# Bell correlations, teleportation, and dense coding {#protocols}

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

## 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 [@nocloning].

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.

## 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 [@chsh; @hensen].

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.

::: {.lab #lab-chsh}
**Laboratory L08 — CHSH experiment.** Change all four measurement angles, visibility, shots per setting, and seed. Compare exact and sampled correlations and S. Visibility models $\rho=v|\Phi^+\rangle\langle\Phi^+|+(1-v)I/4$, not a hardware calibration.
:::

## 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 $Z^aX^b$.

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

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

![Teleportation with two classical feed-forward bits.](figures/teleport.svg)

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 [@teleport].

::: {.lab #lab-teleport}
**Laboratory L09 — Teleportation.** Prepare an arbitrary input, inspect all four branches, disable either correction, and compare the output fidelity with the original state.
:::

## 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 [@dense].

::: {.lab #lab-dense}
**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.
:::

## Exercises

:::: {.exercise #ex07-1}
**7.1.** For the noisy Bell model, determine the visibility needed to violate CHSH with optimal angles.

::: {.solution}

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

::::

:::: {.exercise #ex07-2}
**7.2.** Why is Alice's teleportation record independent of $\alpha,\beta$?

::: {.solution}

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

::::

:::: {.exercise #ex07-3}
**7.3.** Teleport $|+i\rangle$ and condition on ab=01. Show the correction.

::: {.solution}

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

::::

:::: {.exercise #ex07-4}
**7.4.** Dense-code message 11. Find the encoded state and final output.

::: {.solution}

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

::::

:::: {.exercise #ex07-5}
**7.5.** Does an arbitrarily large sample of one local half of Bell pairs reveal the distant party's basis choice?

::: {.solution}

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

::::

