state on a non-unital $C^*$ algebra Suppose $\tau$ is a state on a non-unital $C^*$ algebra $A$.There is a well-known inequality:
$$\tag{$*$}|\tau(a)|^2\leq\tau(a^*a),\ \text{ for all } a\in A.$$


*

*Does there exist some  nonzero element $a_0$ such that $\tau(a_0^*a_0)-|\tau(a_0)|^2$ is small enough?

*Does there exist a nonzero elements $a_0$ such that equality holds in $(*)$?
 A: For any state and any approximate unit $\{e_j\}$ of $A$, you have $\lim_j\tau(e_j)=\|\tau\|$ (cfr. Davidson's C$^*$-Algebras by Example, Lemma I.9.5). So here $\tau(e_j)\to1$. Since one can take $e_j\geq0$ and $\|e_j\|\leq1$ for all $j$, you have 
$$
\tau(e_j^*e_j)=\tau(e_j^2)\leq\|e_j\|\,\tau(e_j)\leq1. 
$$
Thus
$$
1=\lim_j|\tau(e_j)|^2\leq\limsup_j\tau(e_j^*a_j)\leq1,
$$
giving us equality. This gives the existence of your $a_0$. 
The answer to your second question is no when $\tau$ is faithful. Given your $\tau$, you can always extend it to the unitization $\tilde A$ of $A$, and the extension is still faithful. So if $\|a\|\leq1$ and $|\tau(a)|^2=\tau(a^*a)$, this is equality in the Cauchy-Schwarz inequality, so you get that $a=\lambda\,1$. 
When $\tau$ is not faithful, equality can occur. For instance take $A=M_2(\mathbb C)\oplus K(\ell^2(\mathbb N))$, and $\tau(a\oplus k)=a_{22}$. Then the matrix unit $E_{22}\oplus 0$ satisfies the equality
A: Regarding your question about the $a$'s for which $$\tau(a^\ast a) = |\tau(a)|^2$$: It seems that it can be solved using Choi's Theorem on multiplicative domains of cp maps, see [BO; Proposition 1.5.7]. In particular $a$ satisfies the identity above iff for every $b \in A$, $\tau(b a) = \tau(b) \tau(a) = \tau(a) \tau(b) = \tau(a b)$.
[BO] Brown, Nathanial P.; Ozawa, Narutaka, $C^*$-algebras and finite-dimensional approximations, Graduate Studies in Mathematics 88. Providence, RI: American Mathematical Society (AMS). xv, 509 p. (2008). ZBL1160.46001.
