Can distinct elements of a $C^*$-algebra be separated by a maximal left ideal?

Question

Let $A$ be a $C^*$-algebra, and let $f\neq g\in A$. Does there exist a maximal ideal $J\trianglelefteq A$ with $f+J\neq g+J$?

I'm particularly interested in the case of $A=B(\mathcal H)$, and why things aren't obvious. In this case, $f\neq g$ implies that there is some $h\in\mathcal H$ with $f(h)\neq g(h)$, so the left ideal of the annihilator of $h$ certainly separates them. However, $\text{Ann}_{B(\mathcal H)}(h)$ is not maximal, and I don't immediately see how to upgrade this to a maximal ideal.

An alternative approach is to attempt to find an irreducible representation. We have a good family of representations from GNS: for each positive linear functional $\rho$ on $A$, we have a representation $\pi_\rho$ on a Hilbert space, with the image of $1$ being cyclic for the representation. If $f\neq g$, then at least one of these representations has $f\cdot [1]\neq g\cdot [1]$, because the GNS representation is faithful. I know that the annihilator of a simple module is the intersection of maximal left ideals, so if I believe that if I can find a simple $A$-module on which $f$ and $g$ act differently, there is such an ideal.

Answer 1

$\forall a\neq 0$, there is a pure state $\tau$ on $A$ such that $\tau(a^*a)=\|a^*a\|\neq 0$(Theorem 5.1.11, [1]). Let $N_\tau=\{a|\tau(a^*a)=0\},$ then $N_\tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $a\notin N_\tau$.

[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.

Answer 2

Your ideal is maximal when $A=B(H)$. Let $J=\{T:\ Th=0\}$. Let $S\in B(H)\setminus J$. Then $Sh\ne0$. Choose $k$ with $\langle Sh,k\rangle=1$, and let $Wx=\langle x,k\rangle\,h$. Put $R=I-WS$. Then $$ Rh=h-WSh=h-h=0, $$ so $R\in J$. Then $$ I=R+WS\in J+\mathbb C\,WS,$$ and $J$ is maximal (its codimension is 1).