# Clebsch–Gordan coefficients: An Identity

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## Summary:

I have a problem getting a Clebsch–Gordan Identity

## Main Question or Discussion Point

Hi, everyone. I'm trying to get the next identity It is in the format <j1, j2; m1, m2 |j, m>. I hope you can help me

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DrClaude
Mentor
What is your starting point? Wigner 3j-symbols?

What is your starting point? Wigner 3j-symbols?
Hi, DrClaude.
I start with Clebsch–Gordan coefficients to arise angular momemtum coupling i.e. express the addition of two angular momenta in terms of a third.

samalkhaiat
Summary:: I have a problem getting a Clebsch–Gordan Identity

Hi, everyone. I'm trying to get the next identity

View attachment 263536

It is in the format <j1, j2; m1, m2 |j, m>. I hope you can help me
The general expression for the CG coefficients is \begin{align*}C_{j_{3}m_{3}}(j_{1}j_{2};m_{1}m_{2}) &= \delta_{m_{3},m_{1}+m_{2}} \left( \frac{(2j_{3}+1) (j_{1} + j_{2} -j_{3})!(j_{1} + j_{3} - j_{2})!(j_{2} + j_{3} - j_{1})!}{(j_{1}+j_{2}+j_{3}+1)!}\right)^{1/2} \\ & \times \left( \prod_{i = 1}^{3}[(j_{i} - m_{i})!] \prod_{i = 1}^{3}[(j_{i} + m_{i})!]\right)^{1/2} \\ & \times \sum_{n} \frac{(-1)^{n}}{n! (j_{1} + j_{2} - j_{3} - n)!(j_{1} - m_{1} - n)!(j_{2} - m_{2} - n)!(j_{3} - j_{1} - m_{2} + n)!(j_{3} - j_{2} + m_{1} + n)!} , \end{align*} where we take the sum over $n$ when non-of the arguments of factorials are negative. The case you are considering: $j_{1} = j_{3} \equiv j, \ j_{2} = 1, \ m_{1} = m_{3} \equiv m, \ m_{2} = 0$, you sum over $n = 0 , 1$.

• DrClaude