# Is the derivative with respect to a fermion field Grassmann-odd?

Fermion fields anticommute because they are Grassmann numbers, that is, $$$$\psi \chi = - \chi \psi.$$$$ I was wondering whether derivatives with respect to Grassmann numbers also anticommute, as in

$$$$\frac{\partial}{\partial \psi} (\bar \psi \psi) \stackrel{?}{=} - \bar \psi \frac{\partial}{\partial \psi} (\psi) = - \bar \psi.$$$$

1. Since$$^1$$ $$(\frac{\partial}{\partial z}z)~=~1\tag{1}$$ for any supernumber-valued variable $$z$$, the Grassmann-parity of the partial derivative $$\frac{\partial}{\partial z}$$ should be the same as the Grassmann-parity of $$z$$ in order for eq. (1) to preserve Grassmann-parity.

2. In superspace $$\mathbb{R}^{n|m}\ni(x,\theta)$$ a functional derivative $$\frac{\delta}{\delta z(x,\theta)}$$ and its superfield $$z(x,\theta)$$ carry the same (opposite) Grassmann parity if the number $$m$$ of $$\theta$$'s is even (odd), respectively: $$(\frac{\delta}{\delta z(x,\theta)}z(x^{\prime},\theta^{\prime}))~=~\delta^n(x\!-\!x^{\prime})\delta^m(\theta\!-\!\theta^{\prime}) \tag{2}.$$

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$$^1$$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $$\frac{\partial}{\partial z}$$ does not act past $$z$$.

In a Grassmann algebra (or more pedantic, a $$\mathbb{Z}_2$$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,

$$\frac{\partial^L}{\partial \psi} \left(\bar{\psi}\psi\right) = (-)^{\epsilon(\bar{\psi})\epsilon(\psi)} \bar{\psi}$$

$$\frac{\partial^R}{\partial \psi} \left(\bar{\psi}\psi\right) = \bar{\psi}$$,

where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).