Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 50 / 4

Yes: use superdense coding, with a shared Bell state. Let the sender and receiver initially hold the first and second qubits of $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. To encode bits $(a,b)$, the sender applies $Z^a{X}^b$ to her qubit. These Pauli operators produce four orthogonal states:She sends this qubit through the noiseless quantum channel. The receiver then possesses both qubits. He applies a controlled-NOT gate from the received qubit to his original qubit, followed by a Hadamard gate on the received qubit. This maps the four states in the table to $|ab⟩$. A measurement in the computational basis therefore recovers the two bits with certainty. No classical message is sent.

The resource assumption deserves care: one-qubit superdense coding needs a Bell pair already shared before encoding. If none is initially available, the sender can prepare a Bell pair locally and send one half through the same noiseless quantum channel in advance. The protocol then uses two transmitted qubits in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two computational basis qubits directly also communicates two bits. With a pre-shared Bell pair, one transmitted qubit suffices; without it, the entanglement-distribution cost must be counted. An unassisted single transmitted qubit cannot perfectly encode four classical messages: its Holevo bound is at most ${\log }_{2}2=1$ bit.

Yes, the conversion can be deterministic using local operations and classical communication. Put $s=\sin \theta$ and $c=\cos \theta$. The initial state's squared Schmidt coefficients are $(1/2,1/2)$, while the target's decreasingly ordered squared Schmidt coefficients are $(c^2,s^2)$. Since $0<\theta <\pi /4$, one has $1/2<c^2$ and both pairs sum to one. ThusThe direction of this majorization is the allowed one in Nielsen's pure-state conversion theorem: maximal entanglement can be reduced by deterministic LOCC.

Here is an explicit protocol, so no conversion theorem need be assumed. The sender first applies a Pauli Z gate to remove the initial Bell-state minus sign. Starting from $|{\Phi }^{+}⟩$, she performs a two-outcome local measurement with Kraus operatorsThey define a valid measurement because $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. The two unnormalized states after measurement areEach outcome has probability $1/2$. For outcome zero, the normalized state is already $s|00⟩+c|11⟩$. For outcome one, both parties apply Pauli X gates, obtaining the same state. The sender communicates only which outcome occurred. Finally the receiver applies a Pauli X gate, giving $s|01⟩+c|10⟩$ in either branch. This is deterministic two-qubit entanglement dilution, with total success probability one; it does not rely on postselection.