MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the context of (numerically) calculating reduced density matrices in the Lipkin-Meshkov-Glick model (a model introduced to describe atomic nuclei, which has however found many other applications as well), I need to carry out a sum over Clebsch-Gordan coefficients.

If it is possible to find a simplification that replaces the sum below by a (simple) explicit expression, the numerical effort would be greatly reduced (the angular momenta in question are large, therefore the sum over $m_1$ has many terms). The sum in question is:

$\sum_{m_1} \langle J M|j_1 m_1 j_2 m_2\rangle \langle j_1 m_1 j_2 m_2'|J M'\rangle$

I have found sum formulas for Clebsch-Gordan coefficients, but none applicable to this one. Any insight would therefore be greatly appreciated!

share|cite|improve this question

I doubt this is helpful, but on page 26 of "The Classical and Quantum 6j-symbols" we give a closed formula for the $sl_2(C)$ coefficients in a particular normalizations that is compatible with the quantum case.

share|cite|improve this answer
Thank you, but while we were aware of the closed form of the CG coefficients, we could not see a way to use it. That might just be our lack of ability of course! – Johannes Feb 3 '11 at 15:10

If you sum on both $m_1$ and $m_2=m_2'$ then it does simplify into $\delta_{M,M'}$. Without $m_2$ summation, I don't think one can get much better than that. It all depends also on what larger tensor contraction your identity is inserted into. Is this only a small part of a larger formula with a product of a lot of CG coefficients and many sums over $m$'s?

share|cite|improve this answer
Thank you, this simple orthogonality relation was what made us hope that also our sum could be somehow simplified. There is a further tensor contraction, but no more CG coefficients appear in it, which is why I only posted the part which we had hoped to find a simpler expression for. – Johannes Feb 3 '11 at 15:14
This is probably a long shot but you might try to use the Wilf-Zeilberger theory of combinatorial sums of hypergeometric type. Each CG is given by a 3F2 single index sum. So your combination is triple hypergeometric sum. There may be some software implementing the multivariate WZ theory (Inventiones vol 108, p 575) which tells you if such a sum can be simplified. In the univariate case (sum over one index), there is a package to do that in almost every computer algebra software like Maple, Mathematica etc. I don't know about triple sums. – Abdelmalek Abdesselam Feb 3 '11 at 16:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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