Berezin integral of a Grassmann field

Question

Consider a time dependent Grassmann field i.e. $\theta(t)$. Now, consider the following Berezin integral $$\int [\mathcal{D}\theta] ~\prod_{t}\dot{\theta}\tag{1}$$ where $\dot{\theta}$ time derivative of $\theta$ and $[\mathcal{D}\theta]$ is a functional measure over the Grassmann field, something like a path integral measure and a product integral of $\dot{\theta}$. Now, what does the above integral evaluate to? One way I can think of evaluating the integral is that by recognizing that $\dot{\theta} = \delta(\dot{\theta})$ as $\dot{\theta}^2 = 0$. Therefore, we have $$\int [\mathcal{D}\theta] ~\prod_{t}~\dot{\theta} = \int [\mathcal{D}\theta]~\prod_{t} ~\delta(\dot{\theta}) = \det({\partial_t}).\tag{2}$$ Is my evaluation of the integral correct?

Answer 1

TL;DR: Yes, OP is correct.

Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j,$$ so that $$\frac{\partial}{\partial\theta^j} ~=~\sum_{i=1}^n A^i{}_j\frac{\partial}{\partial\theta^{\prime i}}.$$ Then a straightforward calculation shows that OP's path-integral (1) is replaced with$^1$ $$\begin{align} \int\!d\theta^n \ldots d\theta^1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~ \int\! d\theta^n \ldots d\theta^1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr ~=~&\ldots~=~\det(A),\end{align} $$ which indeed is a discrete version of OP's claim (2).

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$^1$Recall that Berezin integration $$ \int\!d\theta^n \ldots d\theta^1 ~=~\frac{\partial}{\partial\theta^n}\ldots \frac{\partial}{\partial\theta^1}$$ is the same as differentiation!