Bound States and Supersymmetry | M.E. Irizarry-Gelpí

Sat 04 October 2014
Physics
#bound states

If a theory is supersymmetric, then the states should form supersymmetric multiplets. In four spacetime dimensions, the simplest massive multiplets are the chiral multiplet,

$$\mathbf{2} \oplus 2 \cdot \mathbf{1},$$

and the $\mathcal{N} = 1$ vector multiplet,

$$\mathbf{3} \oplus 2 \cdot \mathbf{2} \oplus \mathbf{1}.$$

Here I have denoted the states by the dimension of the representation under spatial rotations. Now, consider taking the direct product of two chiral multiplets. A chiral multiplet has two bosonic and two fermionic states for a total of four states, so the direct product of two chiral multiplets should have eight bosonic and eight fermionic states for a total of sixteen. You find:

$$\left( \mathbf{2} \oplus 2 \cdot \mathbf{1} \right) \otimes \left( \mathbf{2} \oplus 2 \cdot \mathbf{1} \right) = \mathbf{3} \oplus 4 \cdot \mathbf{2} \oplus 5 \cdot \mathbf{1}.$$

The chiral multiplet is an $\mathcal{N} = 1$ multiplet. The direct product of two chiral multiplets looks like an $\mathcal{N} = 2$ massive vector multiplet without central charges, or an $\mathcal{N} = 4$ massive vector multiplet with central charges.

Now consider the direct product of two $\mathcal{N} = 1$ massive vector multiplets. One massive vector multiplet has four bosonic and four fermionic states for a total of eight states. The direct product of two $\mathcal{N} = 1$ massive vector multiplets should have 32 bosonic and 32 fermionic states, for a total of 64:

$$\left( \mathbf{3} \oplus 2 \cdot \mathbf{2} \oplus \mathbf{1} \right) \otimes \left( \mathbf{3} \oplus 2 \cdot \mathbf{2} \oplus \mathbf{1} \right) = \mathbf{5} \oplus 4 \cdot \mathbf{4} \oplus 7 \cdot \mathbf{3} \oplus 8 \cdot \mathbf{2} \oplus 6 \cdot \mathbf{1} = \left( \mathbf{5} \oplus 4 \cdot \mathbf{4} \oplus 6 \cdot \mathbf{3} \oplus 4 \cdot \mathbf{2} \oplus \mathbf{1} \right) \oplus \left( \mathbf{3} \oplus 4 \cdot \mathbf{2} \oplus 5 \cdot \mathbf{1} \right).$$

In the second form, this looks like the direct sum of an $\mathcal{N} = 2$ massive vector multiplet and an $\mathcal{N} = 2$ massive graviton multiplet, both without central charges. Each of these $\mathcal{N} = 2$ massive multiplets without central charges can be reinterpreted as $\mathcal{N} = 4$ massive multiplets with central charges.

A similar remark holds for the direct product of an $\mathcal{N} = 1$ massive chiral multiplet and an $\mathcal{N} = 1$ massive vector multiplet. One finds that the $4 \times 8 = 32$ states can be arranged into an $\mathcal{N} = 2$ massive gravitino multiplet without central charges, or an $\mathcal{N} = 4$ massive gravitino multiplet with central charges.

Why do these direct products of $\mathcal{N} = 1$ multiplets admit interpretations as extended multiplets? I arrived at this question after thinking about how would it look like to have the states of supermultiplets form bound states.