# Fermionic measure in path integral

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?

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

## 2 Answers

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.

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

References:

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