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:

Screen Shot 2018-05-16 at 15.34.47.png

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}}

 

 

 

Advertisements