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CHAPTER 07

# Bell correlations, teleportation, and dense coding

Learning goals. Derive the CHSH bound, distinguish correlation from signalling, and track the resources and corrections in two communication protocols.

## 7.1 No-cloning and no-signalling

Suppose a unitary cloned arbitrary states: U|ψ⟩|0⟩=|ψ⟩|ψ⟩U|\psi\rangle|0\rangle=|\psi\rangle|\psi\rangle and likewise for |ϕ⟩|\phi\rangle. Preserving the input overlap requires

⟨ϕ|ψ⟩=(⟨ϕ|ψ⟩)2.\langle\phi|\psi\rangle=(\langle\phi|\psi\rangle)^2.

An overlap strictly between zero and one cannot satisfy this. Thus an unknown arbitrary state cannot be copied perfectly. Orthogonal labels can be copied; the theorem does not prohibit ordinary classical copying [5].

Entanglement also cannot transmit a chosen bit without communication. For a local trace-preserving channel on B with Kraus operators K j K_j,

ρ′A=Tr B∑j(I⊗K j)ρ A B(I⊗K j†)=ρ A,\rho'_A=\operatorname{Tr}_B \sum_j(I\otimes K_j)\rho_{AB}(I\otimes K_j^\dagger)=\rho_A,

because ∑j K j†K j=I\sum_jK_j^\dagger K_j=I. A conditioned state at A may depend on B’s result, but A needs B’s classical record to identify that subensemble.

## 7.2 CHSH from four local answers

Alice chooses setting a a or a′a', Bob chooses b b or b′b', and each outputs ±1\pm1. In a local hidden-variable model, fix the hidden variable and write predetermined responses A,A′,B,B′∈{±1}A,A',B,B'\in\{\pm1\}. Then

A B+A B′+A′B−A′B′=A(B+B′)+A′(B−B′).AB+AB'+A'B-A'B'=A(B+B')+A'(B-B').

Either B=B′B=B' or B=−B′B=-B'. Exactly one parenthesis is zero and the other is ±2\pm2, so the expression is always ±2\pm2. Averaging gives

|S|=|E(a,b)+E(a,b′)+E(a′,b)−E(a′,b′)|≤2.|S|=|E(a,b)+E(a,b')+E(a',b)-E(a',b')|\le2.

Stochastic local responses can be included in the hidden variable, so the bound remains. The assumptions include independent setting choices and an appropriate separation of local responses; a classical simulation programmed with quantum probabilities is not an experimental test of these assumptions [6, 7].

For |Φ+⟩|\Phi^+\rangle and observables A(θ)=cos θ Z+sin θ X A(\theta)=\cos\theta Z+\sin\theta X, one obtains E(a,b)=cos(a−b)E(a,b)=\cos(a-b). Choose a=0,a′=π/2,b=π/4,b′=−π/4 a=0,a'=\pi/2,b=\pi/4,b'=-\pi/4:

S=1 2+1 2+1 2−(−1 2)=2 2.S=\frac1{\sqrt2}+\frac1{\sqrt2}+\frac1{\sqrt2} -\left(-\frac1{\sqrt2}\right)=2\sqrt2.

The joint probabilities are p(s,t∣a,b)=(1+s t cos(a−b))/4 p(s,t\mid a,b)=(1+st\cos(a-b))/4. Summing over t gives p(s)=1/2 p(s)=1/2, independent of Bob’s setting. Strong correlations coexist with no signalling.

Laboratory L08 — CHSH experiment. Change all four measurement angles, visibility, shots per setting, and seed. Compare exact and sampled correlations and S. Visibility models ρ=v|Φ+⟩⟨Φ+|+(1−v)I/4\rho=v|\Phi^+\rangle\langle\Phi^+|+(1-v)I/4, not a hardware calibration.

## 7.3 Teleportation: move a state using shared entanglement

Alice owns input |ψ⟩=α|0⟩+β|1⟩|\psi\rangle=\alpha|0\rangle+\beta|1\rangle and one half of a Bell pair; Bob owns the other half. Order the wires as input, Alice’s Bell half, Bob’s half. The initial state is

1 2(α|000⟩+α|011⟩+β|100⟩+β|111⟩).\frac1{\sqrt2} (\alpha|000\rangle+\alpha|011\rangle+ \beta|100\rangle+\beta|111\rangle).

