Show a relation for the state on $C^*$ algebra

Question

Let $\varphi$ be a state on $C^*$-algebra $A$. Assume $\varphi(a^2)=\varphi(a)^2$ for some self-adjoint elements $a\in A$. Show that $\varphi(ab)=\varphi(ba)=\varphi(a)\varphi(b)$ for any element $b\in A$.

Any element $b\in A$ can be written as the sum of two self-adjoint elements, i.e. $b=a_1+ia_2$ where $a_j=a_j^*, j=1,2$. Then I did sort of naive calculation by just fulfilling the given conditions for these self-adjoints, but it seems like still far away to reach the final desired.

What I am trying to do: suppose $\varphi(ab)=\varphi(a)\varphi(b)$ holds. Then it's equivalent to $\varphi(ab)^2=\varphi(a)^2\varphi(b)^2=\varphi(a^2)\varphi(b)^2$. Plug in $b=a_1+ia_2$ and show that some partial identities hold respectively, where however I got stuck to move on. So did for a few other attempts.

Thanks for illuminating me any pivotal steps or hints/tricky.

Answer 1

By Cauchy-Schwarz, and working on the unitization of $A$, $$ |\varphi(tb+a)|^2\leq \varphi((tb+a)^*(tb+a)),\ \ t\in\mathbb R. $$ Expanding both sides, we get $$ t^2|\varphi(b)|^2+2t\,\operatorname{Re}\,\overline{\varphi(b)}\varphi(a)+\varphi(a)^2\leq t^2\varphi(b^*b)+2t\operatorname{Re}\,\varphi(b^*a)+\varphi(a)^2. $$ The hypothesis then reduces the above inequality to $$ 0\leq t^2(\varphi(b^*b)-|\varphi(b)|^2)+2t\operatorname{Re}\,(\varphi(b^*a)-\overline{\varphi(b)}\varphi(a)). $$ For this to hold for all $t$, we need to have $$ \operatorname{Re}\,(\varphi(b^*a)-\overline{\varphi(b)}\varphi(a)). $$ As the argument also applies to $ib$, we obtain $$ \operatorname{Im}\,(\varphi(b^*a)-\overline{\varphi(b)}\varphi(a)). $$ Thus $$ \varphi(b^*a)-\overline{\varphi(b)}\varphi(a)=0 $$ Replacing $b$ with $b^*$, we get $$ \varphi(ba)=\varphi(b)\varphi(a). $$

I know the above argument from Ozawa's QWEP paper. There is a much nicer argument, using block matrices, that appears in Paulsen's book. This is not limited to scalar-valued maps. Ozawa's argument requires the map to be 2-positive, while Paulsen's requires it to be 4-positive. Any positive linear functional is completely positive.

Answer 2

Suppose $A$ is a C*-algebra, $a\in A$ and $\varphi$ is a state on $A$. Then $A\times A\to \mathbb{C}, (a,b)\mapsto \varphi(b^*a)$ is a pre-inner product, by Cauchy Schwarz Inequality, we have

$\varphi(ba)=0$ for any $b\in A$ iff $\varphi(a^*a)=0$.

Back to your question, and work on the unitization of $A$. Since \begin{align*} & \varphi((a-\varphi(a))^*(a-\varphi(a)))\\= & \varphi((a-\varphi(a))^2)\\= & \varphi(a^2-2\varphi(a)a+\varphi(a)^2)\\= & \varphi(a^2)-2\varphi(a)\varphi(a)+\varphi(a)^2\\= & \varphi(a)^2-2\varphi(a)^2+\varphi(a)^2=0,\end{align*} applying the above lemma, one obtains $\varphi(ba-b\varphi(a))=0$, i.e., $\varphi(ba)=\varphi(b)\varphi(a)$ for any $b\in A$. Hence $\varphi(ab)=\overline{\varphi(b^*a)}=\overline{\varphi(b^*)\varphi(a)}=\varphi(a)\varphi(b).$