Past exam of the quantum information course of the University of Cambridge / 2013 / Paper 1 / 1

Outcome determinism means that a complete hidden state $\lambda$, together with the chosen measurement settings, fixes each outcome. The observed probabilities can still arise from an ensemble of unknown $\lambda$. This differs from deterministic unitary evolution of a wavefunction, which does not by itself specify a unique measurement outcome. Quantum nonlocality means failure of a Bell-local account of the correlations, under assumptions including measurement independence; it is stronger than the existence of entanglement and does not imply controllable faster-than-light signaling.

Use the first ket for the Positron and the second for the electron. Before either final beam splitter, the first pair of splitters and annihilation produceThe photon state $|\gamma ⟩$ is orthogonal to the surviving charged-particle sector. Its probability is $1/4$, while the remaining sector has squared norm $3/4$. The states below retain that photon amplitude and are normalized over all trials, without conditioning on nonannihilation. For compactness write $|cc⟩=|c^{+}{c}^{-}⟩$, and similarly for $|cd⟩,|dc⟩,|dd⟩$.

With both final splitters removed, the path labels map directly into detector labels:In particular $P_{RR}(C^{+},C^{-})=0$, because joint occupation of the two intersecting arms has already annihilated. Each of the other three charged-particle coincidences has unconditional probability $1/4$.

Apply the final splitter only to the first ket. The $|dd⟩$ amplitudes from $i|uv⟩$ and $|vv⟩$ cancel, givingThus detection at $D^{+}$ occurs only in the coincidence $D^{+},C^{-}$, with probability $1/8$. The other two nonzero coincidence probabilities are $P(C^{+},C^{-})=1/8$ and $P(C^{+},D^{-})=1/2$. The cancellation is quantum interference between indistinguishable surviving alternatives.

Interchanging the two wings givesNow detection at $D^{-}$ occurs only together with $C^{+}$, with probability $1/8$. The remaining nonzero coincidence probabilities are $P(C^{+},C^{-})=1/8$ and $P(D^{+},C^{-})=1/2$.

Applying both final splitters givesThe charged-particle coincidence probabilities are $9/16,1/16,1/16,1/16$ in the displayed order, and the annihilation probability remains $1/4$. In particularConditioning on surviving pairs would change this number to $(1/16)/(3/4)=1/12$; it would not change the amplitudes of the unconditioned all-trials state above.

In a deterministic local hidden-variable model, write $C^{+}(R,\lambda )$ or $D^{+}(I,\lambda )$ without an argument for the remote splitter setting. That omission is the parameter independence assumption: the local response for a fixed complete hidden state does not change when the other wing changes its apparatus. Measurement independence also requires the hidden-state ensemble to be independent of the freely selected settings, allowing the same $\lambda$ assignments to be compared across configurations.

The zero probabilities derived above then constrain those assignments, apart from sets of hidden states of zero measure:

For a trial with $D^{+}(I,\lambda )D^{-}(I,\lambda )=1$, the last two implications force $C^{+}(R,\lambda )=C^{-}(R,\lambda )=1$, contradicting the first condition. Thus these deterministic local assignments predict zero probability for the double-$D$ event, whereas quantum theory predicts $1/16$. This is Hardy's paradox, an inequality-free ideal contradiction. With imperfect apparatus the zero-event conditions become small nonzero rates and require a statistical inequality rather than declaring a contradiction from one imperfect detection.^[ref]

The logical contradiction combines counterfactual hidden-variable assignments; it does not assign simultaneously observed path properties to a quantum state. Nor does it allow a message to be sent by changing a distant splitter: the observable local marginal obeys quantum no-signalling. For example, fixing the plus-side splitter to $I$ gives $P(D^{+})=1/8$ for either minus-side setting in the results above.