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In my course, the teacher wrote the Dirac Lagrangian as :

$$ \mathcal{L}=\frac{i}{2} \bar{\psi}\gamma^{\mu}\overleftrightarrow{\partial_\mu} \psi -m \bar{\psi} \psi $$

I just would like to understand what the operator $\overleftrightarrow{\partial}$ mean ? I couldn't find the answer on the internet by myself because they almost always give the complex Lagrangian and not the real one (I know they differ from a surface term).

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    $\begingroup$ It is shorthand for the direction the derivative is operating in: see physicsforums.com/threads/… about halfway down, for an example. $\endgroup$
    – user176049
    Nov 25, 2017 at 10:15

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As Countto10 said, this is a shorthand notation: $$ f\stackrel{\leftrightarrow}{\partial_\mu}g \equiv f\,(\partial_\mu g)-(\partial_\mu \,f)\,g $$ see e.g. Srednicki's QFT book, page 40. To arrive at the usual Dirac Lagrangian,
\begin{align*} \mathcal L &= \frac{\text i}{2} \bar\psi\gamma^\mu\!\!\stackrel{\leftrightarrow}{\partial_\mu} \psi -m \bar\psi \psi\\ &= \frac{\text i}{2} \big( \bar\psi \gamma^\mu\partial_\mu \psi - \partial_\mu\bar\psi\gamma^\mu\psi \big)-m \bar\psi \psi\\ &= \frac{\text i}{2} \bar\psi \gamma^\mu\partial_\mu \psi - \frac{\text i}{2} \big[ \underbrace{\partial_\mu(\bar\psi\gamma^\mu\psi)}_{\text{surface term}}-\bar\psi \gamma^\mu\partial_\mu \psi \big]-m \bar\psi \psi\\ &= \text i \bar\psi \gamma^\mu\partial_\mu \psi-m \bar\psi \psi\\ &= \bar\psi(\text i \gamma^\mu\partial_\mu - m)\psi. \end{align*}

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