Prerequisite Angular Momentum Addition
Phase Convention
A set of normalized eigenbases of any operator are each multiplied by an arbitrary phase factor $ \mathrm{e} ^{ \mathrm{i} \phi_I}$, which is still a set of normalized eigenbases (the eigenvalues remain unchanged). Therefore, the basis transformation matrix (such as the CG matrix), each row or each column is multiplied by a phase factor, and still represents the same basis transformation. How to take these phases is called Phase Convention. The general convention is to make all CG coefficients real numbers, and the matrix elements of the first row and the first column (corresponding to the largest $m_1$ and the largest $L$) are greater than zero (Wikipedia, Griffiths).
Analytic expression is
According to this phase convention, the analytical expression of the CG coefficient is
\begin{equation}
\begin{aligned}
&\quad \left\langle l_1, m_1, l_2, m_2 \middle| l_1, l_2, L, M \right\rangle = \begin{bmatrix}l_1 & l_2 & L\\ m_1 & m_2 & M\end{bmatrix} \\
&= \sqrt{\frac{(2L+1)(L+l_1-l_2)!(L-l_1+l_2)!(l_1+l_2-L)!}{(l_1+l_2+L+1)!}}\\
&\times\sqrt{(L+M)!(L-M)!(l_1-m_1)!(l_1+m_1)!(l_2-m_2)!(l_2+m_2)!}\\
&\times\sum_{k = k_{min}}^{k_{max}} \frac{(-1)^k}{k!(l_1+l_2-L-k)!(l_1-m_1-k)!(l_2+m_2-k)!}\\
&\times \frac{1}{(L-l_2+m_1+k)!(L-l_1-m_2+k)!}
\end{aligned} \end{equation}
The upper and lower limits of the summation need to ensure that each factorial argument containing $k$ is greater than or equal to 0, so there is
\begin{equation}
\begin{aligned}
&k_{min} = \max\{0, \ \ l_2 - m_1 - L,\ \ l_1 + m_2 - L\} \\
&k_{max} = \min\{l_1+l_2-L,\ \ l_1 - m_1, \ \ l_2 + m_2\}
\end{aligned} \end{equation}
Basic selection rules
It can be obtained from "Angular Momentum Addition"
\begin{equation}
\left\lvert l_1 - l_2 \right\rvert \leqslant L \leqslant l_1 + l_2 \qquad
\end{equation}
From a physical understanding, two vectors are added to get the third vector $ \boldsymbol{\mathbf{v}} _1 + \boldsymbol{\mathbf{v}} _2 = \boldsymbol{\mathbf{v}} $, then their modulus length $v_1, v_2, v$ must satisfy the triangle inequality $ \left\lvert v_1 - v_2 \right\rvert \leqslant v \leqslant v_1 + v_2$.
We can also get
\begin{equation}
m_1 + m_2 = M
\end{equation}
\begin{equation}
-L \leqslant M \leqslant L \qquad
-l_1 \leqslant m_1 \leqslant l_1 \qquad
-l_2 \leqslant m_2 \leqslant l_2
\end{equation}
When there is a possibility of a half-integer, it is also required to note that all numbers may take a half-integer in the derivation process. If both numbers are half-integers or both are integers, they add up to an integer, otherwise they add up to a half-integer. A half integer multiplied by 2 equals an odd number, and an integer multiplied by 2 equals an even number. Analogy: If both numbers are odd or even, their addition equals an even number, otherwise the addition equals an odd number. }
\begin{equation}
M + L \in \mathbb{N} \qquad
m_1 + l_1 \in \mathbb{N} \qquad
m_2 + l_2 \in \mathbb{N}
\end{equation}
If and only if a CG coefficient satisfies the above selection rule, it will appear in the CG table (may be 0).
