Let $\mathcal{A}$ be a unital $C^*$-subalgebra of $B(H)$. Then the definition $\phi(a):=\langle ah,h \rangle$ for a fixed $h\in H, \|h\|=1$ and for all $a\in\mathcal{A}$ defines a state on $\mathcal{A}$, (i.e. a positive functional with norm 1).
Now if $\mathcal{A}$ is a non-unital $C^*$-subalgebra of $B(H)$, then does the same example hold true for a state? If not, what could be a simple example of a state in the case of non-unital $C^*$-subalgebra of $B(H)$?
Edit 1: If we assume that $\mathcal{A}$ is separable, can we prove that the functional $\phi$ defined above is a state?
Let $a \in \mathcal A$ be positive, i.e. $\forall x \in H : \langle a x, x \rangle \ge 0$. Then of course also $\langle ah, h \rangle \ge 0$. Hence, $\phi$ is positive. Furthermore, $$ \left| \langle ah, h \rangle \right| \le \|ah\| \|h\| \le \|a\| \|h\| \|h\| = \|a\|, $$ which is why our functional is continuous, and it's norm is bounded by $1$. Furthermore, it is linear due to $$ \langle (a + \lambda b)h, h \rangle = \langle ah, h \rangle + \lambda \langle bh, h \rangle. $$ But, the functional need not have norm $1$. For, if we take $H = \ell^2$ and $\mathcal A$ to be the linear span of the completion of the left-shift operator by composition, then if $h = e_1$, then $ah = 0$ for all $a \in \mathcal A$. Hence, $\phi$ is the zero functional. Note that $\ell^2$ is separable.