Moshinsky
The Talmi-Moshinsky braket can be evaluated as following [1]：
\[\begin{aligned} &\langle n_3l_3,n_4l_4;\Lambda|n_1l_1,n_2l_2;\Lambda\rangle = \prod_{k=1}^4 i^{-l_k} \sqrt{\dfrac{n_k![2(n_k+l_k)+1]!!}{\sqrt{2^{l_k}}}} \\ \times &\sum_{n_an_bn_cn_dl_a l_b l_c l_d} \Bigg[(-1)^{l_d} (\sin\beta)^{2n_a + l_a + 2n_d + l_d}(\cos\beta)^{2n_b+l_b + 2n_c + l_c} \begin{Bmatrix}l_a & l_b & l_1 \\ l_c & l_d & l_2 \\ l_3 & l_4 & \Lambda \end{Bmatrix} \\ \times & C_{l_a0l_b0}^{l_10} C_{l_a0l_c0}^{l_30} C_{l_b0l_d0}^{l_40} C_{l_c0l_d0}^{l_20} \prod_{p = a}^{d} \dfrac{(-1)^{l_p}\sqrt{2^{l_p}}(2l_p+1)}{n_p![2(n_p+l_p) + 1]!!} \Bigg]. \end{aligned}\]
where
\[\begin{aligned} 2n_a + l_a + 2n_b + l_b = 2n_1 + l_1 \equiv f_1, \\ 2n_a + l_a + 2n_c + l_c = 2n_3 + l_3 \equiv f_3, \\ 2n_b + l_b + 2n_d + l_d = 2n_4 + l_4 \equiv f_4, \\ 2n_c + l_c + 2n_d + l_d = 2n_2 + l_2 \equiv f_2. \end{aligned}\]
Here we define $\chi = f_1 + f_2 = f_3 + f_4$。
Similar as Wigner symbols, we try to rewrite it as binomials. First, we have:
\[ (2k+1)!! = \dfrac{(2k+1)!}{(2k)!!} = \dfrac{(2k+1)!}{2^k k!}.\]
We can rewrite the following term:
\[\begin{aligned} t_p &\equiv \dfrac{(-1)^{l_p}\sqrt{2^{l_p}}(2l_p+1)}{n_p![2(n_p+l_p) + 1]!!} \\ &= \dfrac{(-1)^{l_p}\sqrt{2^{l_p}}(2l_p+1)2^{n_p+l_p}}{(2n_p+l_p+1)!}\begin{pmatrix}2n_p+l_p+1 \\ n_p\end{pmatrix} \begin{pmatrix}2(n_p+l_p)+1 \\ n_p + l_p\end{pmatrix}^{-1}. \end{aligned}\]
Similiarly,
\[\begin{aligned} r_k &\equiv \dfrac{n_k![2(n_k+l_k)+1]!!}{\sqrt{2^l_k}} \\ &= \dfrac{(2n_k+l_k+2)!}{\sqrt{2^{l_k}}2^{n_k+l_k}(n_k+l_k+2)}\begin{pmatrix}2(n_k+l_k)+1 \\ n_k + l_k\end{pmatrix}\begin{pmatrix}2n_k+l_k+2 \\ n_k\end{pmatrix}^{-1}. \end{aligned}\]
Next,
\[\begin{aligned} t &\equiv \dfrac{\prod_{k=1}^4 \sqrt{(2n_k+l_k+2)!}}{\prod_{p=a}^d (2n_p+l_p + 1)!} \\ &= \sqrt{\begin{pmatrix}2n_1+l_1+2 \\ 2n_a + l_a + 1\end{pmatrix}\begin{pmatrix}2n_2 + l_2 + 2 \\ 2n_c + l_c + 1\end{pmatrix} \begin{pmatrix}2n_3 + l_3 + 2 \\ 2n_a + l_a + 1\end{pmatrix}\begin{pmatrix}2n_4 + l_4 + 2 \\ 2n_b + l_b + 1\end{pmatrix}} \end{aligned}\]
In the prod of $t_p$, there is
\[ \prod_{p=a}^{d} \sqrt{2^{l_p}}2^{n_p+l_p} = \prod_{p=a}^d 2^{\frac{2n_p+l_p}{2}} 2^{l_p} = 2^{\chi/2} \prod_{p=a}^{d} 2^{l_p}.\]
In the pord of $r_k$, there is
\[ \sqrt{\prod_{k=1}^4 \dfrac{1}{\sqrt{2^{l_k}}2^{n_k+l_k}}} = 2^{-\chi/2} \prod_{k=1}^4 2^{-l_k/2}\]
Finally, we have
