= Solution
In an <ontological model of a quantum system>, a preparation of $|\psi\rangle$ gives a <probability distribution> $\mu_\psi$ over a physical state $\lambda$. A fixed <quantum measurement> has response probabilities $\xi_j(\lambda)\geq0$ with $\sum_j\xi_j=1$, reproducing the <Born rule> after averaging over $\mu_\psi$. The <PBR theorem> says that, assuming <preparation independence> and the quantum predictions, distributions for distinct <pure states> cannot overlap with positive probability. Thus the <quantum state> is determined by the physical state in the <psi-ontic model> sense.
<Preparation independence> says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded <psi-epistemic model> hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.
For the two given <qubit> preparations, consider the following <PBR exclusion measurement for zero and plus>, written in the ordered basis $00,01,10,11$:
$$
\begin{aligned}
|\xi_{00}\rangle&=(|01\rangle+|10\rangle)/\sqrt2,\\
|\xi_{0+}\rangle&=(|00\rangle-|01\rangle+|10\rangle+|11\rangle)/2,\\
|\xi_{+0}\rangle&=(|00\rangle+|01\rangle-|10\rangle+|11\rangle)/2,\\
|\xi_{++}\rangle&=(|00\rangle-|11\rangle)/\sqrt2.
\end{aligned}
$$
Direct <inner products> show that these four vectors are an <orthonormal basis>. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated <projective measurement> satisfies
$$
\boxed{\Pr(xy\mid|x\rangle|y\rangle)=|\langle\xi_{xy}|x,y\rangle|^2=0,\qquad x,y\in\{0,+\}.}
$$
This is an example of <antidistinguishable quantum states>: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.
To prove the contradiction without requiring deterministic <quantum measurement> responses, choose a common dominating measure for $\mu_0,\mu_+$ and write their densities $m_0,m_+$. If they overlap, $\nu(\lambda)=\min(m_0,m_+)$ has mass $\varepsilon>0$. By <preparation independence>, every one of the four product densities dominates $\nu(\lambda_1)\nu(\lambda_2)$. Its total mass is $\varepsilon^2$. The zero <Born rule> probability for outcome $xy$ implies that its nonnegative response $\xi_{xy}(\lambda_1,\lambda_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore \b[$\mu_0$ and $\mu_+$ are mutually singular].
For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|\Psi_i\rangle=|\psi_i\rangle^{\otimes n}$. The <tensor-power reduction of PBR overlap> gives
$$
\langle\Psi_1|\Psi_2\rangle=\langle\psi_1|\psi_2\rangle^n=1/\sqrt2.
$$
Each block therefore has an effective two-dimensional <Hilbert space>. Explicitly,
$$
|e_0\rangle=|\Psi_1\rangle,\qquad
|e_1\rangle=\sqrt2|\Psi_2\rangle-|\Psi_1\rangle
$$
are orthonormal and $|\Psi_2\rangle=(|e_0\rangle+|e_1\rangle)/\sqrt2$. Embed the four-vector exclusion basis above into the <tensor product> of these two block spaces. To obtain a complete <quantum measurement> on all $2n$ <qubits>, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.
If the single-copy preparation distributions overlap with common mass $\varepsilon>0$, their $2n$-fold products all dominate the common measure $\nu^{\otimes2n}$ of mass $\varepsilon^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. Thus
$$
\boxed{\mu_{\psi_1}\perp\mu_{\psi_2}.}
$$
The argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.
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