Fermionic ghost path integral results in $\delta$ function?

Question

This is related to a statement in pg 20 of hep-th/9408074 formula (2.39).

Suppose $$\mathcal{L}\sim\frac{i}{\lambda^{\prime}}\bar{\eta}^xg_{ij}U_x{}^i\psi^j+\cdots \tag{2.35}$$where $\bar{\eta}$ to my guess is ghost field as it is non-dynamical and assume $\cdots$ does not contain contribution of $\bar{\eta}$. Consider $\int d\eta e^{\mathcal{L}}$.

The paper says integration of $\eta$ gives $$\left(\frac{-i}{\lambda^{\prime}}\right)^t\delta(g_{ij}U_x{}^i\psi^j)\tag{2.39}.$$

$\textbf{Q:}$ How do I see this does give rise to Dirac delta function? I had expanded the exponential function to perform fermionic integral but this does not give me $\delta(g_{ij}U_x^i\psi^j)$ but $\sim U_x^i\psi^j$. Normally, Faddeev-Popov ghosts involve 2 fermions (ghost and anti-ghost) to perform gauge fixing. What is the argument here?

Answer 1

Eq. (2.39) is a $t$-dimensional Grassmann-odd delta function$^1$

$$\prod_{x=1}^t \delta(\frac{i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j)~=~\prod_{x=1}^t \int \! d\eta^x~\exp\left\{ \frac{i}{\lambda^{\prime}}\eta^x g_{ij}U_x{}^i\psi^j\right\}~=~\prod_{x=1}^t \frac{\pm i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j .\tag{2.39}$$

Recall that if $f$ is an arbitrary function of a Grassmann-odd variable $\theta$, then Berezin integration yields$^2$ $$\int d\theta~(\pm\theta)~f(\theta)~=~f(0), $$ which formally satisfy the defining property $$\int d\theta~\delta(\theta)~f(\theta)~=~f(0) $$ of a Dirac delta distribution! So we can identify $$\delta(\theta)~=~\pm\theta.$$

References:

1. C. Vafa & E. Witten, arXiv:hep-th/9408074; eqs. (2.35) & (2.39).

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$^1$ There is a typo in the action (2.35) of Ref. 1: The Grassmann-odd $\bar{\eta}$ variable in the fifth term should be an $\eta$ variable.

$^2$ The $\pm$ sign denotes different Berezin conventions $\int d\theta~\theta~=\pm 1 $.