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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?

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  • $\begingroup$ You need Racah coefficietns' technology. Messiah v II Ch XIII §29 and Appendix C , § II . Online here. $\endgroup$ – Cosmas Zachos Feb 7 at 16:59
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Feb 7 at 17:04
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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).

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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.

(This is explored in a bit more depth in my answer to Adding 3 electron spins, particularly regarding the question of whether those final representations can be chosen in a way that has definite symmetry under particle exchange.)

I'm unsure quite how you go from there to a full formal specification of the total addition in terms of a Clebsch-Gordan recursion. But it's nevertheless clear that any formalism that doesn't account for this degeneracy is bound to fail.

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