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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?

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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 $\bar{\Psi}\Psi$.

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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 ($3.2.17$).

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    $\begingroup$ yes, it is exactly the last part of my answer where I say that it's a part of $\bar{\Psi}\Psi$. $\endgroup$ – TwoBs Sep 8 '14 at 10:47

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