I've learned how to add two 1/2-spins, which you can do with C-G-coefficients. There are 4 states (one singlet, three triplet states). States are symmetric or antisymmetric and the quantum numbers needed are total spin and total z-component.

But how do you add three 1/2-spins? It should yield 8 different eigenstates. Which quantum numbers do you need to characterise the 8 states?

It is not as easy as using C-G-coefficients and the usual quantum numbers as for the total momentum the doubly degenerate 1/2 state and quadruple degenerate 3/2 state can describe only 6 or the 8 states. You will need an additional quantum number for the degeneracy.

So how do you get the result?

(I actually tried out myself with a large 8x8 matrix. The total spin 1/2 is each doubly degenerate. For the additional quantum number I chose the cyclic permutation. Spin 1/2 states are neither symmetric nor antisymmetric. But what is the usual way to derive this?)

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The straightforward attempt is to first add two 1/2-spins and to this result add another 1/2-spin. – qoqosz Jun 3 '12 at 20:27
Does a look in Edmonds (books.google.co.uk/books/about/…) help? – Emilio Pisanty Jun 4 '12 at 0:52
@qoqosz: That comment is quite unhelpful. Read the whole question where I already write some objections, that puzzled other theorists who thought "it is straighforward". – Gerenuk Jun 4 '12 at 7:21
@Emilio: I'll keep that as a reference. Supposedly that book offers a solution, but I don't have it and only wanted to know a short outline :) – Gerenuk Jun 4 '12 at 7:22
@Gerenuk when you add two 1/2-spins with C-G coefficients than you won't get pure $|1,0\rangle , \ldots$ states but linear combinations e.g. $|\frac{1}{2}, -\frac{1}{2}\rangle \oplus |\frac{1}{2}, \frac{1}{2}\rangle = \frac{1}{\sqrt{2}} |1,0\rangle - \frac{1}{\sqrt{2}}|0,0\rangle$. Adding 3 spins yields 8 such linear combinations. – qoqosz Jun 4 '12 at 8:01

I don't understand the answers given, nor the reference to an 8x8 matrix. When combing 3 doublets (aka spin 1/2), the guiding principle is that:

2 x 2 x 2 = 4 + 2 + 2

that is, the tensor product can be decomposed into a tensor sum of composite states that are a quartet (spin 3/2) and 2 doublets (spin 1/2).

The quartet is symmetric under interchange and is (up to normalization):

|3/2, 3/2> = uuu

|3/2, 1/2> = uud + udu + duu

|3/2,-1/2> = ddu + dud + udd

|3/2,-3/2> = ddd

The 2 doublets are combinations of:

u(ud-du) = uud-udu

d(ud-du) = dud-ddu

and

(ud-du)u = udu-duu

(ud-du)d = udd-dud

and appear to have mixed symmetry. For example, 1 doublet is:

|1/2, 1/2> = uud + udu - 2duu

|1/2,-1/2> = ddu + udd - 2udd

while an (unnormalized) orthogonal combination is

|1/2, 1/2> = 2uud - udu - duu

|1/2,-1/2> = 2ddu - udd - udd

Any other combination w/ zero spin-3/2 is a linear combo of these 2.

The 8x8 matrix indicates a misunderstanding of the problem: while we can make product states where we know each particles spin-there by justifying an 8x8 operator-- those states are not eigenstates of total angular momentum, hence we don't want to consider them.

We consider the combinations that are eigenstates of total angular momentum, and the way to find them is as the 1st answer stated: pair the 1st and 2nd spin into a spin 1 triplet and a spin 0 singlet, and then using CG coefficients, take their products with a doublet:

spin-1 times spin 1/2:

3 x 2 = 4 + 2

spin-0 times spin 1/2:

1 x 2 = 2

(which is how the doublets broke down explicitly, as shown above).

So recapping, given the product of 3 doublets, break it down pairwise:

2 x 2 x 2 = (2 x 2) x 2

(2 x 2) x 2 = (3 + 1) x 2

(3 + 1) x 2 = (3 x 2) + (1 x 2)

(3 x 2) + (1 x 2) = (4 + 2) + (2) =

4 + 2 + 2

Also: the desire for symmetry is great, but in general, only the extrema case |J,J> is symmetric, and the other states have mixed symmetry. There may or may not be an antisymmetric case. See:

for more on that.

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