How to use Clebsch-Gordan coefficients for 3 particles? I have a Hamiltonian for 3 particles of spin 1 that I boiled down to:
\begin{equation}
k(\textbf{S}^2+\cdots),
\end{equation}
where:
\begin{equation}
\textbf{S}=\textbf{S}_1+\textbf{S}_2+\textbf{S}_3.
\end{equation}
I read somewhere on the internet that the Clebsch-Gordan coefficients were the thing to use to figure out the value of $\textbf{S}^2$, but I'm not seeing it. Could someone explain to me how to use the Clebsch-Gordan coefficients for this case? - In particular, I understand (more or less) Clebsch-Gordan coefficients for adding two angular momenta, but here, I have 3 particles. Is there an equivalent of Clebsch-Gordan coefficients for three angular momenta $j_1,j_2,j_3$?
 A: Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system.
For 2 particles with total angular momentum eigenvalues  $ l_1,l_2 $, such that for example $ m_1=-l_1,-l_1+1,...,+l_1 $ the composite system is - 
$$ (l_1) \otimes (l_2) \sim (l_1+l_2) \oplus ...\oplus(|l_1-l_2|) $$
2 spin half particles reduce to - 
$$ (\frac{1}{2})\otimes(\frac{1}{2})\sim (1)\oplus(0) $$
the left size decribes 2 particles, each has spin 1/2 ,and you can decribe the sany system configuration by combination of those spins - 
$$ |++\rangle , |+-\rangle, |-+\rangle, |--\rangle $$
the right side describes the system as composite system, which can be described by it's total angular momentum and it's total $z$ component - 
$$ |S_{tot}=1,m_s=1\rangle ,|S_{tot}=1,m_s=0\rangle , |S_{tot}=1,m_s=-1\rangle , |S_{tot}=0,m_s=0\rangle ,$$
As I said, the Clebsch-Gordan coefficients are the basis transformation matrix elements, for example, one of them is - 
$$ \langle++|S_{tot}=1,m_s=1\rangle $$ 
The way to add 3 spins is first to add 2 of the and then to add the third one to the composite system - 
$$ [(\frac{1}{2})\otimes (\frac{1}{2})]\otimes (\frac{1}{2})\sim [(1)\oplus(0)]\otimes(\frac{1}{2})\sim [(1)\otimes(\frac{1}{2})]\oplus[(0)\otimes(\frac{1}{2})]\sim  (\frac{3}{2})\oplus (\frac{1}{2})\oplus(\frac{1}{2})$$
(notice the first addition. the later one is "conditional" - if S1+S2=1 then ... if S1+S2=0 then)
so the new basis, in order of $ |S_{tot},S_{1+2},m_{tot}\rangle $ is
$ |\frac{3}{2},1,-\frac{3}{2}\rangle,|\frac{3}{2},1,-\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{3}{2}\rangle,|\frac{1}{2},1,\frac{1}{2}\rangle,|\frac{1}{2},1,-\frac{1}{2}\rangle,|\frac{1}{2},0,-\frac{1}{2}\rangle,|\frac{1}{2},0,\frac{1}{2}\rangle $
this is enough for most applications (finding energy spectrum and such)
A: Be aware that the order in which you do the couplings matters, in the sense that $(j_1\otimes j_2)\to j_{12}$ followed by $j_{12}\otimes j_3\to J$, usually written as $(j_1j_2)j_{12}j_3 J$, will not produce the same basis states as coupling $j_2\otimes j_3\to j_{23}$ first, and then coupling
$j_1\otimes j_{23}\to J$, or $j_1(j_2j_3)j_{23}J$.  The two sets of basis states are related by a unitary transformation that is used to define Racah coefficients as overlaps between the two bases.
