Dear Murat,

you can think of this like for any other conserved quantity. Heavy particles will always decay, with a rate at least proportional to their mass, except for if one of the following two conditions is fulfilled:

1. The decay is kinematically not allowed: This is like for an electron, which could in principle decay into a muon and neutrinos, since the necessary interaction does exist (i.e., the matrix element is non-zero). However, given that the mass of the muon is larger than that of the electron, energy conservation forbids such a decay (i.e., the phase space is zero).

2. The decay violates some conservation low: For kinematical reasons, there is no problem with an electron decaying into, say, two photons (i.e., the phase space is non-zero). However, such a hypothetical decay would violate both, conservation of elelectric charge and of angular momentum, which is why one cannot draw any Feynman diagram transmitting that decay (i.e., the matrix element is zero).

For any sparticle (i.e., the supersymmetric partners of the Standard Model particles) we have such a conserved charge, which is called R-parity. R-parity is to some extend "artificial", in the sense that supersymmetry does not per se require it, but if we do not impose R-parity, then protons would decay very quickly and the resulting theory would just not resemble our real world. If we do impose R-parity, however, there is no contradiction to what we see in Nature (assuming SUSY is broken).

Thus, unless R-parity is violated, all sparticles (which are heavy) would "like" to decay, because of their masses being so large that the decay rates are in fact quite high. However, they can only decay into other sparticles, since any decay conserves R-parity. Still, as long as we can draw a diagram which involves a lighter sparticle, then the decay will ultimately happen.

Now the situation is simple: for any sparticle (except for the LSP), we can draw such a diagram involving the LSP as only sparticle in the final state (just try it in case you do not believe it right away). Unless the initial sparticle is extremely degenerate in mass to the LSP (in which case we would effectively have two LSPs), this decay is kinematically allowed and violates no conservation law - it will thus happen. Only the LSP is absolutely stable, since for any possible diagram transmitting its decay either R-parity would be violated (i.e., the matrix element would be zero) or the decay would be kinematically not allowed (i.e., the phase space would be zero).

Does this clear up the situation for you?

Best regards,

Alexander

R-parity is defined as (-1)3B+L+2S where B is baryon number L is lepton number and S is spin. With this definition particle and corresponding s-particle has opposite R parity. For example electron has B=0,L=1,S=1/2 that is R=+1, whereas slepton has B=0,L=1,S=0, that is R=-1. You can go on in this way and check that for all quarks leptons  gauge bosons and scalars the particle will have R=1 and super partner will have R=-1. All allowed vertices in the MSSM Lagrangian (by construction) conserve R parity, which is a multiplicative quantum number.

As R-parity is conserved at all orders in MSSM, superparticles must decay in a channel containing superparticles. LSP is the lightest superparticle; as it cannot find a kinametically allowed decay channel, it becomes stable.

R-parity violating couplings can also be included in a supersymmetric Lagrangian, but then one has to go beyond the MSSM.

I thought sparticles have conservation lows other than R-parity. For example, squark should not decay into sleptons, like quarks do not decay into leptons. Because supermultiplets only difference is their spins.

Dear Murat,

you are of course right that  sparticles also obey other conservation laws. However, they are typically quite simple to fulfill, since you can always produce an SM particle which is comaratively light and carries the same quantum numbers.

Best regards,

Alexander

Dear Alexander

Thank you, now i got it.

Best regards

Murat Abdughani