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.

  • $\begingroup$ Just expand out $\psi^T M \psi$ in terms of the matrix elements, and use the formulae of grassman integration. The Determinant should come from the permutations of the matrix elements which can be written in terms of the levi-civita symbol. $\endgroup$
    – user7757
    Commented Jan 26, 2014 at 12:36

1 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.


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