## Winger 6-j and 9-j symbol

The meaning of 3-j symbol is same as Clebsch-Gordan coefficient. So, we skip in here.

I am not going to construct the 6-j symbol from 3-j symbol. In here, I just state the meaning and usage in Mathematica.

The 6-j symbol is the coupling between 3 angular momenta, $j_1, j_2, j_3$.

There are 2 ways to couple these 3 angular momenta. First,

$j_1 + j_2 + j_3 \rightarrow j_{12} + j_3 \rightarrow J$

the other way is

$j_ 1 + j_2 + j_3 \rightarrow j_1 + j_{23} \rightarrow J$

The 6-j symbol is

$\begin{pmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{pmatrix}$

We can see that there are 4 vector-sum must satisfy.

$\Delta(j_1, j_2, j_{12})$

$\Delta(j_2, j_3, j_{23})$

$\Delta(j_1, j_{23}, J)$

$\Delta(j_{12}, j_3, J)$

If we draw a line to connect these 4 vector-sum, we have:

In Mathematica, there is a build in function

$\textrm{SixJSymbol}[ \left\{j_1, j_2, j_{23} \right\}, \left\{j_3, J , j_{23} \right\}]$

The 9-j symbol is the coupling between 4 angular momenta, $j_1, j_2, j_3, j_4$.

The 9-j symbol can be used in coupling 2 nucleons, $l_1, s_1, l_2, s_2$.

The 9-j symbol is

$\begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix}$

We can see, each row and column must satisfy the vector-sum.

Unfortunately, there is no build in function in Mathematica. The formula for 9-j symbol is

$\displaystyle \begin{pmatrix} l_1 & s_1 & j_1 \\ l_2 & s_2 & j_2 \\ L & S & J \end{pmatrix} \\ = \sum_{g} (-1)^{2g} (2g+1) \begin{pmatrix} l_1 & s_1 & j_1 \\ j_2 & J & g \end{pmatrix} \begin{pmatrix} j_2 & s_2 & j_2 \\ s_1 & g & S \end{pmatrix} \begin{pmatrix} L & S & J \\ g & l_1 & l_2 \end{pmatrix}$

Where $g$ sum all possible value, which can be calculate using the 6 couplings inside the 3 6-j symbols.To check your result, the coupling between $d_{5/2}$ and $f_{7/2}$ to from a $L = 5, S = 0, J = 0$ state is

$\begin{pmatrix} 2 & 1/2 & 5/2 \\ 3 & 1/2 & 7/2 \\ 5 & 0 & 5 \end{pmatrix} = \frac{1}{2\sqrt{770}}$

## Evaluation of mutual interaction

We are going to evaluation the integral

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

Recalling the multi-pole expansion,

$\displaystyle \frac{1}{r_{12}} = \sum_{l=0}^{\infty} \frac{4 \pi}{2l+1} \frac{r_{<}^l}{r_{>}^{l+1}} \sum_{m=-l}^{l} Y_{lm}^{*}(\Omega_1) Y_{lm}(\Omega_2)$

and the basis

$b_{nlm}(\vec{r}) = R_{nl}(r) Y_{lm}(\Omega)$

Set an averaged basis

$\displaystyle b_{nl}(\vec{r}) = R_{nl}(r) \frac{1}{\sqrt{2l+1}} \sum_{m=-1}^{l}Y_{lm}(\Omega)$

$\displaystyle \Gamma_{ij}^{hk}(l) = \frac{4 \pi}{2l+1} \sum_{m=-l}^{l} \frac{1}{\sqrt{(2l_i+1)(2l_j+1)(2l_h+1)(2l_k+1)}} \\ \sum_{m_i,m_j}\int Y_{l_i m_i}^{*}(\Omega_1) Y_{l m}^{*}(\Omega_1) Y_{l_j m_j}(\Omega_1) d \Omega_1 \\ \sum_{m_h,m_k} \int Y_{l_h m_k}^{*}(\Omega_2) Y_{l m}(\Omega_2) Y_{l_k m_k}(\Omega_2) d \Omega_2$

In the angular integrals, using Wigner 3-j symbol and the integral

$\displaystyle\int Y_{l_1 m_1}(\Omega) Y_{l m}(\Omega) Y_{l_2 m_2}(\Omega) d \Omega \\= \frac{\sqrt{(2l_1+1)(2l+1)(2l_2+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_1 & l & l_2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix}$

$Y_{lm}^{*}(\Omega) = (-1)^m Y_{l(-m)}(\Omega)$

$\displaystyle \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix} = (-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l & l_2 \\ -m_1 & -m & -m_2 \end{pmatrix} \\ =(-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l_2 & l \\ m_1 & m_2 & m \end{pmatrix}$

we have

$\displaystyle\int Y_{l_i m_i}^{*}(\Omega) Y_{l m}^{*}(\Omega) Y_{l_j m_j}(\Omega) d \Omega \\= (-1)^{m_j} \frac{\sqrt{(2l_i+1)(2l+1)(2l_j+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_i & l & l_j \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l & l_j \\ m_i & m & -m_j \end{pmatrix}$

