Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

146 submissions , 123 unreviewed
3,953 questions , 1,403 unanswered
4,889 answers , 20,762 comments
1,470 users with positive rep
507 active unimported users
More ...

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

+ 3 like - 0 dislike
106 views

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

\[<j'm'|T^{(k)}_q|jm>=<jk;mq|jk;j'm'>\frac{<j'||T^{(k)}||j>}{\sqrt{2j+1}},\]

where the coefficient

\[<jk;mq|jk;j'm'>\]

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

\[<w,B_kw'>=c<w,A_kw'>\]

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

\[<w,B_kw'>=c<w,A_kw'>.\]

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

\[<w,A_kw'>=c(A)<w,T_kw'>.\]

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...