Right vs Left Derivatives

Question

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!

Answer 1

I) Let us first review the definition of left and right derivatives:

1. A left derivative means a derivative that acts from the left. A right derivative means a derivative that acts from the right.

2. In more detail, an infinitesimal variation of a function $f(z)$ is of the form $$ \delta z~ \frac{\partial_L f}{\partial z} ~=~\delta f~=~ \frac{\partial_R f}{\partial z}~ \delta z, \tag{LR1}$$ where the left and right derivative satisfy $$\frac{\partial_L f}{\partial z} ~=~(-1)^{|z|(|f|+1)} \frac{\partial_R f}{\partial z}.\tag{LR2}$$ Here $|\cdot|$ denotes the Grassmann-parity.

II) Now let us address OP's questions:

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 $|\epsilon|=|X_R|$.