# Clebsch-Gordan coefficients for more than 2 particles

I need to couple arbitrary spins together and need Clebsch-Gordan coefficients for them. This should be just coupling the last two particles, then couple the next until the first particle is coupled.

Given that we have $$\langle J, M | j_1, m_1, j_2, m_2 \rangle$$, I think that the higher ones should be computable with this recursion relation:

\begin{aligned} &\langle J, M | j_1, m_1, j_2, m_2, j_3, m_3, \ldots \rangle \\&\qquad= \sum_{\tilde J = |j_2 - j_3|}^{j_2 + j_3} \sum_{\tilde M = - \tilde J}^{\tilde J} \langle J, M | j_1, m_1, \tilde J, \tilde M \rangle \langle \tilde J, \tilde M | j_2, m_2, j_3, m_3, \ldots \rangle \,. \end{aligned}

Is that correct?

• You need Racah coefficietns' technology. Messiah v II Ch XIII §29 and Appendix C , § II . Online here. Commented Feb 7, 2019 at 16:59
• Commented Feb 7, 2019 at 17:04

You seem to be thinking that there will be one single $$J$$ representation in the resulting space. That's not correct - you can get multiple independent representations.
As an example, adding two spin-1 particles gives you $$\mathbf{1}\otimes \mathbf{1} = \mathbf{0}\oplus \mathbf{1} \oplus \mathbf{2}$$, and adding a third spin-1 particle produces \begin{align}(\mathbf{1}\otimes \mathbf{1})\otimes\mathbf{1} & = (\mathbf{0}\otimes\mathbf{1})\oplus (\mathbf{1}\otimes \mathbf{1}) \oplus (\mathbf{2}\otimes\mathbf{1}) \\& = \mathbf{1}\oplus (\mathbf{0}\oplus \mathbf{1} \oplus \mathbf{2}) \oplus (\mathbf{1}\oplus \mathbf{2} \oplus \mathbf{3}) \\& = \mathbf{0}\oplus \mathbf{1}\oplus\mathbf{1} \oplus \mathbf{1}\oplus \mathbf{2} \oplus \mathbf{2} \oplus \mathbf{3}, \end{align} i.e. with three independent spin-1 spaces, and two independent spin-2 spaces.
Your answer is correct but is not unique: it is perfectly possible to first combine $$j_1$$ and $$j_2$$ to get $$j_{12}$$, and then combine $$j_{12}$$ to $$j_3$$ to get $$J$$. This ordering will produce different states than if you were to first combine $$j_2$$ and $$j_3$$ to $$j_{23}$$, and then combine $$j_1$$ last to get $$J$$, as you are suggesting.
The set of states $$\{\vert j_1j_2j_3;j_{12};JM\rangle\}$$ is a linear combination of the set $$\{\vert j_1j_2j_3;j_{23};JM\rangle\}$$. This is because there will in general be more than one state with specific $$JM$$ values. The coefficients in the linear combinations are actually recoupling coefficients (Racah $$U$$ coefficients although one often uses $$6j$$ symbols, which are just proportional to the $$U$$s).