Symmetry
Since the 3j symbol has a more concise symmetry, we can use it to derive the symmetry of the CG coefficient1
\begin{equation}
\begin{aligned}
\begin{bmatrix}l_1 &l_2 &L\\ m_1 &m_2 &M\end{bmatrix}
&= (-1)^{l_1+l_2-L} \begin{bmatrix}l_1 &l_2 &L\\ -m_1 &-m_2 &-M\end{bmatrix} \\
&= (-1)^{l_1+l_2-L} \begin{bmatrix}l_2 &l_1 &L\\ m_2 &m_1 &M\end{bmatrix} \\
&= (-1)^{l_1-m_1}\sqrt{\frac{2L+1}{2l_2+1}} \begin{bmatrix}l_1 &L &l_2\\ m_1 &-M &-m_2\end{bmatrix} \\
&= (-1)^{l_2+m_2}\sqrt{\frac{2L+1}{2l_1+1}} \begin{bmatrix}L &l_2 &l_1\\ -M & m_2 &-m_1\end{bmatrix}
\end{aligned} \end{equation}
Symmetry selection rule
The CG coefficient that meets the basic selection rule may also be zero. We can use more selection rules to find these coefficients. According to the choice rule of 3j symbol (eq. 7 , eq. 8 ), when the sum of the three numbers in the first row is odd,2
\begin{equation}
\begin{bmatrix}l_1 & l_2 & L\\ 0 & 0 & 0\end{bmatrix}
= \begin{bmatrix}l & l & L\\ m & m & M\end{bmatrix}
= \begin{bmatrix}l_1 & l & l\\ m_1 & m & -m\end{bmatrix}
= \begin{bmatrix}l & l_2 & l\\ m & m_2 & -m\end{bmatrix}
= 0
\end{equation}
Note that even if these selection rules are added, the CG coefficient may still be zero. In fact, it is very difficult to find all the zero coefficients
3.
Special circumstances
In some cases eq. 1 has a simpler expression
\begin{equation}
\begin{bmatrix}l_1 & l_2 & 0 \\ m_1 & m_2 & 0\end{bmatrix} = \frac{(-1)^{l_1-m_1}}{\sqrt{2l_1+1}}
\end{equation}
Orthonormality
Since $ \left\lvert l_1, l_2, L, M \right\rangle $ are all orthogonal and normalized, there are
\begin{equation}
\begin{aligned}
&\sum_{m_1, m_2} \left\langle l_1', l_2', L', M' \middle| l_1, m_1, l_2, m_2 \right\rangle \left\langle l_1, m_1, l_2, m_2 \middle| l_1, l_2, L, M \right\rangle \\
&= \delta_{l_1, l_1'} \delta_{l_2, l_2'} \delta_{L, L'} \delta_{M, M'}
\end{aligned} \end{equation}
which is
\begin{equation}
\sum_{m_1, m_2} \begin{bmatrix}l'_1 & l'_2 & L'\\ m_1& m_2 &M'\end{bmatrix} \begin{bmatrix}l_1 & l_2 & L\\ m_1& m_2 &M\end{bmatrix} = \delta_{l_1, l_1'} \delta_{l_2, l_2'} \delta_{L, L'} \delta_{M, M'}
\end{equation}
According to the selection rule, only need to meet
eq. 4 and
eq. 5 $m_1$ and $m_2$ (double summation becomes a single summation).
Relationship with spherical harmonic function
The integral of the product of three spherical harmonics can be expressed as two CG coefficients or 3j symbols multiplication4
\begin{equation}
\begin{aligned}
&\quad \int Y_{l_1 m_1} ( \hat{\boldsymbol{\mathbf{r}}} ) Y_{l_2 m_2} ( \hat{\boldsymbol{\mathbf{r}}} ) Y_{l_3 m_3}( \hat{\boldsymbol{\mathbf{r}}} ) \,\mathrm{d}{\Omega} \\
&= (-1)^{m_3} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l_3+1)}} \begin{bmatrix}l_1& l_2& l_3\\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix}l_1 & l_2 & l_3\\ m_1 & m_2 & -m_3\end{bmatrix} \\
&= \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}
\end{aligned} \end{equation}
1. ^ In the following formula, if $L$ is an integer, then the sign in front can be taken, but if it is half If it is an integer, you can only take the negative sign. Pay attention to the derivation.
2. ^ is derived from the symmetry of the CG coefficient The condition will be much more complicated, but it is equivalent to the condition.
3. ^ see TA Heim, J Hinze and ARP Rau, Some classes of `'nontrivial zeroes' of angular momentum addition coefficients, an example is $ \begin{bmatrix}3 & 3 & 2 \\ -2 & 2 & 0\end{bmatrix} $, note that the sum of the first line is an even number.
4. ^ see Bransden Appendix A4, and Wikipedia's 3j/CG coefficients page