tr}(MC)\mathrm{tr}(NC)-\mathrm{tr}(MCNC)$. The disconnected contribution has two fermion loops; the connected contribution has one, giving its minus sign. This covariance is an algebraic normalized Berezin integral, rather than a positivity statement about a probability measure.

Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.

For a free Dirac field with action $\int \bar{\psi }K\psi$ and independent odd sources coupled as $\bar{\psi }J+\bar{J}\psi$, translation invariance of the Berezin integral and completion of the square give $Z[J,\bar{J}]=Z[0,0]\exp (-i\bar{J}{K}_F^{-1}J)$. Here $K_F^{-1}$ has Feynman i-epsilon prescription boundary conditions, and each bilinear includes its spacetime integrations. A left Grassmann derivative with respect to $\bar{J}$, followed by a right Grassmann derivative with respect to $J$, has normalized value $-i{K}_F^{-1}$ at zero sources. Multiplying by $-1$ removes the two insertion factors $i^2$ and yields the Dirac propagator $S_F=i{K}_F^{-1}$. Keeping source order fixed prevents a spurious fermionic sign.

A Grassmann field takes values in a Grassmann algebra, so its components anticommute. A closed loop of Grassmann propagators acquires a minus sign in perturbation theory.