 Solution

For a normalized quantum state, $|\alpha {|}^2+|\beta {|}^2=1$, and the Born rule gives $\boxed{p(z+)=|\alpha {|}^2,p(z-)=|\beta {|}^2}$. If the displayed coefficients are not normalized, divide both by $|\alpha {|}^2+|\beta {|}^2$. A quantum superposition is a coherent linear combination: relative phase affects other spin measurements, unlike an incoherent statistical mixture with the same $z$-measurement probabilities.

Every pure qubit quantum state is an eigenstate of spin along some axis. Its Bloch vector isFor real normalized $\alpha ,\beta$, choose $\theta$ modulo $2\pi$ with $\alpha =\cos (\theta /2)$ and $\beta =\sin (\theta /2)$, allowing an irrelevant global sign. ThenEquivalently, $S_{\mathbf{n}}=(\hslash /2)\left(\begin{array}{cc}{\alpha }^2-{\beta }^2& 2\alpha \beta \\ 2\alpha \beta & {\beta }^2-{\alpha }^2\end{array}\right)$. A Stern-Gerlach spin measurement on this axis yields its positive result with certainty, so the state is not a nontrivial quantum superposition in that measurement basis.