Formula

CG coefficient

Ref ^[1], P240, Section 8.2.4, Formula (20).

\[C_{j_1m_1j_2m_2}^{j_3m_3} = \delta_{m_3, m_1+m_2} \left[ \dfrac{\begin{pmatrix}2j_1 \\ J-2j_2\end{pmatrix}\begin{pmatrix}2j_2\\J-2j_3\end{pmatrix}}{\begin{pmatrix}J+1\\J-2j_3\end{pmatrix}\begin{pmatrix}2j_1\\j_1-m_1\end{pmatrix}\begin{pmatrix}2j_2 \\ j_2 - m_2\end{pmatrix}\begin{pmatrix}2j_3 \\ j_3-m_3\end{pmatrix}} \right]^{1/2} \\ \times \sum_{z} (-1)^{z}\begin{pmatrix}J-2j_3 \\ z\end{pmatrix}\begin{pmatrix}J-2j_2 \\ j_1-m_1-z\end{pmatrix}\begin{pmatrix}J-2j_1 \\ j_2 + m_2 - z\end{pmatrix}\]

where, $J = j_1+j_2+j_3$. It is already combination of binomials.

3j symbol

Ref ^[1], P236, Section 8.1.2, Formula (11).

\[\begin{pmatrix}j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3\end{pmatrix} = (-1)^{j_3+m_3+2j_1}\dfrac{1}{\sqrt{2j_3+1}}C_{j_1-m_1j_2-m_2}^{j_3m_3}\]

This package use the CG coefficient above to calculate 3j symbol.

6j symbol

Ref ^[1], P293, Section 9.2.1, Formula (1).

\[\begin{Bmatrix}j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6\end{Bmatrix} = \Delta(j_1j_2j_3)\Delta(j_4j_5j_3)\Delta(j_1j_5j_6)\Delta(j_4j_2j_6) \\ \times \sum\limits_x\dfrac{(-1)^x(x+1)!}{(x-j_{123})!(x-j_{453})!(x-j_{156})!(x-j_{426})!(j_{1245}-x)!(j_{1346}-x)!(j_{2356}-x)!}\]

Here, I use $j_{123} \equiv j_1+j_2+j_3$ for simplicity. The symbol $\Delta(abc)$ is defined as

\[\Delta(abc) = \left[\dfrac{(a+b-c)!(a-b+c)!(-a+b+c)!}{(a+b+c+1)!}\right]^{\frac{1}{2}}.\]

We can find that

\[\begin{pmatrix}j_1+j_2-j_3 \\ x - j_{453}\end{pmatrix} = \dfrac{(j_1+j_2-j_3)!}{(x-j_{453})!(j_{1245}-x)!} \\ \begin{pmatrix}j_1-j_2+j_3 \\ x - j_{426}\end{pmatrix} = \dfrac{(j_1-j_2+j_3)!}{(x-j_{426})!(j_{1346}-x)!} \\ \begin{pmatrix}j_2+j_3-j_1 \\ x - j_{156}\end{pmatrix} = \dfrac{(j_2+j_3-j_1)!}{(x-j_{156})!(j_{2356}-x)!} \\ \begin{pmatrix}x+1 \\ j_{123}+1\end{pmatrix} = \dfrac{(x+1)!}{(x-j_{123})(j_{123}+1)!}.\]

So, we have

\[\begin{Bmatrix}j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6\end{Bmatrix} = \dfrac{\Delta(j_4j_5j_3)\Delta(j_1j_5j_6)\Delta(j_4j_2j_6)}{\Delta(j_1j_2j_3)} \\ \times \sum\limits_x (-1)^x \begin{pmatrix}x+1 \\ j_{123}+1\end{pmatrix} \begin{pmatrix}j_1+j_2-j_3 \\ x - j_{453}\end{pmatrix} \begin{pmatrix}j_1-j_2+j_3 \\ x - j_{426}\end{pmatrix} \begin{pmatrix}j_2+j_3-j_1 \\ x - j_{156}\end{pmatrix}.\]

Rewrite $\Delta(abc)$ with binomials,

\[\Delta(abc) = \left[\dfrac{1}{\begin{pmatrix}a+b+c+1 \\ 2a + 1\end{pmatrix} \begin{pmatrix}2a \\ a + b - c\end{pmatrix}(2a+1)}\right]^{\frac{1}{2}}.\]

So

\[\dfrac{\Delta(j_4j_5j_3)\Delta(j_1j_5j_6)\Delta(j_4j_2j_6)}{\Delta(j_1j_2j_3)} \\ = \dfrac{1}{2j_4+1} \left[\dfrac{\begin{pmatrix}j_{123}+1 \\ 2j_1+1\end{pmatrix}\begin{pmatrix}2j_1 \\ j_1 + j_2 - j_3\end{pmatrix}}{\begin{pmatrix}j_{156}+1 \\ 2j_1+1\end{pmatrix}\begin{pmatrix}2j_1 \\ j_1+j_5-j_6\end{pmatrix}\begin{pmatrix}j_{453}+1\\2j_4+1\end{pmatrix}\begin{pmatrix}2j_4\\j_4+j_5-j_3\end{pmatrix}\begin{pmatrix}j_{426}+1 \\ 2j_4+1\end{pmatrix}\begin{pmatrix}2j_4 \\ j_4+j_2-j_6\end{pmatrix}}\right]^{\frac{1}{2}}\]

