Is there a simple $C^\*$ algebra $A$, not isomorphic to $M\_2(\mathbb{C})$, such that for every commutator element $x=ab-ba$, $x^2$ is an scalar element?



Simple $C^*$ algebras whose all commutator elements have scalar square
The answer is no:



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> A simple C\*-algebra in which $[x,y]^2$ is central for all $x,y$ is isomorphic to $M\_n(\mathbf{C})$ for $n\in\{1,2\}$.
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Let $A$ be a C\*-algebra. Let $H$ be an irreducible representation of $A$. By irreducibility its commutant is reduced to scalars, and hence its double commutant is equal to the whole algebra $B(H)$, and the double commutant is knows ([von Neumann's bicommutant theorem](https://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem)) to be the strong operator topology ($\mathsf{SOT}$) closure.


Recall that an associative algebra $A$ is a PI-algebra if it satisfies some nontrivial identity $\forall x\in A^n,$ $F(x\_1,\dots,x\_n)=0$, where $F$ is nonzero in the free associative algebra $\mathbf{C}\langle X\_1,\dots,X\_n\rangle$ - $F$ can be then chosen homogeneous. 


In the above setting, we have a continuous algebra homomorphism with dense image $A\to (B(H),\mathsf{SOT})$. If $A$ satisfies the identity $F$ (say of degree $d$) so is $B(H)$. This implies (by a theorem of Kaplansky, see §2.1 in Procesi's book) that $H$ has dimension $\le d/2$. Note that in the precise case in consideration, we have $F(x,y,z)=[[x,y]^2,z]$, which is clearly not an identity in $B(H)$ when $\dim(H)\ge 3$. 


Now if $A$ is a simple C\*-algebra, not reduced to scalars, then $A$ admits a nontrivial irreducible representation (cf §2.5 in Dixmier's book), in a Hilbert space $H$. Since $A$ is simple, this is a faithful representation. If $A$ is a PI-algebra, this forces, by the above, $H$ to be finite-dimensional, so $A$ is a matrix algebra.


This yields the complement result: 



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> A simple C\*-algebra is a PI-algebra if and only if it is isomorphic to a matrix algebra.
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This follows from Kaplansky's theorem, but also directly follows if one can embed the free associative $\mathbf{C}$-algebra on countably many variables into $B(H)$ for $H$ separable infinite-dimensional.


Actually from Dixmier's book, §2.9, the intersection of kernels of irreducible representations is reduced to 0. Hence one gets a stronger result:



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> A C\*-algebra that is a PI-algebra admits a family of finite-dimensional representations of bounded dimension, whose intersection of kernels is trivial.
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(In the case of the identity $[[x,y]^2,z]$ these are of dimension $\le 2$. I'm not sure if one can strengthen the latter conclusion to a more explicit description, but in any case this is beyond what is asked, since the simple case is clear-cut now.)


