Chapter 7: Quantum Teleportation and Communication
In the previous chapter, we discovered entanglement - the strange quantum correlations that Einstein called "spooky action at a distance." Now we put entanglement to work. This chapter presents four landmark protocols that demonstrate what entanglement can actually do: teleport quantum states, double the capacity of a communication channel, and prove that nature itself violates the assumptions of classical physics.
These protocols are not just theoretical curiosities. Quantum teleportation is a fundamental building block of quantum networks and quantum error correction. Superdense coding demonstrates that entanglement is a genuine physical resource. And the Bell and GHZ games settle a deep philosophical question about the nature of reality - with a definitive experimental answer.
7.1 Quantum Teleportation: Moving Qubits Without Moving Particles
Suppose Alice has a qubit in some unknown state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ and she wants to transfer this exact state to Bob, who is far away. She faces two fundamental obstacles:
- The no-cloning theorem (Chapter 6) forbids copying the state, so she cannot just "read it and send the information."
- Measurement destroys the state. If she measures the qubit to learn something about $\alpha$ and $\beta$, the superposition collapses.
It seems impossible. Yet in 1993, Charles Bennett and colleagues discovered that if Alice and Bob share an entangled Bell pair, Alice can transfer $|\psi\rangle$ to Bob using only two classical bits of communication - without ever physically sending a qubit. The original quantum state is destroyed at Alice's end and recreated at Bob's end. This is quantum teleportation.
The setup
The protocol requires three qubits:
- q[0] - Alice's data qubit, in the unknown state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$
- q[1] - Alice's half of a shared Bell pair
- q[2] - Bob's half of the shared Bell pair
Before the protocol begins, qubits q[1] and q[2] are prepared in the Bell state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. The full three-qubit state is:
$$|\psi\rangle \otimes |\Phi^+\rangle = (\alpha|0\rangle + \beta|1\rangle) \otimes \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ $$= \frac{1}{\sqrt{2}}\big(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle\big)$$Step 1: Alice applies CNOT and Hadamard
Alice applies a CNOT gate with q[0] as control and q[1] as target, then a Hadamard gate on q[0]. Let us trace through the math carefully.
After CNOT(q[0], q[1]):
$$\frac{1}{\sqrt{2}}\big(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle\big)$$After H on q[0], using $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$:
$$\frac{1}{2}\big[\alpha(|0\rangle + |1\rangle)|00\rangle + \alpha(|0\rangle + |1\rangle)|11\rangle + \beta(|0\rangle - |1\rangle)|10\rangle + \beta(|0\rangle - |1\rangle)|01\rangle\big]$$Regrouping by the state of the first two qubits (Alice's qubits):
$$\frac{1}{2}\big[|00\rangle(\alpha|0\rangle + \beta|1\rangle) + |01\rangle(\alpha|1\rangle + \beta|0\rangle) + |10\rangle(\alpha|0\rangle - \beta|1\rangle) + |11\rangle(\alpha|1\rangle - \beta|0\rangle)\big]$$Look at what happened. Bob's qubit (the third one) now contains the original state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ in the $|00\rangle$ branch, and simple transformations of it in the other three branches. Alice's measurement will tell her which branch she is in, and then Bob can apply the right correction.
Step 2: Alice measures and sends classical bits
Alice measures q[0] and q[1], obtaining one of four equally likely outcomes. She sends these two classical bits to Bob over an ordinary communication channel (phone, email, carrier pigeon - anything).
Step 3: Bob applies corrections
Based on Alice's two-bit message, Bob applies a correction to his qubit q[2]:
| Alice measures | Bob's qubit state | Bob's correction |
|---|---|---|
| $|00\rangle$ | $\alpha|0\rangle + \beta|1\rangle$ | Do nothing ($I$) |
| $|01\rangle$ | $\alpha|1\rangle + \beta|0\rangle$ | Apply $X$ |
| $|10\rangle$ | $\alpha|0\rangle - \beta|1\rangle$ | Apply $Z$ |
| $|11\rangle$ | $\alpha|1\rangle - \beta|0\rangle$ | Apply $ZX$ |
After Bob's correction, his qubit is always in the state $\alpha|0\rangle + \beta|1\rangle = |\psi\rangle$. The teleportation is complete.
