The discussion centers on deriving a specific Clebsch–Gordan (CG) identity, particularly in the format
Quantum physicists, students of quantum mechanics, and researchers focusing on angular momentum theory will benefit from this discussion.
Hi, DrClaude.DrClaude said:
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.victor01 said: