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The following is an excerpt from K. Varga's paper, Precise solution of few-body problems with stochastic variational method on correlated Gaussian basis:

...The function $θ_{LM_L}(\mathbf{x})$ in Eq. (2), which represents the angular part of the wave function, is a generalization of $\mathcal{Y}$ and can be chosen as a vector-coupled product of solid spherical harmonics of the Jacobi coordinates $$ θ_{LM_L}(\mathbf{x}) = [[[\mathcal{Y}_{l_1}(\mathbf{x}_1) \mathcal{Y}_{l_2}(\mathbf{x}_2)]_{L_{12}} \mathcal{Y}_{l_3}(\mathbf{x}_3)]_{L_{123}}, \ldots]_{LM_L}.\tag{5} $$ Each relative motion has a definite angular momentum...

What is the RHS of the above equation? I've never seen this nested-bracket notation before, and as far as I can tell, it isn't defined in the paper. My first guess would be something like a commutator $$[A, B] = AB - BA$$ but that doesn't explain the subscripts on the closing brackets.

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This is pretty niche notation, and it is indeed not defined in the paper, but the name "vector-coupled product" does seem to be used by a few people beyond Varga and Suzuki. In essence, $$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} $$ is a coupled wavefunction with total angular momentum $L$ that's made up of the single-particle wavefunctions $\mathcal Y_{l_1}(\mathbf x_1)$ and $\mathcal Y_{l_2}(\mathbf x_2)$, which have angular momentum $l_1$ and $l_2$ respectively. This means that you need to couple them via Clebsch-Gordan coefficients as usual.

Thus, the product above is given by $$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} = \sum_{m_1,m_2} ⟨l_1m_1,l_2m_2|LM⟩ \mathcal Y_{l_1m_1}(\mathbf x_1)\mathcal Y_{l_2m_2}(\mathbf x_2) $$ where the sum is over all permissible magnetic quantum numbers for the individual particles.

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  • $\begingroup$ This definitely sounds like the right answer! But could you give a reference as to where you found this definition? I'd be interested in knowing where else this notation is used. $\endgroup$ – David Zhang Aug 15 '15 at 16:22
  • $\begingroup$ This was pieced together from the Google results for 'vector-coupled product', but it makes natural sense as it's the only wavefunction with the required quantum numbers (well defined angular momentum for each individual particle, total angular momentum, and total magnetic quantum number). $\endgroup$ – Emilio Pisanty Aug 15 '15 at 18:45

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