\[\begin{aligned} &\langle n_3l_3,n_4l_4;\Lambda|n_1l_1,n_2l_2;\Lambda\rangle\\ = &\prod_{k=1}^4 i^{-l_k} \sqrt{\dfrac{1}{2^{l_k}(n_k+l_k+2)}\begin{pmatrix}2(n_k+l_k)+1 \\ n_k + l_k\end{pmatrix}\begin{pmatrix}2n_k+l_k+2 \\ n_k\end{pmatrix}^{-1}} \\ \times & \sum_{n_an_bn_cn_dl_a l_b l_c l_d} \Bigg[(-1)^{l_d} (\sin\beta)^{2n_a + l_a + 2n_d + l_d}(\cos\beta)^{2n_b+l_b + 2n_c + l_c} \begin{Bmatrix}l_a & l_b & l_1 \\ l_c & l_d & l_2 \\ l_3 & l_4 & \Lambda \end{Bmatrix} \\ \times & \sqrt{\begin{pmatrix}2n_1+l_1+2 \\ 2n_a + l_a + 1\end{pmatrix}\begin{pmatrix}2n_2 + l_2 + 2 \\ 2n_c + l_c + 1\end{pmatrix} \begin{pmatrix}2n_3 + l_3 + 2 \\ 2n_a + l_a + 1\end{pmatrix}\begin{pmatrix}2n_4 + l_4 + 2 \\ 2n_b + l_b + 1\end{pmatrix}} \\ \times & C_{l_a0l_b0}^{l_10} C_{l_a0l_c0}^{l_30} C_{l_b0l_d0}^{l_40} C_{l_c0l_d0}^{l_20} \prod_{p=a}^d (-1)^{l_p}2^{l_p}(2l_p+1)\begin{pmatrix}2n_p+l_p+1 \\ n_p\end{pmatrix} \begin{pmatrix}2(n_p+l_p)+1 \\ n_p + l_p\end{pmatrix}^{-1} \Bigg] \end{aligned}\]
Consider $m_1 = m_2,\;\omega_1 = \omega_2$, thus $\tan\beta = 1, \; \cos\beta = \sin\beta = 1/\sqrt{2}$, then
\[\begin{aligned} (\sin\beta)^{2n_a + l_a + 2n_d + l_d}(\cos\beta)^{2n_b+l_b + 2n_c + l_c} &= 2^{-\frac{2n_a + l_a + 2n_b + l_b + 2n_c + l_c + 2n_d + l_d}{2}} \\ &= 2^{-\frac{f_1+f_2}{2}} \\ &= 2^{-\chi/2} \end{aligned}\]
Exact evaluation
Still start from the origin equation, define
\[X_{abcd} = \left(\prod_{k=1}^4 i^{-l_k}\right)\left(\prod_{p = a}^d (-1)^{l_p}\right)\begin{Bmatrix}l_a & l_b & l_1 \\ l_c & l_d & l_2 \\ l_3 & l_4 & \Lambda \end{Bmatrix} C_{l_a0l_b0}^{l_10} C_{l_a0l_c0}^{l_30} C_{l_b0l_d0}^{l_40} C_{l_c0l_d0}^{l_20}.\]
According to Ref [2], P251, Sec 8.5.2, Formula(32).
\[C^{c0}_{a0b0} = \begin{cases} 0, & a + b + c = 2g + 1, \\ \dfrac{(-1)^{g-c}\hat{c}g!}{(g-a)!(g-b)!(g-c)!}\Delta(abc), & a +b +c = 2g. \end{cases}\]
Define
\[\Omega(abc) \equiv \dfrac{g!}{(g-a)!(g-b)!(g-c)!} = \frac{1}{(g+1)\Delta^2(\frac{a}{2}\frac{b}{2}\frac{c}{2})} = \binom{g}{c}\binom{c}{g-a}.\]
It is obvious a integer. Then
\[X_{abcd} = (-1)^{f_x}\hat{l}_1\hat{l}_2\hat{l}_3\hat{l}_4\Omega(l_al_bl_1)\Delta(l_al_bl_1)\Omega(l_al_cl_3)\Delta(l_al_cl_3)\Omega(l_bl_dl_4)\Delta(l_bl_dl_4)\Omega(l_cl_dl_2)\Delta(l_cl_dl_2)\begin{Bmatrix}l_a & l_b & l_1 \\ l_c & l_d & l_2 \\ l_3 & l_4 & \Lambda \end{Bmatrix},\]
where
\[\begin{aligned} f_x &= -\frac{l_1+l_2+l_3+l_4}{2} + l_a+l_b+l_c+l_d + \dfrac{l_a+l_b-l_1}{2} + \dfrac{l_c+l_d-l_2}{2} + \dfrac{l_a+l_c-l_3}{2} + \dfrac{l_b+l_d-l_4}{2} \\ &= 2(l_a+l_b+l_c+l_d) - (l_1+l_2+l_3+l_4). \end{aligned}\]
