# Why must SUSY be broken?

Background

One usually claims that supersymmetry must be spontaneously broken. The reasoning is roughly the following:

Since $M^2=P^{\mu}P_{\mu}$ is a casimir operator of the supersymmetry algebra, all the particles in a supermultiplet will have the same mass. Therefore the electron and the selectron will have same mass, and we would be able to produce selectrons at the accelerators, which nowdays operate at an energy scale of $1 \ Tev >> m_e\approx 0.5 \ MeV$. But clearly no selectrons are ever seen at the $MeV$ scale.

The standard way to avoid this is to introduce some kind of supersymmetry breaking mechanism, of a similar kind of the Higgs mechanism in the standard model.

Question

Is it possible that susy is exact and not broken, but still supersymmetric particles can not produced in reaction of ordinary particles, basically because these reactions would violate the conservation of an extra (not yet known) quantum number?

Why is such a scenario discarded a priori, an so much effort is put into the study of the susy breaking mechanism?

• Usually you can get around conservation of quantum numbers by creating a pair of particles, one with 1 for the quantum number and the other with -1 for it. Jan 27, 2014 at 18:54
• Wouldn't the existence of such a conserved quantum number break SUSY? For example, lets say you assigned "Carta charge" 1 to all the superpartners and Carta charge 0 to all the SM particles and said that only charge neutral interactions could occur at TeV energies, for some reason. But now if I do a supersymmetry transformation, my fields will no longer have definite values of Carta charge. On the other hand, if Carta charge were a good symmetry then I should be able to pick my fields to be eigenstates of the Carta charge. So this is inconsistent. Jan 27, 2014 at 21:02
• @JohnRennie, it seems to me, that Fayet-Iliopoulus mechanism, where they add a term like $\xi D$, breaks SUSY spontaneously due to VEV of the $D$-field. Jan 27, 2014 at 21:18
• @Andrew, the existence of the charge - it is called R-parity and is equal to $1$ for SM particles and $-1$ for their superpartners (it adds up multiplicatively) - doesn't break SUSY in any way. The supervariation of a state $\psi$ i.e. $Q\psi$ simply has the opposite parity than the original state $\psi$. "Finite SUSY transformations" require Grassmannian parameters ("angles") and those may be formally assigned a negative R-charge, too, so that "finite supersymmetry transformations" of states have the same R-parity as the original states. Jan 28, 2014 at 12:48
• @Andrew - your (wrong) argument could be more generally used to argue that the generators of symmetries can't carry any charges. But that's just always wrong for non-Abelian symmetries in which case the generators are always charged under some other generators - they carry some charges. It's against nothing. Jan 28, 2014 at 12:49