Representation of SUSY Algebra

[karlzr]

I have some questions about representations of SUSY algebra.
(1) Take $N=1$ as an example. Massive supermultiplet can be constructed in this way:
$$|\Omega>\\
Q_1^\dagger|\Omega>, Q_2^\dagger|\Omega>\\
Q_1^\dagger Q_2^\dagger|\Omega>$$ I understand the z-components $s_z$ of the last state and the first state are the same, but why do they also have the same total spin $s$?

(2)How do we get to know whether the fermions are Weyl or majorana? For instance $N=2$ hypermultiplet
$$|\Omega_{-\frac{1}{2}}>: \chi_\alpha\\
Q^\dagger|\Omega_{-\frac{1}{2}}>: \phi\\
Q^\dagger Q^\dagger |\Omega_{-\frac{1}{2}}>: \psi^{\dagger \dot{\alpha}}$$ Is this representation CPT invariant? if so, I guess $\chi$ or $\psi$ should be majorana
Or we might need to supplement the states with their CPT conjugates when the two fermion fields are weyl?

 

[AndyW]

Hello, thank you for your questions. Let me try to address them one by one.

(1) In SUSY algebra, the supercharges $Q_1$ and $Q_2$ have opposite spinor indices, which means they have different spin projections. However, when acting on the vacuum state $|\Omega>$, they create states with the same total spin $s$. This is because the supercharges are defined as a sum of fermionic and bosonic operators, and the overall spin of a state is determined by the number of fermions and bosons it contains. In this case, the states created by the supercharges have one fermion and one boson, giving them the same total spin.

(2) To determine whether the fermions in a representation are Weyl or Majorana, we need to look at the transformation properties under CPT. In the $N=2$ hypermultiplet example, the fermions transform as a Weyl spinor under CPT, which means they are Weyl fermions. This can also be seen from the fact that the fermions have only one spinor index, indicating that they are left- or right-handed. If the fermions were Majorana, they would have two spinor indices and would transform differently under CPT.

I hope this helps clarify your questions. If you have any further questions, please don't hesitate to ask.