= Solution
The desired output is $|\psi_d\rangle=((\alpha+\beta)|0\rangle+(\alpha-\beta)|1\rangle)/\sqrt2$. To evaluate its <squared quantum fidelity>, write the input <Bloch vector> as $(x,y,z)$, with $x=c$, $y=2\operatorname{Im}(\alpha^*\beta)$ and $z=d$. The desired vector after the Hadamard is $(z,-y,x)$. From part (a), the actual vector is
$$
r_{\mathrm{out}}=\left(r_k z+\frac{s_k-1}{2}x,
-\frac{s_k+1}{2}y,r_kx\right).
$$
For $k\ge2$ this simplifies to $(1-1/k)(z,-y,x)-(x/k,0,0)$. The <pure state> overlap is one half of one plus the dot product with the desired unit vector, so
$$
\boxed{\langle\psi_d|\rho|\psi_d\rangle
=1-\frac{1+xz}{2k},\qquad k\ge2.}
$$
Since $x^2+y^2+z^2=1$, $|xz|\le1/2$. Therefore the infidelity is uniformly bounded by $3/(4k)$, and \b[the <entanglement> effect on the implemented gate is negligible for large $k$], uniformly over all initial pure states. The shifted field packets then have overlaps tending to one, so they retain negligible information about the <qubit> transitions. A large fixed <photon> number without this broad number coherence would not suffice.
The printed overlap silently requires $k\ge2$. If the allowed packet has $k=1$, the exact result instead is
$$
\boxed{\langle\psi_d|\rho|\psi_d\rangle
=\frac12+\frac{y^2-xz}{4}.}
$$
For instance, $\alpha=\cos(\pi/8)$ and $\beta=\sin(\pi/8)$ give actual overlap $3/8$, whereas substituting $k=1$ into the printed expression gives $1/4$. This boundary correction follows directly from the two-shift overlap above.
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