What is the importance of unitary (in-)equivalent representations? Say we have two representations of the observables from an abstract $C^*$-algebra $\mathcal A$ on two Hilbert spaces $H_1$ and $H_2$, i.e. consider the maps $\pi_1,\pi_2: \mathcal A \longrightarrow \mathcal B(H_1), \mathcal B(H_2)$.
If these representations are unitary equivalent, this means that there exists a unitary $U$ such that $\pi_2(A) = U\pi_1(A)U^{-1}$ for all $A\in \mathcal A$. Similarly, if an abstract state is normal with respect to both of these representations, i.e. can be represented by a density matrix on the respective Hilbert space, we find $\rho_2 = U\rho_1 U^{-1}$. And so all predictions we make are the same, irrespective of the Hilbert space on which we represent these observables and states.
However, even for unitary inequivalent representations we can find a unitary $U$ (as the Hilbert spaces are isomorphic) such that $\rho^\prime=U\rho_1 U^{-1}$ and $A^\prime = U\pi_1(A)U^{-1}$ on $H_2$, but with $\rho_2\neq \rho^\prime$ and $\pi_2(A)\neq A^\prime$.
But all expectation values and hence all predictions agree with our first representation. And isn't this the only important thing? So what is the significance of unitary (in-)equivalent representations? It seems that I am missing a crucial thing here.
 A: When the algebra has a non trivial center, then it is easy to understand how inequivalent representations pop out and what is the physical difference between them.
Consider a $C^*$ (or simply $*$) algebra $\cal A$ with non trivial center and two unitary representations $\pi_i: {\cal A} \to \mathfrak{B}(H)$, $i=1,2$, on the same Hilbert space $H$. Finally assume that the two representations are irreducible. If $a=a^* \in \cal A$ is an element of the center (e.g. a Casimir operator in the universal enveloping algebra associated to a Lie group), the operators  $\pi_i(a)$ must be proportional to the unit element $\pi_i(a)= c_iI$ for a pair of reals $c_i$ (Schur's lemma). These values are physical observables and can be measured. The choice of these values distinguishes between the two representations which are unitarily inequivalent: Every unitary intertwiner operator should transform $c_1I$ to $c_2I$ and this is not possible if the two constants are different. However both operators represent the same central element $a$. Both representations are physically admissible, in principle, but physics chooses the value of $a$ and declares which representation is the right one. If $\pi_1$ produces the wrong value of $c_1$, every unitary map from $H$ to $H$ gives rise to a new representation $\pi'_1= U\pi_1U^{-1}$ as you say, but it is  not able to change the wrong value of $c_1$. The new representation is physically admissible, but it is not the right one.
The values of the constants representing the center are fixed by the algebraic pure  states. (Pure states gives rise to irreducible GNS representations.)
Indeed, if $\omega: {\cal A}\to \mathbb{C}$ is an algebraic pure state and $\pi_\omega$ is its GNS representation and $a$ stays in the center, then
$$\omega(a)= \langle \Psi_\omega, \pi_\omega(a)\Psi_\omega\rangle=  \langle \Psi_\omega, c_aI \Psi_\omega\rangle= c_a.$$
The nature of the difference of inequivalent unitary representations of algebras with trivial center (as an infinite dimensions Weyl algebra) is more subtle. A recent book on that is the one  by Schmudgen.
