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

Denote the normalized initial field by $|F⟩$, and its shifted packets byHere the photon-number kets are orthogonal Fock states. Counting shared number labels givesAfter the interaction, linearity gives the joint stateTake the partial trace over the field. Put $d=|\alpha {|}^2-|\beta {|}^2$ and $c=\alpha {\beta }^{\ast }+{\alpha }^{\ast }\beta$. In the ordered qubit basis $(|0⟩,|1⟩)$, the reduced density matrix isFor example, its upper off-diagonal entry is half the overlap of the field accompanying $|1⟩$ with the field accompanying $|0⟩$, fixing the conjugation order. The trace is one, and positivity follows because this is the partial trace of a normalized pure state. For $k\ge 2$, $s_k=1-2/k$; at $k=1$ it is zero, not the continuation $-1$. This is the photon-number reference for a Hadamard gate channel, not an actual coherent-state packet.

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.

Use the rotation-invariant distribution over pure initial states, namely the uniform pure-qubit average with area element $\sin \theta d\theta d\phi /(4\pi )$. Reflection symmetry gives $\mathbb{E}[xz]=0$, and rotational symmetry together with $x^2+y^2+z^2=1$ gives $\mathbb{E}[y^2]=1/3$. Averaging the squared quantum fidelity from part (b) therefore yieldsIf the single-number packet $k=1$ is included, its corrected average isUniform polar angle without the sine weight would not be the Haar-uniform average over all pure qubit states.