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What do the supercharges in extended supersymmetry do?

In ${\cal N}=1$ supersymmetry there are a certain number of fermions and and equal number of bosons. You can transform all fermions to the bosons (and vice versa) in a 1 to 1 fashion using a single supercharge, $Q$.

So what happens when you have, for example, ${\cal N}=2$ supersymmetry with 8 supercharges? Since $Q$ is a generator of supersymmetry transformations, is it a linear combination of these supercharges that act on the particles? In which case could one particle be acted on by two separate linear combinations of $Q$? Or is it strictly one linear combination of $Q$ per fermion/boson?

Also, what does ${\cal N}$ mean physically? What difference does ${\cal N}=2$ have to ${\cal N}=1$ other than more supercharges? Or is that the only difference between the two theories?

Qmechanic
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Siraj R Khan
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1 Answers1

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You need to be a bit careful about the counting of supercharges. In four dimensions, the smallest spinor representation is four-dimensional (over the real numbers), so $\mathcal{N}=1$ supersymmetry has four supercharges.

When there are more supercharges, there are simply more states combined into a single 'multiplet'. For example, in $\mathcal{N}=1$ supersymmetry, a theory with a massless vector boson like a photon necessarily also contains a massless spin-$\frac{1}{2}$ particle, the 'superpartner' of the photon. In an $\mathcal{N}=2$ theory, a photon has more superpartners: two massless spin-$\frac{1}{2}$ particles, and a massless complex scalar. It's still always true that there are the same number of bosonic and fermionic degrees of freedom.

You can find all the details in chapters 2 and 3 of Wess and Bagger's classic supersymmetry textbook.

Rhys
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