# Angular commutation relations

1. Jul 10, 2009

### facenian

There must something wrong with my understanding of this relations because I think the usual way they are derived in many textbooks makes no sense. It goes like this, first assume that to every rotation O(a) in euclidean space there exists a rotation operator R(a) in Hilbert space,second: the relation stated first is an homomorphism T,that is T(O)=R
So far, so good the problem is that after verifying the relation O_x(da)O_y(db)-O_y(db)Ox(da)=O_z(dadb)-I, for infinitesimal rotations da and db in euclidean space the authors conclude that a similar relation holds for infinitesimal rotations in Hilbert space. This last step requieres that besides being T(O_1O_2)=T(O_1)T(O_2) which is ok, the relation
T(O_1+O_2)=T(O_1)+T(O_2) must also hold.
is it correct what I'm saying? in which case, why should T(O_1+O_2)=T(O_1)+T(O_2) hold?

2. Jul 10, 2009

### George Jones

Staff Emeritus
Note that the rotation group is not closed under addition, so O_1 + O_2 is not, in general, a member of the group, and O_1 + O_2 is not in the domain of the group homomorphism T.

The infinitesimal elements (the Lie algebra) of a (Lie) group, however, do form a vector space, and a homomorphism between two Lie groups (in this case the rotation group and a group of operators) gives rise to a (linear) homomorphism between corresponding Lie algebras of infinitesimals (vector spaces).

3. Jul 11, 2009

### facenian

So you say that in the last sentece I posted T should be replaced by another homomorphism P this time between to vector spaces so P(O_1 + O_2)=P(O_1) + P(O_2) holds. Now it makes sence however a don't think it is trivial and should be explained properly in Texts.
Thank you.