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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.