# Weyl Spinors and Lorentz Invariance

Let $\phi_a$ and $\chi_{\dot{a}}$ be two component commuting spinors, where $\chi$ is an anti-spinor.

In terms of some spinor basis, these can both be written in some arbitrary frame as $$\phi_a(P) = \sqrt{E - |p|}\xi_a^-,~~~~~\chi_{\dot{a}}(P) = \sqrt{E+|p|}\tilde{\xi}^-_{\dot{a}}$$ Where $\xi_a^- \neq \tilde{\xi}_{\dot{a}}^-$ and $(\xi^-)^2 = (\tilde{\xi})^2 = 1$

I want to construct a Lorentz scalar from these guys, such as $$\phi_a(P_1)\phi^a(P_2) = \sqrt{(E_1-|p_1|)(E_2 - |p_2|)},~~~~~\chi_\dot{a}(P_1)\chi^\dot{a}(P_2) = \sqrt{(E_1+|p_1|)(E_2 + |p_2|)}$$ If these are Lorentz invariant, then it should not matter what frame I choose to evaluate them in. Thus, choosing the rest frame where $P_1 = P_2 = 0$, I find that these two are equal, since now $$\phi_1\phi_2 = \sqrt{m_1m_2} = \chi_1\chi_2$$ However, as soon as I boost to another frame, these no longer seem to be equal, which would imply that they are not Lorentz invariants.

Any ideas?

EDIT: If the particles both have the same mass, then this can be seen to hold in the COM frame too. Imagining that both particles are travelling along the $z$ axis, then $E_1 = E_2 = E$ and $|p_1| = -|p_2| = |p|$. In that case, we have $$\phi_1\phi_2 = \sqrt{(E-|p|)(E + |p|)} = m = \chi_1\chi_2$$

• Why the downvote? – Akoben Dec 7 '17 at 11:57