$\displaystyle\int Y_{l_h m_h}^{*}(\Omega) Y_{l m}(\Omega) Y_{l_k m_k}(\Omega) d \Omega \\= (-1)^{m_h} \frac{\sqrt{(2l_h+1)(2l+1)(2l_k+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_h & l & l_k \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l & l_k \\ -m_h & m & m_k \end{pmatrix}$

Thus,

$\displaystyle \Gamma_{ij}^{hk}(l) = \sum_{m=-l}^{l} \sum_{m_i,m_j } (-1)^{m_j} \begin{pmatrix} l_i & l_j & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l_j & l \\ m_i & -m_j & m \end{pmatrix} \\ \sum_{m_h,m_k} (-1)^{m_h} \begin{pmatrix} l_h & l_k & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l_k & l \\ -m_h & m_k & m \end{pmatrix}$

The 3-j symbol restricted that

$l_i+l_j+l = even$

$l_h+l_k+l = even$

$m_i-m_j + m = 0, |m_i| \leq l_i , |m_j| \leq l_j$

$-m_h + m_k +m = 0, |m_h| \leq l_h , |m_k| \leq l_k$

I guess it is the most simplified formula for the angular part.

The total integral is

$\displaystyle G_{ij}^{hk} = \sum_{l=0}^\infty \Gamma_{ij}^{hk}(l) \langle R_i(x) R_h(y) | \frac{r_<^l}{r_>^{l+1}} |R_j(x) R_k(y) \rangle$

The angular integral imposes condition for $l$.

I am not sure this is a correct way to treat the problem.

First, the averaged basis is still an energy eigen state. It is not the eigen state for the angular part. So, this averaging could introduces an error and we should reminded that this is an approximation. But in the perturbation view point, this averaged basis is still valid.

Second thing is, the sum

$\displaystyle P_{(l_i m_i) (l_j m_j)}^{(l_h m_h)(l_k m_k)}(l) = \sum_{m=-l}^{l} \langle Y_{l_i m_i}|Y_{lm}^*|Y_{l_j m_j}\rangle \langle Y_{l_h m_h}|Y_{lm}|Y_{l_k m_k}\rangle$

is not symmetry for exchange of $i, j, h,k$ in general. For example,

$\displaystyle \frac{-1}{3}=P_{(1-1) (00)}^{(11)(00)}(1) {\neq} P_{(00)(1-1)}^{(11)(00)}(1) = 0$

This is a very uprising result that the mutual interaction dependent on the magnetic quantum number. Thus, in detail, we should use $n l m m_s$ as a basis.

Third, the sum $P_{ij}^{hk}(l)$ is depend on $l$. The mutual interaction require us to sum all possible $l$.

Fourth, the coupling between 1s2p triplet state, the total spin is $S = 1$, total L is $L = 1$, and the total angular momentum can be  $J = 0, 1, 2$. In our treatment, we did not coupled the angular momentum in the calculation explicitly. In fact, in the integral of the spherical harmonic, the coordinate are integrated separately, and the coupling seem to be calculated implicitly. I am not sure how to couple two spherical harmonics with two coordinates.

## Integration formulas of spherical harmonic

There are several important and useful formulas for the integration of spherical harmonic. We simplify the notation,

$\displaystyle \int_0^{\pi} \sin(\theta)d\theta\int_0^{2\pi}d\phi = \int d\Omega$

The first one is the average of spherical harmonic.

$\displaystyle \int Y_{lm} d\Omega = \sqrt{4\pi} \delta_{l0}\delta_{m0}$

The 2nd one is the orthonormal  condition.

$\displaystyle \int Y^{*}_{l'm'}Y_{lm} d\Omega = \delta_{l'l}\delta_{m'm}$

The 3rd one is triplet integral, we use the product of spherical harmonic,

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y^*_{l_3m_3} d\Omega \\ = \int \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm} Y^*_{l_3m_3} d\Omega \\= \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l_3+1)}} C_{l_10l_20}^{l_30} C_{l_1m_1l_2m_2}^{l_3m_3}$

The 4th one is another triple integral,

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega \\ = \int \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm} Y_{l_3m_3} d\Omega \\ = \int \sum_{lmLM} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} \\ \sqrt{\frac{(2l+1)(2l_3+1)}{4\pi(2L+1)}} C_{l0l_30}^{L0} C_{lml_3m_3}^{LM}Y_{LM}d\Omega$

$\displaystyle = \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} C_{l0l_30}^{00} C_{lml_3m_3}^{00}$

Notice that

$C_{lmLM}^{00} = (-1)^{L+M} \sqrt{\frac{1}{2L+1}} \delta_{Ll}\delta_{-m,M}$

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi (2l_3+1)}} C_{l_10l_20}^{l_30} C_{l_1m_1l_2m_2}^{l_3,-m_3} (-1)^{m_3}$

using Wigner 3-j symbol,

$C_{l_1m_1l_2m_2}^{l_3m_3} = (-1)^{l_1-l_2+m_3} \sqrt{2l_3+1} \begin{pmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & -m_3 \end{pmatrix}$

$\displaystyle \int Y_{l_1m_1}Y_{l_2m_2} Y_{l_3m_3} d\Omega \\= \sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} \begin{pmatrix} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \end{pmatrix}$

For other integral, we can use

$Y^*_{lm}(\theta, \phi) = (-1)^{m}Y_{l(-m)}(\theta,\phi) = Y_{lm}(\theta, -\phi)$