6.2.3 Bell Inequality#

Prompts

What is an entanglement witness, and why does a single measured value \(\langle\hat{W}\rangle < 0\) certify entanglement? In what sense is a Bell inequality a special case of a witness, and why can no single witness detect every entangled state?
What did Einstein, Podolsky, and Rosen argue was incomplete about quantum mechanics, and how does a local hidden variable theory formalize their claim that measurement outcomes are predetermined “elements of reality”?
Why does locality force the CHSH correlation sum to stay within \(\pm 2\) for every local hidden variable theory? What is it about entangled-state correlations that lets quantum mechanics push the sum up to the Tsirelson bound \(2\sqrt{2}\)?
Explain the no-communication theorem: why can’t entanglement transmit information faster than light, even though measurements are correlated?
Walk through the quantum teleportation protocol in detail. Why does Bob need exactly 2 classical bits from Alice to recover the unknown state?

Lecture Notes#

Overview#

How can one certify that a state is entangled without reconstructing it in full? The answer is an entanglement witness: a single Hermitian observable whose expectation value exceeds a classical bound only if the state is entangled. The Bell inequalities are the most famous example — classical bounds on correlation sums that every separable state must obey. Quantum violation of a Bell inequality therefore certifies entanglement, and more: it rules out any local hidden variable description of nature. We develop the witness framework, then turn to the Einstein-Podolsky-Rosen incompleteness argument and the local hidden variable program it motivates, before specializing to the CHSH inequality, its Tsirelson bound \(2\sqrt{2}\), and the loophole-free experimental confirmations. Finally we establish the no-communication theorem and the teleportation protocol — showing that entanglement powers nonlocal correlations without enabling faster-than-light signaling.

Entanglement Witnesses#

Full state tomography — reconstructing \(\hat{\rho}_{AB}\) from measurements — is exponentially expensive in system size. Entanglement witnesses give a shortcut: a single Hermitian operator \(\hat{W}\) whose expectation value distinguishes separable states from (some) entangled ones.

Definition: Entanglement Witness

A Hermitian operator \(\hat{W}\) is an entanglement witness if

\(\mathrm{Tr}(\hat{W}\hat{\rho}_{\mathrm{sep}}) \geq 0\) for every separable state \(\hat{\rho}_{\mathrm{sep}} = \sum_i p_i\, \hat{\rho}_A^{(i)} \otimes \hat{\rho}_B^{(i)}\);
\(\mathrm{Tr}(\hat{W}\hat{\rho}_{\mathrm{ent}}) < 0\) for at least one entangled state.

A measured value \(\langle\hat{W}\rangle < 0\) therefore certifies that the state is entangled.

Example: Witness for a Bell State

Problem. Construct a witness that detects \(\vert\Phi^+\rangle\langle\Phi^+\vert\).

Solution. Take \(\hat{W} = \tfrac{1}{2}\hat{I} - \vert\Phi^+\rangle\langle\Phi^+\vert\). For any separable two-qubit state \(\hat{\rho}_{\mathrm{sep}}\), one can show \(\langle\Phi^+\vert\hat{\rho}_{\mathrm{sep}}\vert\Phi^+\rangle \leq \tfrac{1}{2}\), so \(\mathrm{Tr}(\hat{W}\hat{\rho}_{\mathrm{sep}}) \geq 0\). On \(\vert\Phi^+\rangle\) itself, \(\mathrm{Tr}(\hat{W}\vert\Phi^+\rangle\langle\Phi^+\vert) = -\tfrac{1}{2} < 0\), so \(\hat{W}\) detects this Bell state.

Witnesses Are Complete but Not Universal

By the Hahn–Banach separation theorem, every entangled state admits some witness. But no single witness detects all entangled states — a given \(\hat{W}\) only detects those states lying on the “wrong” side of the separable hyperplane it defines.

