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

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

Now consider an axial U(1) transformation of the Dirac field, but with a spacetime dependent parameter $\alpha(x)$: \begin{equation} \Psi(x) \rightarrow e^{-i\alpha (x)\gamma_{5}} \Psi(x) \tag{77.12} \end{equation} \begin{equation} \overline{\Psi}(x) \rightarrow \overline{\Psi}(x)e^{-i\alpha (x)\gamma_{5}} \tag{77.13} \end{equation} 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 \begin{equation} J(x, y) = \delta^{4}(x-y) e^{-i\alpha (x)\gamma_{5}} \tag{77.17} \end{equation} 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
\begin{equation} \mathcal{D}\Psi \mathcal{D}\overline{\Psi}\rightarrow (\det J)^{-2}\mathcal{D}\Psi \mathcal{D}\overline{\Psi} \tag{77.18} \end{equation}

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

  • $\begingroup$ 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. $\endgroup$
    – knzhou
    Jun 23, 2019 at 10:05
  • $\begingroup$ ... chapter 44, eq. 44.18. $\endgroup$ Jun 23, 2019 at 14:03
  • $\begingroup$ 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 $\int dz d{\bar z} e^{- a z {\bar z} } = \frac{2\pi}{a}$. $\endgroup$
    – Prahar
    Jun 24, 2019 at 22:09

1 Answer 1


Well, the full explanation comes from Berezin integration of Grassmann-odd variables.

  1. One argument is that the Jacobian matrix is a supermatrix, so that the Jacobian determinant is given by the superdeterminant.

  2. Another argument is that Grassmann-odd integration is the same as Grassmann-odd differentiation! The Jacobian factor can therefore be understood from the chain rule of differentiation (rather than integration by substitution).

Both arguments lead to that the Jacobian appears inverted.


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