There is a quotation below:
Definition 1.7.2. Two maps $\pi: A\rightarrow B(H)$ and $\sigma: A\rightarrow B(K)$ are called approximately unitarily equivalent if there is a sequence of unitary operators $U_{n}: H \rightarrow K$ such that $$||\sigma(a)-U_{n}\pi(a)U_{n}^{*}||\rightarrow 0$$
for all $a\in A$. If it also happens that $\sigma(a)-U_{n}\pi(a)U_{n}^{*}$ is a compact operator, for all $a\in A$ and $n\in N$, then we say that $\pi$ and $\sigma$ are approximately unitarily equivalent relative to the compacts.
Note that approximate unitary equivalence relative to the compacts is a much stronger notion as it implies that after passing to the Calkin algebra, the representations $\pi $ and $\sigma$ are actually unitarily equivalent.
My comprehension on the note above is: the $\sigma(a)$ and $\pi(a)$ is unitarily equivalent after passing to Calkin algebra implies $$(u^{*}+K(H))(\sigma(a)+K(H))(u+K(H))=\pi(a)+K(H)$$
here, $K(H)$ denotes all the compacts on $H$ and $u+K(H)$ is a unitary in Calkin algebra. So, there exist a compact $K$ such that $u^{*}\sigma(a)u-\pi(a)=K$, this $K$ may unequal to zero. Hence, approximate unitary equivalence relative to the compacts is much stronger notion. Is there anything wrong in my viewpoint?
What the comment is saying is that when you assume that $\sigma $ and $\pi$ are approximately unitarily equivalent relative to the compacts, you have $\|\sigma(a)-U_n\pi(a)U_n^*\|\to0$, but also that $\sigma(a)-U_n\pi(a)U_n^*\in K(H)$. So, in the Calkin algebra, $\sigma(a)=U_n\pi(a)U_n^*$ for all $a$ and all $n$ (note that the image of a unitary through the quotient map is again a unitary).
But it is not obvious that approximate unitary equivalence relative to the compacts is "much stronger": when $A$ is separable, the two notions are equivalent. See Theorem II.5.8 in Davidson's book.