It is easy to prove that $l_1+l_2+l_3+l_4$ is even, so the phase factor of $X$ is $1$. In the Wigner symbols section, we have obtain
\[\begin{Bmatrix}l_a & l_b & l_1 \\ l_c & l_d & l_2 \\ l_3 & l_4 & \Lambda \end{Bmatrix} = P_0\sum_t P(t) \left(\sum_x A(t,x)\right) \left(\sum_y B(t,y)\right) \left(\sum_z C(t,z)\right).\]
Where, $A(t,x), B(t,y), C(t,z)$ are all integers, and $P(t)$ is a rational. Here
\[P_0 = \frac{\Delta(l_al_cl_3)\Delta(l_bl_dl_4)\Delta(l_1l_2\Lambda)}{\Delta(l_al_bl_1)\Delta(l_cl_dl_2)\Delta(l_3l_4\Lambda)}.\]
So
\[\begin{aligned} X_{abcd} &= \hat{l}_1\hat{l}_2\hat{l}_3\hat{l}_4\Omega(l_al_bl_1)\Omega(l_al_cl_3)\Omega(l_bl_dl_4)\Omega(l_cl_dl_2) \Delta^2(l_al_cl_3)\Delta^2(l_bl_dl_4)\frac{\Delta(l_1l_2\Lambda)}{\Delta(l_3l_4\Lambda)} \\ &\times\sum_t P(t) \left(\sum_x A(t,x)\right) \left(\sum_y B(t,y)\right) \left(\sum_z C(t,z)\right). \end{aligned}\]
In this formula, the only sqrt occurs in
\[\hat{l}_1\hat{l}_2\hat{l}_3\hat{l}_4\frac{\Delta(l_1l_2\Lambda)}{\Delta(l_3l_4\Lambda)},\]
which is only related to $l_1,l_2,l_3,l_4,\Lambda$, and is equal in all the summations. If we only deal with $\tan\beta = 1$, that's $(\sin\beta)^{2n_a + l_a + 2n_d + l_d}(\cos\beta)^{2n_b+l_b + 2n_c + l_c} = 2^{-\chi/2}$, then the Moshinsky braket is a square root of a rational.
Define
\[\begin{aligned} M_{9j}(l_al_bl_cl_dl_1l_2l_3l_4\Lambda) &= \Omega(l_al_bl_1)\Omega(l_al_cl_3)\Omega(l_bl_dl_4)\Omega(l_cl_dl_2) \Delta^2(l_al_cl_3)\Delta^2(l_bl_dl_4)\\ &\times \sum_t P(t) \left(\sum_x A(t,x)\right) \left(\sum_y B(t,y)\right) \left(\sum_z C(t,z)\right). \end{aligned}\]
Then we can finally write down
\[\begin{aligned} &\langle n_3l_3,n_4l_4;\Lambda|n_1l_1,n_2l_2;\Lambda\rangle\\ = &\frac{1}{(2n_1+l_1+2)(2n_2+l_2+2)}\sqrt{\frac{\binom{\chi+2}{f_1+1}\binom{l_3+l_4+\Lambda+1}{2\Lambda+1}\binom{2\Lambda}{\Lambda + l_3-l_4}}{\binom{\chi+2}{f_3+1}\binom{l_1+l_2+\Lambda+1}{2\Lambda+1}\binom{2\Lambda}{\Lambda+l_1-l_2}}} \prod_{k}^4 \sqrt{(2l_k+1)\frac{\binom{2(n_i+l_i)+1}{n_k+l_k}}{\binom{2n_k+l_k+1}{n_k}}} \\ &\times \sum_{n_an_bn_cn_dl_a l_b l_c l_d} \Bigg[(-1)^{l_d} (\sin\beta)^{2n_a + l_a + 2n_d + l_d}(\cos\beta)^{2n_b+l_b + 2n_c + l_c} M_{9j}(l_al_bl_cl_dl_1l_2l_3l_4\Lambda) \\ &\times \binom{2n_1+l_1+2}{2n_a+l_a+1}\binom{2n_2+l_2+2}{2n_c+l_c+1} 2^{l_a+l_b+l_c+l_d-\frac{l_1+l_2+l_3+l_4}{2}} \prod_{p=a}^d (2l_p+1) \frac{\binom{2n_p+l_p+1}{n_p}}{\binom{2(n_p+l_b)+1}{n_p+l_p}}\Bigg] \end{aligned}\]