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I want to combine three spin half particles and this is what I have so far.

I used the lowering operator $J_{-}$ on the top states and found the following states fine:

$$|\frac{3}{2},\frac{3}{2}\rangle , |\frac{3}{2},\frac{1}{2}\rangle , |\frac{3}{2},0\rangle , |\frac{3}{2},-\frac{1}{2}\rangle , |\frac{3}{2},-\frac{3}{2}\rangle $$

So that is the combination of three spin $\frac{1}{2}$ particles is equivalent to a spin $\frac{3}{2}$ particle, right?

In the case where I did it for combining two spin $\frac{1}{2}$ particles I found this was equivalent to a spin 1 particle and an additional spin zero particle.

So my question is, is there a singlet or anymore that accompany what I found for the case where I combined three spin $\frac{1}{2}$?

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    $\begingroup$ I didn't read your question carefully when I posted my answer (thanks for the upvote) but now I notice you list five states, including one with m=0. I'm quite sure that's wrong...there's always going to be an odd half-spin left over with three electrons. $\endgroup$ – Marty Green Mar 21 '14 at 21:14
  • $\begingroup$ cheers just spotted that $\endgroup$ – user32462 Mar 21 '14 at 21:32
  • $\begingroup$ physics.stackexchange.com/questions/29443/… $\endgroup$ – Gerenuk Dec 22 '16 at 6:17
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It's funny how hard this question turns out to be. I wrote a blogpost about it last year which you can read here. But here's the short version.

There are six possible states you can create in terms of total spin (l) and z-axis spin (m), which are:

  1. (3/2, 3/2)
  2. (3/2, 1/2)
  3. (1/2, 1/2)
  4. (1/2, -1/2)
  5. (3/2, -1/2)
  6. (3/2, -3/2)

It's easy to make states 1 and 6 from three electrons: you just take the states

1 = uuu (all up); and,

6 = ddd (all down)

Its the m=1/2 that get tricky: how do you know which is the l=3/2 state, and which is the l=1/2 state? The problem is there are six ways to mix three electrons:

A= uud B= udu C= duu

and three more with m = -1/2, but we don't need to write them out because everthing is the same as the spin-up case. How do we mix the A, B and C states to give us alternately (3/2, 1/2) and (1/2, 1/2)?

As I explain in my blogpost, it's not to hard to see that the symmetric combination A+B+C gives you the (3/2, 1/2) state. But what about the other state? When we had two electrons, it was easier: we just took the symmetric combination to give you the triplet state and the antisymmetric combination to give you the singlet state. But how do you do it when there are three? It turns out there are different ways, but maybe the nicest way is to use the cube roots of unity, w^3 = 1. We can then write:

(1/2, 1/2) = A + wB + (w^2)C

or, alternately:

(1/2, 1/2) = A + (w^2)B + wC

So there are actually two independent ways of making the (1/2, 1/2) state.

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We know 2 x 2 = 3 + 1. (A symmetric vector and an antisymmetric scalar state). Add one more:

2 x 2 x 2 = 3 x 2 + 1 x 2 = 4 + 2 + 2.

"4" is symmetric, with

uuu = 3/2,

uud + udu + duu = 1/2,

ddu + dud + udd = -1/2,

ddd = -3/2 .

The scalar times a spinor is a mixed symmetry j=1/2:

+1/2 = (ud-du)u

-1/2 (ud-du)d

The remaining "2" is mixed symmetry, and orthogonal to the above:

+1/2 = udu + duu - 2duu

-1/2 = dud + udd - 2udd

(with the appropriate normalization applied where needed).

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No, it is not equivalent. The Spin $s = S = 3/2$ particle will have spin projections between $S_3 = 3/2$ and $-3/2$, as you have worked out. That is it, it will just be a multiplet with 5 members.

The three particles with spin $s = 1/2$ can also have a combined spin with $S = 3/2$ which will form the same 5-multiplet. However, they can also form a $S = 1/2$ multiplet, which is a triplet. With two electrons for instance, you will have a symmetric and an antisymmetric comination of up and down, one will the the $|1, 0\rangle$ from the triplet, where the other will be the $|0, 0\rangle$ from the singlet.

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protected by Qmechanic Sep 29 '15 at 5:14

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