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So, in doing some numerical computations in QFT, I've run into the following Wigner $6j$-Symbol:

$$ \left\{ \begin{array}{ccc} x & J_1 & J_2 \\ \frac{N}{2} & \frac{N}{2} & \frac{N}{2} \\ \end{array} \right\}. $$

In the regime where $x \ll J_1,J_2,N$ and $J_1 \approx J_2 \approx N$, and $N$ is large. I would like to know if there is an asymptotic formula for such a symbol, or if one can be derived. Using symmetries we can get

$$ \left\{ \begin{array}{ccc} x & \frac{1}{2} \left(J_1+J_2\right) & \frac{1}{2} \left(J_1+J_2\right) \\ \frac{N}{2} & \frac{1}{2} \left(N+J_1-J_2\right) & \frac{1}{2} \left(N-J_1+J_2\right) \\ \end{array} \right\}. $$

Perhaps this could help, I'm really not sure.

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2 Answers 2

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The source for this is the book of Varshalovich et al, Quantum Theory of angular momentum. In section 9.8 one can find the following: $$ \left\{\begin{array}{ccc} a&b&c\\ d+R&e+R& f+R\end{array}\right\} \approx \frac{(-1)^{a+b+d+e}}{\sqrt{2R(2c+1)}}C^{c\gamma}_{a\alpha;b\beta} $$ where $C^{c\gamma}_{a\alpha;b\beta}$ is a Clebsch Gordan coefficient, and where $\alpha=f-e, \beta=d-f, \gamma=d-e$. This expression is valid in the limit where $R\gg 1$.

(I have never personally used this but with a few simple test using Mathematica gives a pretty good estimate. For instance, with $(a,b,c,d,e,f)=(3,3,2,2,4,4)$ and $R=75$, the $6j\approx -0.01763$ while the approximate expression gives $-0.01781$.)

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For the asymptotic behavior of the Wigner 6j-Symbol when all the coefficients but one grow, you can use the Edmonds formula. In your case it reads as:

$$ \left\lbrace\begin{matrix} x & J_1 & J_2 \\ \frac{N}{2} & \frac{N}{2} & \frac{N}{2}\end{matrix}\right\rbrace \approx \dfrac{(-1)^{J_2+N+x}}{\sqrt{(2J_2+1)(N+1)}}d^x_{J1-J2, 0}(\phi), $$

where $d^x_{J1-J2, 0}(\phi)$ is the small Wigner d-matrix and

$$ \cos(\phi)=\frac{1}{2}\sqrt{\dfrac{J_1(J_1+1)}{N/2(N/2+1)}}. $$

Here it is a good reference https://aip.scitation.org/doi/10.1063/1.532474 .

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  • $\begingroup$ Welcome !Could you elaborate the key points of this publication here? This would make your answer within this thread a lot more useful. $\endgroup$
    – engineer
    Commented Oct 23, 2020 at 8:57

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