Racha coefficient

Ref ^[1], P291, Section 9.1.2, Formula (11)

\[W(j_1j_2j_3j_4, j_5j_6) = (-1)^{j_1+j_2+j_3+j_4} \begin{Bmatrix} j_1 & j_2 & j_5 \\ j_4 & j_3 & j_6 \end{Bmatrix}\]

This package uses the 6j symbol above to calculate Racha coefficient.

9j symbol

Ref ^[1], P340, Section 10.2.4, Formula (20)

\[\begin{Bmatrix}j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6 \\ j_7 & j_8 & j_9\end{Bmatrix} = \sum\limits_{t}(-1)^{2t}(2t+1)\begin{Bmatrix}j_1 & j_2 & j_3 \\ j_6 & j_9 & t\end{Bmatrix} \begin{Bmatrix}j_4 & j_5 & j_6 \\ j_2 & t & j_8\end{Bmatrix} \begin{Bmatrix}j_7 & j_8 & j_9 \\ t & j_1 & j_4\end{Bmatrix}\]

Use the 6j symbol result above, we get

\[\dfrac{\Delta(j_1j_9t)\Delta(j_6j_9j_3)\Delta(j_6j_2t)}{\Delta(j_1j_2j_3)} \dfrac{\Delta(j_4tj_8)\Delta(j_2tj_6)\Delta(j_2j_5j_8)}{\Delta(j_4j_5j_6)} \dfrac{\Delta(j_7j_1j_4)\Delta(tj_1j_9)\Delta(tj_8j_4)}{\Delta(j_7j_8j_9)} \\ = \dfrac{\Delta(j_3j_6j_9)\Delta(j_2j_5j_8)\Delta(j_1j_4j_7)}{\Delta(j_1j_2j_3)\Delta(j_4j_5j_6)\Delta(j_7j_8j_9)} \Delta^2(j_1j_9t)\Delta^2(j_2j_6t)\Delta^2(j_4j_8t).\]

Define

\[P_0 \equiv \dfrac{\Delta(j_3j_6j_9)\Delta(j_2j_5j_8)\Delta(j_1j_4j_7)}{\Delta(j_1j_2j_3)\Delta(j_4j_5j_6)\Delta(j_7j_8j_9)} \\ = \left[\dfrac{\begin{pmatrix}j_{123} + 1 \\ 2j_1+1\end{pmatrix}\begin{pmatrix}2j_1\\j_1+j_2-j_3\end{pmatrix}\begin{pmatrix}j_{456}+1\\2j_5+1\end{pmatrix}\begin{pmatrix}2j_5 \\ j_4+j_5-j_6\end{pmatrix}\begin{pmatrix}j_{789}+1 \\ 2j_9+1\end{pmatrix}\begin{pmatrix}2j_9 \\ j_7 + j_9 - j_8\end{pmatrix}}{\begin{pmatrix}j_{147} + 1\\ 2j_1 + 1\end{pmatrix}\begin{pmatrix}2j_1 \\ j_1+j_4-j_7\end{pmatrix}\begin{pmatrix}j_{258}+1 \\ 2j_5+1\end{pmatrix}\begin{pmatrix}2j_5 \\ j_2+j_5 - j_8\end{pmatrix}\begin{pmatrix}j_{369}+1 \\ 2j_9+1\end{pmatrix}\begin{pmatrix}2j_9 \\ j_3+j_9 - j_6\end{pmatrix}}\right]^{1/2}.\]

and then define

\[P(t) \equiv (-1)^{2t}(2t+1)\Delta^2(j_1j_9t)\Delta^2(j_2j_6t)\Delta^2(j_4j_8t) = \dfrac{(-1)^{2t}}{(2t+1)^2} \times \\ \dfrac{1}{\begin{pmatrix}j_1+j_9+t+1 \\ 2t+1\end{pmatrix}\begin{pmatrix}2t \\ j_1+t-j_9\end{pmatrix}\begin{pmatrix}j_2+j_6+t+1 \\ 2t+1\end{pmatrix}\begin{pmatrix}2t \\ j_2+t-j_6\end{pmatrix}\begin{pmatrix}j_4+j_8+t \\ 2t+1\end{pmatrix}\begin{pmatrix}2t \\ j_4+t-j_8\end{pmatrix}}.\]

