Lagrangian for a massless Dirac field a total derivative? I've confused myself about a rather trivial point. I could write the Lagrangian of the Dirac equation as
$\cal L = {\rm i}/2 \left( \bar \psi \gamma^\nu \partial_\nu \psi + {\rm cc} \right)$
which, for all I can tell is the same as
${\rm i}/2 \left( \partial_\nu (\bar \psi \gamma^\nu \psi) \right) $
So, assuming the current vanishes sufficiently fast at infinity, the volume integral should always vanish, regardless of what $\psi$ is. But that can't be because that would mean that, according to the principle of least action, literally all wave-functions would be solutions (and they'd all be stable under variation too).
 A: Your Lagrangian is not real valued. In fact, it is imaginary valued!
The term in your parentheses is real valued, when it is multiplied by $i$, it is imaginary valued.
The correct Lagrangian should be
$$\mathcal{L} = \frac{i}{2}\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi + {\rm cc} = \frac{i}{2}\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - \frac{i}{2}\partial_{\mu}\bar{\psi}\gamma^{\mu}\psi = i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - i\partial_{\mu} (\bar{\psi}\gamma\psi)$$
If we neglect the total divergence, we have
$$\mathcal{L} = i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$$
This is the most common form of Lagrangian of Dirac field in most textbooks.
A: The Dirac Lagrangian contains only the first term with no complex conjugate, so you cannot integrate by parts and obtain ONLY a total derivate. 
If you add the complex conjugate, you are not working with the Dirac Lagrangian.
Edit: The Lagrangian does not have to be real valued, it is operator valued anyways. What's important is that the lagrangian is hermitian, so all observables are real valued. 
Don't forget the i in the lagrangian, when you add the complex conjugate I don't think you can integrate by parts to get a total derivative alone. Be careful with the complex conjugate. 
