Quantum Communication Protocols

Step 1: Initial state decomposition.

The initial three-qubit state is:

Step 2: Expansion in Bell basis for qubits 1,2.

The four Bell states form a complete orthonormal basis for two qubits:

We rewrite the total state by changing the basis of qubits 1 and 2:

This expansion is exact: the four terms correspond to the four possible measurement outcomes.

Step 3: Bell measurement outcome.

When Alice measures qubits 1 and 2 in the Bell basis, the combined system \(|\Psi\rangle_{123}\) projects onto one of the four terms above, each with equal probability \(1/4\).

Step 4: Bob's correction.

Based on Alice's measurement result (communicated classically), Bob applies the appropriate unitary to qubit 3:

In all cases, after Bob's correction, qubit 3 is in the state \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\), which matches the original state on \(S^2_1\). \(\blacksquare\) □

Before Alice's measurement, qubits 2 and 3 form a maximally entangled Bell pair with density matrix \(\rho_{23}^{\text{initial}} = |\Phi^+\rangle\langle\Phi^+|_{23}\).

After Alice's Bell measurement (a projective measurement on the 1-2 subsystem), the combined state \(|\Psi\rangle_{123}\) is projected onto one of four terms in Equation above. The density matrix of the full system after measurement (before Bob's correction) is a mixture of the four projection outcomes.

Tracing out qubit 2 after Alice's measurement yields:

where the sum reflects the four possible measurement results. This is a maximally mixed state on qubit 3: it contains no entanglement with qubit 2 (which is now in a definite state).

Thus the \(S^2_2 \times S^2_3\) entanglement structure is “consumed” or “transferred”: it is used to establish the correlation between Alice's measurement outcome and Bob's correction, and cannot be reused. \(\blacksquare\) □

Step 1: Fidelity definition.

For a specific input state \(|\psi\rangle\) and output \(\rho_{\text{out}}\), the fidelity is:

The average fidelity over all input states (using the uniform (Haar) measure on \(S^2\)) is:

Step 2: Entanglement fidelity relation.

For the standard teleportation protocol with shared entangled state \(\rho_{23}\), there is a direct relationship between the average fidelity and the entanglement fidelity \(F_e\) of the shared state:

where \(d\) is the dimension of each qubit space. For qubits, \(d = 2\).

Step 3: Evaluate for qubits.

Substituting \(d = 2\):

For a maximally entangled Bell state, \(F_e = 1\), so \(\bar{F} = 1\) (perfect fidelity).

Step 4: Physical interpretation on \(S^2\).

The fidelity \(F = |\langle\psi_{\text{in}}|\psi_{\text{out}}\rangle|^2\) measures the overlap of two quantum states as points on the Bloch sphere \(S^2\). Specifically, if \(\theta\) is the angle between the input and output states on \(S^2\), then:

Perfect teleportation (\(F = 1\)) means \(\theta = 0\): the output state coincides exactly with the input state on \(S^2\). \(\blacksquare\) □

Step 1: Setup.

Alice and Bob share a maximally entangled Bell state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\). Alice holds qubit 1, Bob holds qubit 2.

Step 2: Encoding by Alice.

Alice wants to send 2 classical bits (a message in \(\{00, 01, 10, 11\}\)). She applies one of four unitary operations to her qubit 1:

Each message maps to one of the four maximally entangled Bell states. Note: these operations are applied only to Alice's qubit; the entanglement structure changes the global state.

Step 3: Transmission.

Alice sends her qubit to Bob through a quantum channel (1 qubit transmitted).

Step 4: Decoding by Bob.

Bob now possesses both qubits and performs a Bell state measurement (projecting onto the four Bell basis states). The measurement outcome is guaranteed to be one of the four Bell states with certainty (no randomness), so Bob can distinguish all four states and recover Alice's 2-bit message with probability 1.

Step 5: Capacity analysis.

Alice transmits 1 qubit but communicates 2 bits of information (distinguished among \(2^2 = 4\) possible messages). \(\blacksquare\) □

Step 1: Holevo formula.