The complete circuit
Here is the full teleportation circuit. We prepare Alice's data qubit in a specific state (here, $|1\rangle$ via an X gate) so we can verify the teleportation worked. Alice's measurement outcomes feed into classically controlled corrections on Bob's qubit:
Stage 1: Create Bell Pair
Alice and Bob share an entangled Bell pair. H on q[1] followed by CNOT(q[1],q[2]) creates $|\Phi^+\rangle$. Meanwhile, Alice's data qubit q[0] is in state $|1\rangle$ (prepared with X gate).
Stage 2: Alice's Bell Measurement
Alice applies CNOT(q[0],q[1]) then H on q[0]. This entangles her data qubit with her half of the Bell pair, effectively performing a Bell-state measurement on her two qubits.
Stage 3: Bob's Correction
Alice measures q[0] and q[1], sending two classical bits to Bob. Based on the results, Bob applies corrections: X if q[1]=1, Z if q[0]=1. After correction, Bob's qubit is in state $|1\rangle$ - the teleportation is complete!
Run this circuit. Since we prepared Alice's qubit as $|1\rangle$, Bob's qubit should always measure $1$ after the corrections. The first two bits (Alice's measurements) will be random - each of the four outcomes $00$, $01$, $10$, $11$ occurs with equal probability - but Bob's bit will always be $1$.
Try changing the initial state. Replace x q[0] with h q[0] to teleport the state $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. Bob's measurement should then give $0$ and $1$ with equal probability.
What teleportation is NOT
Before Alice sends her classical bits, Bob's qubit is in a completely random mixed state - it looks like a coin flip regardless of what $|\psi\rangle$ was. Bob cannot extract any information about $|\psi\rangle$ until he receives Alice's classical message, which travels at most at the speed of light. Teleportation moves quantum information, but it does not allow faster-than-light communication. The classical channel is essential.
Teleportation transfers the quantum state - the full complex amplitudes $\alpha$ and $\beta$ - not the physical particle. Alice's original qubit is destroyed in the process (it collapses when she measures it). This is consistent with the no-cloning theorem: the state is moved, not copied. One copy is destroyed and another is created.
Why teleportation matters beyond communication
Quantum teleportation is not just a communication protocol. It is a fundamental primitive in quantum computing and quantum networking:
- Quantum error correction (Part VI) uses teleportation-like circuits to move quantum information between encoded blocks without exposing it to noise.
- Gate teleportation is a technique where a quantum gate is applied to a qubit by teleporting it through a specially prepared entangled state. This allows "offline" preparation of difficult gates and is a key ingredient in fault-tolerant quantum computing.
- Quantum networks (Chapter 34) use teleportation to transmit qubits between distant nodes. Since photons (the natural carriers for long-distance communication) are easily lost, teleportation with entanglement swapping allows quantum information to traverse networks without requiring a single photon to travel the full distance.
Predict, Observe, Explain: Can we send information faster than light?
7.2 Superdense Coding: Two Bits Through One Qubit
Teleportation uses 2 classical bits plus 1 shared entangled pair to transmit 1 qubit. Superdense coding is its mirror image: it uses 1 qubit plus 1 shared entangled pair to transmit 2 classical bits. These two protocols are dual to each other, and together they reveal the deep relationship between entanglement and communication.
The protocol was proposed by Charles Bennett and Stephen Wiesner in 1992, one year before the teleportation paper.
The protocol
Alice and Bob again start by sharing a Bell pair $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, with Alice holding q[0] and Bob holding q[1].
