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
\begin{equation}
\frac{\partial^{R}}{\partial \theta_{i}} \theta_{k} \theta_{l} = \theta_{k} \delta_{il} - \delta_{ik}\theta_{l}.\tag{3.195}
\end{equation}
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 $\left\{ \frac{\partial^{R}}{\partial \theta_{i}} , \theta_{k} \right\} = \delta_{ik}$.
 A: 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 $\theta$.
We can correct eq. (3.197) in at least 2 ways:

*

*Introduce left and right multiplication operators:
$$ m^L_f(g)~:=~fg \quad\text{and}\quad m^R_f(g)~:=~gf. $$
Then
$$ \left\{\frac{\partial^L}{\partial\theta^i}, m^L_{\theta^k} \right\}_+~=~\delta^k_i
\quad\text{and}\quad
\left\{\frac{\partial^R}{\partial\theta^i}, m^R_{\theta^k} \right\}_+~=~\delta^k_i.\tag{3.197'}$$


*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
$$ \left\{\frac{\stackrel{\rightarrow}{\partial}}{\partial\theta^i}, \theta^k \right\}_+~=~\delta^k_i
\quad\text{and}\quad
\left\{\frac{\stackrel{\leftarrow}{\partial}}{\partial\theta^i}, \theta^k \right\}_+~=~\delta^k_i.\tag{3.197"}$$
References:

*

*R.A. Bertlmann, Anomalies in QFT, 1996; section 3.3.1 p. 150.

