 Solution

The angular-momentum commutators giveFor example $[L_x+i{L}_y,L_z]=-i\hslash L_y+i(i\hslash L_x)=-\hslash L_{+}$. In the displayed basis $|1,1⟩,|1,0⟩,|1,-1⟩$, the only nonzero upper entries of $L_{+}$ connect adjacent magnetic quantum numbers. Subtracting their column and row $L_z$ eigenvalues multiplies each by $-\hslash$, proving the first matrix commutator; the lower matrix similarly gives the second. Direct multiplication giveswhose difference is $2\hslash L_z$. Also $L^2=L_z^2+(L_{+}{L}_{-}+L_{-}{L}_{+})/2=2{\hslash }^{2}I$, confirming $\ell =1$.

A rotation changes the quantization axis without changing the spectrum, so $L_{\mathbf{n}}$ has eigenvalues $+\hslash ,0,-\hslash$. Its dimensionless matrix isSolving $(L_{\mathbf{n}}-\hslash I)v=0$ and normalizing gives the spin-one maximum-projection eigenvectorFor instance the first equation is $(\cos \theta -1){\cos }^2(\theta /2)+{\sin }^2\theta /2=0$, and the squared components sum to one. An overall phase is immaterial.

At $\theta =\pi /3$, $v_{+}=(3/4,\sqrt{6}/4,1/4{)}^{\mathsf{T}}$. The Born rule therefore predicts, in the requested order,This treats the Stern-Gerlach experiment as the stipulated ideal measurement of orbital projection. In an actual atom, other magnetic moments and spin-orbit coupling can change which angular momentum the apparatus resolves.