Just to preface things, this exact question has been asked before here, but I don't feel like the answer really clarifies things for me.

The core issue is that we want to construct a 4-vector that we can contract with $\partial_\mu$ to form a Lorentz invariant kinematic term we can introduce in the Lagrangian. Schwartz finds that, for a right-handed Weyl spinor,

$$ V^\mu_R =\ \left(\psi^\dagger_R\psi_R,\psi^\dagger_R\sigma^i\psi_R\right) $$

transforms like a 4-vector. Clearly this means that $\partial_\mu V^\mu_R$ is Lorentz invariant, but this isn't useful because a total derivative makes no meaningful contribution to the Lagrangian. Schwartz then claims that the fact that $V^\mu_R$ transforms like a 4-vector must mean that

$$ \psi_R^\dagger\partial_t\psi_R+\psi^\dagger_R\partial_j\sigma^j\psi_R $$

is also Lorentz invariant. This isn't at all obvious to me, for example, in principle, why couldn't, the terms in parenthesis from the following product rule

$$ \partial_\mu V^\mu_R = \left(\psi_R^\dagger\partial_t\psi_R+\psi^\dagger_R\partial_j\sigma^j\psi_R\right)+\left(\partial_t(\psi_R^\dagger)\psi_R+\partial_j(\psi_R^\dagger)\sigma^j\psi_R\right) $$

transform oppositely such that $\partial_\mu V^\mu_R$ remains invariant? That is, why must each individual term above be invariant?

To further clarify, the reason why the answer from the post I linked doesn't help me is because helicity has not yet been introduced. In principle there should be a self contained way of understanding this.

  • $\begingroup$ Just write down the Lorentz generators and work it out. $\endgroup$ Mar 22, 2023 at 23:00
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    $\begingroup$ @ConnorBehan I get that that works, but the way Schwartz phrases it, it sounds like it should be an immediate consequence of the fact that $V^\mu_R$ is a 4-vector. This is what bothers me. $\endgroup$
    – FranDahab
    Mar 22, 2023 at 23:03
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    $\begingroup$ I suggest you read the original in Weyl's "Group theory and Quantum mechanics" His account is just so neat! $\endgroup$
    – mike stone
    Mar 22, 2023 at 23:17
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    $\begingroup$ It sounds like you're asking a yes or no question. Namely whether Schwarz gave a proper explanation. I vote no. $\endgroup$ Mar 23, 2023 at 12:51
  • $\begingroup$ @ConnorBehan Yeah I guess that's fair. Maybe I'm looking for a formal justification where there isn't one... $\endgroup$
    – FranDahab
    Mar 23, 2023 at 14:39


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