3
$\begingroup$

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

\begin{equation} \frac{\partial}{\partial \psi} (\bar \psi \psi) \stackrel{?}{=} - \bar \psi \frac{\partial}{\partial \psi} (\psi) = - \bar \psi. \end{equation}

$\endgroup$
7
$\begingroup$
  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: $$ (\frac{\delta}{\delta z(x,\theta)}z(x^{\prime},\theta^{\prime}))~=~\delta^n(x\!-\!x^{\prime})\delta^m(\theta\!-\!\theta^{\prime}) \tag{2}.$$

--

$^1$ 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 $z$.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.