Past exam of the quantum theory course of the University of Oxford / 2021 / 3 / a

The angular momentum eigenstates have $J^2$ eigenvalue ${\hslash }^{2}j(j+1)$, where the angular momentum quantum number is $j=0,\frac{1}{2},1,\frac{3}{2},\dots$. For each $j$, the magnetic quantum number runs in unit steps from $-j$ to $j$. Here that magnetic quantum number is $+1/2$, so only half-integer $j$ are compatible:Integer $j$ cannot contain a $J_3$ eigenvalue $\hslash /2$. Applying the angular momentum lowering operator and using $[J_3,J_{-}]=-\hslash J_{-}$ givesThe lowered vector is nonzero, since the adjoint operator identity $J_{-}^{\ast }=J_{+}$ givesFinally, $J_3$ is a self-adjoint operator, so its eigenstates for distinct eigenvalues are orthogonal. Explicitly,and hence $\boxed{⟨\phi \mid \psi ⟩=0}$.

For $\lambda =3/4$, the adjoint operator relation $J_{+}^{\ast }=J_{-}$ and the given product identity implyA vector of zero norm is zero, so $J_{+}\psi =0$. Part (i) givesThe commutator $[J^2,J_{-}]=0$ also gives $J^2\phi =(3/4){\hslash }^2\phi$, while $J_3\phi =-(\hslash /2)\phi$. Applying the other product identity to this unnormalized vector,ThereforeThese are the two angular momentum ladder endpoints of a spin one-half pair. In particular, ${\psi }_2=\phi /\hslash$ has unit norm, and $J_{+}{\psi }_2=\hslash {\psi }_1$ because $J_{+}{J}_{-}\psi ={\hslash }^2\psi$.