the situation i want to talk about is the following:
$(H_1,\varphi_1),(H_2,\varphi_2)$ irreducible representation of a $C^*$-algebra $A$. A bounded operator $T:H_1\rightarrow H_2$ such that $T\varphi_1(a)=\varphi_2(a)T$ for all $a\in A$
I am asked to prove the following implication: If $\varphi_1$ and $\varphi_2$ are NOT unitarily equivalent, then $T=0$
My ideas for so far: Proof by contradiction: Suppose $T\neq 0$, then $T^*$ exists and is bounded. I want to prove that $T$ is unitary because then we have the situation that both representations are equivalent (unitarily). But therefore i have to show that $T$ is surjective and that $(Tx,Ty)=(x,y)$ for all $x,y$, but this seems to be very difficult or is there something which i don't see and which makes the proof easier? I only know more that non-zero vectors are cyclic vectors since the representations are irreducible but i have no idea how to use it. Or: is the direct implication the better choise?