 Solution

Taking $n=e_i$ gives ${\sigma }_i^2=I$. Taking $n=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ and expanding the square gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$. Thus the anticommutator and commutator together implySince $n\cdot \sigma$ is Hermitian and $\theta$ is real, the exponent is anti-Hermitian. Its exponential has inverse equal to its adjoint and is therefore unitary. Splitting the exponential series into even and odd powers, using $(n\cdot \sigma {)}^2=I$, givesPut $N=n\cdot \sigma ,M=m\cdot \sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot \sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. ConsequentlyThe spin-half angular momentum is $S_i=\hslash {\sigma }_i/2$. The commutators then become $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, and $(n\cdot S{)}^2={\hslash }^{2}I/4$ expresses the two possible directional spin values $\pm \hslash /2$. Also $S^2=3{\hslash }^{2}I/4$, appropriate to spin one half. The exponential represents a spatial rotation on spin states.

Choose $n=(0,1,0)$ and initially ${\sigma }_3|\chi ⟩=|\chi ⟩$. The conjugation formula sends ${\sigma }_3$ to ${\sigma }_3\cos \theta +{\sigma }_1\sin \theta$. ThereforeThus $U|\chi ⟩$ is spin up along $(\sin \theta ,0,\cos \theta )$, as required.