Let $A$ be a $C^{\ast}$-algebra and $I$ be a closed subspace of $A$. Then $I$ is called $\textbf{modular}$ if $A/I$ is unital $C^{\ast}$-algebra and $I$ is called $\textbf{primitive}$ if $I=\operatorname{ker}(\pi) $ for some irreducible representation $\pi$ of $A$. It is known that maximal modular ideal is primitive.
How does one constructs modular ideals which are not primitive and also the primitive ideals which are not modular?
In $B(\mathcal{H}) $, I could not find any such example. Moreover, in $C_{0}(X) $, I know that modular ideals corresponds to compact subsets.
$\def\cA{\mathcal A}$ $\def\cB{\mathcal B}$ $\def\cC{\mathcal C}$ $\def\cI{\mathcal I}$ $\def\CC{\mathbb C}$ $\def\RR{\mathbb R}$ $\def\h{\mathcal H}$ $\def\k{\mathcal K}$ $\def\tr{\operatorname{tr}}$
You cannot expect to obtain many examples from $\cB(\h)$, since it has a single nontrivial ideal.
Note that you have $A/\ker\pi\simeq\pi(A)$. So you can get non-modular ideals by representing onto non-unital algebras. You can control primitivity by giving a "big" complement to the ideal that cannot have an irrep.
For instance if you take $\cA=\k(\h)$ and $\pi$ the identity representation, you have that $\cA/\ker \pi=\cA$ which is non-unital. And $\pi$ is irreducible. So $\ker\pi$ is primitive but not modular.
To find a modular ideal that is not primitive, we need to start with a unital C$^*$-algebra (so the quotient will be unital) and consider a non-irreducible representation. For example let $\cA=\CC^3$ and $\pi(a,b,c)=a$. Then $\pi(\cA)$ is abelian and two-dimensional, so $\pi$ is not irreducible. That is, the ideal $$I=\ker\pi=\{(0,b,c):\ b,c\in\CC\}$$ is modular but not primitive.