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Given the fermionic generating functional $$Z[\eta]=\ det^{\frac{1}{2}}(K_{ij})e^{-\frac{i}{2}\eta_{i}G^{ij}\eta_{j}},\tag{1}$$ where $$G^{ij}=K^{-1}_{ij}$$ is the Green function of our theory, then we know that the propagator can be written as $$\langle\psi^{i}\psi^{j}\rangle = \frac{1}{Z[0]}\bigg(\frac{1}{i}\bigg)^2\frac{\delta^2Z[\eta]}{\delta\eta_{i}\delta\eta_{j}}\bigg|_{\eta=0}=-iG^{ij}.$$ I can’t understand how to derive this result. Indeed $$\langle\psi^{i}\psi^{j}\rangle=\frac{\delta}{\delta\eta_{i}}\bigg(\frac{i}{2}G^{jk}\eta_{k} - \frac{i}{2}\eta_{k}G^{kj}\bigg),$$where I used the fact that for Grassmann variables the Leibniz rule should read as $$\frac{\partial}{\partial\theta}(\theta\theta)=\frac{\partial\theta}{\partial\theta}\theta-\theta\frac{\partial\theta}{\partial\theta}$$ and assuming that the Green function is Grassmann even. Hence, eventually we can write the propagator as $$\langle\psi^{i}\psi^{j}\rangle=\frac{i}{2}G^{ji}-\frac{i}{2}G^{ij}.$$ Exploiting the symmetry of the Green function, the result is zero. I am missing a sign but I don’t know where I am wrong.

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  • $\begingroup$ Use \langle and \rangle for $\langle$ and $\rangle$, respectively. $\endgroup$ Apr 15 at 9:11

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Hint: For eq. (1) to be non-trivial we must assume that $G^{ij}$ is antisymmetric in $i\leftrightarrow j$.

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  • $\begingroup$ Oh yeah, now I see the error. Thank you! $\endgroup$
    – Michael
    Apr 15 at 7:57
  • $\begingroup$ @Michael Consider to "accept" the answer, by clicking on the check-mark. $\endgroup$ Apr 15 at 9:11

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