Jan22-14, 01:31 AM
From supersymmetry, gauge particles have superpartners, gauginos. Supersymmetry breaking will make all the gauginos massive, since none have been observed. But that has certain problems.
A gauge field is a multiplet in its gauge group where each member corresponds to a generator of that group. That puts the field in the adjoint representation of that group. There's a theorem that states that that rep is always a real rep, so a gauge field can be real-valued instead of complex-valued without loss of generality. A gauge field is a massless vector field in the absence of the Higgs mechanism or anything similar, meaning that it has degrees of freedom corresponding to helicities +1 and -1.
By supersymmetry, each gauge-field generator has a corresponding gaugino mode with only two degrees of freedom. That makes gauginos Majorana fields, with helicities +1/2 and -1/2. Is that right about them?
If they get mass from SUSY breaking, that would make them massive Majorana fields.
Massive Majorana fields follow the Majorana equation - Wikipedia:
i*D(ψ) = m*ψc
ψ is the field, ψc is its charge conjugate, m is the mass, and D is the derivative operator γμ.Dμ
ψc = i*C.ψ*
where C is some matrix, the identity matrix in the Majorana basis.
At first sight, it seems as if a massive Majorana field cannot be in a nontrivial rep of a gauge group. But if the rep's group-element matrices are all real, then it becomes possible.
So a Majorana particle can be in any real rep of a gauge group. That includes the adjoint rep, meaning that massive Majorana gauginos are possible. Is that correct?
|Register to reply|
|gauge invariance requires gauge bosons, why not for neutral fermions?||Quantum Physics||19|
|When will we know whether neutrinos are Majorana fermions?||High Energy, Nuclear, Particle Physics||1|
|48 Majorana fermions in type II?||General Physics||0|
|Massive Fermions in Four Spatial Dimensions||General Physics||10|
|Why gauge bosons, but no gauge fermions||Quantum Physics||47|