Given a simple lie algebra (over ${\mathbb C}$ or ${\mathbb R}$). What is the number of operators such that their eigenvalues sufficiently label all state vectors in the algebra's representation space.

According to this paper http://arxiv.org/abs/quant-ph/0409209 (page 24), if $d$ is the dimension of the algebra and $n$ is its rank then the number is $(d+n)/2$.

I am not familiar with this result; a reference or explanation would be appreciated.

(adding additional background information from the paper above; this was too large to include in a comment) :

Although the setting in the paper is the lie group $SO(4,2) \otimes SU(2)$ I moved it to a lie algebra setting : $so(6) \oplus su(2)$ or $A_3 \oplus A_1$ for simplicity.

What the author is saying is that it takes (15+3)/2=9 operators to label an $A_3$ state and (3+1)/2=1 operators to label an $A_1$ state. A state here would be any element in the basis of a finite dimensional irrep. For $A_3$ the 9 operators would include 3 casimir elements and 3 cartan generators. Quoting the author : "According to a (not very well-known) result popularised by Racah, we need to find $(d-3n)/2=3$ additional operators in order to complete the set of the six preceding operators". This is what motivated the question. So it seems that these 9 operators would correspond to 9 commuting elements of the universal eneveloping algebra of $A_3$; I would expect that they are linearly independant.

ofrepresentations, or a space whichisa representation)? I'm also not sure what "the number of operators such their eigenvalues sufficiently label all state vectors" is supposed to mean. If you have a collection of operators whose joint spectrum on a finite dimensional space is simple (my best guess for what "sufficiently label all state vectors" means), then any generic linear combination of them also has simple spectrum. – Ben Webster♦ Feb 20 '14 at 9:24