# Converting two component product to four component notation

Consider the product of two left Weyl spinors in the notation commonly found in supersymmetry, \begin{equation} \chi ^\alpha\eta_\alpha = \chi ^\alpha \epsilon _{ \alpha \beta } \eta ^\beta \end{equation} This is equal to, \begin{equation} \left( \begin{array}{c} \chi ^\alpha \\ 0 \end{array} \right) ^T\left( \begin{array}{cc} \epsilon _{ \alpha \beta } & 0 \\ 0 & \epsilon ^{ \dot{\alpha} \dot{\beta} } \end{array} \right) \left( \begin{array}{c} \eta ^\beta \\ 0 \end{array} \right) = \bar{\eta} _L ^\ast \gamma _0 C \chi _L \end{equation} where I have used some common spinor identites and defined, $\eta _L \equiv P _L \eta, \chi _L \equiv P _L \chi$ ($\eta$ and $\chi$ are now four component spinors). I also use the defintion, $C \equiv i \gamma_0 \gamma _2$. While I don't think anything is particularly wrong with this derivation, I have never seen a term like this in normal quantum field theory. It there a simpler way to reformulate this to correspond to common expression for such mass terms or is my uncomfort with this term due to my ignorance?

Just realize that you can form ordinary Dirac spinors from 2-spinors by using charge conjugation, $i\sigma_2\eta^*$, that gives a right- handed field that can fit in the right-handed slot (forming a 4 component Majorana field) $$\Psi_1=\left(\begin{array}{c}\eta \\ i\sigma_2\eta^*\end{array}\right)$$ And analogous for $\Psi_2$ in terms of $\chi$. Then you just look at the 'mass terms' $\bar\Psi_1 \Psi_2$ to get your term (well in fact you need to insert also a $P_L$). I think the textbook by Ramond shows this kind of things.
Actually, if you add also the hermitian conjugate to your expression, you can even fit all in a single Dirac spinor $$\Psi=\left(\begin{array}{c}\eta \\ i\sigma_2\chi^*\end{array}\right)$$ and look at $\bar{\Psi}\Psi$.
Following this ref, one sees that, in some basis where the current is diagonal ($3.2.16$), then a term like $\chi \eta$ is just a part of the mass term ($3.2.17$).
• yes, it is exactly the last part of my answer where I say that it's a part of $\bar{\Psi}\Psi$. Sep 8 '14 at 10:47