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  generating Clebsch-Gordan coefficients by applying Wigner-Eckart theorem to some "standard" tensor operator

+ 3 like - 0 dislike

Hi. This question came to me when I was reading Wigner-Eckart theorem. This theorem appears both on quantum mechanics textbooks, e.g., J.J. Sakurai's Modern QM, and on books about Lie groups such as Brian Hall, yet in slightly different forms.

In Sakurai's book, the theorem looks like


where the coefficient


is precisely the Clebsch-Gordan coefficient. 

In Brian Hall's book on Lie groups, Appendix D, the theorem states

Let A and B be two vector operators on a representation space V and W1, W2 be two irreducible invariant subspaces of V. Then there is a constant c such that


for all \(w\in W_1\),\(w'\in W_2\) and \(k\in\{1,2,3\}\).

Hall also mentioned that in light of this theorem, we can write

\[<u_l, B_k u_{l'}>=c\alpha(m,n,k,l,l')\]

in which the \(\alpha\) is also called the Clebsch-Gordan coefficient by Hall. 

With a little bit of effort, I managed to figure out that the two "versions" of Wigner-Eckart theorem are essentially the same thing.However, I wonder how to show that the \(\alpha\) here is really the CG coefficient defined in the inner product form \(<jk;mq|jk;j'm'>.\)

Standard proof given on QM books like Sakurai did its job by computing

\(<j'm'|[J_{\pm}, T^{(k)}_q]|j,m>,\) and showing that this inner product has the same recursive relation as the one in computing the Clebsch-Gordan coefficients. I'm pretty convinced, but I wonder whether there is a more direct way of doing this. 

My guess is that in Hall's proof, we have


So every tensor operator's matrix coefficients are proportional to those of a standard operator T, meaning, once this T is fixed, for any tensor operator A, we have


When we choose w and w' to be the standard weight vectors \(|j,m>\) and \(|j'm'>\), the inner product \(<jm|T_k|j'm'>\) could just give us the Clebsch-Gordan coefficients.

If such a tensor operator T exists, the equivalence of the two "versions" of Wigner-Eckart theorem will appear more transparent to me, and I'd appreciate it if someone helps me construct it.

P.S., I'm more closely following Brian Hall's notation and define a spherical tensor operator as an equivariant map

\[T: V_k\rightarrow End(V),\]

a map that intertwines the spin-k action on \(V_k\) and the standard adjoint action on End(V). If anyone wants to use a different notation, I think I can do the conversion myself, but it would be nice if more details could be given on its definition.Sorry for any inconvenience.

Thanks a lot in advance.

asked Jul 7, 2016 in Theoretical Physics by LIU ZHENG (15 points) [ no revision ]

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