# Closed form for 3j-symbol ratios

I am working on the spherical harmonic decomposition of cosmic microwave background maps, therefore I often deal with functions that are proportional to Wigner 3J symbols/Clebsch–Gordan coefficients.

I would be very grateful if you could share with me a closed form of the ratio between

$$\begin{pmatrix} l_1 &l_2 &l_3\\ 0&2&-2 \end{pmatrix}$$

and

$$\begin{pmatrix} l_1 &l_2 &l_3\\ 0&0&0 \end{pmatrix} \;,$$

for $l_1+l_2+l_3$ even, if it exists. Ideally, I would like an expansion of the first 3j-symbol in terms of the second one, i.e.

first_3j = second_3j * ( 1 + ... )

$$C(m_2+1,m_3-1)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2+1&m_3-1\end{pmatrix}+ D(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2&m_3\end{pmatrix}+$$ $$C(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2-1&m_3+1\end{pmatrix}=0, \tag{1}$$ where $$C(m_2,m_3)=\sqrt{(l_2-m_2+1)(l_2+m_2)(l_3+m_3+1)(l_3-m_3)},\tag{2}$$ and $$D(m_2,m_3)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1)+2m_2m_3,\tag{3}$$ can be found (formula 9) in http://scitation.aip.org/content/aip/journal/jmp/16/10/10.1063/1.522426?ver=pdfcov (Exact recursive evaluation of 3j‐ and 6j‐coefficients for quantum‐mechanical coupling of angular momenta, by Klaus Schulten and Roy G. Gordon). For $m_1=m_2=m_3=0$, we get from (1): $$C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+ D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}+ C(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=0. \tag{4}$$ Using the symmetry relation $$\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=(-1)^{l_1+l_2+l_3} \begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix},$$ (4) can be rewritten as follows $$[C(1)+(-1)^{l_1+l_2+l_3}C(0)]\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0.\tag{5}$$ Here, according to (2) and (3), $$C(0)=\sqrt{l_2l_3(l_2+1)(l_3+1)},\;\; C(1)=C(0),$$ and $$D(0)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1).$$ Now let us assume $m_1=0,m_2=1,m_3=-1$ in (1). We get $$\tilde C(2)\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}+ \tilde D(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+ \tilde C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0.$$ so that $$\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}=-\frac{\tilde C(2)} {\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix} -\frac{\tilde C(1)} {\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}.$$ Substituting this into (5), we get after a little algebra $$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}} {\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)\tilde D(1)}{C(1)+(-1)^{l_1+l_2+l3}C(0)}-\tilde C(1)\right ].$$ It remains to take into account that $l_1+l_2+l_3$ is even and calculate from (2) and (3) (for $m_3=-1$) $$\tilde C(1)=C(1),\;\;\tilde C(2)=\sqrt{(l_2-1)(l_2+2)(l_3-1)(l_3+2)},\;\;\; \tilde D(1)=D(0)-2.$$ Finally, $$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}} {\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)[D(0)-2]}{2C(0)}-C(0)\right ].$$