Chapter 24 Quiz: Entanglement, Bell's Theorem, and Foundations
Instructions: This quiz covers the core concepts from Chapter 24. For multiple choice, select the single best answer. For true/false, provide a brief justification (1-2 sentences). For short answer, aim for 3-5 sentences. For applied scenarios, show your work.
Multiple Choice (10 questions)
Q1. The EPR argument concludes that:
(a) Quantum mechanics gives incorrect predictions for entangled particles (b) Quantum mechanics is incomplete — hidden variables must exist to restore local realism (c) The uncertainty principle is wrong (d) Entangled particles communicate faster than light
Q2. In the CHSH inequality $|S| \leq 2$, the bound of 2 follows from the assumption(s) of:
(a) The uncertainty principle (b) Locality alone (c) Locality and realism (outcomes $\pm 1$ determined by hidden variables and local settings) (d) Energy conservation and Lorentz invariance
Q3. For the singlet state measured along axes separated by angle $\theta$, the quantum correlation function is:
(a) $E(\theta) = \cos\theta$ (b) $E(\theta) = -\cos\theta$ (c) $E(\theta) = -\cos^2\theta$ (d) $E(\theta) = 1 - 2\theta/\pi$
Q4. The Tsirelson bound states that the maximum quantum violation of the CHSH inequality is:
(a) $|S| = 2$ (b) $|S| = 2\sqrt{2} \approx 2.828$ (c) $|S| = 4$ (d) $|S| = \pi$
Q5. Aspect's 1982 experiment was particularly important because it:
(a) Was the first to generate entangled photon pairs (b) Used rapid switching of measurement settings to close the locality loophole (c) Achieved 100% detection efficiency (d) Demonstrated quantum teleportation
Q6. In the quantum teleportation protocol, the resources consumed are:
(a) 1 Bell pair only (b) 1 Bell pair + 1 classical bit (c) 1 Bell pair + 2 classical bits (d) 2 Bell pairs + 2 classical bits
Q7. The no-cloning theorem states that:
(a) Entangled states cannot be created (b) No quantum operation can copy an arbitrary unknown quantum state (c) Quantum states cannot be measured (d) Quantum information cannot be transmitted classically
Q8. Which interpretation of quantum mechanics is explicitly non-local (meaning it includes faster-than-light influences in its formalism)?
(a) Copenhagen (b) Many-worlds (c) Bohmian mechanics (d) QBism
Q9. The detection loophole in Bell tests arises because:
(a) Detectors cannot distinguish between different Bell states (b) Imperfect detection efficiency could allow a local hidden variable model to fake a Bell violation if detected pairs are a biased sample (c) Photon detectors cannot measure spin (d) Detectors disturb the quantum state
Q10. In superdense coding, Alice transmits to Bob:
(a) 1 classical bit using 1 qubit (no entanglement needed) (b) 2 classical bits using 1 qubit + 1 shared Bell pair (c) 1 qubit using 2 classical bits + 1 shared Bell pair (d) 2 qubits using 1 classical bit
True/False (4 questions)
For each statement, indicate TRUE or FALSE and provide a brief justification (1-2 sentences).
Q11. "Bell's theorem proves that quantum mechanics is non-local — that measurements on one particle instantaneously affect the other."
Q12. "The EPR argument assumes that quantum mechanics gives correct predictions for the singlet state."
Q13. "All four Bell states have the same amount of entanglement (1 ebit), even though they are different states."
Q14. "The many-worlds interpretation and the Copenhagen interpretation make different experimental predictions, which is why the debate continues."
Short Answer (5 questions)
For each question, answer in 3-5 sentences.
Q15. What is the difference between Bell's original inequality and the CHSH inequality? Why is the CHSH form preferred for experiments?
Q16. Explain in physical terms why quantum teleportation does not violate the no-signaling principle (the principle that information cannot travel faster than light). What specific aspect of the protocol enforces this?
Q17. The 2015 loophole-free Bell tests closed three main loopholes simultaneously. Name and briefly explain each loophole.
Q18. State the threshold concept of this chapter — "Entanglement is not classical correlation" — and explain what it means in terms of Bell's theorem. Why can't a shared random variable (like a hidden variable $\lambda$) reproduce entangled correlations?
Q19. Why is the measurement problem considered a genuine unsolved problem in physics, rather than just a philosophical quibble? What specific question does it ask that the quantum formalism does not answer?
Applied Scenarios (3 questions)
Show your work for full credit.
Q20. Alice and Bob share the Bell state $|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$. Both measure spin along the $z$-axis.
(a) What are the possible joint outcomes and their probabilities? (b) What is the correlation $E(\hat{z}, \hat{z})$? (c) Now Alice measures along $\hat{z}$ and Bob along $\hat{x}$. What is $E(\hat{z}, \hat{x})$? (d) Is $|\Psi^+\rangle$ rotationally invariant like the singlet $|\Psi^-\rangle$? Explain.
Q21. A Werner state is $\rho_W = 0.8|\Psi^-\rangle\langle\Psi^-| + 0.2 \cdot \frac{1}{4}\hat{I}$.
