# Why is the Jacobian factor for fermionic variables different from that for bosonic ones?

In Srednicki's textbook Quantum Field Theory, Section 77 discusses anomalies and the path integral for fermions. The path integral over the Dirac field is defined to be

$$$$Z(A) \equiv \int \mathcal{D}\Psi\mathcal{D}\overline{\Psi} e^{iS(A)} \tag{77.8}$$$$ where $$$$S(A) \equiv \int d^{4}x \overline{\Psi}i\displaystyle{\not}{D}\Psi \tag{77.9}$$$$ is the Dirac action, $$i\displaystyle{\not}{D}= i\gamma^{\mu}D_{\mu}$$ is the Dirac wave operator, and $$$$D_{\mu} = \partial_{\mu} -igA_{\mu}$$$$ is the covariant derivative.

Now consider an axial U(1) transformation of the Dirac field, but with a spacetime dependent parameter $$\alpha(x)$$: $$$$\Psi(x) \rightarrow e^{-i\alpha (x)\gamma_{5}} \Psi(x) \tag{77.12}$$$$ $$$$\overline{\Psi}(x) \rightarrow \overline{\Psi}(x)e^{-i\alpha (x)\gamma_{5}} \tag{77.13}$$$$ We can think of eqs. (77.12) and (77.13) as a change of integration variable in eq.(77.8); ... ...

The change of variable in eqs.(77.12) and (77.13) is implemented by the functional matrix $$$$J(x, y) = \delta^{4}(x-y) e^{-i\alpha (x)\gamma_{5}} \tag{77.17}$$$$ Because the path integral is over fermionic variables (rather than bosonic), we get a $$(\det J)^{-1}$$ (rather than $$\det J$$) for each of the transformations in eqs. (77.12) and (77.13), so that we have
$$$$\mathcal{D}\Psi \mathcal{D}\overline{\Psi}\rightarrow (\det J)^{-2}\mathcal{D}\Psi \mathcal{D}\overline{\Psi} \tag{77.18}$$$$

I don't understand this. Why is the Jacobian factor for fermionic variables $$(\det J)^{-1}$$ while that for bosonic ones $$det J$$?

• That just comes down to how Grassmann variables are defined; they're not at all like ordinary numbers. This should be covered in detail early in the book. – knzhou Jun 23 '19 at 10:05
• ... chapter 44, eq. 44.18. – AccidentalFourierTransform Jun 23 '19 at 14:03
• It can all be boiled down to the following fact about Grassmannian integrals $\int d\eta d{\bar \eta} e^{ - a \eta {\bar \eta} } = a$. The same formula for bosonic variables is $\int dz d{\bar z} e^{- a z {\bar z} } = \frac{2\pi}{a}$. – Prahar Mitra Jun 24 '19 at 22:09