Bell inequalities as witnesses. The CHSH inequality below is a canonical witness: the CHSH observable \(\hat{S}_{\mathrm{CHSH}}\) satisfies \(\langle\hat{S}_{\mathrm{CHSH}}\rangle \leq 2\) for every local hidden variable theory and every separable quantum state, while the singlet attains \(2\sqrt{2}\). Equivalently, \(\hat{W}_{\mathrm{CHSH}} = 2\,\hat{I} - \hat{S}_{\mathrm{CHSH}}\) is a witness for states that violate local realism. Bell inequalities are thus a historically important — and experimentally accessible — family of entanglement witnesses, detecting every state whose nonlocal correlations exceed the classical bound.

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 \(\hat{Z}\) and gets \(+1\), then Bob’s particle is instantly in \(\vert 1\rangle\). If she instead measures \(\hat{X}\) 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):

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

Derivation: CHSH Bound

Setup. In a local hidden variable theory, a shared hidden state \(\lambda\) is drawn from some distribution \(\rho(\lambda)\). For every realization of \(\lambda\), Alice’s outcome \(A_i(\lambda) \in \{+1,-1\}\) depends only on her setting \(i \in \{1,2\}\) and on \(\lambda\), and Bob’s outcome \(B_j(\lambda) \in \{+1,-1\}\) depends only on his setting \(j\) and on \(\lambda\). Expectation values are averages over \(\rho(\lambda)\): \(\langle A_i B_j\rangle = \int \rho(\lambda)\,A_i(\lambda)B_j(\lambda)\,\mathrm{d}\lambda\).

Step 1 — bound the integrand pointwise. If the quantity

\[ S(\lambda) \;=\; A_1 B_1 + A_1 B_2 + A_2 B_1 - A_2 B_2 \]

satisfies \(\vert S(\lambda)\vert \leq 2\) for every single realization of \(\lambda\), then averaging preserves the bound: \(\vert\langle S\rangle\vert \leq 2\). So hold \(\lambda\) fixed and treat \(A_1, A_2, B_1, B_2\) as four specific numbers in \(\{+1,-1\}\).

Step 2 — regroup. Factor out Alice’s outcomes:

\[ S = A_1(B_1 + B_2) + A_2(B_1 - B_2). \]

Step 3 — Bob’s values force one parenthesis to zero. Because \(B_1, B_2 \in \{+1,-1\}\), only two cases occur:

If \(B_1 = B_2\), then \(B_1 + B_2 = \pm 2\) and \(B_1 - B_2 = 0\).
If \(B_1 \neq B_2\), then \(B_1 + B_2 = 0\) and \(B_1 - B_2 = \pm 2\).

Exactly one of the two parentheses equals \(\pm 2\); the other vanishes.

Step 4 — the sum is \(\pm 2\). In either case the non-vanishing term is \(A_i\,(\pm 2)\) with \(A_i \in \{+1,-1\}\), so \(S \in \{+2, -2\}\) — in particular \(\vert S\vert \leq 2\).

Step 5 — average. Since \(\vert S(\lambda)\vert \leq 2\) holds for every \(\lambda\),

\[ \left\vert \langle A_1 B_1\rangle + \langle A_1 B_2\rangle + \langle A_2 B_1\rangle - \langle A_2 B_2\rangle \right\vert \leq 2. \]

Where locality enters. The grouping in Step 2 treats \(A_1, A_2, B_1, B_2\) as four numbers coexisting for the same \(\lambda\). This is exactly the assumption that each outcome depends only on the local setting and the shared \(\lambda\) — not on the distant setting. Quantum mechanically, \(A_1\) and \(A_2\) correspond to non-commuting observables (e.g. \(\hat{Z}\) and \(\hat{X}\)), so no single state assigns simultaneous values to both. The CHSH inequality is built precisely to detect this: any violation rules out the existence of such pre-assigned joint values.

Quantum Violation#

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

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

where \(\theta\) is the angle between measurement directions. Choosing optimal angles, the singlet achieves:

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

Derivation: Quantum CHSH Violation

For the singlet state, the spin correlation between measurement directions \(\hat{\boldsymbol{a}}\) and \(\hat{\boldsymbol{b}}\) is \(\langle \hat{\boldsymbol{a}}\cdot\hat{\boldsymbol{\sigma}} \otimes \hat{\boldsymbol{b}}\cdot\hat{\boldsymbol{\sigma}} \rangle = -\hat{\boldsymbol{a}} \cdot \hat{\boldsymbol{b}}\).

