# Simplifying a sum of products of Clebsch-Gordan Coefficients

I'm trying to simplify this sum involving the products of four Clebsch Gordan coefficients:

\begin{align} &\sum_{m_1 m_2 m_3 K K' M} (-1)^{j_1+j_2+j_3+G+m_1+m_2+m_3-M} \langle j_1\ m_1; j_2\ m_2|J\ K\rangle \\ &\times \langle j_2\ -m_2;j_3\ m_3|J'\ K'\rangle \times \langle j_3\ -m_3;j_1\ -m_1|G\ -M\rangle \times \langle J\ K;J'\ K'|G\ M\rangle \end{align} where each of $j_1$, $j_2$, and $j_3$ are half-integers, and the sums are taken over all allowed values for each parameter ($-j_1$ to $j_1$, $-j_2$ to $j_2$, etc.). I looked through several identities in Wolfram's list here, but I couldn't find any among the two-term identities that I could apply, and I'm not familiar (and unfortunately don't have time to learn at the moment) with the 6-j and 9-j symbols used in the more complicated ones. I expect the sum to simplify nicely (as all of the others that I've encountered in this context so far have), but I can't see how.

• unless there is a surprise with the phases or with the signs of your projections (which by cursory inspection don't look quite right btw) very very likely a $6j$ symbol. – ZeroTheHero Apr 17 '17 at 23:27
• What makes you say that the signs don't look correct? – jawheele Apr 18 '17 at 2:06
• I've looked back at my source calculations that led to this sum a few times, and I'm confident that these are the correct signs. Your help is very much appreciated! – jawheele Apr 18 '17 at 9:31
• If ZeroTheHero's expression doesn't work, it would be helpful to have a bit more context about where this came from, as that can often help point out the right direction. Clebsch-Gordan manipulations are messy and coefficienty, but ultimately you're just doing geometry, and forgetting the geometry doesn't help often. – Emilio Pisanty Apr 18 '17 at 15:35
• @jawheele Please check my edited expression as I had an overall phase issue in my previous expression. – ZeroTheHero Apr 18 '17 at 20:14

You should double-check but the summation appears to be given by $$(2G+1)\sqrt{(2J'+1)(2J+1)} \left\{\begin{array}{ccc} j_1&j_2&J\\ J'&G&j_3 \end{array}\right\}\tag{1}$$

To get there the simplest way is to start from the definition of the $6j$ symbol: \begin{align} &\sum_{\bar{m}_1\bar{m}_2\bar{m}_3\bar{m}_{12}\bar{m}_{23}} \langle \bar{j}_{12}\bar{m}_{12};\bar{j}_3\bar{m}_3\vert \bar{j}\bar{m}\rangle \langle \bar{j}_{1}\bar{m}_{1};\bar{j}_2\bar{m}_2\vert \bar{j}_{12}\bar{m}_{12}\rangle\, ,\\ &\qquad\qquad \times \langle \bar{j}_{1}\bar{m}_{1};\bar{j}_{23}\bar{m}_{23}\vert \bar{j'}\bar{m'}\rangle \langle \bar{j}_{2}\bar{m}_{2};\bar{j}_3\bar{m}_3\vert \bar{j}_{23}\bar{m}_{23}\rangle\\ &=\delta_{\bar{j}\bar{j'}} (-1)^{\bar{j}_1+\bar{j_2}+\bar{j}_3+\bar{j}} \sqrt{(2\bar{j}_{12}+1)(2\bar{j}_{23}+1)} \left\{\begin{array}{ccc} \bar{j}_1&\bar{j}_2&\bar{j}_{12}\\ \bar{j}_3&\bar{j}&\bar{j}_{23} \end{array}\right\} \end{align} This is equation (9.1.8) from D.A. Varshalovich et al, Quantum Theory of angular momentum (1988 English edition by WorldScientific; in the Russian edition some of the material is in different places).

There are some CG's to manipulate to the right form but basically the identification is $$\bar{j}_1\to j_1\, ,\quad \bar{j}_2\to j_2\, ,\quad \bar{j}_3\to J' \, ,\quad \bar{j}_{12}\to J\, ,\quad \bar{j}_{23}\to j_3\, ,\quad \bar{j}=\bar{j'}\to G\, ,$$ Your expression has a final sum on $M$ which provides an additional $(2G+1)$ factor, giving (1) as final expression.

I've checked it with about half-a-dozen values and it seems to work, but please double-check this as I could have made a typesetting error.

Edit: after comments I double checked and found that my original expression had the incorrect overall phase. I believe the current Eq.(1) is correct, i.e. the overall phase is $+1$. I've checked the result for various half-integer and integer values of $j_1,j_2$ and $j_3$ using the built-in ClebschGordan and SixJSymbol routines of Mathematica.

• Hi ZTH, I've checked your result using Mathematica, and it works fine except for a sign, so you either use a different convention for the $6j$ symbol, or there is a factor of $(-1)^\mathrm{something}$ missing somewhere (or I messed up my MMA code). Are you using the same def. for $6j$ as in wikipedia? – AccidentalFourierTransform Apr 18 '17 at 17:17
• @AccidentalFourierTransform Thanks for checking. I had trouble with the sign and will recheck later. I checked using Mma as you did so the problem will not be in the definition of the $6j$; I could have lost a phase somewhere, made error in the subtitutions when I keyed it in Mma etc.. My original phase was $(-1)^{(2j_2+J')}$ and didn't quite work with the examples I tried, and I somehow found another factor of $(-1)^{J'}$ but I could have messed it up. – ZeroTheHero Apr 18 '17 at 17:31
• @AccidentalFourierTransform I think I fixed the phase in the calculation. See my edit. – ZeroTheHero Apr 18 '17 at 20:10
• Yep, now it works :-) – AccidentalFourierTransform Apr 18 '17 at 20:20
• @AccidentalFourierTransform thanks for checking on your own. – ZeroTheHero Apr 18 '17 at 20:24