# Lorentz-invariant Lagrangian for spinor field

Schwartz book on QFT (page 167), Zee book on group (page 461) and Maggiore book on QFT (page 55), prove that $$\psi_R^{\dagger}\sigma^{\mu}\psi_R$$ is a 4-vector, where $$\psi_R$$ is a right-handed spinor field and $$\sigma^{\mu} \equiv (I,\sigma^i)$$.

So far so good, then from that result they trivially states that $$\psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R$$ is Lorentz-invariant, and here I am lost.

I know that $$\partial_{\mu}(\psi_R^{\dagger}\sigma^{\mu}\psi_R)$$ is Lorentz-invariant, but how trivially comes that $$\psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R$$ is Lorentz-invariant?

Moreover, I believe it would help the reader of that books if they states how the product $$\psi_R^{\dagger}\psi_R$$ and the $$\partial_{\mu}$$ is defined when applied to a spinor. I guess the product is the usual scalar product of two complex vectors, and the derivative merely acts on each element of the vector.

I might haver found a trivial proof, if it's correct:

$$\partial_{\mu}(\psi_R^{\dagger}\sigma^{\mu}\psi_R) = \partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R + \psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R$$

$$[\partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R]^{\dagger} = \psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R$$

$$\partial_{\mu}(\psi_R^{\dagger}\sigma^{\mu}\psi_R) = \partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R + [\partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R]^{\dagger}$$

$$\partial_{\mu}(\psi_R^{\dagger}\sigma^{\mu}\psi_R) = f_{\psi_R}(x) + f_{\psi_R}^{\dagger}(x)$$

where $$f_{\psi_R}:\Bbb R^4\to \Bbb C$$

We know that $$\partial_{\mu}(\psi_R^{\dagger}\sigma^{\mu}\psi_R)$$ is a Lorentz-invariant, so for every right-handed spinor field (and then for every $$f_{\psi_R}:\Bbb R^4\to \Bbb C$$) we have (Lorentz transformation $$\Lambda$$ is linear):

$$\Lambda(f_{\psi_R} + f_{\psi_R}^{\dagger}) = \Lambda(f_{\psi_R}) + \Lambda(f_{\psi_R}^{\dagger}) = f_{\psi_R} + f_{\psi_R}^{\dagger}$$

and then necessarily:

$$\Lambda(f_{\psi_R}) = f_{\psi_R}, \,\Lambda(f_{\psi_R}^{\dagger}) = f^{\dagger}_{\psi_R}$$

which means

$$\Lambda (\psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R) = \psi_R^{\dagger}\sigma^{\mu}\partial_{\mu}\psi_R$$

$$\Lambda (\partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R) = \partial_{\mu}(\psi_R^{\dagger})\sigma^{\mu}\psi_R$$

• $\sigma^aX_a=X_{AA'}\in SL(2,C)$ is invariant under orthochronous Lorentz transformation of $X^a$ (see: jstor.org/stable/20520687 ). Which means $\sigma^{\mu}\partial_{\mu}$ 2x2 matrix operator is invariant. Further, the parameters appearing in Lorentz trnaformation $\Lambda=\exp(\theta_{ab}S^{ab})$ are constants,... that would explain the lorentz invariance of $\psi^{\dagger}_R\sigma^{\mu}\partial_{\mu}\psi_R$
– KP99
Nov 28, 2021 at 16:27