I was reading the article "Dynamical Supersymmetry Breaking" by Yael Shadmi and Yuri Shirman. In particular i was studing the relation between the presence of a global continus spontaneusly broken symmetry and SUSY breaking. All it's clear, except for this statement (on page 28-29):

"If the global symmetry is spontaneously broken, there is a massless scalar field, the Goldstone boson, with no potential. With unbroken supersymmetry, the Goldstone boson is part of a chiral supermultiplet that contains an additional massless scalar, again with no potential. ... "

I don't get who is the scalar companion of the Goldstone boson: if i think of the simple N=1 susy the massless chiral supermultiplet contains only one scalar and one Weyl Fermion.

Any suggestion?


1 Answer 1


The Goldstone boson of a spontaneously global symmetry is a real scalar. This is in contrast to the complex scalar that is part of a chiral supermultiplet.

When a global symmetry is broken in the supersymmetric limit, the real scalar Goldstone boson has another real scalar partner so that they can together form a complex scalar. As you note, there is also a Weyl fermion.


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