When Alice encodes a classical variable \(X \in \{x_1, x_2, \ldots, x_m\}\) into a quantum state \(\rho_X\), the accessible classical information is bounded by:

where \(B\) is Bob's measurement outcome, and \(\chi\) is the Holevo quantity.

Step 2: For two input states.

In superdense coding, Alice chooses between 4 messages (2 bits), which map to 4 pure states (the four Bell states). Each Bell state is maximally entangled and has zero entropy: \(S(|\Phi^+\rangle) = S(|\Phi^-\rangle) = S(|\Psi^+\rangle) = S(|\Psi^-\rangle) = 0\).

The average state (mixture) is:

which is the completely mixed state on the 2-qubit system: \(\rho_{\text{avg}} = I/4\), with entropy \(S = 2\) bits.

Step 3: Holevo quantity.

Since Bob transmits 1 qubit (entropy at most 1 bit), and Alice sends 2 bits of classical information, the Holevo bound allows exactly 2 bits. Superdense coding achieves this maximum. \(\blacksquare\) □

Step 1: No-cloning theorem from \(S^2\) linearity.

Suppose there exists a universal cloning machine \(U\) that copies any quantum state: \(U|\psi\rangle|0\rangle = |\psi\rangle|\psi\rangle\) for all \(|\psi\rangle\) on \(S^2\).

Consider two states: \(|0\rangle\) and \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\).

By assumption, the cloning machine must work on both:

However, by linearity of quantum mechanics:

But \(\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) \neq |+\rangle|+\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)\).

This is a contradiction. Therefore, no universal cloning machine exists. \(\blacksquare\)

Step 2: Eavesdropping introduces detectable errors.

Alice sends qubits prepared in random bases (Z or X, each with probability 1/2). Eve does not know the correct basis, so she must guess: she measures each intercepted qubit in either Z or X basis, also with probability 1/2.

Case 1: Eve uses the correct basis. She gets the correct bit value and resends it undisturbed. No error.

Case 2: Eve uses the wrong basis (probability 1/2). She measures and obtains one of two possible outcomes with equal probability. Then she resends her measured state.

Example: Suppose Alice prepared \(|+\rangle\) (X-basis). Eve incorrectly measures in Z-basis and gets either \(|0\rangle\) or \(|1\rangle\) with probability 1/2. Eve resends what she measured.

If Alice later measures her transmitted qubit in X-basis and gets \(|+\rangle\), but Eve had measured in Z-basis and found \(|0\rangle\) (resending \(|0\rangle\)), then Bob receives \(|0\rangle\) instead of \(|+\rangle\). When Bob measures in X-basis (to extract the key bit), he gets \(|+\rangle\) or \(|-\rangle\) with equal probability, introducing a 50% error rate among the bits where Eve used the wrong basis.

Step 3: Overall error rate.

Among the bits where Alice and Bob use the same basis (which they later compare):

Alice and Bob publicly compare a random subset of their sifted bits (those measured in the same basis). If the error rate exceeds 25%, they abort and assume an eavesdropper is present. If the error rate is below the threshold, they assume no eavesdropping and use the remaining bits as a secret key. □

Step 1: Connection to Part 7 Bell inequality.

Part 7, §57.6 established that the Bell correlation for a singlet state is:

where \(\sigma_{\vec{a}}\) denotes a spin-1/2 measurement along direction \(\vec{a}\), and the singlet \(|\Psi^-\rangle\) is the maximally entangled state.

The CHSH Bell parameter (from Part 7) is:

For a perfect singlet and optimal angle choices, \(S = 2\sqrt{2} \approx 2.828\), violating the classical bound \(S \leq 2\).

Step 2: Eve's access to information.

Suppose Eve has performed a measurement on the source before Alice and Bob receive the qubits. If Eve has obtained information about the outcomes, she has effectively “reduced” the entanglement of the state Alice and Bob receive. The reduced state would satisfy a local hidden variable model with \(S \leq 2\).

