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Can we have supersymmetry using real scalar instead of complex scalars?

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I am aware that a suersymmetric theories containing a complex scalar a Weyl fermion and an auxiliary field exist.

I was wondering if we can have something analogous using real and not complex scalar fields, and some kind of real Weyl fermions (if such a thing does indeed exist).

asked Aug 15, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

A spinor satisfying a reality condition is a Majorana spinor and the question is about the existence of a Majorana-Weyl spinor. But in four spacetime dimensions (in Lorentzian signature), no Majorana-Weyl spinor exists, the Weyl spinor is the smallest and so the minimal supersymmetry in four dimensions has four real supercharges and the scalar superpartner of a Weyl spinor has to be a complex scalar.

It is possible to find something analogous in three spacetime dimensions. In this case, the smallest spinor is a Majorana spinor, with two real components, and so the minimal supersymmetry in three dimensions has two real supercharges (half the number of the minimal supersymmetry in four dimensions) and the scalar superpartner of a Majorana spinor in three dimensions is a real scalar.

answered Aug 15, 2015 by 40227 (4,660 points) [ no revision ]

A hands-on (but not rigorous and definitive) approach would be to investigate the supermultiplets (irreducible representations) arising in the $N=1$ superfield. There, the chiral and antichiral superfields completely cover the scalar+fermion degree of freedom and what is left is only the vector+bispinor degrees of freedom (which is what you get from the "real superfield" under the Wess-Zumino gauge). I am not sure, however, whether it is not possible to get a longitudinal component of the vector field ($\sim$ spin 0 because $\partial_\mu \phi \leftrightarrow A_\mu$) from an appropriately constructed Lagrangian of the "real superfield". @silvrfuck 

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