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I read that the chiral nature of SM fields is an indication that they must be realized in a N=1 supermultiplet (and not N=2). I don't quite understand how so. Please enlighten.

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There are no chiral multiplets in the $N=2$, $d=4$ supersymmetry.

To clarify this claim, one must distinguish two meanings of the word "chiral". One of them is more general – any field or interaction that is left-right-asymmetric. But there's one more specific meaning in supersymmtry: a "chiral supermultiplet" is a particular supermultiplet including a complex scalar and a Majorana or Weyl fermion – in some sense, two real bosonic and two real fermionic polarizations. This is to be contrasted with the "vector supermultiplet" and perhaps others.

A chiral field is one that only depends on $\theta^\alpha$ but not their complex conjugates $\bar\theta^\alpha$: $$ \phi\equiv \phi (x^\mu, \theta^\alpha) $$ In the Taylor decomposition over the fermionic coordinates, one finds out that this includes the complex scalar boson component and a Weyl fermion (which is better rephrased as a Majorana fermion if the field is completely neutral and allows the oscillations between the matter and antimatter). At any rate, the Weyl or Majorana fermion is 1/2 of a Dirac field and the full Dirac field is needed for a left-right-symmetric theory so the fermionic content of the chiral multiplet is chiral in the other sense.

The vector supermultiplet contains two transverse polarizations of a gauge boson (like a photon) and its Majorana superpartner; if it's charged, like the W-boson vector supermultiplet, the two Weyl fermions from the oppositely charged states combine into a Dirac particle.

In $N=2$ supersymmetry contains a vector supermultiplet as well – which decomposes as a vector plus chiral multiplet under the $N=1$ subalgebra – and a hypermultiplet – which decomposes to two chiral multiplets. However, the fermionic content of the hypermultiplet (two Weyl spinors) combines into a Dirac spinor which is non-chiral in the old sense. So the $N=2$ matter multiplet – the hypermultiplet – is non-chiral.

To some extent, we could say that the $N=2$ vector multiplet is chiral. However, the quantum numbers of these particles are determined – they have to transform as the adjoint of the gauge group (just like the gauge bosons) – so they're no good for the quarks and leptons. Moreover, the fermionic content may be combined to a Dirac spinor, too.

The minimal $N=1$ SUSY is sort of the maximum that allows chirality because this low degree of SUSY allows multiplets with a very small number of the fermionic degrees of freedom, smaller than a Dirac spinor, which are chiral. All greater supersymmetries imply that the fermionic content always comes in full Dirac spinors and is therefore non-chiral. So at least at the level of field theory, $N=1$ SUSY is the maximum to produce realistic models. However, one has to point out that this is true for quarks and leptons; the gauge bosons subsector of physics may have a greater, especially $N=2$, supersymmetry.

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I like your last remark about the gauge subsector. – arivero Jun 14 '13 at 22:08

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