= Solution
{c}
On Bob's polarization <qubit>, take
$$
X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad
XZ=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=-iY.
$$
Suitable conditional <unitary operators> are
$$
\boxed{U_{\Phi^+}=XZ,\quad U_{\Phi^-}=X,\quad
U_{\Psi^+}=Z,\quad U_{\Psi^-}=I.}
$$
Indeed $(I\otimes XZ)|\Phi^+\rangle=|\Psi^-\rangle$, $(I\otimes X)|\Phi^-\rangle=|\Psi^-\rangle$, and $(I\otimes Z)|\Psi^+\rangle=-|\Psi^-\rangle$. The last minus sign is a harmless <global phase>.
For an ideal <half-wave plate> whose fast axis makes <angle> $\theta$ with horizontal polarization, discard the common <phase> and use the <Jones matrix>
$$
W(\theta)=\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}.
$$
It implements $Z$ at $0^\circ$ and $X$ at $45^\circ$. Thus for outcome $\Phi^+$ use two <half-wave plates>, first $0^\circ$ then $45^\circ$ in the propagation direction; for $\Phi^-$ use one at $45^\circ$; for $\Psi^+$ one at $0^\circ$; for $\Psi^-$ use no plate. Ideal retardance is assumed at the <photon> <wavelength>, and the setting is chosen after receiving the detector's classical result.
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