 Solution

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 $\boxed{⟨\phi \mid \psi ⟩=0}$.