 Solution

Putting the unit vector along a coordinate axis gives ${\sigma }_i^2=I$. Putting it along $(e_i+e_j)/\sqrt{2}$ gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$ for $i\ne j$. Combining the resulting anticommutators with the given commutators yields ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, and henceThe operators $S_i=(\hslash /2){\sigma }_i$ obey angular-momentum commutators and ${\mathbf{S}}^2=3{\hslash }^{2}I/4$, the spin-one-half value. In a spin-half irreducible sector they are the intrinsic spin operators; the algebra also permits direct sums of such sectors.

Since $Q=\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian with $Q^2=I$, its operators $P_{\pm }=(I\pm Q)/2$ are Hermitian and satisfy $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Moreover $Q{P}_{\pm }=\pm P_{\pm }$. Thus $|\chi ⟩=P_{+}|\chi ⟩+P_{-}|\chi ⟩$ is its orthogonal eigenstate decomposition, omitting any zero component. The Born probabilities for a normalized state areFor spin up along $\mathbf{m}$, the anticommutator $\{\mathbf{m}\cdot \mathbf{\sigma },{\sigma }_j\}=2m_{j}I$ gives $⟨{\sigma }_j⟩=m_j$. Therefore the probability of spin up along $\mathbf{n}$ is $(1+\mathbf{n}\cdot \mathbf{m})/2$.