6.2.3 Bell Inequality#

Prompts

What was the EPR argument, and how does Bell’s theorem refute it?
Derive the CHSH inequality and show that the singlet state achieves the Tsirelson bound \(2\sqrt{2}\).
Explain the no-communication theorem: why can’t entanglement transmit information faster than light?
Walk through the quantum teleportation protocol. Why does Bob need Alice’s 2 classical bits?

Lecture Notes#

Overview#

Bell’s theorem (1964) settles the EPR debate: no local hidden variable theory can reproduce quantum predictions. The CHSH inequality provides the sharpest test — classical theories are bounded at 2, but quantum mechanics achieves \(2\sqrt{2}\). We also establish the no-communication theorem and quantum teleportation, showing that entanglement powers nonlocal correlations without enabling faster-than-light signaling.

The EPR Paradox#

Einstein, Podolsky, and Rosen (1935) argued that quantum mechanics must be incomplete. Consider an entangled pair in the singlet state, with Alice measuring particle 1 and Bob measuring particle 2 far away. If Alice measures \(\hat{\sigma}^z\) and gets \(+1\), then Bob’s particle is instantly in \(\vert\downarrow\rangle\). If she instead measures \(\hat{\sigma}^x\) and gets \(+1\), Bob’s particle is in \(\vert-\rangle\). Since the particles are far apart, Bob’s outcome cannot depend on Alice’s choice — so EPR concluded the outcomes must be predetermined by hidden variables \(\lambda\) that each particle carries.

A local hidden variable theory assumes outcomes \(A(a, \lambda)\) and \(B(b, \lambda)\) depend only on the local measurement setting and the shared hidden state. Bell showed that any such theory satisfies constraints — Bell inequalities — that quantum mechanics violates.

CHSH Inequality#

The Clauser–Horne–Shimony–Holt (CHSH) form is the most commonly tested Bell inequality:

CHSH Inequality

For any local hidden variable theory with measurement settings \(a_1, a_2\) (Alice) and \(b_1, b_2\) (Bob):

(127)#\[ \left\vert \langle A_1 B_1 \rangle + \langle A_1 B_2 \rangle + \langle A_2 B_1 \rangle - \langle A_2 B_2 \rangle \right\vert \leq 2 \]

Derivation: CHSH Bound

For any realization of \(\lambda\), all outcomes are \(\pm 1\). Write:

\[ A_1(B_1 + B_2) + A_2(B_1 - B_2) \]

Since \(B_1, B_2 \in \{\pm 1\}\), either \(B_1 + B_2 = \pm 2\) and \(B_1 - B_2 = 0\), or vice versa. In either case the expression equals \(\pm 2\). Averaging over \(\lambda\) gives the bound.

Quantum Violation#

For the singlet state \(\vert\Psi^-\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle - \vert 10\rangle)\), the spin correlation is:

(128)#\[ \langle A(\boldsymbol{a})\, B(\boldsymbol{b}) \rangle = -\boldsymbol{a} \cdot \boldsymbol{b} = -\cos\theta \]

where \(\theta\) is the angle between measurement directions. Choose the optimal angles \(a_1 = 0°\), \(a_2 = 45°\), \(b_1 = 22.5°\), \(b_2 = 67.5°\) to obtain:

(129)#\[ \text{CHSH}_\text{quantum} = 2\sqrt{2} \approx 2.828 > 2 \]

Tsirelson Bound

The maximum CHSH value achievable by any quantum state is \(2\sqrt{2}\), achieved by maximally entangled states with optimal measurement directions:

\[ \text{Classical}: \leq 2 \qquad \text{Quantum}: \leq 2\sqrt{2} \qquad \text{No-signaling (PR box)}: \leq 4 \]

Bell’s Theorem

No local hidden variable theory can reproduce all predictions of quantum mechanics. Any theory of nature must abandon at least one of: (1) locality — distant events cannot influence each other instantaneously, or (2) realism — physical properties exist prior to measurement.

Discussion

The CHSH inequality gives classical maximum 2 and quantum maximum \(2\sqrt{2}\), but hypothetical “PR boxes” achieve 4. Why doesn’t nature allow PR boxes? What physical principle sets the quantum bound?

Experimental Confirmation#

Early Bell tests suffered from loopholes (detection efficiency, locality of measurement choices). Since 2015, loophole-free Bell tests have closed all major loopholes simultaneously, confirming quantum violation with \(> 5\sigma\) confidence. Local hidden variable theories are decisively ruled out.

Detection and Locality Loopholes

The detection loophole arises when not all particles are detected — undetected events could conspire to satisfy the CHSH bound. The locality loophole requires that measurement settings be chosen while the particles are in flight, with spacelike separation ensuring no signal can travel between the detectors. Modern experiments use high-efficiency detectors and random number generators to close both loopholes in a single run.

No-Communication Theorem#

Despite nonlocal correlations, entanglement cannot transmit information:

No-Communication Theorem

For any bipartite state \(\hat{\rho}_{AB}\), Bob’s reduced state \(\hat{\rho}_B = \mathrm{Tr}_A(\hat{\rho}_{AB})\) is independent of Alice’s measurement choice. Local statistics reveal no information about distant operations.

For the singlet, regardless of what Alice measures, Bob’s reduced state is always \(\hat{\rho}_B = \hat{I}/2\) (maximally mixed). Correlations become visible only when Alice and Bob compare results via classical communication, which is limited by the speed of light.

