Yoran91 Posted April 12, 2013 Posted April 12, 2013 Hi everyone, I'm reading Edmonds' Angular momentum in quantum physics and I'm stuck on one expansion of 9j symbols in terms of 6j symbols. Technically, I'm stuck on the expansion of recoupling coefficients concerning four angular momenta in terms of recoupling coefficients related to three angular momenta. The relation I don't understand is found on page 100-101, section 6.4, the very first equality in equation 6.4.1. Edmonds states that there are two adding schemes of four angular momenta. Denoting them by J1, J2 ,J3,J4, these are 1. J1+J2=J12 and J3+J4=J34 and J=J12+J34 2. J1+J3=J13 and J2+J4=J24 and J=J13+J24. These two adding schemes produce basisvectors of the total Hilbert space (tensor product HIlbert space) and there is a unitary transformation between them. Denoting the basisvectors by |j12 j34 jm> and |j13j24 jm> respectively, the coefficients in the unitary transformation are <(j1j2)j12 (j3j4) j34 j m | (j1j3) j13 (j2j4) j24 jm>. Now Edmonds states that this transformation can be performed in three steps with coupling of only three angular momentum operators in each step. His formula is <(j1j2) j12 (j3j4) j34 j m| (j1j3) j13 (j2j4) j24 j m> = Sum < (j1j2) j12 j34 j | j1 (j2 j34) j' j> <j2 (j3j4) j34 j' | j3 (j2j4) j24 j' > <j1 (j3j24) j' j | (j1j3) j13 j24 j> but I really don't see how the unitary transformation can be performed in these three steps involving only addition of three angular momenta. Can anyone shed some light on this?
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