A criterion for vector states to be in the same irreducible representation A little wish...: is there a theorem that corresponds or implies the following 
Let $A$ be a $C^*$ algebra with the data of a representation in $B(H)$. Let $x,y$ be two vectors and call $S(x,y)$ the set of states induced by vectors of $P(span(x,y))$, where $P$ stands for projective space. 
The map from $P(span(x,y))$ to $S(x,y)$ is injective iff $x$ and $y$ belong to the unitarily equivalent irreducible subrepresentation of $A$.
or maybe it is not a "iff" but only  one direction.
I want to write such a result as a motivation to introduce superselection sector as irreducible representation of the algebra of observables in physics.
 A: Given a vector, a state is defined as
$$\omega_{\lambda\mathbf{x}+\mu\mathbf{y}}: A\ni a\mapsto \langle    \lambda\mathbf{x}+\mu\mathbf{y}|\pi(a)|\lambda\mathbf{x}+\mu\mathbf{y}\rangle / \lVert  \lambda\mathbf{x}+\mu\mathbf{y} \rVert^2\\
 =\lvert \lambda\rvert^2 \omega_{\mathbf{x}}+ \lvert \mu\rvert^2 \omega_{\mathbf{y}} + \lambda^* \mu \langle\mathbf{x}|\pi(a) |\mathbf{y}\rangle + \lambda \mu^*\langle\mathbf{y}|\pi(a) |\mathbf{x}\rangle $$
Assuming the representation decomposes into direct sum (a sufficient condition: semi-simple algebra, are there other conditions?) one sees that the two last term vanish if x,y are in different irreducible subrepresentations. And hence one cannot observe the argument of $\lambda^* \mu$
There is not to worry about unitarily equivalent irreducible representation of anything. The usual situation is that we have a direct sum of irreducible representation of the gauge group (first kind) and in that case the algebra can contain intertwiners, it is not the case here.
However that is not a proof, we've just seen that we definitely don't have an injective map from the projective plane to the subset of states, if $\mathbf{x}$ and $\mathbf{y}$ are in different representations.
A proof can be found in Dixmier's book on $C^*$ algebra, proposition 2.5.6