Alice wants to send one of four two-bit messages: 00, 01, 10, or 11. She applies one of four single-qubit gates to her qubit only:
| Message | Alice's gate | Resulting Bell state |
|---|---|---|
| 00 | $I$ (do nothing) | $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ |
| 01 | $X$ (bit flip) | $|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle)$ |
| 10 | $Z$ (phase flip) | $|\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$ |
| 11 | $ZX$ | $|\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ |
The key insight is that Alice's single-qubit operation transforms the shared Bell pair into one of the four orthogonal Bell states. Since orthogonal states can be perfectly distinguished by measurement, Bob can recover both bits.
Why do these gates produce different Bell states?
Let us verify one case in detail. Starting from $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, Alice applies $X$ to her qubit (q[0]):
$$X \otimes I \cdot \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle) = |\Psi^+\rangle$$The $X$ gate flips the first qubit of each term, transforming $|00\rangle \to |10\rangle$ and $|11\rangle \to |01\rangle$. Similarly, the $Z$ gate changes the sign of the $|1\rangle$ component: $Z \otimes I \cdot \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) = |\Phi^-\rangle$. Each single-qubit gate on Alice's side produces a distinct, orthogonal Bell state.
Bob's decoding
After receiving Alice's qubit, Bob has both halves of the (now modified) Bell pair. He applies the reverse of the Bell pair creation circuit: first a CNOT with q[0] as control and q[1] as target, then a Hadamard on q[0]. This maps each Bell state back to a unique computational basis state:
$$|\Phi^+\rangle \to |00\rangle, \quad |\Psi^+\rangle \to |01\rangle, \quad |\Phi^-\rangle \to |10\rangle, \quad |\Psi^-\rangle \to |11\rangle$$Let us trace one example. For message 01, the state is $|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle)$. After CNOT:
$$\frac{1}{\sqrt{2}}(|11\rangle + |01\rangle) = \frac{1}{\sqrt{2}}(|1\rangle + |0\rangle) \otimes |1\rangle$$After Hadamard on q[0]: since $H \cdot \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |0\rangle$, we get $|01\rangle$. Bob measures and reads the message 01.
Bob then measures both qubits to read the two-bit message.
A single qubit can normally carry at most one bit of information (Holevo's theorem). Superdense coding achieves two bits only because Alice and Bob share a pre-existing entangled pair. The entanglement acts as a resource that was "pre-distributed" before communication began. You can think of one bit as stored in the qubit Alice sends and the other bit as stored in the entanglement itself.
Interactive: Superdense Coding
Teleportation consumes 1 Bell pair + 2 classical bits to send 1 qubit. Superdense coding consumes 1 Bell pair + 1 qubit to send 2 classical bits. In both cases, entanglement serves as a catalyst that bridges the gap between quantum and classical information. This duality is a deep feature of quantum information theory.
The resource accounting of entanglement
These two protocols give us a precise "exchange rate" between quantum and classical communication when entanglement is available:
$$1 \text{ ebit} + 2 \text{ cbits} \geq 1 \text{ qubit} \quad \text{(teleportation)}$$ $$1 \text{ ebit} + 1 \text{ qubit} \geq 2 \text{ cbits} \quad \text{(superdense coding)}$$Here "ebit" means one shared Bell pair and "cbit" means one classical bit. These inequalities are tight - you cannot do better with less. Together they imply a beautiful identity at the heart of quantum information theory:
$$1 \text{ qubit} = 1 \text{ ebit} + 1 \text{ cbit}$$A qubit of communication is worth exactly one bit of entanglement plus one classical bit. This relationship, formalized by the quantum reverse Shannon theorem, shows that entanglement is a quantifiable physical resource with a precise value.
7.3 Bell's Inequality and the CHSH Game
In Chapter 6, we mentioned the EPR paradox: Einstein, Podolsky, and Rosen argued that quantum mechanics must be incomplete because entangled particles seem to exhibit "spooky action at a distance." They proposed that there must be local hidden variables - pre-existing instructions that each particle carries, determining the outcomes of any measurement. Under this view, entanglement is just classical correlation: each particle "knows" in advance what it will do when measured.
