 Solution

Taking $\mathbf{n}$ along a coordinate axis gives ${\sigma }_i^2=I$. Taking $\mathbf{n}=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ gives $\{{\sigma }_i,{\sigma }_j\}=0$. Combining this anticommutator relation with the specified commutator givesand contraction with $a_i{b}_j$ provesThe operators for spin angular momentum, $S_i=\hslash {\sigma }_i/2$, satisfy $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ and $S^2=3{\hslash }^{2}I/4$, identifying spin $1/2$. The usual irreducible spin space is two-dimensional; the assumptions also allow repeated copies of it if no Hilbert-space dimension is specified.

Since $\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian and squares to $I$, $P_{\pm }=(I\pm \mathbf{n}\cdot \mathbf{\sigma })/2$ are Hermitian, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. They are its spectral projectors. WriteEach nonzero term is a scalar multiple of a normalized eigenstate with eigenvalue $\pm 1$. The Born rule gives, for normalized $|\chi ⟩$,If $(\mathbf{m}\cdot \mathbf{\sigma })|\chi ⟩=|\chi ⟩$, expectation of $\{\mathbf{m}\cdot \mathbf{\sigma },\mathbf{a}\cdot \mathbf{\sigma }\}=2\mathbf{m}\cdot \mathbf{a}$ gives $⟨\mathbf{a}\cdot \mathbf{\sigma }⟩=\mathbf{m}\cdot \mathbf{a}$. Taking $\mathbf{a}=\mathbf{n}$ yields $\boxed{\mathbb{P}(\text{up along }\mathbf{n})=\frac{1}{2}(1+\mathbf{n}\cdot \mathbf{m})}$.