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: $U|\psi\rangle|0\rangle=|\psi\rangle|\psi\rangle$ and likewise for $|\phi\rangle$. Preserving the input overlap requires

$$\langle\phi|\psi\rangle=(\langle\phi|\psi\rangle)^2.$$

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 $K_j$,

$$\rho'_A=\operatorname{Tr}_B \sum_j(I\otimes K_j)\rho_{AB}(I\otimes K_j^\dagger)=\rho_A,$$

because $\sum_jK_j^\dagger K_j=I$. 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 $a$ or $a'$, Bob chooses $b$ or $b'$, and each outputs $\pm1$. In a local hidden-variable model, fix the hidden variable and write predetermined responses $A,A',B,B'\in\{\pm1\}$. Then

$$AB+AB'+A'B-A'B'=A(B+B')+A'(B-B').$$

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

$$|S|=|E(a,b)+E(a,b')+E(a',b)-E(a',b')|\le2.$$

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

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

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

7.3 Teleportation: move a state using shared entanglement

Alice owns input $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ 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

$$\frac1{\sqrt2} (\alpha|000\rangle+\alpha|011\rangle+ \beta|100\rangle+\beta|111\rangle).$$

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

$$\frac12\big[ |00\rangle(\alpha|0\rangle+\beta|1\rangle) +|01\rangle(\alpha|1\rangle+\beta|0\rangle)$$ $$\qquad+ |10\rangle(\alpha|0\rangle-\beta|1\rangle) +|11\rangle(\alpha|1\rangle-\beta|0\rangle) \big].$$

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

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

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 $I/2$. The original protocol is due to Bennett and collaborators [8].

7.4 Superdense coding: the complementary resource trade

Alice and Bob first share $|\Phi^+\rangle$. Alice encodes two classical bits a,b by applying X when $b=1$, then Z when $a=1$ 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].

7.5 Exercises