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Asymptotics of the Wigner $6j$-Symbol

Physics Asked by Mesoscopic C on May 13, 2021

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

$$
left{
begin{array}{ccc}
x & frac{1}{2} left(J_1+J_2right) & frac{1}{2} left(J_1+J_2right)
frac{N}{2} & frac{1}{2} left(N+J_1-J_2right) & frac{1}{2} left(N-J_1+J_2right)
end{array}
right}.
$$

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

2 Answers

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: $$ left{begin{array}{ccc} a&b&c d+R&e+R& f+Rend{array}right} approx frac{(-1)^{a+b+d+e}}{sqrt{2R(2c+1)}}C^{cgamma}_{aalpha;bbeta} $$ where $C^{cgamma}_{aalpha;bbeta}$ is a Clebsch Gordan coefficient, and where $alpha=f-e, beta=d-f, gamma=d-e$. This expression is valid in the limit where $Rgg 1$.

(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 $6japprox -0.01763$ while the approximate expression gives $-0.01781$.)

Answered by ZeroTheHero on May 13, 2021

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:

$$ leftlbracebegin{matrix} x & J_1 & J_2 frac{N}{2} & frac{N}{2} & frac{N}{2}end{matrix}rightrbrace approx dfrac{(-1)^{J_2+N+x}}{sqrt{(2J_2+1)(N+1)}}d^x_{J1-J2, 0}(phi), $$

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 .

Answered by Milo Viviani on May 13, 2021

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