# Right vs Left Derivatives

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!

• The problem also doesn't appear if $|f|=|\theta|$. Thus, there is a sign difference if and only if $|f|=|\delta\theta|=|\theta|+1$ – 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 $$|\epsilon|=|X|$$.

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

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