Root vectors of Lie algebras
I need some help with understanding the basics of Lie algebra theory. I suspect my problems are due to a fundamental misunderstanding somewhere so apologies in advance for the naivety of the questions and for the restatement of elementary mathematics.
As I understand it elements of Lie groups can be represented in terms of a set of generators by exponentiation. The set of generators form a Lie algebra whose properties are determined by the commutators of the generators. A subset of the generators (the Cartan subalgebra or CSA ) commute and are simultaneously diagonalisable. The eigenvectors of the CSA generators are used to represent physical states. The remaining generators, referred to as step or ladder operators, when applied to the eigenvectors of the CSA generators transform them into other eigenvectors with corresponding eigenvalues that differ by 2 from the eigenvalue corresponding to the original eigenvector. The algebra can be largely defined in terms of the eigenvalues of the CSA generators which must be integer and are referred to as roots or weights. Is this correct and is there any difference between a root and a weight?
The specific problem I have involves sl(3). I have seen a representation of this in terms of 8 3*3 real matrices (related to the Gell-Mann matrices) of which 2 are diagonal with diagonals (1, -1,0) and (0,1,-1). Iíve also been told that the root vectors of sl(3) are α(1)=(2,1) α(2)=(-1,2) and θ=(2,1) apparently regardless of the representation used. However the eigenvalues of real diagonal matrices are the diagonal elements themselves so why are the elements of the root vectors different, why do they not differ by 2 and why are there three of them? Clearly I have misunderstood something basic here but despite referring to several texts I am unable to resolve. Any help appreciated.
Re: Root vectors of Lie algebras
Root is employed (instead of weight) when you are in the adjoint representation of the algebra.
Where have you found eigenvalues differ by 2? do not forget that the states in the CSA are eigenvectors with nul eigenvalues (commutation in the CSA).
Ie for su(3) you have 8 eigenvectors (2 in the CSA) and 3 pairs of ladders
|All times are GMT -5. The time now is 02:56 AM.|
Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums