Suppose we have an object X composed of three spin-1/2 fermions, R, G, and B. I should mention that “spin-1/2” means SU(2) here, in particular the usual 2-dimensional Pauli spin. The three fermions can each have spin +- 1/2, what spin states can object X have? This is a problem learned in undergraduate quantum mechanics; the answer is that 2x2x2 = 4+2+2, that is, one obtains a spin-3/2 quadruplet, and two spin-1/2 doublets. If that didn’t make any sense, then the wikipedia article on 2×2 = 3+1 spin might help, but this explanation is longer and better.

One easy way to derive the decomposition for n spin-1/2 particles uses Pascal’s triangle, and the fact that the simple representations of SU(2) have quantum numbers that run from -n/2 to +n/2, and have multiplicity 1 at each location. This fact is proved on Wikipedia in the usual method of the place (short and difficult to follow), but is probably well known to most of the readers. If we are to graph the quantum numbers of the various representations, we need one dimension, i.e. a number line, which we will draw horizontally. Then the various simple representations of SU(2) have quantum numbers as follows:

In the above, symbols “1” give the multiplicity of the eigenvalue. The spin-3/2 states, which are the “4” of 2x2x2 = 4+2+2, are circled in red. For the simple representations of SU(2) these are all the same so we could have used a dot, but farther down we’ll be dealing with higher multiplicities. Each representation is labeled by its highest spin state. The various representations have been assembled this way so that they are at least a little reminiscent of Pascal’s triangle.

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