\[A(t,x) \equiv (-1)^x\begin{pmatrix}x+1 \\ j_{123} + 1\end{pmatrix}\begin{pmatrix}j_1+j_2-j_3 \\ x - (j_6+j_9+j_3)\end{pmatrix} \begin{pmatrix}j_1+j_3-j_2 \\ x - (j_6+j_2+t)\end{pmatrix}\begin{pmatrix}j_2+j_3-j_1 \\ x - (j_1 + j_9 + t)\end{pmatrix}.\]

\[B(t,y) \equiv (-1)^y\begin{pmatrix}y+1 \\ j_{456} + 1\end{pmatrix}\begin{pmatrix}j_4+j_5-j_6 \\ y - (j_2+t+j_6)\end{pmatrix}\begin{pmatrix}j_4+j_6-j_5 \\ y - (j_2+j_5+j_8)\end{pmatrix}\begin{pmatrix}j_5+j_6-j_4 \\ y - (j_4+t+j_8)\end{pmatrix}.\]

\[C(t,z) \equiv (-1)^z \begin{pmatrix}z+1 \\ j_{789} + 1\end{pmatrix}\begin{pmatrix}j_7+j_8-j_9 \\ z - (t+j_1+j_9)\end{pmatrix}\begin{pmatrix}j_7+j_9-j_8 \\ z - (t+j_8+j_4)\end{pmatrix}\begin{pmatrix}j_8+j_9- j_7 \\ z - (j_7 + j_1 + j_4)\end{pmatrix}.\]

At last, we get

\[\begin{Bmatrix}j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6 \\ j_7 & j_8 & j_9\end{Bmatrix} = P_0\sum_t P(t) \left(\sum_x A(t,x)\right) \left(\sum_y B(t,y)\right) \left(\sum_z C(t,z)\right)\]

It deserves to be mentioned that, although the formula has 4 $\sum$s, the $\sum$ of $x,y,z$ are decoupled. So we can do the three for loops respectively, which means the depth of for loop is not 4 but 2.

Estimate the capacity

Assume we are doing a calculation for a system, we will not calculate the Winger Symbols with very large angular momentum, because usually we will take some trunction. If in such truncated system, the max angular momentum is $J_{max}$, now we can estimate how many binomial coefficients we need to store to compute those Wigner Symbols.

With maximum $J_{max}$

CG & 3j

According to the formula, the max possible binomial is $\begin{pmatrix} J+1 \\ J-2j_3\end{pmatrix}$, where $J = j_1+j_2+j_3$. So we need at least store binomials to $\begin{pmatrix} n_{min} \\ k\end{pmatrix}$ where

\[n_{min} \geq 3J_{max}+1\]

6j & Racha

For binomials in sqrt，the maximum possible $n = 3J_{max}+1$, and for those in the summation, we have to calculate the boundary of $x + 1$. According to the formula

\[x \leq \min\{j_1+j_2+j_4+j_5, j_1+j_3+j_4+j_6, j_2+j_3+j_5+j_6\} \leq 4J_{max}\]

So we have to at least store to

\[n_{min} \geq 4J_{max} + 1\]

9j

Similarly, according to $P_0$, we need $3J_{max}+1$. According to $P(t)$,

\[t \leq \min\{j_2 + j_6, j_4 + j_8, j_1 + j_9\} \leq 2J_{max}\]

so we need $j_1 + j_9 + t + 1 \leq 4J_{max} + 1$. According to $A(t, x),\; B(t, y),\; C(t, z)$,

\[x \leq \min\{j_1+j_2+j_6+j_9, j_1+j_3+j_6+t, j_2+j_3+j_9+t\} \leq 5J_{max}\]

so we need to at least store to

\[n_{min} \geq 5J_{max} + 1\]

With maximum single particle $j_{max}$

Considering a quantum many body calculation, we can get a more optimistic estimation of the calculation capacity. In a quantum many body calculation, we often truncate single particle orbits, and in this condition we assume that we only need to consider two-body coupled angular momentum.

The maximum angular momentum of the single particle orbits defined as $j_{max}$. If for each function, there are at least one of the parameters is single particle angular momentum, then we can get the following estimation.

CG & 3j

In this condition, the $j_1, j_2, j_3$ contains two single particle angular momentum, and the rest one is a coupled two body angular momentum. So

\[n_{min} \geq 4j_{max} + 1\]

6j & Racha

Similarly

\[x \leq \min\{j_1+j_2+j_4+j_5, j_1+j_3+j_4+j_6, j_2+j_3+j_5+j_6\}\]

These summations of four angular momentum can be split into two kinds, all single particle angular momentum, and two single particle with two two body angular momentum. Consider the worst condition

\[n_{min} \geq 6j_{max} + 1\]

9j

We assume the 9j symbol contains at least one single particle angular momentum, then according to the coupling rules, the 9j symbol can be split into kinds

\[\begin{Bmatrix} 1 & 1 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 2 \end{Bmatrix} \quad \begin{Bmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{Bmatrix}\]

where $1$ represents single particle and $2$ for two body angular momentum. In both cases, we can deduce that

\[n_{min} \geq 8j_{max} + 1\]

Reference