Spin Statistics
Most recent answer: 06/09/2011
- Bob (age 19)
Seattle, WA, USA
Wow, it doesn't take many words to ask some very tough questions.
On spin: many particles have some internal property which is often described as having two possible states. These two states have angular momentum of ± (h/2π)/2 along some chosen axis. To be more careful, what we really mean is that any of the possible internal states can be made up out of a combination of these two states. Formally, it's a vector space of dimension 2. Usually the two states also have different (and opposite) magnetic moments. Saying that the spin is 1/2 (in units of h/2π) means that rotation of the state by 360° multiplies by -1, not by +1 as you might expect for a complete rotation. I think I've told you what spin is, but know that's not a very intuitively satisfying answer.
On Pauli's exclusion principle: There's something called the spin-statistics theorem. It says among other things that for spin-1/2 particles, only 0 or 1 particle can go in any quantum state. The argument is not trivial. You start by representing the possible states for some type of particle in terms of things called creation and annihilation operators, which either add a particle of some type or remove it. You then ask how to represent the change in the state as you go to a moving reference frame in terms of those operators. You then make a loop of those changes by adding little motions to your reference frame until you come back to the starting reference frame's motion. You can show that this is equivalent to rotating the starting frame a little. So now you have an expression for how the state changes as you rotate it a little in terms of the creation and annihilation operators. You manipulate this a bit and find that for s=1/2 the product of two creation operators for the same state is zero. So you can't have two of these in the same state. You call that the Pauli exclusion principle. See? Easy!
That strains my mind enough for today. Let me steer you to another site for lepton number: .
Mike W.
I should add that lepton number doesn't always have to be conserved. Observed neutrino oscillations violate lepton number conservation. Also, the "black holes have no hair" theorem impies that lepton number would be lost in black hole formation and evaporation.
(published on 06/09/2011)
Follow-Up #1: spin statistics
- Anonymous
As my extremely sketchy outline of the spin-statistics theorem indicates, whether a particle is a boson or a fermion depends only on whether its spin is an integer or a half-integer. So 0 is boson, 3/2 is fermion.
There are plenty of composite S=0 and S=3/2 particles around, so we aren't just relying on theory here, although the theory is extremely solid. There are numerous indications that S=0 fundamental particles exist quite aside from supersymmetry.
I don't know much about which particles are good candidates for dark matter. The key traits needed are to have some rest mass (to allow clumping with galaxies) and to lack electric charge and QCD "charge" so as not to interact much with ordinary matter.
Mike W.
(published on 07/08/2011)