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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!

edited Feb 2 2011 at 19:02

Steve Huntsman
8,139●1●22●56

asked Feb 2 2011 at 18:28

Johannes
11●1

Scott Carter · Answer 1 · 2011-02-02 22:33:20Z UTC

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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.

answered Feb 2 2011 at 22:33

Scott Carter
3,448●1●9●23

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 2011 at 15:10

Abdelmalek Abdesselam · Answer 2 · 2011-02-03 00:20:44Z UTC

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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?

edited Feb 3 2011 at 0:20

answered Feb 3 2011 at 0:14

Abdelmalek Abdesselam
2,251●4●17

	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 2011 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 2011 at 16:34

Sum relation for Clebsch-Gordan-Coefficients?

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