The CHSH combination \(E_{11} + E_{12} + E_{21} - E_{22}\) is largest in magnitude when all four terms add with the same sign. Since the singlet correlator \(E = -\cos\theta\) depends only on the angle \(\theta\) between directions, this means the three \(+\) correlators should be equal and the \(-\) correlator their supplement. The configuration achieving this places all four directions in the \(x\)-\(z\) plane with Alice’s two settings mutually orthogonal (\(\hat{\boldsymbol{a}}_1\) along \(x\), \(\hat{\boldsymbol{a}}_2\) along \(z\)) and Bob’s two settings at \(\pm 45^\circ\) from \(\hat{\boldsymbol{a}}_1\), so three of the four Alice-Bob pairs are separated by \(45^\circ\) and the pair \((\hat{\boldsymbol{a}}_2,\hat{\boldsymbol{b}}_2)\) by \(135^\circ\):

\[\begin{split} \begin{split} \hat{\boldsymbol{a}}_1 &= (1, 0, 0)\\ \hat{\boldsymbol{a}}_2 &= (0, 0, 1) \end{split} \end{split}\]

\[\begin{split} \begin{split} \hat{\boldsymbol{b}}_1 &= \frac{1}{\sqrt{2}}(1, 0, 1)\\ \hat{\boldsymbol{b}}_2 &= \frac{1}{\sqrt{2}}(1, 0, -1) \end{split} \end{split}\]

These give \(\hat{\boldsymbol{a}}_1 \cdot \hat{\boldsymbol{b}}_1 = 1/\sqrt{2}\), \(\hat{\boldsymbol{a}}_1 \cdot \hat{\boldsymbol{b}}_2 = 1/\sqrt{2}\), \(\hat{\boldsymbol{a}}_2 \cdot \hat{\boldsymbol{b}}_1 = 1/\sqrt{2}\), \(\hat{\boldsymbol{a}}_2 \cdot \hat{\boldsymbol{b}}_2 = -1/\sqrt{2}\). With the singlet correlator \(E(\hat{\boldsymbol{a}},\hat{\boldsymbol{b}}) = -\hat{\boldsymbol{a}}\cdot\hat{\boldsymbol{b}}\), the four correlations are \(E_{11} = E_{12} = E_{21} = -1/\sqrt{2}\) and \(E_{22} = +1/\sqrt{2}\). The CHSH sum (with its standard \(+,+,+,-\) signs on the four correlators) becomes:

\[\begin{split} \begin{split} \mathrm{CHSH} &= E_{11} + E_{12} + E_{21} - E_{22}\\ &= -\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}}\\ &= -\frac{4}{\sqrt{2}} = -2\sqrt{2}, \end{split} \end{split}\]

so \(\vert\mathrm{CHSH}\vert = 2\sqrt{2}\) — the Tsirelson bound, the maximal violation allowed by quantum mechanics. The overall sign of \(\mathrm{CHSH}\) is a labeling convention — reversing Bob’s \(\pm 1\) outcome labels (\(\hat{\boldsymbol{b}}_i \to -\hat{\boldsymbol{b}}_i\)) sends it to \(+2\sqrt{2}\) — so the body result (232) and the CHSH inequality both bound the magnitude \(\vert\mathrm{CHSH}\vert\).

Tsirelson Bound

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

\[\begin{split} \begin{split} \text{Classical} &: \leq 2 \\ \text{Quantum} &: \leq 2\sqrt{2} \end{split} \end{split}\]

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 Nature of Bell’s Theorem

Bell’s theorem asks a deep question: What does it mean for a property to “exist” before we measure it? Einstein believed in realism — that distant particles carry predetermined values. Quantum mechanics says no: the universe is fundamentally nonlocal, or fundamentally indeterminate, or both. Which view do you find more compelling? Does Bell’s theorem suggest that reality depends on observation, or that the universe is “spooky” (nonlocal), or something else entirely? How do you reconcile quantum nonlocality with Einstein’s principle that nothing travels faster than light?

Poll: Bell inequality and local realism

The CHSH inequality \(\vert\langle A_1 B_1 \rangle + \langle A_1 B_2 \rangle + \langle A_2 B_1 \rangle - \langle A_2 B_2 \rangle\vert \leq 2\) is violated by certain quantum states (e.g., maximally entangled states violate it by a factor of \(\sqrt{2}\)). What does this violation demonstrate?

(A) Quantum mechanics is incomplete; hidden variables are needed.

(B) Local hidden variable theories cannot reproduce all quantum predictions.

(D) Measurements on entangled pairs always give correlated results.

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 \(\vert\psi\rangle_B\).

Derivation: Teleportation Protocol

Setup. Alice holds qubit C with unknown state \(\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\) and one half of a shared Bell pair (qubit A). Bob holds the other half (qubit B). The shared state is \(\vert\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\).

Step 1: Write initial state. The combined state is:

\[ \vert\Psi_\text{init}\rangle = \vert\psi\rangle_C \otimes \vert\Phi^+\rangle_{AB} = (\alpha\vert 0\rangle_C + \beta\vert 1\rangle_C) \otimes \frac{1}{\sqrt{2}}(\vert 00\rangle_{AB} + \vert 11\rangle_{AB}) \]

Step 2: Expand in the Bell basis of qubits C and A. The CA Bell basis is \(\vert\Phi^\pm\rangle = \tfrac{1}{\sqrt{2}}(\vert 00\rangle \pm \vert 11\rangle)\) and \(\vert\Psi^\pm\rangle = \tfrac{1}{\sqrt{2}}(\vert 01\rangle \pm \vert 10\rangle)\); inverting these relations expresses each computational-basis pair of qubits C and A as a combination of Bell states:

\[\begin{split} \begin{split} \vert 00\rangle_{CA} &= \tfrac{1}{\sqrt{2}}\bigl(\vert\Phi^+\rangle_{CA} + \vert\Phi^-\rangle_{CA}\bigr)\\ \vert 11\rangle_{CA} &= \tfrac{1}{\sqrt{2}}\bigl(\vert\Phi^+\rangle_{CA} - \vert\Phi^-\rangle_{CA}\bigr)\\ \vert 01\rangle_{CA} &= \tfrac{1}{\sqrt{2}}\bigl(\vert\Psi^+\rangle_{CA} + \vert\Psi^-\rangle_{CA}\bigr)\\ \vert 10\rangle_{CA} &= \tfrac{1}{\sqrt{2}}\bigl(\vert\Psi^+\rangle_{CA} - \vert\Psi^-\rangle_{CA}\bigr) \end{split} \end{split}\]

First regroup the tensor factors of \(\vert\Psi_\text{init}\rangle\) so qubits C and A sit together and qubit B stands alone (qubit A is the first half of the shared pair, B the second):

\[ \vert\Psi_\text{init}\rangle = \frac{1}{\sqrt{2}}\bigl(\alpha\vert 00\rangle_{CA}\vert 0\rangle_B + \alpha\vert 01\rangle_{CA}\vert 1\rangle_B + \beta\vert 10\rangle_{CA}\vert 0\rangle_B + \beta\vert 11\rangle_{CA}\vert 1\rangle_B\bigr) \]

Substitute the four identities — each carries a factor \(1/\sqrt{2}\) that combines with the \(1/\sqrt{2}\) above into an overall \(1/2\) — and expand the products:

\[\begin{split} \begin{split} \vert\Psi_\text{init}\rangle = \frac{1}{2}\bigl[&\alpha\bigl(\vert\Phi^+\rangle_{CA} + \vert\Phi^-\rangle_{CA}\bigr)\vert 0\rangle_B + \alpha\bigl(\vert\Psi^+\rangle_{CA} + \vert\Psi^-\rangle_{CA}\bigr)\vert 1\rangle_B\\ &+ \beta\bigl(\vert\Psi^+\rangle_{CA} - \vert\Psi^-\rangle_{CA}\bigr)\vert 0\rangle_B + \beta\bigl(\vert\Phi^+\rangle_{CA} - \vert\Phi^-\rangle_{CA}\bigr)\vert 1\rangle_B\bigr] \end{split} \end{split}\]

