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Why must the superpotential be analytic?

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In the MSSM, it is necessary to have two Higgs doublets. The reason often given is that, in the standard model one uses the Higgs field to give mass to up-type fermions, and its complex conjugate for the down-type ones. However, in the MSSM this is not possible because the superpotential must be an analytic (holomorphic) function of the fields, and taking the c.c. is not analytic.

My question is, why must the superpotential $W$ be analytic? In the plain SM, the Lagrangian does not seem to have this restriction. Is the analycity of $W$ necessary for the Lagrangian to be invariant under SUSY?

asked Jun 25, 2014 in Theoretical Physics by jdm (50 points) [ no revision ]

The superpotential must be holomorphic function (and hence analytic) of chiral superfields -- this follows from writing supersymmetric actions using superfields. See related discussion here: http://physicsoverflow.org/19122/using-covariant-derivative-when-building-superpotential

1 Answer

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This is transparent if you work in a manifestly supersymmetric formalism such as superspace.

There are two kinds of superfields in a four-dimensional $N=1$ supersymmetric field theory such as the MSSM : chiral superfields and vector superfields.  The physical degrees of freedom in a chiral superfield are a fermion, a scalar and a pseudoscalar; whereas in the vector superfield it's a vector field and a fermion.  Hence the Higgs fields are component fields of a chiral superfield.

The most general renormalisable N=1 supersymmetric action involving chiral superfields has two kinds of terms:

  1. a kinetic term, which involves the Berezin integral  over all of superspace: $$ \int d^4x d^2\theta d^2\bar\theta \Phi\bar\Phi$$  or, if you abandon renormalisability, then more generally $$ \int d^4x d^2\theta d^2\bar\theta K(\Phi,\bar\Phi)$$ where $K$ is a real function (which can be interpreted geometrically as a Kähler potential); and
  2. superpotential term, which is the Berezin integral over "half of the superspace": $$\int d^4x d^2\theta W(\Phi) + c.c.$$

For this to be supersymmetric, the function $W(\Phi)$ must itself be a chiral superfield, which means that it must be a function of $\Phi$ alone and not of $\bar\Phi$.  Renormalisability further constrains $W$ to be at most cubic.

So in summary, supersymmetry says that the superpotential has to be a chiral superfield and this in turn means that is a function of $\Phi$, which is tantamount to analyticity.

answered Jun 26, 2014 by José Figueroa-O'Farrill (2,135 points) [ revision history ]

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