In 1964, John Bell proved that any local hidden variable theory makes specific predictions that differ from quantum mechanics. In 1969, Clauser, Horne, Shimony, and Holt (CHSH) reformulated Bell's result as a precise inequality that can be tested experimentally. We will present this as a cooperative game.
The CHSH game
Alice and Bob are in separate rooms with no communication. A referee sends each of them a random input bit:
- Alice receives bit $x \in \{0, 1\}$
- Bob receives bit $y \in \{0, 1\}$
Each player must respond with an output bit:
- Alice outputs $a \in \{0, 1\}$
- Bob outputs $b \in \{0, 1\}$
They win if and only if:
$$a \oplus b = x \wedge y$$In other words: their output bits should be the same (both 0 or both 1), except when $x = y = 1$, in which case they should be different. Let us write the winning condition for each input pair:
| $x$ | $y$ | $x \wedge y$ | Winning condition |
|---|---|---|---|
| 0 | 0 | 0 | $a = b$ (same outputs) |
| 0 | 1 | 0 | $a = b$ (same outputs) |
| 1 | 0 | 0 | $a = b$ (same outputs) |
| 1 | 1 | 1 | $a \neq b$ (different outputs) |
Classical strategies: at most 75%
Without quantum resources, the best deterministic strategy is simple: both players always output 0. This wins on inputs $(0,0)$, $(0,1)$, and $(1,0)$ but loses on $(1,1)$. That is 3 out of 4 input pairs, giving a 75% win rate.
You might wonder if a cleverer strategy could do better. Let us check systematically. Alice has a function $a(x)$ and Bob has a function $b(y)$. There are $2^2 = 4$ possible strategies for each player (since each function maps one bit to one bit), giving $4 \times 4 = 16$ total deterministic strategies. For each, count the wins:
| Alice's strategy | Bob's strategy | Wins |
|---|---|---|
| Always 0 | Always 0 | 3/4 (lose on $x{=}y{=}1$) |
| Always 0 | Always 1 | 1/4 (win only $x{=}y{=}1$) |
| Output $x$ | Output $y$ | 2/4 |
| ...and so on for all 16 combinations | ||
No deterministic strategy wins more than 3 out of 4. Can a randomized classical strategy do better? No. Any classical strategy (including shared randomness) can be decomposed into a probabilistic mixture of deterministic strategies. Since each deterministic strategy wins at most 3 out of 4 cases, no mixture can exceed 75%. This is the content of Bell's inequality (in the CHSH form): classical correlations are bounded.
Define the CHSH value $S = E_{00} + E_{01} + E_{10} - E_{11}$, where $E_{xy}$ is the correlation between Alice's and Bob's outputs for inputs $x, y$. For any local hidden variable theory, $|S| \leq 2$. This bound is equivalent to the 75% maximum classical win rate.
The quantum strategy: approximately 85%
Now suppose Alice and Bob share an entangled Bell pair $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. They use the following strategy:
- Alice: If $x = 0$, measure in the standard basis ($Z$ basis). If $x = 1$, measure in a basis rotated by $\pi/2$ on the Bloch sphere (apply $R_y(-\pi/2)$ then measure in $Z$).
- Bob: If $y = 0$, measure in a basis rotated by $\pi/4$ (apply $R_y(-\pi/4)$ then measure). If $y = 1$, measure in a basis rotated by $-\pi/4$ (apply $R_y(\pi/4)$ then measure).
