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).

• Comment to the post (v1): There is an important $i$ missing compared to the usual Dirac Lagrangian. – Qmechanic Jun 10 '17 at 15:58
• I've added the prefactor, but it's not relevant to my question. – WIMP Jun 11 '17 at 4:54
• An overall non-zero constant factor in the action is of course irrelevant for the classical EOM, and not what I meant. – Qmechanic Jun 11 '17 at 6:41

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.

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.

• The Lagrangian has to be real valued. This means the complex conjugate is either the same as the first term, in which case it doesn't matter that I wrote it, or it's not, in which case it has to be there. – WIMP Jun 11 '17 at 4:47
• I checked this in a standard textbook and it's correct. It's normally written with a double-sided arrow over the derivative, but I couldn't figure out how to make this. There should be a factor 1/2 in front of the Lagrangian, but that doesn't matter for my question. – WIMP Jun 11 '17 at 4:51
• My answer has been edited – CStarAlgebra Jun 11 '17 at 4:57
• Explain the downvotes? – CStarAlgebra Jun 11 '17 at 15:57
• ? I haven't downvoted anything. – WIMP Jun 12 '17 at 5:26