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.

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Apr 17, 2017 at 23:27
  • $\begingroup$ What makes you say that the signs don't look correct? $\endgroup$
    – jawheele
    Commented Apr 18, 2017 at 2:06
  • $\begingroup$ 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! $\endgroup$
    – jawheele
    Commented Apr 18, 2017 at 9:31
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 18, 2017 at 15:35
  • $\begingroup$ @jawheele Please check my edited expression as I had an overall phase issue in my previous expression. $\endgroup$ Commented Apr 18, 2017 at 20:14

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Apr 18, 2017 at 17:17
  • $\begingroup$ @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. $\endgroup$ Commented Apr 18, 2017 at 17:31
  • $\begingroup$ @AccidentalFourierTransform I think I fixed the phase in the calculation. See my edit. $\endgroup$ Commented Apr 18, 2017 at 20:10
  • 1
    $\begingroup$ Yep, now it works :-) $\endgroup$ Commented Apr 18, 2017 at 20:20
  • $\begingroup$ @AccidentalFourierTransform thanks for checking on your own. $\endgroup$ Commented Apr 18, 2017 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.