Is There a Symmetric Form of the Dirac Lagrangian?

[ChrisVer]

Is there any way to write the Dirac lagrangian to have symmetric derivatives (acting on both sides)? Of course someone can do that by trying to make the Lagrangian completely hermitian by adding the hermitian conjugate, and he'll get the same equations of motion (a 1/2 must exist in that case)...However same equations of motion, imply that there should be a connection between the two Lagrangians via a 4divergence, right? Unfortunately I cannot see what that could be.
L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi
In particular choosing
L_{D}= \bar{\psi} (i γ^{\mu}∂_{\mu}-m) \psi + ∂_{\mu}K^{\mu}
seems to bring some result if I choose K^{\mu}=i\bar{\psi}\gamma^{\mu}\psi
but the initial lagrangian seems to get doubled...

 

[ChrisVer]

L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi -m \bar{\psi}\psi

L_{D}=\frac{1}{2} \bar{\psi} i γ^{\mu}∂_{\mu} \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi) - \frac{1}{2}i (∂_{\mu}\bar{\psi})\gamma^{\mu}\psi -m \bar{\psi}\psi

L_{D}=\frac{1}{2} \bar{\psi} (i γ^{\mu}[∂_{\mu}^{→}-∂_{\mu}^{←}] -2m) \psi + \frac{1}{2} i ∂_{\mu}(\bar{\psi}γ^{\mu}\psi)

Should the - exist?

 

[The_Duck]

Yes, your last expression is right.