Unitary Equivalence of Two Irreducible $ * $-Representations of a GCR $ C^{*} $-Algebra that Have the Same Kernel. In general, if two irreducible $ * $-representations of a $ C^{*} $-algebra $ A $ have the same kernel, then we can say that they are approximately unitarily equivalent. When $ A $ is GCR, how can we prove that they are actually unitarily equivalent?
 A: In order to prove the OP’s claim, we need the following two basic propositions concerning the representation theory of $ C^{*} $-algebras. Their proofs can be found in any good graduate textbook on $ C^{*} $-algebras (I highly recommend William Arveson’s An Invitation to $ C^{*} $-Algebras).

Proposition 1: Let $ (\pi,\mathcal{H}_{\pi}) $ be an irreducible $ * $-representation of a $ C^{*} $-algebra $ A $. Let $ I $ be a closed two-sided ideal of $ A $.
  
  
*
  
*If $ \pi[I] \neq \{ 0_{\mathcal{B}(\mathcal{H})} \} $, then $ \left( \pi|_{I},\mathcal{H}_{\pi} \right) $ is an irreducible $ * $-representation of $ I $.
  
*Every irreducible $ * $-representation of $ I $ extends uniquely to an irreducible $ * $-representation of $ A $.
  
*If two irreducible $ * $-representations of $ I $ are unitarily equivalent, then so are their unique irreducible extensions.
  
  
  Proposition 2: Let $ \mathcal{H} $ be a Hilbert space. Then any irreducible $ * $-representation of the $ C^{*} $-algebra $ \mathcal{K}(\mathcal{H}) $ of compact operators on $ \mathcal{H} $ is unitarily equivalent to the identity $ * $-representation
  $$
(
\iota_{\mathcal{H}}: \mathcal{K}(\mathcal{H}) \hookrightarrow \mathcal{B}(\mathcal{H}),
\mathcal{H}
).
$$

Let $ A $ be a $ C^{*} $-algebra that is GCR, and suppose that $ (\pi,\mathcal{H}_{\pi}) $ and $ \left( \rho,\mathcal{H}_{\rho} \right) $ are two irreducible $ * $-representations of $ A $ that have the same kernel. Consider the following $ * $-homomorphism:
\begin{align}
\lambda: \pi[A] & \to \rho[A] \subseteq \mathcal{B} \! \left( \mathcal{H}_{\rho} \right), \\
         \pi(a) & \mapsto \rho(a).
\end{align}
Observe that $ \lambda $ is well-defined and injective (because $ \text{Ker}(\pi) = \text{Ker}(\rho) $). Furthermore, $ \left( \lambda,\mathcal{H}_{\rho} \right) $ is an irreducible $ * $-representation of $ \pi[A] $ because $ \text{Range}(\lambda) = \rho[A] $. As $ A $ is GCR by hypothesis, $ \pi[A] $ contains $ \mathcal{K}(\mathcal{H}_{\pi}) $ as a closed two-sided ideal, and as $ \lambda[\mathcal{K}(\mathcal{H}_{\pi})] \neq \{ 0_{\mathcal{B}(\mathcal{H}_{\rho})} \} $ (because $ \lambda $ is injective), Proposition 1 says that $ \left( \lambda|_{\mathcal{K}(\mathcal{H}_{\pi})},\mathcal{H}_{\rho} \right) $ is an irreducible $ * $-representation of $ \mathcal{K}(\mathcal{H}_{\pi}) $. Then by Proposition 2,
$$
\left( \lambda|_{\mathcal{K}(\mathcal{H}_{\pi})},\mathcal{H}_{\rho} \right)
\sim_{u}
\left( \iota_{\mathcal{H}_{\pi}},\mathcal{H}_{\pi} \right).
$$
By Proposition 1 again,
$$
\left( \lambda,\mathcal{H}_{\rho} \right)
\sim_{u}
(i: \pi[A] \hookrightarrow \mathcal{B}(\mathcal{H}_{\pi}),\mathcal{H}_{\pi});
$$
in other words, there exists a unitary operator $ U: \mathcal{H}_{\rho} \to \mathcal{H}_{\pi} $ such that
$$
\forall a \in A: \quad
U \circ \rho(a) \circ U^{-1} = U \circ \lambda(\pi(a)) \circ U^{-1} = \pi(a).
$$
Therefore, $ (\pi,\mathcal{H}_{\pi}) \sim_{u} \left( \rho,\mathcal{H}_{\rho} \right) $ as required to be shown.
