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Symmetry groups

by Dreak
Tags: groups, symmetry
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Aug19-13, 02:08 PM
P: 52

PF have helped me a lot understanding a lot of important things in physics, I hope you guys can help me with this too :).

I have problems understand the symmetry groups.

I know there are groups like SU(2), O(3).. etc. But I have no idea how they represent certain particles.

So particles are characterized by spin (0, 1, 1/2...) at which the spin of a particle gets represented by the SO(3) subgroup which 'forms' the corresponding field.

A) Where does 'SO(3)' comes from?

B) Each group represents a certain spin; for example: for spin 1/2. We need to rotate over 720 to get the 'same' particle (can't find the exact English term right now :X). But how do I know which rotation corresponds to which spin?

C) How does this all connect to symmetry breaking?
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Aug21-13, 09:40 PM
P: 4
The elementary particles can be classified according to space-time isometry group representations. In case of the Minkovsky space-time it's the Poincare group: translations and Lorentz transformations. This translations can be thought of as transitions to another reference systems. If a particle is thought of as elementary it's reasonable to expect that a corresponding representation is "in one piece" -- irreducible. So the classification of elementary particles comes to classification of irreducible representations.

For the Poincare group they are determined by two numbers(excluding "an infinite/continuous spin" representations and mixed symmetry cases(the latter corresponds to several "spin numbers"; they arise if space-time dimension if more than four)). One of them changes smoothly. It's mass(a square of mass, to be precise). There are two different cases: m = 0 and m > 0. Another one changes discretely. We call it spin.

Why 0, 1/2, 1, ...? Because it can be shown that the representation to be irreducible must be the functions defined on an orbit of special orthochronous Lorentz subgroup. In the simplest massive case they are poles(one of two) of a two sheeted hyperboloid p^2 = p0^2 - p1^2 - p2^2 - p3^2 = m^2. The action of an aforementioned Lorentz subgroup is transitive of the pole. So we have to describe a representation of a subgroup which leaves some point invariant. It's easy to see(in a coordinate system where the point have coordinates (m, 0, 0, 0)) that this subgroup is SO(3)(O(3) to be precise, but let's forget about it right now).

So here comes SO(3). The hard problem(to find irreducible representation of an non-compact group) have been reduced to the easy one(rep of a compact -- SO(3)). The standard way is an infinitesimal way -- to find a Lie algebra representation and then exponentiate it to get a corresponding group representation. so(3) algebra irreps are classified by one non-negative integer number. But only a half of them(even ones) corresponds to some irreps of SO(3)(and exhaust them). Odd numbers correspond to spin representations. Roughly speaking, their exponentiations are not SO(3) representations(but can be said that they're projective ones) but it's universal covering. Which happens to be SU(2).
Aug30-13, 03:36 AM
Sci Advisor
P: 5,437
Dreak, are you talking about spin of particles according to SU(2) representations? Or are you talking about other symmetry groups like isospin, color, ... as well?

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