Collecting the eight terms by their CA Bell state — for instance \(\vert\Psi^-\rangle_{CA}\) picks up \(+\alpha\vert 1\rangle_B\) from the \(\vert 01\rangle_{CA}\) term and \(-\beta\vert 0\rangle_B\) from the \(\vert 10\rangle_{CA}\) term — gives:

\[\begin{split} \begin{split} \vert\Psi_\text{init}\rangle = \frac{1}{2}\bigl[&\vert\Phi^+\rangle_{CA}(\alpha\vert 0\rangle_B + \beta\vert 1\rangle_B) + \vert\Phi^-\rangle_{CA}(\alpha\vert 0\rangle_B - \beta\vert 1\rangle_B)\\ &+ \vert\Psi^+\rangle_{CA}(\beta\vert 0\rangle_B + \alpha\vert 1\rangle_B) + \vert\Psi^-\rangle_{CA}(\alpha\vert 1\rangle_B - \beta\vert 0\rangle_B)\bigr] \end{split} \end{split}\]

Or more compactly: \(= \frac{1}{2}\sum_j \vert\Phi_j\rangle_{CA} \hat{U}_j \vert\psi\rangle_B\) where \(\hat{U}_0 = \hat{I}\), \(\hat{U}_1 = \hat{Z}\), \(\hat{U}_2 = \hat{X}\), \(\hat{U}_3 = \hat{X}\hat{Z}\).

Step 3: Alice measures in Bell basis. Alice measures qubits C and A, obtaining one of four outcomes \(j = 0, 1, 2, 3\) with equal probability \(1/4\) each. The measurement projects to \(\vert\Phi_j\rangle_{CA}\), leaving Bob’s qubit in state \(\hat{U}_j\vert\psi\rangle_B\).

Step 4: Bob applies correction. Alice sends the 2-bit outcome \(j\) to Bob. Bob applies \(\hat{U}_j^\dagger\) to his qubit, recovering \(\vert\psi\rangle_B\).

This protocol transmits an unknown quantum state using one Bell pair and 2 classical bits—no direct physical transfer of the qubit itself.

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#

Entanglement witnesses: a Hermitian \(\hat{W}\) with \(\langle\hat{W}\rangle \geq 0\) on all separable states and \(\langle\hat{W}\rangle < 0\) on some entangled state; every entangled state admits a witness, but no one witness detects all.
Bell’s theorem: no local hidden variable theory reproduces quantum predictions; the CHSH bound \(\leq 2\) is violated by quantum mechanics at the Tsirelson bound \(2\sqrt{2}\). Bell inequalities are a specific family of entanglement witnesses.
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. Entanglement Witness. For the Bell state \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\), define the witness \(\hat{W} = \frac{\hat{I}}{2} - \vert\Phi^+\rangle\langle\Phi^+\vert\).

(a) Show that \(\mathrm{Tr}(\hat{W}\hat{\rho}_{\mathrm{sep}}) \geq 0\) for any separable state (it suffices to check product states \(\hat{\rho}_A\otimes\hat{\rho}_B\)).

(b) Compute \(\mathrm{Tr}(\hat{W}\vert\Phi^+\rangle\langle\Phi^+\vert)\) and verify it is negative, so \(\hat{W}\) detects \(\vert\Phi^+\rangle\).

(c) For the Werner state \(\hat{\rho}(p) = p\,\vert\Phi^+\rangle\langle\Phi^+\vert + (1-p)\,\tfrac{\hat{I}}{4}\), find the threshold \(p_c\) above which the witness signals entanglement.

(d) Explain briefly why no single witness can detect all entangled states.

2. Classical CHSH Bound. Let \(A_1, A_2 \in \{\pm 1\}\) and \(B_1, B_2 \in \{\pm 1\}\) be four \(\pm 1\)-valued random variables. Show that \(A_1 B_1 + A_1 B_2 + A_2 B_1 - A_2 B_2 = \pm 2\) for any single realization, and conclude that \(\vert\langle A_1 B_1 \rangle + \langle A_1 B_2 \rangle + \langle A_2 B_1 \rangle - \langle A_2 B_2 \rangle\vert \leq 2\).

