The Dirac Lagrangian (Density) is defined in the text "Quantum Field Theory, An Integrated Approach" by Fradkin as:
$$\mathcal{L}=\bar{\Psi}\left(i\not{\partial}-m\right)\Psi\equiv \frac{1}{2}\bar{\Psi}i\stackrel{\leftrightarrow}{\not{\partial}}\Psi-m\bar{\Psi}\Psi\tag{2.142}$$
Where:
$$\bar{\Psi}\stackrel{\leftrightarrow}{\not{\partial}}\Psi\equiv\bar{\Psi}\left(\not{\partial}\Psi\right)-\left(\partial_\mu\bar{\Psi}\right)\gamma^\mu\Psi\tag{p.26} $$
Ignoring $m$, if the $\partial_\mu$ only acts rightward (on $\Psi$) I have trouble seeing how this relation holds. I've tried using the anticommutation relation: $\left\{\gamma^\mu,\gamma^\nu\right\}=2g^{\mu\nu}I$, but I can't get the correct relationship.
Does the partial act both leftwards and rightwards to give this relation or is there a simple mathematical manipulation that I am missing?
Best Answer
The $\equiv$ symbol in eq. (2.142) allegedly means equality modulo total divergence terms (which don't change the Euler-Lagrange (EL) equations).