# Right derivative of Grassmann number and associated anti-commutation relation

I am reading chapter 3 of Anomalies in quantum field theory by Reinhold Bertlmann and I found one statement that I don't know how to prove. First of all he defined the right derivative on the Grassmann number to satisfy $$$$\frac{\partial^{R}}{\partial \theta_{i}} \theta_{k} \theta_{l} = \theta_{k} \delta_{il} - \delta_{ik}\theta_{l}.\tag{3.195}$$$$ My understanding is that the derivative will act from the right. So we can first take derivative on $$\theta_{l}$$ and this will produce $$\theta_{k} \delta_{il}$$. Note that no minus sign is here since we act the derivative from right. However, to differentiate $$\theta_{k}$$ from the right, we need to move the right derivative next to $$\theta_{k}$$, this will produce a minus sign and then we get $$-\delta_{ik} \theta_{l}$$. Since I can reproduce this formula, I guess I am on the right track.

He then stated that for both the left and right derivatives we have \begin{align} \left\{ \frac{\partial}{\partial \theta_{i}} , \theta_{k} \right\} = \delta_{ik}.\tag{3.197} \end{align} Let us consider the right derivative and let $$\left\{ \frac{\partial^{R}}{\partial \theta_{i}} , \theta_{k} \right\}$$ acts on a Grassmann function $$f$$. We will have \begin{align} \left\{ \frac{\partial^{R}}{\partial \theta_{i}} , \theta_{k} \right\}f & = \left( \frac{\partial^{R}}{\partial \theta_{i}} \left[ \theta_{k}f\right] + \theta_{k} \frac{\partial^{R}}{\partial \theta_{i}}f \right) \\ & = \theta_{k} \frac{\partial^{R}}{\partial \theta_{i}}f - \delta_{ik} f + \theta_{k} \frac{\partial^{R}}{\partial \theta_{i}}f \\ & \neq \delta_{ik} f. \end{align} I am confused, what goes wrong with my attempt? I can show that for left derivatives we have $$\left\{ \frac{\partial^{L}}{\partial \theta_{i}} , \theta_{k} \right\} = \delta_{ik}$$, but it seems to me that when it comes to right derivative we need to be more careful. However, I am not sure what exact thing we should be careful of since I can really get the $$\frac{\partial^{R}}{\partial \theta_{i}} \theta_{k} \theta_{l} = \theta_{k} \delta_{il} - \delta_{ik}\theta_{l}$$ correct and I just apply the same logic but it fails to prove $$\left\{ \frac{\partial^{R}}{\partial \theta_{i}} , \theta_{k} \right\} = \delta_{ik}$$.

OP has a point. Ref. 1 did not expect that OP would evaluate eq. (3.197) on a Grassmann-odd function $$f$$, and thereby effectively insert a Grassmann-odd object in between the $$\theta$$-derivative and $$\theta$$.
1. Introduce left and right multiplication operators: $$m^L_f(g)~:=~fg \quad\text{and}\quad m^R_f(g)~:=~gf.$$ Then $$\left\{\frac{\partial^L}{\partial\theta^i}, m^L_{\theta^k} \right\}_+~=~\delta^k_i \quad\text{and}\quad \left\{\frac{\partial^R}{\partial\theta^i}, m^R_{\theta^k} \right\}_+~=~\delta^k_i.\tag{3.197'}$$
2. Introduce the notation $$(\frac{\stackrel{\rightarrow}{\partial}}{\partial\theta^i}f) ~=~(\frac{\partial^L}{\partial\theta^i}f)\quad\text{and}\quad (f\frac{\stackrel{\leftarrow}{\partial}}{\partial\theta^i}) ~=~(\frac{\partial^R}{\partial\theta^i}f),$$ where a right derivative stands to the right of its argument. Then $$\left\{\frac{\stackrel{\rightarrow}{\partial}}{\partial\theta^i}, \theta^k \right\}_+~=~\delta^k_i \quad\text{and}\quad \left\{\frac{\stackrel{\leftarrow}{\partial}}{\partial\theta^i}, \theta^k \right\}_+~=~\delta^k_i.\tag{3.197"}$$