Past exam of the introduction to quantum information course of the University of Oxford / 2021 / 2

Each local reduced density operator of the Bell state is $I/2$. Alice's unrecorded local measurement leaves Bob's reduced state unchanged: he still sees a fair random bit. Only after learning her outcome through a classical channel can he identify the corresponding conditional state. Alice cannot choose the random outcome to encode a bit, so the correlations obey the no-signalling theorem.

If the shared state is the trusted pure Bell state, measuring both sides in the $Z$ basis gives identical, uniformly random raw bits. A pure joint state of Alice and Bob has no correlations with an external system, so these bits can form a shared secret. In an actual quantum key distribution protocol, the parties must verify the necessary state/device assumptions and authenticate their classical communication; correlation alone, without a security test, does not certify secrecy against an untrusted source.

Expand the Bell state in the standard basis:The transpose is not an adjoint: the formula is complex-linear in both operators. The equality follows since $(A^T{)}_{kj}=A_{jk}$.

The matrices $X,Z$ are real symmetric, $\mathrm{tr}(X^2)=\mathrm{tr}(Z^2)=2$, and $\mathrm{tr}(XZ)=0$. Part b therefore givesSince the local measurement outcomes are $\pm 1$, their product is one exactly on agreement. Thus $E[ab]=2\mathbb{P}(a=b)-1$, and

The first three angle differences have cosine $1/\sqrt{2}$, while the fourth has cosine $-1/\sqrt{2}$. The first three winning conditions require agreement and the fourth requires disagreement. Each consequently succeeds with the same probabilityUniform independent selection of the four setting pairs gives this same average. Equivalently the CHSH inequality expression is $E_{11}+E_{12}+E_{21}-E_{22}=2\sqrt{2}$, attaining the CHSH Tsirelson bound.

For predetermined outcomes, the four products multiply to one:The four desired signs are $(+1,+1,+1,-1)$, whose product is minus one. Thus at least one winning condition must fail for every deterministic assignment. Uniform independent settings give success at most $3/4$ per assignment, and averaging over hidden assignments preserves the bound.

An observed rate near $0.85$, with enough trials to exclude statistical fluctuation, therefore contradicts Eve's predetermined-local-outcome claim, provided the settings are independent of her devices and the devices cannot communicate during a run. It violates the local bound of the CHSH inequality while remaining consistent with the quantum optimum ${\cos }^2(\pi /8)\approx 0.853553$.

The observation is not by itself exact certification of a perfect two-qubit state. Maximal quantum CHSH correlations support CHSH self-testing only with the appropriate quantum/locality assumptions and up to local changes of representation and unused ancillary systems; a near-maximal finite data set requires a quantitative error analysis. An unrestricted communicating pair of black boxes can instead win every run, so the quantum bound is not a bound on arbitrary devices without these assumptions.

No. The independent-setting assumption can fail if Eve controls the random-number generators. Her hidden information can then include the settings that will be selected, and the outcome assignment can depend on that information. For each known setting pair she can prearrange its one tested winning relation, without having to satisfy all four counterfactual relations at once.

For example, arrange equal outputs for the first three setting pairs and opposite outputs for the fourth, conditioning the programmed outputs on the known pair. This can give success one with predetermined records. It does not contradict the $3/4$ proof, which assumes the same hidden-assignment distribution for every independently selected pair. This is the measurement-independence loophole, so the claimed security conclusion is not established by such a compromised test.