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

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

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

  • 1
    $\begingroup$ $\sigma^aX_a=X_{AA'}\in SL(2,C)$ 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 $\psi^{\dagger}_R\sigma^{\mu}\partial_{\mu}\psi_R$ $\endgroup$
    – KP99
    Nov 28 '21 at 16:27
  • $\begingroup$ Unfortunately I don't have access to that link. $\endgroup$
    – Andrea
    Nov 28 '21 at 20:32
  • $\begingroup$ You’re making it far too complicated. You can show that your second term is Lorentz invariant just like you showed your first term is. $\endgroup$
    – Prahar
    Nov 29 '21 at 4:17
  • 1
    $\begingroup$ Oh. This one should help: scipp.ucsc.edu/~haber/archives/physics251_13/… $\endgroup$
    – KP99
    Nov 29 '21 at 7:22

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