If $X$ is a compact metrisable space, a metric $d$ on $X$ can be take as an element of $C(X\times X)$ such that

(1) $ev_x\otimes ev_y (d)=d(x,y)\geq 0$ for all $x,y\in X$ (Non-negativity).

(2) $ev_x\otimes ev_y (d)=0$ iff $x=y$ (Identity of indiscernibles).

(3) $ev_x\otimes ev_y (d)=ev_y\otimes ev_x (d)$ for $x,y\in X$ (Symmetry).

(4) For all $x,y,z\in X$, $ev_x\otimes ev_y (d)\leq ev_x\otimes ev_z (d)+ev_z\otimes ev_y (d)$ (Triangle inequality).

Motivated by this, we can give the following definition of quantum compact metrisable space.

A unital nuclear $C^*$-algebra $A$ is called a quantum compact metrisable space if there exists an element $d\in A\otimes A$ such that

(1) $d\geq 0$(non-negativity).

(2) $\psi\otimes \phi(d)=0$ iff $\psi=\phi$ for $\psi,\phi\in P(A)$ where $P(A)$ is the pure state space of $A$(Identity of indiscernibles).

(3) $\psi\otimes \phi(d)=\phi\otimes \psi(d)$ for $\psi,\phi\in P(A)$(Symmetry).

(4) $\psi\otimes \phi(d)\leq \psi\otimes \varphi(d)+\varphi\otimes \phi(d)$ for all $\psi,\phi$ and $\varphi\in P(A)$(Triangle inequality).

I check that for $M_n(C)$, the $C^*$-algebra of $n$ by $n$ complex matrices and found that if a $d$ satisfies non-negativity, identity of indiscernibles and symmetry, then $d$ does not satisfy triangle inequality. So this means that $M_n(C)$ can only admit a semi-metric in this sense.

My question: is there any genuine quantum compact metrisable space(i.e. noncommutative $C^*$-algebra $A$ with such a $d\in A\otimes A$ satisfying the above properties)?