Past exam of the laser science and quantum information processing course of the University of Oxford / 2015 / 7

The quantum bit-flip channel has Kraus operatorsThusThe orthonormal diagonal linear polarizations $|D⟩=(|H⟩+|V⟩)/\sqrt{2}$ and $|A⟩=(|H⟩-|V⟩)/\sqrt{2}$ are the Pauli X eigenstates. Each is transmitted perfectly as a density operator for every $p$; the possible minus sign on $|A⟩$ is global. Encode a classical zero and one as $D$ and $A$, then measure in that basis at the receiver. This sends one classical bit per transmitted qubit without error. General superpositions of $D$ and $A$ need not be preserved, because their relative phase is randomized.

Label the unknown input $1$, Alice's resource photon $2$, and Bob's resource photon $3$. First distribute photon $3$ through the channel while keeping photon $2$ with Alice; at $p=1$ this leaves the perfect shared Bell state $|{\Phi }^{+}{⟩}_{23}=(|00⟩+|11⟩)/\sqrt{2}$. When the input arrives, the three-qubit state isAlice applies ${\mathrm{CNOT}}_{1\to 2}$, givingShe next applies $H_1$, producingThese two gates followed by computational measurement implement a perfect Bell state analyzer. For outcome $(m,n)$ the normalized conditional state is $|mn{⟩}_{12}{X}^n{Z}^m|\psi {⟩}_3$, with probability $1/4$. Alice sends the two classical bits $m,n$; Bob applies $Z^m{X}^n$, recovering $|\psi ⟩$ exactly. Averaging before reception of the bits gives Bob $I/2$, independent of the input, so this protocol provides no faster-than-light communication.

For $p=1$, directly sending the original qubit is already perfect. Quantum teleportation therefore has no fidelity advantage in this ideal case and consumes a shared entangled pair plus classical communication. Its operational advantage is that entanglement can be distributed beforehand: once the unknown input appears, only classical information need travel, and the original input photon need not itself propagate to Bob. The Bell measurement consumes the input; it does not create a second copy.

Use the same protocol: distribute one resource photon through $\mathcal{C}$, and send Alice's two classical bits error-free in the $D,A$ alphabet from part a. At $p=1/2$ the distributed resource isFor the second component Bob has an extra $X$ before his conditional correction. Conjugation by that correction changes $X$ at most by a sign, which cancels in the density operator. This bit-flipped Bell resource teleportation therefore induces the bit-flip map,Using the squared pure-target quantum fidelity convention,The maximum is $1$ precisely for the Pauli X eigenstates $D,A$. The minimum is $1/2$ whenever $\mathrm{Re}({\alpha }^{\ast }\beta )=0$, the great circle with zero $x$ component of the Bloch vector, including $H,V$ and both circular polarizations. Uniform averaging over pure qubits gives $\overline{F}=2/3$.

The resource-distribution and classical-encoding assumptions are essential. If a perfect shared pair had already been supplied independently and this channel were used only for the robustly encoded classical bits, teleportation would remain perfect even at $p=1/2$. Conversely, unprotected $H,V$ transmission of Alice's correction bits would add classical errors. The stated expression applies to the resource sent through the given channel and the error-free classical encoding explicitly available in part a.