# Fining Clebsch-Gordan coeffs in special cases, alternate method

by Jesssa
Tags: alternate, cases, clebschgordan, coeffs, fining, method, special
 P: 51 Hello, I was wondering, is there a way to derive the expression for these coefficients without the use of the general CGC formula? For example, the J=0 case (this is taken from wikipedia) The 1/sqrt term is clear but the (-1)^.. term not so much, is there a way to find this coefficient? I have read it requires the use of spinors and so on but I was wondering if there was a more simple way? I found a derivation of the CGC using only binomial expansions, but it was difficult to see exactly how they saw the starting point, it seemed like it must have been educated guesses until worked. Other derivations of the coefficients rely on the J=0 coefficients or more advanced mathematics. Thanks,
 Sci Advisor P: 3,448 The usual way to derive Clebsch-Gordan coefficients in special cases is to use stepping operators. For the example you give, if you let |0,0> = ∑ Cm|j,m>|j,-m>, you know that J+|0,0> = 0. J+(∑ Cm|j,m>|j,-m>) = ∑ Cm√(j-m)(j+m+1) (|j,m+1>|j,-m> + |j,m>|j,-m+1>) = ∑√(j-m)(j+m+1)(Cm + Cm+1)|j,m+1>|j,-m> In order for this sum to vanish, the Cm's must all have the same value and alternate in sign.
 P: 51 Oh I see, Thanks for your answer! However I am still having trouble figuring out how the negative sign of the coefficients depends on j1 and m1 in this example. (-1)j1 + m1 From application of either of the ladder operators you get Cm = -Cm+1 But from this it's not clear which coefficient is actually negative, only that they have opposite signs. Is it possible to simply find things like this? What I mean is, in the case of this example, the dependence of the negative sign on j1 and m1?