# Transformation of Weyl spinors

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.

• 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. – Stephan Nov 27 '17 at 2:07

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.