 Solution

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.