Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 66 / 1

Put $|0⟩=|{\uparrow }_z⟩$ and $|1⟩=|{\downarrow }_z⟩$. The total-spin sector with spin zero consists of the spin singlet state $|{\Psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$, not the whole subspace with zero $z$ component. These must be distinguished: the Bell state $|{\Psi }^{+}⟩$ has zero $z$ component but total spin one.

There are two meanings of verification to separate here. A one-pair entanglement-assisted statistical singlet verification is possible: choose a shared random axis from $x,y,z$, measure that axis's two-spin parity by the meter circuit below, and accept anticorrelation. An $x$ or $y$ test uses local basis changes before and after the computational-axis circuit. The singlet always passes and is unchanged. Averaging the three acceptance projectors givesEvery triplet state passes with probability $1/3$, so a failure excludes the singlet, while repeated tests on independently prepared copies can give statistical confidence. A single pass does not certify the singlet with certainty.

Exact single-shot verification cannot be implemented with just one shared Bell pair. Here exact verification means a yes outcome with probability one on the spin singlet state, zero on its orthogonal complement, and preservation of the singlet on yes. If that stronger meaning is intended, the question needs an extra resource. The Bell-pair cost of exact nondemolition singlet verification gives a short proof. For a fully resolved tuple $\mu ,\nu$ of local measurement records, contraction of the shared $|{\Phi }^{+}⟩$ gives a system Kraus operatorLocal ancillas and locally adaptive operations are included in these operators. Refining any unobserved local environment gives the same form, so the operator Schmidt rank is at most two. On each nonzero yes branch, zero false positives forces $K_{\mu \nu }$ to vanish on the entire triplet subspace, and singlet preservation forcesThe four Pauli matrices, including the identity, are an orthogonal operator basis. Thus $P_s$ has operator Schmidt rank four, contradicting the rank-two bound. At least one yes branch is nonzero because the singlet must pass with certainty. Shared classical randomness cannot evade this branchwise argument.

With two shared Bell pairs, an explicit corrected protocol is available. On the first meter pair, apply local system-to-meter CNOT gates and measure both meters in the computational basis. If their records are $u,v$, the Kraus operator is $P_z/\sqrt{2}$, where $z=(-1{)}^{u\oplus v}$ and $P_z=(I+z{Z}_A{Z}_B)/2$. On the second pair, measure $X_A{X}_B$ by the same circuit conjugated with local Hadamard gates on the system. The two parities commute, and their joint projectors are the four Bell-state nondemolition measurement projectors. The singlet is exactly the outcome $(z,x)=(-1,-1)$ and remains unchanged. Local interactions and readouts fit within the stated interval; the combined verdict becomes available only after the records are compared by classical communication.

This corrected exact protocol resolves the triplet into three Bell states. It preserves the singlet and every Bell state, but generally destroys triplet superpositions. A binary Lüders rule measurement preserving all triplet coherence is a different operation: it violates the singlet-triplet measurement causality obstruction. The one-pair protocol used below measures a single parity and supplies the statistical test above; it does not give exact single-shot singlet verification.

Use the nonnegative representative of reduction modulo four. The four product eigenstates and eigenvalues arebecause $-2\equiv 2(\mathrm{mod}4)$. Consequently the observable is $I+Z_A{Z}_B$. It has the even sector $\mathrm{span}\{|00⟩,|11⟩\}$ with value two and the odd sector $\mathrm{span}\{|01⟩,|10⟩\}$ with value zero.

The entanglement-assisted nondemolition parity measurement needs just the single shared $|{\Phi }^{+}⟩$ provided in the question. Apply ${\mathrm{CNOT}}_{A\to d_1}$ and ${\mathrm{CNOT}}_{B\to d_2}$. For a computational-basis input $|ab⟩$, the meter becomesMeasuring the meters gives $u\oplus v=a\oplus b$. For an arbitrary coherent system input, its conditional Kraus operator isIt preserves every superposition within the measured sector, so this realizes the Lüders rule for the two degenerate eigenvalues. The answer is two for equal meter bits, zero for unequal meter bits. Each individual meter bit is uniform; only later comparison reveals the parity, respecting quantum no-signalling. This construction is independent of the defective one-pair singlet-verification request in part (a).

Use one shared $|{\Phi }^{+}⟩$ pair to measure $Z_A{Z}_B$, and the other to measure $X_A{X}_B$. The latter entanglement-assisted nondemolition parity measurement is obtained by applying Hadamard gates to both system qubits before and after the $Z$-parity circuit. Since $ZX=-XZ$ at each site, the two minus signs cancel, giving $[Z_A{Z}_B,X_A{X}_B]=0$.

The Bell states have joint eigenvaluesand are therefore distinguished uniquely by the two meter parities. For records $(u,v)$ and $(r,s)$, the combined Kraus operator isEach $P_{z,x}$ is the rank-one projector onto the corresponding Bell state. This is a Bell-state nondemolition measurement: an input Bell state remains exactly that state for every possible local record. The local circuits can be scheduled without communication; the identity of the Bell state is known only when the classical records are brought together. Two parity bits distinguish all four Bell states without disturbing them.

Interpret the requested measurement as distinguishing all four stated orthogonal eigenstates, hence as their rank-one projective measurement. If they have indistinguishable degenerate eigenvalues, this assumption need not hold; the identity observable, for example, cannot reveal the basis and gives no contradiction.

Let Bob initially prepare $|0{⟩}_B$. Alice encodes a bit by preparing $|0{⟩}_A$ or $|1{⟩}_A$, without changing Bob's initial reduced density matrix. If Alice prepares zero, the global input is the first eigenstate, so nondemolition leaves Bob in $|0⟩$ with certainty. If Alice prepares one, the rank-one Lüders rule, after discarding the outcome, dephases Bob in the rotated basisHis output is ${\rho }_B={\cos }^2\theta |u⟩⟨u|+{\sin }^2\theta |v⟩⟨v|$. A computational-basis measurement then givesIt is positive for every $0<\theta \le \pi /4$. Repeating the experiment would transmit Alice's choice across a spacelike interval, violating quantum no-signalling. The controlled-basis measurement causality obstruction therefore excludes the entire nonzero interval, including $\pi /4$.

At $\theta =0$, the basis is the computational product basis, up to an irrelevant sign on the fourth vector. Alice and Bob each measure $Z$ locally, then compare their results later. Each product eigenstate is preserved. Thus being a product basis is insufficient for an instantaneous quantum nondemolition measurement: a remote party's choice of which local basis is measured can still cause signalling.