= Solution
{c}
For $\lambda=3/4$, the <adjoint operator> relation $J_+^*=J_-$ and the given product identity imply
$$
\|J_+\psi\|^2=\langle\psi,J_-J_+\psi\rangle
=\hbar^2\left(\frac34-\frac14-\frac12\right)=0.
$$
A vector of zero <norm> is zero, so $J_+\psi=0$. Part (i) gives
$$
\boxed{\|\varphi\|^2=\hbar^2,\qquad \|\varphi\|=\hbar.}
$$
The <commutator> $[J^2,J_-]=0$ also gives $J^2\varphi=(3/4)\hbar^2\varphi$, while $J_3\varphi=-(\hbar/2)\varphi$. Applying the other product identity to this unnormalized vector,
$$
\|J_-\varphi\|^2=\langle\varphi,J_+J_-\varphi\rangle
=\hbar^2\left(\frac34-\frac14-\frac12\right)\|\varphi\|^2=0.
$$
Therefore
$$
\boxed{J_+\psi=0,\qquad J_-\varphi=0.}
$$
These are the two <angular momentum ladder endpoints> of a <spin one-half> pair. In particular, $\psi_2=\varphi/\hbar$ has unit <norm>, and $J_+\psi_2=\hbar\psi_1$ because $J_+J_-\psi=\hbar^2\psi$.
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