 Solution

Direct matrix multiplication givesThe normalized $S_x$ eigenvectors are $(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$, and the $S_y$ eigenvectors are $(|\uparrow ⟩\pm i|\downarrow ⟩)/\sqrt{2}$, with eigenvalues $\pm \hslash /2$ in each case.

For $N=|a{|}^2+|b{|}^2>0$, allowing for a possibly unnormalized input pair of coefficients,For a normalized state $N=1$. The second result follows because each squared Spin operator matrix equals ${\hslash }^2\hat{I}/4$.

The tilted component is ${\hat{S}}_{z^{\prime }}=(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Its eigenvalues remain $\pm \hslash /2$, with one real phase convention for the normalized eigenvectorsProjecting the initial $|\uparrow ⟩$ onto them gives $P_{+}={\cos }^2(\theta /2)$ and $P_{-}={\sin }^2(\theta /2)$. ThusA smaller expectation does not imply a smaller individual spin eigenvalue; it describes the probabilities of the same two measurement outcomes.