(a) What is the purity $\text{Tr}(\rho_W^2)$ of this state? (b) Is this state entangled? (Recall the entanglement threshold for Werner states is $p > 1/3$.) (c) Does this state violate the CHSH inequality? (Recall the CHSH threshold is $p > 1/\sqrt{2}$.) (d) What is the maximum CHSH parameter $|S|$ achievable with this state?
Q22. Alice wants to teleport the state $|\chi\rangle = |0\rangle$ (the spin-up state) using the Bell pair $|\Phi^+\rangle$. Trace through the teleportation protocol step by step:
(a) Write the initial three-qubit state. (b) Express it in the Bell basis for qubits $C$ and $A$. (c) If Alice's Bell measurement yields $|\Psi^-\rangle_{CA}$, what is Bob's qubit state? (d) What correction must Bob apply? Verify the final state is $|0\rangle$.
Answer Key
Q1: (b). EPR concluded that QM is incomplete — they never claimed it was incorrect. They accepted all QM predictions and argued that hidden variables must exist to explain the correlations locally.
Q2: (c). The CHSH bound requires both locality (Alice's outcome independent of Bob's setting) and realism (outcomes are definite functions of hidden variables and local settings). Neither alone suffices.
Q3: (b). $E(\theta) = -\cos\theta$ for the singlet state. The negative sign reflects the anticorrelation.
Q4: (b). Tsirelson proved $|S|_{\max} = 2\sqrt{2}$ for any quantum state and any measurement settings.
Q5: (b). Aspect's key innovation was acousto-optic switching of polarizer settings during the flight of the photons, addressing the locality loophole.
Q6: (c). Teleportation requires 1 shared Bell pair consumed + 2 classical bits sent from Alice to Bob.
Q7: (b). The no-cloning theorem is about arbitrary unknown states. Known states or states from an orthogonal set can be copied.
Q8: (c). Bohmian mechanics uses a guidance equation where each particle's velocity depends on the wave function, which in turn depends instantaneously on the positions of all particles.
Q9: (b). If only a fraction of pairs are detected, a LHV model could produce violations by having the hidden variables bias which pairs are detected. Closing this requires detector efficiency above ~82.8%.
Q10: (b). Superdense coding: 1 qubit sent + 1 pre-shared Bell pair = 2 classical bits received.
Q11: FALSE. Bell's theorem proves that local realism is false — at least one of locality or realism must go. Different interpretations make different choices. Bohmian mechanics sacrifices locality; many-worlds and QBism arguably sacrifice aspects of realism.
Q12: TRUE. This is crucial and often overlooked. EPR accepted quantum predictions and used them to argue for hidden variables. If the predictions were wrong, the argument would have no force.
Q13: TRUE. Each Bell state has entanglement entropy $E = S(\rho_A) = -\text{Tr}(\rho_A \log_2 \rho_A) = 1$ because the reduced density matrix is $\rho_A = \frac{1}{2}\hat{I}$ for all four.
Q14: FALSE. All mainstream interpretations (Copenhagen, many-worlds, Bohmian, QBism, consistent histories) make identical experimental predictions. The debate continues because the question is about ontology, not empirical outcomes. (Objective collapse theories like GRW are exceptions — they do make different predictions.)
Q15: Bell's original inequality requires perfect correlations and three measurement settings; it is vulnerable to experimental imperfections. CHSH requires only two settings per party and works with imperfect correlations, making it directly testable in real experiments with finite statistics and detector noise.
Q16: Before receiving Alice's 2-bit classical message, Bob's qubit is in the maximally mixed state $\frac{1}{2}\hat{I}$ — it carries zero information about $|\chi\rangle$. Bob cannot extract any useful information until the classical message arrives, which travels at or below the speed of light. The classical communication step is essential and rate-limiting.
Q17: (1) Locality loophole: measurement settings must be chosen after the particles separate and the choice events must be space-like separated. (2) Detection loophole: detector efficiency must be high enough that the detected sample cannot be biased by hidden variables (~82.8% for CHSH). (3) Freedom-of-choice loophole: the random number generators choosing settings must be truly independent of the hidden variables.
Q18: Classical correlations arise from shared randomness ($\lambda$) and are bounded by the CHSH inequality $|S| \leq 2$. Entangled quantum states produce correlations reaching $|S| = 2\sqrt{2}$, exceeding this bound. Bell's theorem proves that no shared random variable, no matter how complex, can reproduce quantum correlations. The "extra" correlation comes from the non-factorizability of the entangled state.
Q19: The measurement problem asks: what physical process causes a superposition $\sum c_n |a_n\rangle$ to yield a definite outcome $|a_k\rangle$ with probability $|c_k|^2$? The Schrödinger equation is linear and deterministic; it never produces "collapse." The formalism tells us the probabilities but gives no physical mechanism for how or when a definite outcome emerges. This is not philosophy — it is a gap in the dynamical description.
