# Acting for a covariant derivative on charged spinor [closed]

For field, theory what i know $i.e$,complex scalar QED \begin{align} D_\mu \phi = \partial_{\mu} \phi - i Q A_{\mu} \phi \end{align} and \begin{align} D_\mu \phi^{\dagger} = \partial_{\mu} \phi^{\dagger} + i Q A_{\mu} \phi^{\dagger} \end{align}

In Geometry, by adapting vielbein formulation for spinor $\psi$, \begin{align} D_\nu \psi = \partial_\nu \psi + \frac{1}{4} w_{\nu ab} \gamma^{[a} \gamma^{b]} \psi, \quad D_\nu \bar{\psi} = \partial_\nu \bar{\psi} - \frac{1}{4} w_{\nu ab} \bar{\psi} \gamma^{[a}\gamma^{b]} \end{align}

Consider charged spinor, then the covariant derivatives

\begin{align} D_a \psi= e^{\mu}{}_a (\partial_\mu - i eQ A_\mu + \frac{1}{2} \sigma^{bc} e_b{}^\nu \partial_\mu e_{c\nu})\psi, \quad \sigma^{ab} = \frac{\gamma^a \gamma^b - \gamma^b \gamma^a}{4} \end{align} Then how about $D_a \bar{\psi}$?

My guess is \begin{align} D_a \bar{\psi}= e^{\mu}{}_a (\partial_\mu\bar{\psi} + i eQ A_\mu \bar{\psi} - \frac{1}{2} \bar{\psi}\sigma^{bc} e_b{}^\nu \partial_\mu e_{c\nu}), \quad \sigma^{ab} = \frac{\gamma^a \gamma^b - \gamma^b \gamma^a}{4} \end{align} Is this right?

## closed as off-topic by ACuriousMind♦, user36790, honeste_vivere, CuriousOne, knzhouJul 7 '16 at 7:27

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• Use the fact that $\bar \psi \psi$ is a scalar, apply the derivative and the Leibniz rule, it should be obvious enough – Slereah Jul 6 '16 at 9:40
• – Qmechanic Jul 7 '16 at 18:36