Interpretations of Quantum Mechanics · Chapter 09
Can distinct pure quantum states represent overlapping ignorance about the same underlying reality?
Starting point: Chapters 1–8; familiarity with density operators and linear algebra.
Anchor example: A four-outcome measurement excluding four nonorthogonal two-qubit preparations.
Method: Calculate a concrete experiment, identify the assumptions, then compare what different accounts explain.
Watch for: ψ-ontic does not mean complete, a material wave in ordinary space, or a proof that every informational interpretation is impossible.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Can distinct pure quantum states represent overlapping ignorance about the same underlying reality?
To discuss whether a state describes knowledge, we first need a model in which that sentence has a precise meaning. Let λ denote an underlying physical state. Preparing a pure state ψ produces a distribution μψ(λ). A measurement M has response probabilities ξ(k|M,λ), normalized over k. Reproducing quantum predictions requires
\[\int d\lambda\,\mu_\psi(\lambda)\xi(k|M,\lambda)=\langle\psi|E_k|\psi\rangle.\]Within this framework, a model is ψ-ontic if distributions for distinct pure states have disjoint support up to measure-zero sets. It is ψ-epistemic if at least two distinct pure states have overlap of positive measure. In the first case, λ determines which pure-state preparation is compatible with it, but λ may contain more information than ψ.
These definitions do not classify every use of “information” in philosophy. They presume an underlying state space, preparation distributions, and response functions. They also concern pure-state distinctions; different decompositions of the same mixed density operator pose additional questions.
Orthogonal states already need disjoint supports in such a model: a measurement that distinguishes them perfectly cannot assign both certainty zero and certainty one at a shared λ of positive weight. Nonorthogonal states are the interesting test because no single-copy measurement perfectly distinguishes them.
Worked check 9.1. If ψ determines only part of λ, can the model still be ψ-ontic?
Yes. Disjoint supports can make ψ recoverable from λ even when λ contains additional variables. Onticity does not imply completeness.
Use the two single-qubit preparations |0⟩ and |+⟩. Suppose their distributions overlap on a region of positive weight. Preparation independence assumes that independently prepared systems have a product ontic distribution \(\mu_{ab}(\lambda_1,\lambda_2)=\mu_a(\lambda_1)\mu_b(\lambda_2)\), with the appropriate composite state space. The shared region then occurs with positive weight for all four preparations 00, 0+, +0, and ++.
Now define four orthonormal measurement vectors, with |±⟩=(|0⟩±|1⟩)/√2:
\[\begin{aligned}|\xi_{00}\rangle&=(|01\rangle+|10\rangle)/\sqrt2,\\|\xi_{0+}\rangle&=(|0-\rangle+|1+\rangle)/\sqrt2,\\|\xi_{+0}\rangle&=(|+1\rangle+|-0\rangle)/\sqrt2,\\|\xi_{++}\rangle&=(|+-\rangle+|-+\rangle)/\sqrt2.\end{aligned}\]The outcome labelled ab is impossible for preparation ab: \(\langle\xi_{ab}|a,b\rangle=0\). This is exclusion, not discrimination. One outcome rules out its associated preparation, but generally does not identify which of the other three was prepared.
It is worth expanding these vectors rather than trusting the labels. In particular, the final vector is (|00⟩−|11⟩)/√2. Its overlap with |++⟩ vanishes by cancellation. The four vectors form a projective measurement on the four-dimensional joint space.
Worked check 9.2. Verify that ξ₀₊ is orthogonal to |0+⟩.
Its overlap is (⟨−|+⟩+0)/√2=0. Its normalization follows from the orthogonal first-qubit labels in its two terms.
Each impossible quantum outcome has zero average response probability. Because the integrand μξ is nonnegative, the corresponding response must vanish almost everywhere on that preparation's support. On the joint overlap region, all four preparations are possible, so every one of the four response probabilities would have to vanish. That contradicts their required sum of one.
This is the two-copy Pusey–Barrett–Rudolph argument for this pair of states. The general result uses suitable numbers of copies for other distinct states. It rules out the relevant overlap under the ontological-model assumptions and preparation independence. Relaxing preparation independence changes the argument and must be evaluated as a concrete alternative assumption.
Notice what has not been shown. We have not proved the wavefunction is a material field in three-dimensional space, that it is a complete physical description, or that a particular interpretation is uniquely correct. We have constrained one way of treating pure states as overlapping distributions over a deeper state.
Experimental noise also matters: exact zeros become small nonzero probabilities, so a robust inference requires quantitative inequalities rather than repeating the zero-probability proof unchanged. This chapter gives a finite exact example and an assumption audit, not a claim to settle all meanings of quantum reality.
Worked check 9.3. Why does nonnegativity matter in the support argument?
A nonnegative function with zero integral vanishes almost everywhere with respect to the preparation measure. Without nonnegativity, cancellations could produce zero while individual responses remained nonzero.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Explain the difference between ψ-ontic and ψ-complete.
Ask whether λ can have additional components.
ψ-ontic means distinct pure states have disjoint underlying supports. ψ-complete additionally treats the wavefunction as a complete state description; extra variables can make an ontic model incomplete in that sense.
Why must perfectly distinguishable states have disjoint supports?
Use one outcome with probabilities one and zero.
On shared positive support, the response would have to be both one and zero almost everywhere to reproduce the two deterministic predictions, an impossibility.
Is any personal-probability interpretation automatically a PBR model?
Check whether it posits λ and μψ.
No. The theorem applies to the specified ontological-model framework. A philosophy of probability need not adopt that underlying representation.
Expand ξ₀₊ in the computational basis.
Expand |−⟩ and |+⟩.
It is (|00⟩−|01⟩+|10⟩+|11⟩)/2.
Expand ξ₊₀ and check its overlap with |+0⟩.
Expand both first-qubit states.
It is (|00⟩+|01⟩−|10⟩+|11⟩)/2. The overlap with (|00⟩+|10⟩)/√2 is zero.
Show ξ₀₊ and ξ₊₀ are orthogonal.
Take the dot product of their four real coefficients.
The numerator is 1−1−1+1=0, divided by four.
If the single-system overlap weight is ω, what is the product-overlap weight under preparation independence?
Factor the two integrals for the shared subdistribution.
It is ω², positive whenever ω>0. This is why a shared two-system region cannot be ignored.
Identify the exact contradiction on the joint overlap.
Sum all response probabilities.
Each outcome has zero response there because some possible preparation forbids it. Their sum becomes zero, contradicting normalization to one.
Does the exclusion measurement identify every preparation perfectly?
Distinguish exclusion from discrimination.
No. Outcome ab rules out preparation ab, but it can be compatible with multiple other preparations. Nonorthogonal states have not become perfectly distinguishable.
What must change in the proof for noisy observations?
Replace exact zero statements with bounds.
Small probabilities no longer force response functions to vanish. One needs a quantitative bound on overlap from the observed forbidden-outcome probabilities and experimental assumptions.