The idea is geometrical. Picture the measurement bases as directions on the Bloch sphere. Alice has two directions (angles $0$ and $\pi/2$) and Bob has two directions ($\pi/4$ and $-\pi/4$). For each input pair, the angle between their chosen measurement bases determines the correlation:
| Inputs $(x, y)$ | Alice's angle | Bob's angle | Angle between | Want |
|---|---|---|---|---|
| $(0, 0)$ | $0$ | $\pi/4$ | $\pi/4$ | Same |
| $(0, 1)$ | $0$ | $-\pi/4$ | $\pi/4$ | Same |
| $(1, 0)$ | $\pi/2$ | $\pi/4$ | $\pi/4$ | Same |
| $(1, 1)$ | $\pi/2$ | $-\pi/4$ | $3\pi/4$ | Different |
When Alice and Bob share the Bell state $|\Phi^+\rangle$ and measure along axes separated by an angle $\theta$ on the Bloch sphere, the probability of obtaining matching outcomes is $\cos^2(\theta/2)$. For the three "same" cases, the Bloch sphere angle between their measurement bases is $\pi/4$, giving a matching probability of $\cos^2(\pi/8) \approx 85.4\%$. For the "different" case $(1,1)$, the angle is $3\pi/4$, giving a matching probability of $\cos^2(3\pi/8) \approx 14.6\%$ - meaning the different outcomes appear with probability $\sin^2(3\pi/8) = \cos^2(\pi/8) \approx 85.4\%$.
Crucially, all four input cases yield the same win probability of $\cos^2(\pi/8)$. The angles are arranged so that the three cases requiring matching outputs and the one case requiring different outputs all benefit equally from the quantum correlations.
The quantum strategy wins with probability:
$$p_{\text{win}} = \cos^2\!\left(\frac{\pi}{8}\right) \approx 0.8536$$This corresponds to a CHSH value of $S = 2\sqrt{2} \approx 2.828$, which violates the classical bound of 2. The value $2\sqrt{2}$ is known as Tsirelson's bound - it is the maximum achievable by quantum mechanics.
Experiments have confirmed the quantum prediction with high statistical significance (most notably the 2015 "loophole-free" Bell tests). This means at least one of the following classical assumptions must be false: (1) locality - physical influences cannot travel faster than light, (2) realism - physical properties have definite values before measurement, or (3) freedom of choice - experimenters can freely choose their measurement settings. Most physicists accept that local realism is violated, while the speed-of-light limit on communication is preserved.
Interactive: the CHSH game
Below is a circuit implementing the quantum strategy for the input case $x = 0, y = 0$. Alice measures in the standard basis and Bob rotates by $\pi/4$ on the Bloch sphere. They should mostly output matching bits and win this round:
The matching outcomes ($|00\rangle$ and $|11\rangle$) should appear approximately $\cos^2(\pi/8) \approx 85\%$ of the time - significantly more than the 75% classical limit.
Now try the critical case $x = 1, y = 1$, where they want different outputs:
Here the different outcomes ($|01\rangle$ and $|10\rangle$) should again appear about 85% of the time. The quantum strategy manages to win approximately 85% in all four input cases, not just three out of four.
Try the other two cases yourself by setting $(x, y) = (0, 1)$ and $(1, 0)$. For $(x=0, y=1)$: no rotation on Alice, $R_y(\pi/4)$ on Bob. For $(x=1, y=0)$: $R_y(-\pi/2)$ on Alice, $R_y(-\pi/4)$ on Bob. In both cases, matching outputs should dominate.
Why not better than $2\sqrt{2}$?
One might wonder: could there be "super-quantum" correlations that win the CHSH game with even higher probability? Mathematically, the winning condition $a \oplus b = x \wedge y$ could be satisfied with up to 100% probability if the players shared a hypothetical "PR box" (named after Popescu and Rohrlich). However, quantum mechanics limits correlations to $S \leq 2\sqrt{2}$ (Tsirelson's bound). This is not just an accident - it follows from the mathematical structure of quantum mechanics: operators on a Hilbert space satisfy specific norm bounds that prevent stronger-than-quantum correlations.
Understanding why nature obeys exactly the Tsirelson bound rather than the algebraic maximum remains one of the deep open questions in quantum foundations. Various "information-theoretic" principles have been proposed (such as "information causality" and "macroscopic locality") that attempt to derive this bound from more basic assumptions.
The quantum strategy works without any communication between Alice and Bob during the game. They only share entanglement that was prepared beforehand. The entanglement does not allow them to send messages - remember the no-communication theorem from our teleportation discussion. Instead, it creates correlations that are stronger than anything achievable with shared classical randomness. This is the essence of quantum nonlocality.
