Gaussian Integral over Grassmann variables

Question

I need to evaluate two Grassmann integrals, one over "real" Grassmann variable another one over complex variables.

Let's start with the real one first:

The prototype we have for $n$ real Grassmann variables :

$$ \tag{1}\int d^n\psi \exp{[\frac{1}{2}\psi^T M\psi]} ~=~ (\det M)^{1/2}. $$

Now we can shift the integration variable and use the shift invariance of the integration. If we replace $$\psi\rightarrow \psi - M^{-1}\eta, \tag{2}$$ then the argument inside exponential becomes:

$$ [\frac{1}{2}\psi^T M\psi] ~\longrightarrow~ \frac{1}{2}(\psi - M^{-1}\eta)^T M(\psi - M^{-1}\eta) $$ $$ =\frac{1}{2}[\psi^T M\psi + \eta^T (M^{-1})^T MM^{-1}\eta \space\underbrace{-\eta^T (M^{-1})^T M\psi - \psi^TMM^{-1}\eta}] $$

$$ =\frac{1}{2}\psi^T M\psi+\frac{1}{2}\underbrace{\eta^T (M^{-1})^T MM^{-1}\eta}_{\text extra}+\eta^T\psi. \tag{3} $$

This "extra" term should vanish, but how? I will run into same problem is do the same Gaussian integral over complex Grassmann variables.

Answer 1

Note that in OP's eq. (1) it is implicitly assumed that the matrix $M$ is antisymmetric

$$\tag{A} M^T~=~-M.$$

[A symmetric part in eq. (1) would not contribute to the integrand (1).] The term $\eta^TM^{-1}\eta$ in eq. (3) does in general not vanish

$$ 2S(\psi) ~:=~ (\psi - M^{-1}\eta)^T M(\psi - M^{-1}\eta) ~\stackrel{(A)}{=}~(\psi^T + \eta^T M^{-1}) M(\psi - M^{-1}\eta)$$ $$~=~\psi^T M\psi+\eta^T\psi -\psi^T\eta-\eta^TM^{-1}\eta . \tag{B}$$

One may check that if we vary the shifted action (B) wrt. the integration variable $\psi$, we unsurprisingly get the classical value $$ \psi~\approx~M^{-1}\eta, \tag{C}$$ which reflect the shift (2) that OP performed in the first place. As a check, note that the classical action vanishes $$S\left(\psi=M^{-1}\eta\right)~\stackrel{(B)}{=}~0 ,\tag{D}$$ as it should.