Originally posted by Ric:
Umm... what? Maybe it's just me that doesn't understand this, but I would surmise from the lack of posts on this thread that im not the only one. Could you possibly restate this in a more "Dumbed down" way? Not that i'm stupid or anything, just don't understand what you're talking about... sorry. 
Little help here maybe? The idea is that supersymmetry could allow for fermions to transform into bosons. Fermions are particles like electrons and quarks. Two feermions must always be in different states. This explains why we have different kinds of atoms (and chemistry, life biology etc.).
Electrons aren't allowed to all occupy the lowest energy state. The lowest allowed energy state for a system of N fermions is a state where one is in the lowest (single particle) state another is in the next lowest etc.
Bosons can all occupy the same state. E.g. a laser beam consists of a beam of photon who are all in the same state.
What happens to a star that has run out of fuel? the gas pressure isn't able to resist gravity. Gravity will compress the gas to higher and higher densities. If the mass isn't too high this process will be halted because the pressure will rise sharply at a certain density. This rise in pressure is caused by the electrons who don't want to be in the same state and are forced to go to higher energy states (the smaller the volume the less states are available below the same energy). The contraction then halts and you end up with a white dwarf star. In such a star the electrons have very high energies.
Suppose that electrons could transorm themselves to bosons. Then, because the bosons are allowed to occupy the same state, all the elctrons could transform themselves to the bosons in the lowest energy state. An enormous amount of energy could thus be released.
Another effect of such a transition would be that the released energy (in the form of gamma rays)would be beamed in one direction. This is because of a similar effect that makes lasers work: Stimulated emission. Bosons can not only occupy the same states, they actually prefer to be in the same state! once you have photons in the same state (moving in the same direction), other atoms will preferentially decay producing more of the same photons.
This last property may sound mysterious (bosons surely don't have free will?), but it can be explained (intuitively) as follows. According to the rules of quantum mechanics, the probability that a state x evolves to a state y is the same as the reverse process of y evolving into x. Let's apply this to atoms decaying producing photons.
An atom in an exited state as some probability of making a transition to the ground state, producing a photon. That photon can travel in any possible direction. Suppose that the atom decays and a photon is produced. Suppose that you have another atom also in an exited state and the photon collides with that atom. The probablity that the atom will decay producing another photon moving in the same direction is now enhanced. To see this let's apply the rule that the reverse process has the same probability.
The reverse process would be a process in which two photons are in the same state and are colliding with an atom in the ground state. Each of these photons can bring the atom to the exited state, so the probability is twice as likely than if you had only one photon. The one photon case would be the reverse of the process in which you started with an atom in the excited state and there were no photons interacting with it.
So, in general, if there are n identical photons interacting with an atom in an excited state the decay probability producing another photon in the same state is enhanced by a factor of (n + 1), because, in the reverse process, you could let each of the n + 1 photons interact with the atom in the ground state, bringing the atom to the same exited state.
Note that a process in which you produce an aditional photon in a different state is not enhanced, because in the reverse process you are then only allowed to let that additional photon interact with the ground state atom bringing it back to the exited state. If you use one of the other photons then, you are left with the ''additionally produced photon'' but this was assumed to be different from the others. So, this is process isn't the reverse of the decay process.
