3
$\begingroup$

Background: I'm a quantum chemist and even once wrote a proggie for 3j and 6j symbols. Imagine my gaping mouth when I read the paper (Reshetikhin/Turaev, I think) who let them pop up in my fave, knot theory. My ultimate goal would be to rewrite my zoo of S matrices with 3j symbols etc. so that checking the invariance under Reidemeister moves would just need a few applications of Biedenharn-Elliot and whatsnot. (AND I don't have to solve 1000 nonlinear equations in 20 variables to get my S in the first place :-)

Namagiri is throwing epiphanies at me lately in torrents: meanwhile I can rewrite most of my zoo into simpler trivalent vertexes (looking like this: =< ) - and maybe the matrix elements of those tensors are just 3j symbols? One day later I tripped over q-alg/9706029v2 and if you look closely, the intertwiners ARE 3j symbols, look like trivalent vertices and fulfil same relations as those in the Kuperberg G2 paper!

But now I'm caught between two contradictory statements. My instinct says that you just draw quantum brackets around any integer in the standard sum formula: http://en.wikipedia.org/wiki/Table_of_Clebsch-Gordan_coefficients and voila, Quantum Clebsch. But in the paper above, a psi for a representation replaces the usual J, and also the Wiki implies Quantum Clebsches depend on the quantum group. (Which makes more sense anyway, because trivalent-vertex-for-G2 != trivalent-vertex-for-A2, for example).

Question: Can you confirm that Quantum Clebsches depend on the quantum group? If yes, why does the intuitive approach (replace all integers by quantum integers) fail here? Or maybe it fails generally, except for a special quantum group, which is? (And does, by chance, Scott Morrisons QuantumKnot package contain a function for computing Quantum Clebsches for which the formula is known?)

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Different bases for your models of the representations will give different Clebsch-Gordon coefficients. $\endgroup$
    – Charlie Frohman
    Commented Aug 2, 2011 at 12:59
  • 7
    $\begingroup$ I haven't voted on your question, but I did find the flow of this question very hard to follow. At the moment it reads very much like a stream of consciousness or personal diary; including more background, defining more of your terms more precisely, and possibly narrowing the scope of your narrative to a core set of facts or observations which are relevant to potential answerers would greatly help. $\endgroup$
    – j.c.
    Commented Aug 2, 2011 at 13:36
  • $\begingroup$ <shrug> Sorry, I know my writing is most annoying to read for a mathematician. My brain is extremely jumpy, I don't speak Math-ese, and I never had a formal education in higher math. I'm already proud when I manage to state an intellegible question. I'm trying to convince Namagiri to direct her insights to a new Ramanujan instead of me, but argue with a goddess :-) $\endgroup$
    – Hauke Reddmann
    Commented Aug 3, 2011 at 9:18

1 Answer 1

Reset to default
11
$\begingroup$

In chemistry the only Lie group that comes up is SU(2). Naively you'd think the relevant group would be SO(3), which is the group of symmetries of space and so acts on the wave functions. But actually because electrons are Fermions you probably also care about SU(2) (which is the double cover of SO(3)).

Thus, traditional "Clebsch-Gordon" coefficients involve the group SU(2). You can generalize this in two independent directions:

  1. Replacing SU(2) with other Lie groups. (This is classical mathematics, but not so relevant to chemistry, unless you somehow ended up doing chemistry in higher dimensions.)

  2. Quantizing. (This is much newer mathematics involving quantum groups.)

I think you're getting confused because you hadn't heard of generalization 1 before.

(Warning: "replace all the integers with quantum integers" will cause some errors in general. For example, the dimension of the fundamental representation of SO(n) is n, but the quantum dimension of the fundamental rep of quantum SO(n) is either $[n-1]_q+1$ or $[n-1]_{q^2}+1$ depending on parity. In neither case is it [n].)

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Your warning is duely noted. :-) (In 99 of 100 cases, I can trust my intuition, but that is rather useless, as I never know which one is the rotten...) $\endgroup$
    – Hauke Reddmann
    Commented Aug 3, 2011 at 9:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .