 Squaring is not order preserving in a C-star algebra

In $M_2(\mathbb{C})$ take $a=\mathrm{diag}(1,2)$ and $b=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$. Then $a-1\ge 0$ and $b-a=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ge 0$, but $b^2-a^2=\left(\begin{array}{cc}4& 5\\ 5& 6\end{array}\right)$ has determinant $-1$ and a negative eigenvalue. In contrast, for commuting positive elements $b^2-a^2=(b-a)(b+a)$ is positive in the commutative function algebra supplied by continuous functional calculus. The obstruction is noncommutativity, not a failure of the scalar inequality.