positive linear functionals and states(C* algebras) Let $\omega$ be a state of an algebra $A$ . 
if $a \in A_{sa} $ (self-adjoint) such that $\omega(a^2)=\omega(a)^2$ then $\omega(ab)=\omega(a)\omega(b)=\omega(ba)$.
I would like to prove this argument for  $b\neq{a}$, but I don't understand how the fact that $\omega(a^2)=\omega(a)^2$ can help me in this case.

Assume that $\omega$ is a positive linear functional on $A$ such that $\omega(a^2)=\omega(a)^2$ for all $a\in A_{sa}$ and $b\in A_{sa}$ is arbitrary.


*

*As @SamM suggested expanding both sides of the equality $\omega((a+b)^2)=\omega(a+b)^2$ you will get 
$$
\text{Re}\,\omega(ab)=\omega(a)\omega(b).
$$

*Cauchy-Schwarz inequality for positive functionals gives 
$$
|\omega(ab)|^2\le\omega(a^2)\omega(b^2)=\bigl(\omega(a)\omega(b)\bigr)^2.
$$

*"1+2" gives that $\text{Re}\,\omega(ab)=|\omega(ab)|$, i.e. $\omega(ab)\in\mathbb{R}$ and, hence, $$
\omega(ba)=\overline{\omega(ab)}=\omega(ab)=\omega(a)\omega(b).
$$


P.S. We have used that a positive linear functional is self-adjoint, i.e. 
$\omega(x)=\overline{\omega(x^*)}$, and have not used that $\omega$ is a state, i.e. $\|\omega\|=1$.
