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?


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?

  • $\begingroup$ This is one of those weird things about Grassmann numbers: the Jacobian determinant factor is the opposite of what you would expect. It should be explained in detail in the chapter of Peskin that introduces path integrals. $\endgroup$
    – knzhou
    Apr 1, 2019 at 18:38
  • $\begingroup$ @knzhou Peskin claims to use Eqs. (9.63) and (9.69) in order to obtain that 'rule' for the jacobian but that are properties of Grassman variables. I've made an edit using them giving a possible way to understand this. What do you think? $\endgroup$
    – Vicky
    Apr 1, 2019 at 19:01
  • 2
    $\begingroup$ Yup, that's essentially the intuition for why it has to come out opposite! Another way to see this is that Grassmann integrals are the same thing as Grassmann derivatives, so you get a chain rule factor, which behaves in the opposite way to a Jacobian factor. $\endgroup$
    – knzhou
    Apr 1, 2019 at 19:02

1 Answer 1

  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.

  • $\begingroup$ I don't see this step: $\frac{\partial\theta^{\prime k_1}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_1}} \ldots \frac{\partial\theta^{\prime k_n}}{\partial\theta^1} \frac{\partial}{\partial\theta^{\prime k_n}}$ $~=~\det M \frac{\partial}{\partial\theta^{\prime 1}} \ldots \frac{\partial}{\partial\theta^{\prime n}}$. Could you do it in a little more detail, I'd really appreciate it $\endgroup$
    – Vicky
    Apr 1, 2019 at 20:01
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 1, 2019 at 20:57

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