7.4 The GHZ Game: Entanglement Wins Again
The CHSH game shows that quantum strategies beat classical ones by roughly 10 percentage points (85% versus 75%). The GHZ game, devised by Daniel Greenberger, Michael Horne, and Anton Zeilinger, provides an even more dramatic separation: a problem where classical strategies can never win with certainty, but a quantum strategy wins every single time.
The rules
Three players - Alice, Bob, and Charlie - are in separate rooms. A referee gives each player an input bit $x$, $y$, $z \in \{0, 1\}$, with the constraint that the total number of 1s is even. That is, the valid input triples are:
$$\{(0,0,0),\; (0,1,1),\; (1,0,1),\; (1,1,0)\}$$Each player outputs a bit ($a$, $b$, $c$ respectively). They win if:
$$a \oplus b \oplus c = x \vee y \vee z$$In other words: if all inputs are 0, the outputs should XOR to 0. If any input is 1 (which means exactly two inputs are 1), the outputs should XOR to 1.
Classical strategies fail
Let us see why no deterministic strategy can win every round. The players agree on a strategy before the game. Since the only thing each player knows is their own input bit, Alice's output is a function $a(x)$, Bob's is $b(y)$, and Charlie's is $c(z)$.
The winning conditions for each input triple give four equations (writing $a_0 = a(0)$, $a_1 = a(1)$, etc.):
- $(0,0,0)$: $a_0 \oplus b_0 \oplus c_0 = 0$
- $(0,1,1)$: $a_0 \oplus b_1 \oplus c_1 = 1$
- $(1,0,1)$: $a_1 \oplus b_0 \oplus c_1 = 1$
- $(1,1,0)$: $a_1 \oplus b_1 \oplus c_0 = 1$
XOR all four equations together. Every variable appears exactly twice on the left, so the left side is $0 \oplus 0 \oplus 0 = 0$. The right side is $0 \oplus 1 \oplus 1 \oplus 1 = 1$. We get $0 = 1$, a contradiction.
The system of equations has no solution, so any deterministic strategy must fail on at least one of the four input triples. The best possible classical success rate is 3 out of 4, or 75%. (For instance, if all three players always output 1, then $a \oplus b \oplus c = 1$, which matches $x \vee y \vee z = 1$ for the three cases where at least one input is 1, but fails the $(0,0,0)$ case where the required output XOR is 0.)
The quantum strategy: 100% win rate
Now suppose the three players share a GHZ state:
$$|\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$$The strategy is elegant. Each player applies the following rule based on their input:
- If input is $0$: apply $H$ (Hadamard), then measure in the $Z$ basis
- If input is $1$: apply $S^\dagger$ (the gate that maps $|1\rangle \to -i|1\rangle$), then $H$, then measure in the $Z$ basis
The $S^\dagger$ followed by $H$ sequence is equivalent to measuring in the $Y$ basis. So the strategy amounts to: measure in $X$ for input 0, measure in $Y$ for input 1.
Let us verify this works. The GHZ state in the $X$/$Y$ measurement basis has the property that:
- $XXX$ measurement (input $000$): outcomes always have an even number of $1$s, so $a \oplus b \oplus c = 0$
- $XYY$ measurement (input $011$): outcomes always have an odd number of $1$s, so $a \oplus b \oplus c = 1$
- $YXY$ measurement (input $101$): outcomes always have an odd number of $1$s, so $a \oplus b \oplus c = 1$
- $YYX$ measurement (input $110$): outcomes always have an odd number of $1$s, so $a \oplus b \oplus c = 1$
This matches the winning condition exactly, for all four input cases. The quantum strategy wins with probability 100%.
