In BV formalism of the gauge theory, we need to compute the right / left functional derivatives of the actions that include fermions.
I do not quite see what it means by that.
For example, let us think of the QCD action given as
\begin{equation}
S[\psi, \overline{\psi}]:= \int d^4x \text{ } \overline{\psi} [\gamma_\mu D^\mu+m]\psi\tag{1}
\end{equation}
where $\psi$ is the quark field, $\overline{\psi}$ is its Dirac adjoint, and $D^\mu$ is the covariant derivative with respect to the given gauge group (which should be $SU(3)$ but not really important here).
Then what does it mean by
\begin{equation}
\frac{\delta_L S}{\delta {\psi}} \text{ or } \frac{\delta_R S}{\delta {\psi}}~?\tag{2}
\end{equation}
In the QFT book by Weinberg, it is simply stated that the derivative acts from the left or right, but I do not understand.. In particular, is it always true that
\begin{equation}
\frac{\delta_L S}{\delta {\psi}}=-\frac{\delta_R S}{\delta {\psi}}~?\tag{3}
\end{equation}
Could anyone please clarify?
Best Answer
TL;DR: A left derivative means a derivative that acts from the left. A right derivative means a derivative that acts from the right.
In more detail, an infinitesimal variation of a functional $F[\phi]$ is of the form $$\int \!d^4x ~ \delta \phi(x)~ \frac{\delta_L F}{\delta \phi(x)} ~=~\delta F~=~ \int \!d^4x ~~ \frac{\delta_R F}{\delta \phi(x)}~ \delta \phi(x), \tag{A}$$ where the left and right functional derivative satisfy $$\frac{\delta_L F}{\delta \phi(x)} ~=~(-1)^{|\phi|(|F|+1)} \frac{\delta_R F}{\delta \phi(x)}.\tag{B}$$ Here $|\cdot|$ denotes the Grassmann-parity.
Eq. (B) implies OP's eq. (3) for Grassmann-odd $\psi$ and Grassmann-even $S$.