Clebsch-Gordan Coefficients, M=m1+m2

Can someone tell me why the Clebsch-Gordan coefficients vanish unless m = m1 + m2 ? I can't undetstand its proof....
- James (age 25)
Graylingt, Mich, USA

Sure, this just says that the total angular momentum component along some axis is the sum of the contributions from the two parts. A more formal way to see it is that the state is the product of the states of the two parts. For orbital wave functions, one has an e^im₁φ and the other an e^im₂φ dependence on φ, so the overall state has an an e^i(m₁+m₂)φ dependence. The same general principle extends to spin as well, though without the simple angular functional dependence picture.

Mike W.

p.s. This reminds me how anachronistic our categories are on this site, since this gets categorized as "new and exciting physics".

(published on 06/01/2011)