Conversely, if \(S = 2\sqrt{2}\) is measured, the entanglement is definitely not reduced by Eve's measurement: she has no information about the outcomes.

Step 3: Quantifying Eve's information.

For a partially entangled or eavesdropped state, the Bell parameter decreases:

where \(V \leq 1\) is the visibility (purity) of the entanglement. Eve's information about the key bits is bounded by \(1 - V\).

Step 4: Key generation and security.

Alice and Bob use the basis-matching measurements (those where they randomly chose the same angle) as key bits. They publicly test the Bell inequality using the non-matching measurements. If the measured \(S\) is close to \(2\sqrt{2}\), they confirm the entanglement is secure and use the key bits. If \(S < 2\sqrt{2}\), Eve has information, and they discard the key. \(\blacksquare\) □

Step 1: Initial state.

The state of all four qubits is:

Expanding explicitly:

Step 2: Rewrite in Bell basis for qubits B and C.

We need to express \(|\Psi\rangle_{ABCD}\) in terms of Bell states on the B-C subsystem. After expanding the product and regrouping:

(The algebra is tedious but straightforward—the point is that each Bell state on B-C entangles with a corresponding Bell state on A-D.)

Step 3: Bell measurement on B,C.

When the middle pair (B,C) is measured in the Bell basis, the result projects the state onto one of the four terms above. Each outcome has probability 1/4.

For example, if the measurement outcome is \(|\Phi^+\rangle_{BC}\), the A-D pair collapses to the state \((|00\rangle + |11\rangle)_{AD} = |\Phi^+\rangle_{AD}\) (up to normalization).

Step 4: Local corrections.

The other three measurement outcomes project A-D onto \(|\Phi^-\rangle_{AD}\), \(|\Psi^+\rangle_{AD}\), or \(|\Psi^-\rangle_{AD}\). Local unitary corrections (Pauli operators applied to either A or D) can convert any of these to \(|\Phi^+\rangle_{AD}\). The required correction is classically communicated from B and C to the outer nodes.

Result: After measurement and correction, A and D are entangled in state \(|\Phi^+\rangle_{AD}\). \(\blacksquare\) □

The idea is to use two noisy entangled pairs to create one higher-fidelity pair:

• Setup: Alice and Bob each hold two noisy singlet pairs with fidelity \(F > 1/2\).
• Measurement: Alice applies a CNOT (controlled-NOT) from one pair to the other. She then measures the “target” qubits of the CNOT in the Z-basis.
• Communication: Alice sends her measurement results (2 classical bits) to Bob.
• Conditioning: Bob applies corresponding operations to his qubits. He measures the targets in Z-basis as well.
• Outcome: The two pairs are processed simultaneously. The measurement outcomes that match between Alice and Bob indicate successful purification. The control qubits from the two pairs are now entangled in a state with higher fidelity than the originals.
• Iteration: By repeating this procedure, the fidelity increases exponentially toward 1.

Fidelity increase: Each round transforms fidelity \(F \to F^{\text{new}}\) such that the purity \((2F - 1)\) approximately doubles. Since purification is iterative and each iteration increases purity, and since fidelity can exceed classical threshold \(F = 1/2\), repeated purification drives \(F \to 1\). The exact rates and overheads depend on the noise model. \(\blacksquare\) □

Step 1: Kraus representation of local operations.

Any CPTP (completely positive, trace-preserving) local operation on Bob's system can be represented in Kraus form:

where the Kraus operators \(K_k\) satisfy:

This ensures trace preservation.

Step 2: Apply \(\mathcal{E}_B\) to the bipartite state.

When \(\mathcal{E}_B\) acts on the B subsystem of \(\rho_{AB}\):

Step 3: Trace out B.

Using the cyclic property of trace: \(\text{Tr}(ABC) = \text{Tr}(CAB)\), and the fact that \(\text{Tr}_B(X_A \otimes Y_B) = X_A \cdot \text{Tr}(Y_B)\):

Thus Alice's marginal density matrix is completely independent of Bob's operation \(\mathcal{E}_B\). \(\blacksquare\) □