I usually see Weyl spinor and Weyl equations as derived from Dirac equation, like in Peskin. Now, I'm following a course where the professor actually builds Weyl spinor lagrangians BEFORE talking about Dirac equation.

So he takes a Weyl field $\psi_L$ (he built Weyl fields from group theory, as objects transforming under a $(1/2,0)$ representation) and he defines $\bar{\sigma}^\mu=(\mathbb{1}, -\vec{\sigma})$ and $\sigma^\mu=(\mathbb{1}, \vec{\sigma})$. He then says that $\psi_L^\dagger\bar{\sigma}^\mu\psi_L$ is a vector while $\psi_L^\dagger\sigma^\mu\psi_L$ is not, and it's immediate to check.

I don't understand how I am supposed to prove these facts without going into pages of calculations, passing to the infinitesimal form and searching for the $[\bar{\sigma}^\mu, \sigma^i]$ commutation relations. It's important for me to understand this point because from there he builds the invariant lagrangian as $\mathcal{L}_L=\psi_L^\dagger\bar{\sigma}^\mu\partial_\mu\psi_L$ and he derives the EOM, etc.

Am I missing something? Maybe even a conceptual point that could prove that easily.

  • $\begingroup$ If you use the transformation of a spinor under a Lorentz transformation and look at $\mu=0$, $\mu=i$ separately, you can derive the transformation of a regular four-vector. This is example 2.3 in Labelle’s “Supersymmetry Demystified” (solution in appendix). On mobile right now so I can’t type every detail, I’m sorry. $\endgroup$
    – ersbygre1
    Nov 27, 2017 at 2:07
  • 1
    $\begingroup$ Although the one who asked this question may have solved it years ago, I still want to add that the answers can be found in chapter 19.1 and 19.2 in Coleman's qft book. explained in a clear way. the Lorentz invariant Lagrangian can be derived quickly from the content in the book. $\endgroup$ Jul 9 at 4:12
  • $\begingroup$ @Lagrange629 Thanks! Indeed, I probably solved it at the time but your comment will be very useful to other people :) $\endgroup$
    – Luthien
    Jul 10 at 11:07

1 Answer 1


Contracting the vectorial indices of the 4-momentum with the sigma matrices as $P = P_{m}\sigma^{m}$, the Lorentz transformations will act on these resulting matrices as $P\rightarrow M P M^{\dagger}$, where $\det(M)=1$. In order to see why $\det(M)=1$, just note that $\det(P)=P_mP^{m}$, and by definition the Lorentz transformations are the ones that preserve $A_{m}A^{m}$. All this matrices $M$ forms a known group called the $SL(2,\mathbb C)$ group. It is very useful to write explicitly the $SL(2,\mathbb C)$ indices, as:

$$ \chi_{\alpha}\rightarrow M_{\alpha}\,^{\beta}\chi_{\beta},\qquad\bar\chi_{\dot\alpha}=(\chi_{\alpha})^{*}\rightarrow M_{\dot\alpha}\,^{\dot\beta}\chi_{\dot\beta} $$

where $M_{\dot\alpha}\,^{\dot\beta}=(M_{\alpha}\,^{\beta})^{*}$ is the complex conjugation of the $SL(2,\mathbb C)$ matrix $M$. With this we have that in order to $P\rightarrow M P M^{\dagger}$ the $\sigma^{m}$ matrix should have the following index structure: $\sigma^{m}_{\alpha\dot\alpha}$. Now, since $\det(M)=1$ we have:

$$ \varepsilon_{\alpha\beta}M_{\gamma}\,^{\alpha}M_{\delta}\,^{\beta}=\varepsilon_{\gamma\delta} $$

and the same for putting dots in each SL(2,C) index. This means that we can raise and lower indices by using the $\varepsilon_{\alpha\beta}$ and its inverse $\varepsilon^{\alpha\beta}$ (same for the dotted indices). This allows you to define:

$$ \bar\sigma^{m\dot\alpha\alpha}=\varepsilon^{\alpha\beta}\varepsilon^{\dot\alpha\dot\beta}\sigma^{m}_{\beta\dot\beta} $$

Now returning to your question, it becomes very easy too see why $\psi_{L}^{\dagger}\sigma^{m}\psi_{L}$ is not Lorentz covariant, since the index structure does not agree, i.e. there are contraction between dotted and undotted indices:

$$ \bar\psi^{\dot\alpha}\sigma^{m}_{\alpha\dot\beta} \psi^{\beta} $$

As you can see the $SL(2,\mathbb C)$ notation prohibits contractions of spinorial indices that are not covariant under Lorentz transformations. Note that $\psi_{L}\sigma^{m}\psi_{L}^{\dagger}$ does transforms covariantly under Lorentz.


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