Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 66 (Quantum Foundations)

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.

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

Bob uses qudit teleportation to teleport his carrier $B_1$ to Clare through the pair $B_{2}C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a generalized Bell basis measurement on $B_1{B}_2$. Contraction of the two initial pairs givesThe probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^r{X}_C^{-s}$. Alice and Clare then share $|{\psi }_{00}{⟩}_{AC}$ with certainty. This is entanglement swapping; neither an additional entangled pair nor a quantum transmission during the protocol is required.

The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible generalized Bell states leaves $AC$ maximally mixed. Thus the conditional entanglement swapping does not supply faster-than-light communication.

The EPR criterion of reality proposes that a quantity has an element of physical reality when its value can be predicted with certainty without disturbing the system. It is a sufficient criterion, not a claim that every uncertain prediction is unreal. In the original locality argument, measuring one member of a separated correlated pair allows such a prediction for the other, whose physical condition is assumed unaffected by the remote choice.

Combining the EPR criterion of reality with perfect predictive correlations and locality motivates pre-existing values for the locally selectable observables. The general local hidden-variable theory additionally assumes a complete shared variable $\lambda$ and conditional factorizationThe second condition is measurement independence. The reality criterion alone does not prove this factorization for arbitrary imperfect correlations: it motivates the local completion whose consequences are being tested.

A deterministic local hidden-variable model may be used without loss of generality by putting local random seeds into $\lambda$. For each $\lambda$, write its setting responses as $A,A^{\prime },B,B^{\prime }\in \{-1,1\}$. Integrating with the same setting-independent distribution and using the triangle inequality givesExactly one of $B-B^{\prime }$ and $B+B^{\prime }$ vanishes and the other has magnitude two. Hence the requested CHSH bound is two. No assumption about the spacelike quantum state enters this local-model derivation.

The equal mixture of $|{\Phi }^{+}⟩$ and $|{\Phi }^{-}⟩$ cancels their off-diagonal coherence, leaving the separable quantum stateA trivial LHV model chooses a shared random bit $j$, uniformly zero or one, and prepares both local qubits in $|j⟩$. Each party uses the local Born rule for its own chosen spin direction, with independent local random seeds if a deterministic description is desired. This reproduces the entire joint measurement distribution, not merely the $z$-axis correlation.

Since the state has an explicit local hidden-variable theory for these local measurements, it cannot violate the CHSH inequality, by the derivation in part (b). Classical correlation of the shared preparation bit is sufficient; the cancellation of coherence removes the entanglement.

This is the dephased Bell-state mixture with coherence parameter $2x$. The Pauli correlation tensor isso spin measurements along unit vectors have $E(\mathbf{a},\mathbf{b})={\mathbf{a}}^{T}T\mathbf{b}$. ChooseThen $E(a,b)-E(a,b^{\prime })=2/\sqrt{1+4x^2}$ and $E(a^{\prime },b)+E(a^{\prime },b^{\prime })=8x^2/\sqrt{1+4x^2}$. The CHSH inequality expression is thereforeThese axes also attain the CHSH optimum from the correlation tensor, since the two largest eigenvalues of $T^{T}T$ are $1$ and $4x^2$. At $x=1/2$ the value is $2\sqrt{2}$; at $x=0$ the explicit LHV model in part (c) applies. Thus exactly $0<x\le 1/2$ has correlations incompatible with a local hidden-variable description.

Arbitrarily small positive coherence suffices in this particular one-parameter family, with optimally chosen axes; the violation above two tends to zero quadratically as $x\to 0$. This does not imply that every entangled mixed state violates a CHSH inequality.

All time-dependent Hamiltonian operators are multiples of the same fixed operator, so they commute and their time-ordered exponential reduces to an ordinary exponential. Let $P_1=(I-Z_S)/2$. With $\hslash =1$ and the given unit coupling integral,Since $X_D^2=I$,ThusIt is exactly the CNOT gate with $S$ controlling $D$, with no residual relative phase. Acting on $(c_0|0⟩+c_1|1⟩{)}_S|0{⟩}_D$ correlates the system basis label with the apparatus label, producing $c_0|00⟩+c_1|11⟩$ as required.

In the Hadamard basis, $X_D$ is diagonal with projectors $Q_{\pm }=(I\pm X_D)/2$. Rewrite the same CNOT gate in the original tensor-factor order $S\otimes D$ asExpanding $P_0\otimes I+P_1\otimes X_D$ verifies this identity directly. Now $Z_S|+⟩=|-⟩$ and $Z_S|-⟩=|+⟩$, so $Z_S$ is ${\tilde{X}}_S$ in the complementary basis. If $D$ is plus, the identity acts on $S$; if $D$ is minus, its sign-basis label flips. Hence $D$ is the control and $S$ the target in this basis.

Equivalently,This CNOT control reversal in the Hadamard basis concerns the matrix in changed local bases, rather than a physical exchange of the two carriers.

The Hadamard basis changes the Pauli matrices by $Z_S={\tilde{X}}_S$ and $X_D={\tilde{Z}}_D$. Thus, keeping tensor factors ordered as $S,D$, the Hamiltonian operator isIf $D$ has new-basis label plus, the second factor vanishes. If $D$ has label minus, the integrated generator on $S$ is $\pi (I-{\tilde{X}}_S)/2$, whose exponential is ${\tilde{X}}_S$ by part (a). The generator therefore has the same controlled form with $D$ as control in the complementary bases.

To use $S$ as the apparatus, prepare it in the ready state $|+{⟩}_S$ and later read it in the Hadamard basis. For an arbitrary input $\alpha |+{⟩}_D+\beta |-{⟩}_D$, part (b) givesThe two apparatus states are orthogonal, so its readout measures the corresponding projectors on $D$ and preserves each input eigenstate. The observable iswith outcomes $+1$ and $-1$. This is the measurement direction of a CNOT interaction in the complementary basis.

The ready state and readout basis matter. Keeping $S$ in the original $|0{⟩}_S$ would make the old-basis controlled gate act as the identity on $D$, producing no record. Thus the same interaction can serve either measurement direction, with different apparatus preparation and pointer readout; the Hamiltonian operator alone does not assign an intrinsic system/apparatus role.