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  Who is the “scalar supersymmetric companion” of a Goldstone boson?

+ 2 like - 0 dislike

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?

Thanks a lot!

asked Mar 27, 2018 in Theoretical Physics by SuperBaba (20 points) [ revision history ]
edited Mar 27, 2018 by SuperBaba

It's a complex scalar so it has two real degrees of freedom.

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