# Forming massive $\mathcal{N} = 2$ supermultiplets

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This question stems from page 14 of John Terning's Modern Supersymmetry.

Suppose we consider $\mathcal{N} = 2$ supersymmetry, and denote the spin-0 Clifford vaccuum by $\Omega_0$. Now, the non-trivial anticommutators of the SUSY algebra for a massive supermultiplet of mass $m$ are

$\{Q_\alpha^a, {Q^\dagger}_{\dot{\alpha}b}\} = 2 m \delta_{\alpha\dot{\alpha}}\delta^a_b$

where $a, b = 1, \ldots, \mathcal{N}$. Let's build the states.

$|\Omega_0\rangle$ spin = 0

${Q^\dagger}_{\dot{\alpha}a} |\Omega_0\rangle$ spin = 0 + 1/2 = 1/2

${Q^\dagger}_{\dot{\alpha}a} {Q^\dagger}_{\dot{\beta}b} |\Omega_0\rangle$ naively, spin = 0 + 1/2 + 1/2 = 1, except this doesn't seem to be true, because you can get both spin 0 and spin 1 in this case.

In the book, the author first provides a schematic construction of the supermultiplet starting from a Clifford vacuum of spin s. Then, he considers a specific example for $\mathcal{N} = 2$.

But in the general case of spin s, he doesn't seem to distinguish between states of spin $s + 1/2$ and $s - 1/2$. Why not?

I am guessing this is more to do with notation than anything else, but doesn't $s$ denote the TOTAL SPIN? A massive state should be labeled then by $|m, s, s_3\rangle$ and not just the total spin.

To be perfectly clear, what happens when $Q^\dagger$ acts on a general state of spin $s$? Does it produce a rung of states with total spin ranging from $-|s-1/2|$ to $|s+1/2|$?

This post imported from StackExchange Physics at 2015-01-01 13:06 (UTC), posted by SE-user leastaction
asked Jan 1, 2015
Try p.172 of Muller-Kirsten/Wiedemann (Introduction to Supersimmetry).

This post imported from StackExchange Physics at 2015-01-01 13:06 (UTC), posted by SE-user Rexcirus

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But in the general case of spin s, he doesn't seem to distinguish between states of spin s+1/2 ands−1/2. Why not?

I am not sure I understand your confusion but think of the following. The reason people do not consider this is because these states are CPT related. Despite that if you take a look at the representation tables  you will find the info you need and the classification of the states (including the CPT conjugates). In specific, this is the article you should be looking into. Check also Witten's lecture on Seiberg-Witten theory from IAS.

We conclude that CPT flips the sign of the helicity. Therefore, unless the helicity is distributed symmetrically around 0, which is not the case in general, a supermultiplet is not CPT-invariant. This means that in order to have a CPT-invariant theory one should in general double the supermultiplet we have just constructed adding its CPT conjugate.  This is not needed if the supermultiplet is itself CPT conjugate but $\mathcal{N}=2$ is not! Note simply that we have two $\mathcal{N}=1$ chiral supermultiplets (plus our vector supermultiplet).

$Q_{\dot{\alpha}a} Q_{\dot{\beta} b} =|\Omega_0 \rangle$ = naively, spin=0+1/2+1/2=1

Now this is not true. In a chiral multiplet you only get up to spin 1/2. Remember that you are acting with one of the $\mathcal{N}=1$ susy's onto the spin 0 vacuum. To get vectors you need to act on the 1/2 vacuums.

answered Jan 1, 2015 by (3,625 points)
Any good references?
@Outlander John Terning is an excellent SUSY reference. I strongly recommend it.

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