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Fining ClebschGordan coeffs in special cases, alternate method 
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#1
Nov1712, 09:49 AM

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, 


#2
Nov1712, 10:36 AM

Sci Advisor
Thanks
P: 4,160

The usual way to derive ClebschGordan coefficients in special cases is to use stepping operators. For the example you give, if you let 0,0> = ∑ C_{m}j,m>j,m>, you know that J_{+}0,0> = 0.
J_{+}(∑ C_{m}j,m>j,m>) = ∑ C_{m}√(jm)(j+m+1) (j,m+1>j,m> + j,m>j,m+1>) = ∑√(jm)(j+m+1)(C_{m} + C_{m+1})j,m+1>j,m> In order for this sum to vanish, the C_{m}'s must all have the same value and alternate in sign. 


#3
Nov1912, 08:38 PM

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 j_{1} and m_{1} in this example. (1)^{j1 + m1} From application of either of the ladder operators you get C_{m} = C_{m+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 j_{1} and m_{1}? 


#4
Nov2012, 05:07 AM

Sci Advisor
Thanks
P: 4,160

Fining ClebschGordan coeffs in special cases, alternate method
Some of the phase factors involved in the definition of the angular momentum states and the ClebschGordan coefficients are not uniquely determined, and must be arbitrarily assigned. For example the state with the highest mvalue is conventionally chosen to have a plus sign. The standard choices are known as the Condon and Shortley phase conventions.



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