Concrete realization of quotient $C^*$-algebra Let $A$ be a $C^*$-subalgebra of the bounded operators $B(H)$ on some Hilbert space $H$. Let $J$ be a proper closed $2$-sided ideal of $A$. Then one knows that $A/J$ is another $C^*$-algebra.
By the Gelfand-Naimark theorem, one also knows that $A/J$ is $*$-isomorphic to a $C^*$-subalgebra of $B(H')$ for some Hilbert space $H'$.
Question: Is there a "natural" choice of $H'$?
This seems like a natural and basic question, but having looked around a bit, it seems that an answer may not be straightforward. In particular, in the case where $A=B(H)$ and $J=K(H)$ for $H$ separable, the Gelfand-Naimark-Segal construction gives a non-separable example of $H'$. But I would be interested if there are any references where this is discussed further.

I guess it depends on what you consider "natural". If $A$ is separable, then you can take $H$ separable, which allows you to take $H'=H$; but I don't think you'll have any relation between those concrete realizations of $A$ and $A/J$. It is probably worth remarking that the concrete presentation of the Hilbert spaces $H$ and $H'$ can very a lot, and while some representations are better than others, there is no obvious "best" one.
For a kind of generic example, let $A=C^*(\mathbb F_\infty)$, the universal C$^*$-algebra of the free group on infinitely many generators. This C$^*$-algebra is universal, in the sense that any separable C$^*$-algebra is a quotient of it. Being separable, $A$ has a faithful state, $\phi$, so we can do GNS and represent $A\subset B(L^2(A,\phi))$. Because $A$ has countably many generators with no relations, any separable C$^*$-algebra $A_0$ is a quotient of $A$. So you can choose any separable C$^*$-algebra $A_0$, represent it on a Hilbert space $H'$, in one of many many ways, and conclude that one cannot expect any natural relation between $H'$ and $H=L^2(A,\phi)$.