Quantum Teleportation#

Alice can transmit an unknown state \(\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\) to Bob using one shared Bell pair and 2 classical bits:

Setup: Alice holds qubit C (the state to send) and qubit A; Bob holds qubit B. Qubits A and B share \(\vert\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\).
Bell measurement: Alice measures qubits C and A in the Bell basis, obtaining 2 classical bits \(m_1 m_2\).
Classical communication: Alice sends \(m_1 m_2\) to Bob.
Correction: Bob applies \(\hat{X}^{m_2}\hat{Z}^{m_1}\) to qubit B, recovering \(\vert\psi\rangle_B\).

Example: Teleportation Algebra

Problem. Verify that after Alice’s Bell measurement yields outcome \(m_1 m_2\), Bob’s correction \(\hat{X}^{m_2}\hat{Z}^{m_1}\) recovers \(\vert\psi\rangle\).

Solution. Expand the initial state \(\vert\psi\rangle_C \otimes \vert\Phi^+\rangle_{AB}\) in the Bell basis of qubits C and A:

\[ \vert\psi\rangle_C \vert\Phi^+\rangle_{AB} = \tfrac{1}{2}\bigl[\vert\Phi^+\rangle_{CA}\vert\psi\rangle_B + \vert\Phi^-\rangle_{CA}\hat{Z}\vert\psi\rangle_B + \vert\Psi^+\rangle_{CA}\hat{X}\vert\psi\rangle_B + \vert\Psi^-\rangle_{CA}\hat{Z}\hat{X}\vert\psi\rangle_B\bigr] \]

Each Bell outcome leaves Bob’s qubit in a Pauli-rotated version of \(\vert\psi\rangle\). The 2 bits \((m_1, m_2)\) tell Bob which Pauli to undo.

Why Teleportation Respects No-Communication

Alice’s Bell measurement produces random outcomes. Without the 2 classical bits, Bob’s qubit is maximally mixed — indistinguishable from noise. The quantum state is reconstructed only after Bob receives the classical message, which travels at most at light speed.

Summary#

Bell’s theorem: no local hidden variable theory reproduces quantum predictions; the CHSH bound \(\leq 2\) is violated by quantum mechanics at \(2\sqrt{2}\).
Loophole-free experiments (2015+) confirm quantum nonlocality with high confidence.
No-communication: entanglement creates correlations, not signals; local statistics are independent of distant operations.
Quantum teleportation: entanglement + 2 classical bits transmit an unknown quantum state without physical transport.

Homework#

1. (Classical bound) Let \(A, A' \in \{\pm 1\}\) and \(B, B' \in \{\pm 1\}\) be four \(\pm 1\)-valued random variables. Show that \(AB + AB' + A'B - A'B' = \pm 2\) for any single realization, and conclude that \(\vert\langle AB \rangle + \langle AB' \rangle + \langle A'B \rangle - \langle A'B' \rangle\vert \leq 2\).

2. (Singlet correlation) For the singlet state \(\vert\Psi^-\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle - \vert 10\rangle)\), show that \(\langle \hat{\boldsymbol{a}}\cdot\hat{\boldsymbol{\sigma}} \otimes \hat{\boldsymbol{b}}\cdot\hat{\boldsymbol{\sigma}} \rangle = -\hat{\boldsymbol{a}}\cdot\hat{\boldsymbol{b}}\).

3. (Optimal angles) Choose measurement directions \(\hat{a} = \hat{z}\), \(\hat{a}' = \hat{x}\), \(\hat{b} = \frac{1}{\sqrt{2}}(-\hat{z} - \hat{x})\), \(\hat{b}' = \frac{1}{\sqrt{2}}(\hat{z} - \hat{x})\). Compute the four correlators for the singlet and show \(\text{CHSH} = 2\sqrt{2}\).

4. (Tsirelson bound) Prove that for any quantum state and any choice of \(\pm 1\)-valued observables, \(\vert\langle \text{CHSH} \rangle\vert \leq 2\sqrt{2}\). (Hint: compute \(\hat{C}^2\) where \(\hat{C} = \hat{A}\otimes\hat{B} + \hat{A}\otimes\hat{B}' + \hat{A}'\otimes\hat{B} - \hat{A}'\otimes\hat{B}'\), and use \(\hat{A}^2 = \hat{I}\).)

5. (EPR reasoning) For the singlet state, if Alice measures \(\hat{\sigma}^z\) and gets \(+1\), what is Bob’s state? If she measures \(\hat{\sigma}^x\) and gets \(+1\)? Explain why EPR argued that “elements of reality” must exist for Bob’s particle before measurement, and how Bell’s theorem refutes this reasoning.

6. (Werner state) The Werner state \(\hat{\rho}_W(p) = p\vert\Psi^-\rangle\langle\Psi^-\vert + (1-p)\frac{\hat{I}}{4}\) interpolates between maximally entangled (\(p=1\)) and maximally mixed (\(p=0\)). Show that its maximum CHSH value is \(2\sqrt{2}\,p\), and find the critical \(p\) above which Bell inequality is violated.

7. (No-signaling) Show that \(\sum_{a} p(a,b\vert x,y) = \sum_{a} p(a,b\vert x',y)\) for quantum probabilities \(p(a,b\vert x,y) = \mathrm{Tr}[(\hat{M}_a^x \otimes \hat{M}_b^y)\hat{\rho}]\). Interpret: Alice’s measurement choice cannot affect Bob’s marginal distribution.

8. (Teleportation) Verify the teleportation protocol: starting from \(\vert\psi\rangle_C \otimes \vert\Phi^+\rangle_{AB}\), show that after Alice’s Bell measurement with outcome \((m_1, m_2)\), Bob’s correction \(\hat{X}^{m_2}\hat{Z}^{m_1}\) recovers \(\vert\psi\rangle_B\). Work through all four Bell outcomes explicitly.