Consider an $\mathcal{N}=1$ left-handed chiral supermultiplet. The particle content is

$$L = (\phi\quad e_L) $$

where $\phi$ is a complex scalar and $e_L$ a left handed Weyl fermion.

People usually denote a right-handed fermion as $\bar{e}$ or $e^c$ (see for example this review). So we can say that $e_R$ is the same as $\bar{e}$ or $e^c$.

If I have $\bar{e}_R$ field, i.e $e_R^\dagger \gamma_0$, is that just the same as $e_L$?


Your Answer

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

Browse other questions tagged or ask your own question.