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

4. Optimal Measurement Angles. Choose measurement directions \(\hat{\boldsymbol{a}}_1 = \hat{\boldsymbol{x}}\), \(\hat{\boldsymbol{a}}_2 = \hat{\boldsymbol{z}}\), \(\hat{\boldsymbol{b}}_1 = (\hat{\boldsymbol{x}}+\hat{\boldsymbol{z}})/\sqrt{2}\), \(\hat{\boldsymbol{b}}_2 = (\hat{\boldsymbol{x}}-\hat{\boldsymbol{z}})/\sqrt{2}\). Using the singlet correlator \(E(\hat{\boldsymbol{a}},\hat{\boldsymbol{b}}) = -\hat{\boldsymbol{a}}\cdot\hat{\boldsymbol{b}}\), compute the four correlators and show \(\vert\mathrm{CHSH}\vert = 2\sqrt{2}\). Verify that flipping the sign of any one \(\hat{\boldsymbol{b}}_{i}\) would give \(\vert\mathrm{CHSH}\vert = 0\) instead — the relative orientation matters.

5. 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 \(\hat{C} = \hat{A}_1\otimes\hat{B}_1 + \hat{A}_1\otimes\hat{B}_2 + \hat{A}_2\otimes\hat{B}_1 - \hat{A}_2\otimes\hat{B}_2\), and use \(\hat{A}_i^2 = \hat{I}\).)

6. EPR Local Realism. For the singlet state, if Alice measures \(\hat{Z}\) and gets \(+1\), what is Bob’s state? If she measures \(\hat{X}\) 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.

7. Werner State Violations. 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.

8. No-Signaling Constraint. 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.

9. Bell Measurement and Teleportation. Consider two qubits \(A\) and \(B\) with the stabilizer observables \(\hat{X}_A\hat{X}_B\) and \(\hat{Z}_A\hat{Z}_B\).

(a) Write \(\hat{X}_A\hat{X}_B\) and \(\hat{Z}_A\hat{Z}_B\) as \(4\times 4\) matrices in the computational basis \(\{\vert 00\rangle, \vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}\).

(b) Show that \([\hat{X}_A\hat{X}_B, \hat{Z}_A\hat{Z}_B] = 0\), so they can be measured simultaneously. Find the four joint eigenvalues \((x, z) \in \{(\pm 1, \pm 1)\}\) and show that the corresponding eigenstates are the four Bell states.

(c) A third qubit \(C\) is in an unknown state \(\vert\psi\rangle_C = \alpha\vert 0\rangle + \beta\vert 1\rangle\). The full state is \(\vert\psi\rangle_C \otimes \vert\Phi^+\rangle_{AB}\), where \(\vert\Phi^+\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)\). Rewrite this state by expanding qubits \(C\) and \(A\) in the Bell basis.

(d) After a Bell measurement on \((C, A)\) yields outcome \((x, z)\), show that Bob can recover \(\vert\psi\rangle\) on qubit \(B\) by applying a single-qubit correction \(\hat{X}^{(1-z)/2}\hat{Z}^{(1-x)/2}\). This is quantum teleportation.

10. Entanglement Swapping. Four qubits \(A, B, C, D\) are prepared in a product of two Bell-singlet pairs: \(\vert\Psi\rangle = \vert\Psi^-\rangle_{AB} \otimes \vert\Psi^-\rangle_{CD}\), where \(\vert\Psi^-\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle - \vert 10\rangle)\).

(a) Expand \(\vert\Psi\rangle\) in the Bell basis on the \((B, C)\) pair (the four Bell states are \(\vert\Phi^\pm\rangle, \vert\Psi^\pm\rangle\)).

(b) A Bell measurement is performed on \((B, C)\). Show that, conditional on each outcome, qubits \(A\) and \(D\) end up in a Bell state — even though they never interacted directly.

(d) Explain why this protocol is called entanglement swapping: entanglement is transferred from the pairs \((A, B)\) and \((C, D)\) to the pair \((A, D)\) via a measurement on \((B, C)\).