Q20: (a) Outcomes: $(|0\rangle_A, |1\rangle_B)$ with probability 1/2 and $(|1\rangle_A, |0\rangle_B)$ with probability 1/2. (b) $E(\hat{z},\hat{z}) = \frac{1}{2}(+1)(-1) + \frac{1}{2}(-1)(+1) = -1$. Wait — for $|\Psi^+\rangle$: $E = \frac{1}{2}(-1) + \frac{1}{2}(-1) = -1$? Let's recompute. $|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$, so $\langle \sigma_z \otimes \sigma_z \rangle = \frac{1}{2}[(+1)(-1) + (-1)(+1)] = -1$. So $E(\hat{z}, \hat{z}) = -1$. (c) $E(\hat{z}, \hat{x}) = \langle \sigma_z \otimes \sigma_x \rangle$. Using the identity for triplet states: $E = \hat{z} \cdot \hat{x} - 2\langle\Psi^+|P_{\text{singlet}}|\Psi^+\rangle(\hat{z}\cdot\hat{x})$... More directly, compute $(\sigma_z \otimes \sigma_x)|\Psi^+\rangle = \frac{1}{\sqrt{2}}(\sigma_z|0\rangle \otimes \sigma_x|1\rangle + \sigma_z|1\rangle \otimes \sigma_x|0\rangle) = \frac{1}{\sqrt{2}}(|0\rangle|0\rangle - |1\rangle|1\rangle) = |\Phi^-\rangle$. So $\langle\Psi^+|\Phi^-\rangle = 0$. Thus $E(\hat{z},\hat{x}) = 0$. (d) $|\Psi^+\rangle$ is NOT rotationally invariant (it transforms as a triplet state with $m=0$). Only $|\Psi^-\rangle$ is rotationally invariant.
Q21: (a) $\text{Tr}(\rho_W^2) = p^2 + (1-p)^2/4 + 2p(1-p)/4$... more carefully: $\rho_W = p|\Psi^-\rangle\langle\Psi^-| + \frac{1-p}{4}\hat{I}$. The eigenvalues are $\frac{1+3p}{4}$ (singlet, multiplicity 1) and $\frac{1-p}{4}$ (triplet, multiplicity 3). So $\text{Tr}(\rho^2) = \left(\frac{1+3p}{4}\right)^2 + 3\left(\frac{1-p}{4}\right)^2 = \frac{(1+3p)^2 + 3(1-p)^2}{16} = \frac{1+6p+9p^2+3-6p+3p^2}{16} = \frac{4+12p^2}{16} = \frac{1+3p^2}{4}$. For $p=0.8$: $\text{Tr}(\rho^2) = \frac{1+3(0.64)}{4} = \frac{2.92}{4} = 0.73$. (b) Yes: $p = 0.8 > 1/3$, so the state is entangled. (c) Yes: $p = 0.8 > 1/\sqrt{2} \approx 0.707$, so CHSH is violated. (d) $|S|_{\max} = 2\sqrt{2} \times 0.8 = 2.263$.
Q22: (a) $|0\rangle_C \otimes \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)_{AB} = \frac{1}{\sqrt{2}}(|000\rangle + |011\rangle)_{CAB}$. (b) Using the Bell decomposition: $|0\rangle_C|0\rangle_A = \frac{1}{\sqrt{2}}(|\Phi^+\rangle + |\Phi^-\rangle)_{CA}$ and $|0\rangle_C|1\rangle_A = \frac{1}{\sqrt{2}}(|\Psi^+\rangle + |\Psi^-\rangle)_{CA}$. So: $|\Psi\rangle = \frac{1}{2}[|\Phi^+\rangle_{CA}|0\rangle_B + |\Phi^-\rangle_{CA}|0\rangle_B + |\Psi^+\rangle_{CA}|1\rangle_B + |\Psi^-\rangle_{CA}|1\rangle_B]$. Wait — let me redo this more carefully. $|0\rangle_C|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}}|0\rangle_C(|0\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B) = \frac{1}{\sqrt{2}}(|00\rangle_{CA}|0\rangle_B + |01\rangle_{CA}|1\rangle_B)$. Substituting the Bell decomposition: $|00\rangle_{CA} = \frac{1}{\sqrt{2}}(|\Phi^+\rangle + |\Phi^-\rangle)$ and $|01\rangle_{CA} = \frac{1}{\sqrt{2}}(|\Psi^+\rangle + |\Psi^-\rangle)$. So: $= \frac{1}{2}[(|\Phi^+\rangle + |\Phi^-\rangle)|0\rangle_B + (|\Psi^+\rangle + |\Psi^-\rangle)|1\rangle_B] = \frac{1}{2}[|\Phi^+\rangle|0\rangle + |\Phi^-\rangle|0\rangle + |\Psi^+\rangle|1\rangle + |\Psi^-\rangle|1\rangle]$. (c) If Alice gets $|\Psi^-\rangle_{CA}$, Bob has $|1\rangle_B$. (d) Bob applies $i\hat{\sigma}_y$ to $|1\rangle$: $i\sigma_y|1\rangle = i\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = i\begin{pmatrix}-i\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} = |0\rangle$. Confirmed.