 Solution

matrix multiplication gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, hence $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. For physical neutron spin, use $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, so $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $S^2=3{\hslash }^2/4$, and measurement along any axis gives $\pm \hslash /2$.

For the specified field, $H=(\alpha B\hslash /2){\sigma }_y$. The Time-dependent Schrödinger equation is $i\hslash \dot{\psi }=H\psi$ with $\psi (0)=(1,0{)}^T$. The uniform-field spin precession follows from ${\sigma }_y^2=I$:

The sign of the precession is contained in the given $\alpha$. The normalized $x$-axis eigenstates are $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$.

The measurement direction is $\mathbf{n}=(0,\sin \theta ,\cos \theta )$, as confirmed by the original diagram. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The spinors ${\chi }_{+}=(c,is{)}^T$ and ${\chi }_{-}=(is,c{)}^T$ have unit norm and inner product $ics-isc=0$. Applying $\mathbf{n}\cdot \mathbf{\sigma }$ gives eigenvalues $+1$ and $-1$. At $\theta =0$ they are the usual $z$ basis; at $\pi /2$ they are the $y$ basis up to a phase; at $\pi$ the $z$ up/down roles are exchanged, with a factor $i$; and at $2\pi$ both return with a minus sign. A spinor changes sign under a full rotation, while its isolated-state measurement probabilities are unchanged.

For spin measurement in the yz plane, projection givesFor $\theta =0$ these are ${\cos }^2\varphi ,{\sin }^2\varphi$; for $\pi /2$ both are $1/2$ at all times; for $\pi$ they are exchanged; and for $2\pi$ they are the same as at zero angle. The constant $y$ probabilities also follow from conservation of $S_y$.