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 + ... )


Thank you for your consideration.


Best wishes,
Guido



Closed form for 3j-symbol ratios
The following recursion relation


$$
C(m\_2+1,m\_3-1)\begin{pmatrix}l\_1 &l\_2 &l\_3\\m\_1&m\_2+1&m\_3-1\end{pmatrix}+
D(m\_2,m\_3)\begin{pmatrix}l\_1 &l\_2 &l\_3\\m\_1&m\_2&m\_3\end{pmatrix}+$$ $$
C(m\_2,m\_3)\begin{pmatrix}l\_1 &l\_2 &l\_3\\m\_1&m\_2-1&m\_3+1\end{pmatrix}=0,
\tag{1}$$
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,
$$\frac{\begin{pmatrix}l\_1 &l\_2 &l\_3\\0&2&-2\end{pmatrix}}
{\begin{pmatrix}l\_1 &l\_2 &l\_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)[D(0)-2]}{2C(0)}-C(0)\right ].$$


