Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 2 / b / Solution

The desired output is $|{\psi }_d⟩=((\alpha +\beta )|0⟩+(\alpha -\beta )|1⟩)/\sqrt{2}$. To evaluate its squared quantum fidelity, write the input Bloch vector as $(x,y,z)$, with $x=c$, $y=2\mathrm{Im}({\alpha }^{\ast }\beta )$ and $z=d$. The desired vector after the Hadamard is $(z,-y,x)$. From part (a), the actual vector isFor $k\ge 2$ 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, soSince $x^2+y^2+z^2=1$, $|xz|\le 1/2$. Therefore the infidelity is uniformly bounded by $3/(4k)$, and 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 $k\ge 2$. If the allowed packet has $k=1$, the exact result instead isFor 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.