I have been reading 'Quantum Field Theory and the Standard Model' by Schwartz and have gotten stuck on a line of reasoning in Section 10.2.2.

I understand that we can construct a (right-handed) four-vector $V_R^\mu$ given by, $$ V_R^\mu = (\psi_R^\dagger\psi_R, \psi_R^\dagger\vec{\sigma}\psi_R). $$ Schwartz then says that (eq. 10.54) $$ \psi_{R}^{\dagger} \partial_{t} \psi_{R}+\psi_{R}^{\dagger} \partial_{j} \sigma_{j} \psi_{R} $$ is therefore a Lorentz invariant. This can be proven using the same methodology as here, but how does this result follow from the construction of $V_R^\mu$?

I expect that $\partial_\mu V_R^\mu$ is Lorentz invariant but this isn't equivalent to the expression given by Schwartz.

Any help is appreciated!

  • $\begingroup$ Are you using the $S=\int L dx$ to determine Lorentz invariance? $\endgroup$ Dec 25, 2019 at 1:18

1 Answer 1


$\partial_{\mu}V^{\mu}$ is a total derivative term in the Lagrangian, hence can be dropped. The term written in Eq.10.54 is the kinetic term in the Lagrangian which satisfies Weyl equation of motion which comes from helicity of the particular representation. \begin{eqnarray} \sigma^{\mu}_{+}p_{\mu}u_{R}=0\\ \sigma^{\mu}_{+}=(I,\vec{\sigma}) \end{eqnarray} $u_{R}$ is the helicity eigenstate of right handed Weyl Fermion.

  • $\begingroup$ In the Schwartz's book, the author uses the Lorentz invariance of $\psi_{R}^{\dagger} \partial_{t} \psi_{R}+\psi_{R}^{\dagger} \partial_{j} \sigma_{j} \psi_{R}$ directly to derive the Dirac Lagrangian from which the Weyl equation can be derived by taking $m\rightarrow 0$. Could you derive this from the fact that $V_R^\mu$ is a 4-vector? $\endgroup$
    – asdf
    Dec 2, 2020 at 17:23
  • $\begingroup$ We need another vector that can contract with $V^{\mu}_R$ to give a scalar. Now, $\partial_{\mu} V^{\mu}$ is a total derivative, but we can use a part of it.\begin{equation}\partial_{\mu} V^{\mu} =\partial_{\mu}\psi^{\dagger}_{R}\sigma^{\mu}\psi_R + \psi^{\dagger}_{R}\sigma^{\mu}\partial_{\mu} \psi_R\end{equation} we can use either of terms in the right hand side of the equation to write down Dirac Lagrangian. $\endgroup$
    Dec 4, 2020 at 6:11
  • $\begingroup$ I'm sorry but I cannot get your point. We can include $\psi^{\dagger}_{R}\sigma^{\mu}\partial_{\mu} \psi_R$ in the Lagrangian if it is Lorentz invariant, and the author insists that it's indeed so. Why on earth is $\psi^{\dagger}_{R}\sigma^{\mu}\partial_{\mu} \psi_R$ Lorentz invariant? $\endgroup$
    – asdf
    Dec 4, 2020 at 10:54
  • $\begingroup$ @asdf Have you ever figured this out? $\endgroup$
    – Yadeses
    Jan 15 at 20:25
  • $\begingroup$ Use lorentz transformation properties for R type helicity spinor and transformation for \delta_\mu. From there the lorentz invariance property would come. $\endgroup$
    Jan 17 at 3:43

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