= Solution
For a <bounded linear functional> on a unital C\*-algebra, the state criterion is
$$
\boxed{x(1_A)=\|x\|=1.}
$$
Equivalently it is positive and takes the unit to one.
In the <Gelfand–Naimark–Segal construction>, set $N_x=\{b:x(b^*b)=0\}$. This is a <left ideal>. Give $A/N_x$ the <inner product> $\langle[b],[c]\rangle=x(c^*b)$, linear in the first variable, and complete it to a <Hilbert space> $H_x$. Left multiplication gives the bounded \*-representation $\pi_x(a)[b]=[ab]$. The vector $\xi=[1_A]$ is cyclic and satisfies $x(a)=\langle\pi_x(a)\xi,\xi\rangle$.
The Hermitian matrix $\rho$ has eigenvalues $\lambda_\pm=1/2\pm|\alpha|$ and <trace> one. In an <orthonormal eigenbasis> $v_+,v_-$,
$$
x_\rho(a)=\lambda_+\langle av_+,v_+\rangle+\lambda_-\langle av_-,v_-\rangle.
$$
If $|\alpha|\leq1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha|>1/2$, evaluating on the positive <rank>-one projection onto the negative-<eigenvalue> eigenspace gives a negative value, so it is not a state. Therefore
$$
\boxed{x_\rho\text{ is a state}\iff|\alpha|\leq1/2.}
$$
To obtain the representation dimension, cyclicity of <trace> gives
$$
x_\rho(b^*b)=\operatorname{Tr}(\rho b^*b)=\|b\rho^{1/2}\|_{\mathrm{HS}}^2.
$$
Thus $A/N_{x_\rho}$ identifies isometrically with the matrix space $\{b\rho^{1/2}:b\in M_2(\mathbb C)\}$, with left multiplication as the representation. If $r=\operatorname{rank}\rho$, this space consists, after a <unitary> change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. <Rank> is one at $|\alpha|=1/2$ and two at $|\alpha|<1/2$, proving
$$
\boxed{\dim H_{x_\rho}=2\text{ at }|\alpha|=1/2,\qquad\dim H_{x_\rho}=4\text{ at }|\alpha|<1/2.}
$$
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