Alice applies CNOT from the input to her Bell half, then H to the input. Grouping by Alice’s two measurement bits gives

1 2[|00⟩(α|0⟩+β|1⟩)+|01⟩(α|1⟩+β|0⟩)\frac12\big[ |00\rangle(\alpha|0\rangle+\beta|1\rangle) +|01\rangle(\alpha|1\rangle+\beta|0\rangle) +|10⟩(α|0⟩−β|1⟩)+|11⟩(α|1⟩−β|0⟩)].\qquad+ |10\rangle(\alpha|0\rangle-\beta|1\rangle) +|11\rangle(\alpha|1\rangle-\beta|0\rangle) \big].

Each bracket has norm one, so every two-bit outcome has probability 1/4 1/4, independent of the input. Bob’s conditioned state is X b Z a|ψ⟩X^bZ^a|\psi\rangle, where a is Alice’s first bit and b her second. He corrects by applying X when b=1 b=1, then Z when a=1 a=1: the total correction is Z a X b Z^aX^b.

Alice’s bits ab | Bob before correction | Operations in time order
--- | --- | ---
00 | |ψ⟩|\psi\rangle | None
01 | X|ψ⟩X|\psi\rangle | X
10 | Z|ψ⟩Z|\psi\rangle | Z
11 | X Z|ψ⟩XZ|\psi\rangle | X, then Z

For branch 11, (Z X)(X Z)=I(ZX)(XZ)=I. Reversing correction order only adds a branch-global minus sign, but consistent conventions make the derivation transparent.

Figure 7.1. Teleportation with two classical feed-forward bits.

No copy remains at Alice. The protocol consumes shared entanglement and two classical bits. Bob cannot use it for instantaneous messaging: before receiving a and b, his averaged state is I/2 I/2. The original protocol is due to Bennett and collaborators [8].

Laboratory L09 — Teleportation. Prepare an arbitrary input, inspect all four branches, disable either correction, and compare the output fidelity with the original state.

## 7.4 Superdense coding: the complementary resource trade

Alice and Bob first share |Φ+⟩|\Phi^+\rangle. Alice encodes two classical bits a,b by applying X when b=1 b=1, then Z when a=1 a=1 to her qubit. This gives four orthogonal Bell states up to phase. She then physically sends her qubit to Bob.

Bob applies CNOT from Alice’s qubit to his, then H to Alice’s qubit. A Z measurement returns ab exactly. One transmitted qubit carries two selected classical bits given an already shared Bell pair. The prior entanglement is a consumed resource; ignoring its distribution would misstate the communication cost [9].

Laboratory L10 — Dense coding. Select a two-bit message, inspect the encoded Bell state, and run the decoding circuit. Compare the sender’s reduced state for all four messages.

## 7.5 Exercises

7.1. For the noisy Bell model, determine the visibility needed to violate CHSH with optimal angles.

Show solution / guidance

S=2 2 v S=2\sqrt2v. Violation requires v>1/2≈0.7071 v>1/\sqrt2\approx0.7071. Failure to violate this particular inequality does not by itself prove separability.

7.2. Why is Alice’s teleportation record independent of α,β\alpha,\beta?

Show solution / guidance

Each branch is a unitary image of the normalized input multiplied by 1/2 1/2, hence has probability 1/4 1/4. Two classical bits cannot reveal the continuous input amplitudes.

7.3. Teleport |+i⟩|+i\rangle and condition on ab=01. Show the correction.

Show solution / guidance

Bob has X|+i⟩=(i|0⟩+|1⟩)/2 X|+i\rangle=(i|0\rangle+|1\rangle)/\sqrt2. Applying X again returns |+i⟩|+i\rangle, since X 2=I X^2=I.

7.4. Dense-code message 11. Find the encoded state and final output.

Show solution / guidance

X on Alice produces (|10⟩+|01⟩)/2(|10\rangle+|01\rangle)/\sqrt2. Z gives (−|10⟩+|01⟩)/2(-|10\rangle+|01\rangle)/\sqrt2. CNOT and H produce |11⟩|11\rangle up to an irrelevant overall phase.

7.5. Does an arbitrarily large sample of one local half of Bell pairs reveal the distant party’s basis choice?

Show solution / guidance

No. Every local copy has density matrix I/2 I/2 regardless of a trace-preserving distant operation. Joint records reveal correlations only after communication.

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