Interpretations of Quantum Mechanics · Chapter 03
Which assumptions about reality are actually excluded by quantum correlations?
Starting point: Chapters 1–2; familiarity with density operators and linear algebra.
Anchor example: CHSH correlations and a two-qubit Peres–Mermin square.
Method: Calculate a concrete experiment, identify the assumptions, then compare what different accounts explain.
Watch for: Bell violations do not exclude every hidden-variable theory, and no-signalling is weaker than Bell locality.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Which assumptions about reality are actually excluded by quantum correlations?
Alice chooses x∈{0,1}, Bob chooses y∈{0,1}, and each records ±1. Introduce a candidate complete variable λ. Bell-local factorization says \(p(a,b|x,y,\lambda)=p(a|x,\lambda)p(b|y,\lambda)\). Measurement independence says the distribution μ(λ) does not depend on x or y. Neither assumption should disappear behind the word “realism.”
Let Aₓ(λ) and Bᵧ(λ) be the conditional mean outcomes, each between −1 and 1. For a fixed λ the CHSH expression is \(A_0(B_0+B_1)+A_1(B_0-B_1)\). Its magnitude is at most \(|B_0+B_1|+|B_0-B_1|\le2\). Averaging over a setting-independent μ preserves the bound:
\[|S|=|E_{00}+E_{01}+E_{10}-E_{11}|\le2.\]Notice that we did not assume deterministic outcomes. They appear only as extreme cases of conditional means. The result constrains a combination of probabilistic and causal assumptions. Rejecting factorization is one response; rejecting measurement independence is another, with its own explanatory responsibilities. An inequality alone does not decide which account to adopt.
Worked check 3.1. For deterministic B₀=B₁=−1, what is the fixed-λ CHSH value?
It is −2A₀, with magnitude two; the A₁ contribution vanishes.
Now use \(|\Phi^+\rangle\), Alice's A₀=Z, A₁=X, and Bob's \(B_0=(Z+X)/\sqrt2\), \(B_1=(Z-X)/\sqrt2\). The nonzero correlations XX and ZZ both equal one; XZ and ZX vanish. Substituting gives E₀₀=E₀₁=E₁₀=1/√2 and E₁₁=−1/√2, hence S=2√2.
Does Bob send a message by choosing his setting? No. For a complete local POVM \(\{F_b^y\}\),
\[\sum_b p(a,b|x,y)=\operatorname{Tr}\!\left[\rho(E_a^x\otimes\sum_bF_b^y)\right]=\operatorname{Tr}[\rho(E_a^x\otimes I)].\]Alice's marginal is independent of y. This is operational no-signalling. It does not imply the stronger factorization conditional on λ. Confusing the two makes it impossible to describe theories that reproduce Bell correlations without controllable faster-than-light communication.
The calculation is ideal. Experimental inference also needs appropriate treatment of detection, setting choice, timing, and statistical uncertainty. Here we derive the theoretical benchmark rather than claiming a particular experiment closes every loophole. Later chapters will use this benchmark to ask what a proposed ontology must reproduce.
Worked check 3.2. For white-noise visibility v, ρ=v|Φ+⟩⟨Φ+|+(1−v)I/4, when is CHSH violated with these settings?
All correlations scale by v, so S=2√2v. A violation requires v>1/√2.
Contextuality asks a different question: could every observable have a fixed value independent of which compatible observables accompany it? For sharp measurements consider the array
\[\begin{pmatrix}X\otimes I&I\otimes X&X\otimes X\\I\otimes Z&Z\otimes I&Z\otimes Z\\X\otimes Z&Z\otimes X&Y\otimes Y\end{pmatrix}.\]Each row and column consists of commuting operators. The three row products and first two column products are +I; the final column product is −I. Suppose each entry has a context-independent value ±1 and the value of a commuting product equals the product of values. Multiplying all six required constraints gives +1 on the left, because every entry occurs twice, but −1 on the right. No such assignment exists.
This is a state-independent contradiction for that value model in dimension four. It does not forbid all descriptions containing additional variables: those descriptions can be contextual. It also is not the complete modern definition of contextuality. Generalized noncontextuality concerns operationally equivalent preparations, transformations, and measurements and does not simply assume predetermined values for every effect.
The lesson is methodological. Identify the compatibility relations and the value rule before invoking a no-go theorem. An interpretation earns its place by explaining how it handles the assumptions, not by declaring the theorem irrelevant.
Worked check 3.3. Verify the last column product.
(X⊗X)(Z⊗Z)=−Y⊗Y because XZ=−iY. Multiplying by Y⊗Y gives −I.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
What role does measurement independence play in the average?
Consider a different μ for each setting pair.
A common μ permits combining four correlations into one bounded integrand. Setting-dependent distributions invalidate that step.
Prove |u+v|+|u−v|≤2 for u,v∈[−1,1].
Order their absolute values.
The expression equals 2 max(|u|,|v|), which is at most two.
Must Bell-local responses be deterministic?
Use conditional means.
No. The bound holds for all means in [−1,1], including stochastic response functions.
Compute E₁₁ for the specified Bell experiment.
Use XX=1 and XZ=0.
The expectation of X⊗(Z−X)/√2 is −1/√2.
For v=0.8 calculate S.
Multiply 2√2 by visibility.
S≈2.2627, above two for these ideal measurements.
Why does postselecting on Bob’s outcome not enable signalling?
Who has access to the outcome label?
Conditional ensembles can differ, but Alice must receive Bob’s outcome information to select them. Her unconditioned statistics do not depend on his setting.
Verify the third row product of the square.
Use XZ=−iY and ZX=iY.
(X⊗Z)(Z⊗X)=Y⊗Y, whose square is I, so the row product is +I.
Why is assigning a new value in each context an escape from the contradiction?
Count repeated variables in the product argument.
The same observable no longer supplies the same variable twice. The squared-value cancellation fails; this is precisely contextuality.
Does the square establish a contradiction for a single qubit?
Count the Hilbert-space dimension.
No. The construction uses two qubits, dimension four. Its nine observables cannot be read as a single-qubit value table.
Contrast Bell locality and generalized noncontextuality.
Identify what is held fixed in each assumption.
Bell locality constrains spatial conditional probabilities given λ. Generalized noncontextuality requires identical representations for operationally equivalent procedures. Their assumptions and experimental structures differ.