Verifying the math
Let us prove the $XXX$ case (input $000$) in detail. The GHZ state is $|\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$. Applying $H$ to each qubit:
$$H^{\otimes 3}|\text{GHZ}\rangle = \frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{2}}\Big[\sum_{a,b,c}|abc\rangle + \sum_{a,b,c}(-1)^{a+b+c}|abc\rangle\Big]$$The terms where $a + b + c$ is even get coefficient $\frac{2}{4} = \frac{1}{2}$, and the terms where $a + b + c$ is odd cancel to 0. So the outcome is uniform over the four states $|000\rangle$, $|011\rangle$, $|101\rangle$, $|110\rangle$ - all having even parity ($a \oplus b \oplus c = 0$), exactly as required.
For the $XYY$ case (input $011$), we apply $H$ to qubit 0 and $S^\dagger H$ to qubits 1 and 2. The $S^\dagger$ introduces a factor of $-i$ for each $|1\rangle$ in the last two qubits. Working through the algebra (or using the stabilizer formalism from later chapters), the outcomes are exactly those with odd parity.
Sandbox: the GHZ game in action
Below is the quantum strategy for the input case $(x, y, z) = (0, 1, 1)$. Alice measures in $X$ (just $H$ then measure), Bob and Charlie measure in $Y$ ($S^\dagger$ then $H$ then measure). The outputs should XOR to 1:
Run this and examine the outcomes. You should see only states where the three bits XOR to 1 (an odd number of 1s): $|001\rangle$, $|010\rangle$, $|100\rangle$, and $|111\rangle$. The states with an even number of 1s ($|000\rangle$, $|011\rangle$, $|101\rangle$, $|110\rangle$) should never appear.
Try the all-zeros case $(0, 0, 0)$ where all three players measure in the $X$ basis (all apply $H$ before measuring). You should see only outcomes where the bits XOR to 0.
Here you should see only outcomes where the bits XOR to 0 (an even number of 1s): $|000\rangle$, $|011\rangle$, $|101\rangle$, and $|110\rangle$.
Finally, try the case $(1, 1, 0)$ where Alice and Bob measure in $Y$ and Charlie measures in $X$:
Again, only odd-parity outcomes should appear. Try modifying the circuit for the remaining case $(1, 0, 1)$ yourself - Alice and Charlie measure in $Y$, Bob in $X$.
The CHSH game shows a statistical gap between classical and quantum strategies (75% versus approximately 85%). The GHZ game shows a logical gap: the classical impossibility is not about probabilities but about the non-existence of any consistent assignment. A single round of the GHZ game, if all outcomes are as predicted by quantum mechanics, rules out all local hidden variable theories. No statistics needed - just one perfect round.
These "games" are not just mathematical abstractions. They correspond to real experiments that have been performed in laboratories around the world. Bell inequality violations using entangled photon pairs were demonstrated as early as the 1980s (by Alain Aspect and colleagues), and GHZ experiments followed in the late 1990s. In 2022, the Nobel Prize in Physics was awarded to Aspect, Clauser, and Zeilinger for their experimental work establishing that Bell inequalities are violated in nature.
Chapter Summary
This chapter explored four protocols that showcase the power of quantum entanglement:
- Quantum teleportation transfers an unknown quantum state from Alice to Bob using a shared Bell pair and two classical bits. The original state is destroyed and recreated, consistent with no-cloning. Classical communication is essential - no faster-than-light signaling is possible.
- Superdense coding is the dual protocol: using a shared Bell pair, Alice sends two classical bits by transmitting only one qubit. These two protocols together show that 1 entangled pair + 2 classical bits = 1 qubit of communication, and vice versa.
- The CHSH game demonstrates that entangled particles produce correlations that cannot be explained by any local hidden variable theory. The classical win rate is bounded at 75%, while the quantum strategy achieves $\cos^2(\pi/8) \approx 85\%$, corresponding to the CHSH value of $2\sqrt{2}$ (Tsirelson's bound).
- The GHZ game provides an even starker demonstration: no classical strategy can win every round, but the GHZ state gives a 100% win rate, ruling out local hidden variables with a single perfect experiment.
These protocols form the conceptual foundation of quantum communication. In Chapter 8, we will see how entanglement and quantum measurement lead to practical applications in cryptography - specifically, quantum key distribution protocols that offer security guaranteed by the laws of physics.