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Question

Is there a quick way to identify the branches in a 3J symbol?

Context

I need to compute Wigner 3J symbols/Clebsch–Gordan coefficients,

$$ \begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&m&-m \end{pmatrix} , $$

for all configurations of ($\ell_1,\ell_2,\ell_3$) up to $\ell_\text{max}\sim 2000$, and $m=\mathcal{O}(1) $. Speed is crucial so rather than computing the 3j a billion times, I would resort to an interpolation scheme.

Interpolating the raw 3j symbol is impossible because it has four branches; for example, consider this plot of the 3j coefficients as a function of $\ell_3$ with $\ell_1=120$, $\ell_2=90$ and $m=2$:

3j symbol with l1=120, l2=90, m=2 (source)

Messy, uh? Let's split it in two plots based on the parity of $\ell_1+\ell_2+\ell_3$. We have two branches for even $\ell_3$:

3j symbol with l1=120, l2=90, m=2 and EVEN l3 (source)

and two branches for odd $\ell_3$:

3j symbol with l1=120, l2=90, m=2 and ODD l3 (source)

Each branch is rather smooth, as long as $m$ stays small; this means that interpolation is possible on a per-branch basis. However, first I need a way to identify which branch corresponds to a given $(\ell_1,\ell_2,\ell_3)$. Hence the question: is there a quick way to identify the branch of a 3j symbol?

Note that this question is a follow-up of my previous question, Sign of 3j symbol, where I asked about the sign of

$$ \begin{pmatrix} \ell_1 &\ell_2 &\ell_3\\ 0&0&0 \end{pmatrix} , $$

which, thanks to @GjergjiZaimi, I now know is $-1^{(\ell_1+\ell_2+\ell_3)/2}$. In that simplified case, knowing the sign was equivalent to identifying the 3J branch. In the $m\neq0$ case, however, the sign is not enough because each branch crosses the zero.

Thank you for your attention, Guido

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  • $\begingroup$ In your example, your "branches" depend periodically on the residue of $\ell_3$ mod 4. I take from your question that this is special to your values $\ell_1 = 120, \ell_2 = 90, m = 2$? Have you worked other examples to see what the pattern is? $\endgroup$
    – Theo Johnson-Freyd
    Commented Oct 31, 2015 at 21:08
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    $\begingroup$ Any reason you are not already using well-known recurrence relations to speed up the calculations and asymptotic formulas instead of interpolation? $\endgroup$
    – Igor Khavkine
    Commented Nov 1, 2015 at 9:33

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