5
$\begingroup$

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?

Improve this question
$\endgroup$
12
  • 4
    $\begingroup$ 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. $\endgroup$
    – Brandon Enright
    Jan 27, 2014 at 18:54
  • 2
    $\begingroup$ 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. $\endgroup$
    – Andrew
    Jan 27, 2014 at 21:02
  • 2
    $\begingroup$ @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. $\endgroup$
    – Edvard
    Jan 27, 2014 at 21:18
  • 2
    $\begingroup$ @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. $\endgroup$
    – Luboš Motl
    Jan 28, 2014 at 12:48
  • 1
    $\begingroup$ @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. $\endgroup$
    – Luboš Motl
    Jan 28, 2014 at 12:49

1 Answer 1

Reset to default
-4
$\begingroup$

If there is a person that writes beautiful historical reviews about subjects related to physics then that person is Ed. Witten... because of this, I quote here these papers where you may find information about SUSY breaking mechanisms, dynamical breaking and mainly how susy is different from "ordinary symmetries". http://inspirehep.net/record/10634?ln=en

If you read that paper carefully, and maybe some papers about origins of susy you may even find hints for why susy is not real at all... but that's up to your ability to see things not stated directly ;)

Improve this answer
$\endgroup$
1
  • $\begingroup$ Downvoted, since I believe you answer and the article you cited don't actually answer the question. Sorry $\endgroup$
    – Federico Carta
    Jan 30, 2014 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.