Let $\theta$ be a fermionic quantity and $f(\theta)=f(0)+\theta\frac{\partial f}{\partial\theta}=f(0)+\frac{\partial_r f}{\partial\theta}\theta$. Under a variation $\theta\mapsto\theta+\delta\theta$ we have $$f(\theta)\mapsto f(\theta)+\delta\theta\frac{\partial f}{\partial\theta},$$ using the first formula, or $$f(\theta)\mapsto f(\theta)+\frac{\partial_r f}{\partial\theta}\delta\theta,$$ using the second one. However, $$\delta\theta\frac{\partial f}{\partial\theta}=(-1)^{|\delta\theta|(|f|+|\theta|)}\frac{\partial f}{\partial\theta}\delta\theta=(-1)^{|\delta\theta|(|f|+|\theta|)+|\theta|(|f|+1)}\frac{\partial_rf}{\partial\theta}\delta\theta$$ which is different from $\frac{\partial_rf}{\partial\theta}\delta\theta$ in general. This yields a contradiction between both variations. Of course problems are avoided if $|\delta\theta|=|\theta|$ but I don't see how this affect the first two equations. I am very confused by this!

  • $\begingroup$ The problem also doesn't appear if $|f|=|\theta|$. Thus, there is a sign difference if and only if $|f|=|\delta\theta|=|\theta|+1$ $\endgroup$ – Iván Mauricio Burbano Jul 13 at 19:18
  1. Yes, by definition the Grassmann parity $|\delta z|$ of a variation $\delta z$ of a supernumber $z$ (of definite Grassmann parity) is the same as the Grassmann parity $|z|$ of the supernumber $z$ itself: $$|\delta|~=~0.\tag{1}$$

  2. Perhaps OP is wondering about the following question.

    Question: How does an infinitesimal variation $\delta$ relate to a left vector-field/linear derivation $X$ of Grassmann-parity $|X|$?

    Answer: In order to relate $X$ to an infinitesimal variation$^1$ $$\delta~=~\epsilon X,\tag{2L}$$ one needs to introduce an infinitesimal parameter $\epsilon$ of the same Grassmann-parity $|\epsilon|=|X|$.


$^1$ For a right vector-field/linear derivation $X_R$, we instead have $$\delta~=~X_R \epsilon ,\tag{2R}$$ with $|\epsilon|=|X_R|$.

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  • $\begingroup$ Of course, I was worried by supersymmetries but in there we have fermionic parameters that make $|\delta\theta|=|\theta|$! thanks $\endgroup$ – Iván Mauricio Burbano Jul 13 at 19:34
  • $\begingroup$ Well, I guess we can do supersymmetry without superspace. In that case $|\delta\theta|\neq|\theta|$. Then my problem still stands. $\endgroup$ – Iván Mauricio Burbano Jul 13 at 19:52
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jul 13 at 20:16

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