quantum field theory - Change of variables in path integral measure

In fermion's path integral we have a measure that you can write, in terms of the Grassmann variables $\psi, \bar{\psi}$ as

$$ D\bar{\psi}D\psi, \quad \psi(x) = \sum_n a_n\phi_n(x), \quad \bar{\psi}(x) = \sum_n \bar{a}_n \bar{\phi}_n(x) $$

Where $a_n, \bar{a}_n$ are Grassmann variables and $\phi_n(x)$ a set of orthonormal functions such that

$$ \int d^3x\ \phi^\dagger_n(x)\phi_m(y) = \delta_{nm} $$

Now if you perform a change of variables in, for instance, axial group $U(1)_A$ with an small parameter $\alpha(x)$, this renders

$$ a'_m = \sum_n(\delta_{mn} + i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x))a_n = \sum_n(1 + C)_{mn}a_n $$ $$ \bar{a}'_m = \sum_n(1 + C)_{mn}\bar{a}_n $$ $$ C_{mn} = i\int d^3x\ \alpha(x)\phi^\dagger_m(x)\gamma^5\phi_n(x)), \quad 1\ {\rm is\ the\ identity} $$

Now, following Peskin (chapter 19.2, Eq. (19.69)), the path integral meassure should change as

$$ D\bar{\psi}'D\psi' = D\bar{\psi}D\psi·(det[1 + C])^{-2}. $$

I don't understand where the -2 power for the Jacobian ($det[1 + C]$) came up since if we were talking about a usual integral with usual variables we would end up with +2 power.

What am I missing?

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EDIT

Thinking about the problem I found a possible explanation. Grassmann variables, let's call it $\eta$, are forced to satisfy

$$ \int d\eta\ \eta = 1 $$

Therefore, a change of variables such that $\eta$ changes to $$\eta' = A\eta\tag{A}$$ and $\eta'$ is still a Grassmann variable should fulfill

$$ \int d\eta'\ \eta' = 1 \tag{B} $$

But if we follow the change of variables given by Eq. (A) and we want Eq. (B) satisfied,

$$ \int d\eta\ \eta = A^{-2}\int d\eta'\ \eta' = 1 \Leftrightarrow \int d\eta'\ \eta' = A^2 $$

Then, we are violating Eq. (B) and $\eta'$ isn't a Grassmann variable. So, if $\eta, \eta'$ are Grassmann variables then the Jacobian ($j = A^{-1}$) among themselves must be introduced in the measure with the opposite power sign, so:

$$ \int d\eta'\ \eta' = \int j^{-1}·d\eta\ j·\eta \equiv 1 $$

Fine or something to complain about?

Answer

1. 
   

   OP is right. Grassmann-odd integrals are the same thing as Grassmann-odd derivatives $\int \!d\theta^j=\frac{\partial}{\partial\theta^j}$, cf. e.g. this Phys.SE post and above comments by user knzhou.

   
   


2. 
   

   For this reason, if $\theta^{\prime k} = M^k{}_j ~\theta^j$ is a linear change of Grassmann-odd variables [where the matrix elements $M^k{}_j$ are Grassmann-even], then $$\begin{align}\int \!d\theta^1 \ldots \int \!d\theta^n &~=~\frac{\partial}{\partial\theta^1} \ldots \frac{\partial}{\partial\theta^n} ~=~\sum_{k_1=1}^n\frac{\partial\theta^{\prime k_1}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_1}} \ldots \sum_{k_n=1}^n\frac{\partial\theta^{\prime k_n}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_n}} \cr &~=~\sum_{\pi\in S_n} M^{\pi(1)}{}_1 \frac{\partial}{\partial\theta^{\prime \pi(1)}} \ldots M^{\pi(n)}{}_n \frac{\partial}{\partial\theta^{\prime \pi(n)}}\cr &~=~\sum_{\pi\in S_n} M^{\pi(1)}{}_1 \ldots M^{\pi(n)}{}_n (-1)^{{\rm sgn}(\pi)} \frac{\partial}{\partial\theta^{\prime 1}} \ldots \frac{\partial}{\partial\theta^{\prime n}}\cr &~=~\det M \frac{\partial}{\partial\theta^{\prime 1}} \ldots \frac{\partial}{\partial\theta^{\prime n}}~=~\det M \int \!d\theta^{\prime 1} \ldots \int \!d\theta^{\prime n}. \end{align}$$

   
   



3. 
   

   More generally, change of super-integration variables transform with the superdeterminant/Berezinian.