Fermionic measure in path integral

Question

When writing the fermionic path integral one arrives at an expression containing $\mathcal{D}\bar{\psi}$ and $\mathcal{D}\psi$:

$$ \int \mathcal{D}\bar{\psi} \mathcal{D}\psi e^{iS} $$

Usual derivations of this expression involve discretizing into time intervals and then inserting the realization of the identity operator using coherent states. Following this procedure, one defines

$$ \mathcal{D}\bar{\psi} \mathcal{D}\psi := \lim_{n}\prod_{i}^n\textrm{d}\bar{\psi}_i\textrm{d}\psi_i $$

I have a question at this point. What would happen if we exchange $\textrm{d}\bar{\psi}_i$ and $\textrm{d}\psi_i$ for all $i$? We would get a minus sign for each term, but what would happen with this minus sign in the limit $n\to\infty$?

My question is based on the fact that on books, notes and many resources $\mathcal{D}\bar{\psi}$ and $\mathcal{D}\psi$ seemed to be exchanged under the integral sign without any justification, and people also write

$$ \int \mathcal{D}\psi \mathcal{D}\bar{\psi} e^{iS} $$

Clearly, for a finite number $n$ of $\textrm{d}\psi_i$'s such an exchange changes the integral by a factor $(-1)^n$, but what happens in the case of the path integral measure?

Suppose we have a Lagrangian $\mathcal{L}$ given with a certain order for $\bar{\psi}$ and $\psi$? Let's say for instance $\mathcal{L}$ is normal ordered. Can we commute the measures as follows?

$$ \int \mathcal{D}\bar{\psi} \mathcal{D}\psi e^{iS} = \int \mathcal{D}\psi \mathcal{D}\bar{\psi} e^{iS} $$

Answer 1

The relevant object in QFT is the generating functional $Z[\eta, \bar{\eta}] = \int \mathcal{D} \bar{\psi} \mathcal{D} \psi \, e^{i\{S[\psi, \bar{\psi}] +\int d^d x(\bar{\psi}(x) \eta(x)+\bar{\eta}(x) \psi(x)\}}$. The (Grassmann) fields $\eta$ and $\bar{\eta}$ are external sources and the generating functional is normalized as $Z[0,0]=1$. Thus all your (potential) concerns are absorbed by this normalization condition. $\mathcal{D} \bar{\psi} \mathcal{D} \psi$ should simply be seen as some symbolic notation for a translation invariant measure for the fermionic path integral.

Answer 2

1. For a finite-dimensional Berezin/Grassmann integral the order of the measure clearly do matter, as OP states. Since the path integral is supposed to be a continuum limit of a finite-dimensional integral, it is in principle also affected. However, we can avoid this sign ambiguity if we always increase the discretization with an even number of pairs $(\psi_i, \bar{\psi}_i)$.

2. Be aware that different authors have

   • different ordering conventions for measure + integrand, and

   • different sign conventions for evaluating a 1-dimensional Berezin/Grassmann integral.

References:

1. M. Srednicki, QFT, 2007; chapter 44. A prepublication draft PDF file is available here.