Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 61 (Quantum Information Physics)

Choose a real unitary matrix and undo it at the second stage:After the first rotation the photon qubit is $\sqrt{q}|0⟩+\sqrt{p}|1⟩$. A dud leaves this quantum superposition unchanged, so the inverse rotation restores $|0⟩$. Its final projective measurement can never return $1$.

For a live bomb the absorbing component explodes with probability $p$. If the photon survives, the normalized state is $|0⟩$, and the second rotation makes it $\sqrt{q}|0⟩-\sqrt{p}|1⟩$. The unconditional probabilities of the three possible live-bomb outcomes are consequentlyThey sum to one. The nonexplosive result $1$ certifies a live bomb because its dud probability is exactly zero. The inconclusive result $0$ resets the qubit to its original state, allowing another identical trial.

The repeated weak-rotation bomb-test efficiency follows by summing the geometric series over the first successful trial:This is conditional on testing a live bomb, not an average over an unspecified prior fraction of duds. To make the finite-run success probability exceed $1/2-\epsilon$, choose positive $p$ so that $p/[2(2-p)]<\epsilon /2$, then choose $N$ so that $q^{2N}<\epsilon /2$. The first condition controls the limiting deficit and the second controls the remaining tail. Exactly $p=0$ yields no explosion but also no information, so the limit must be approached with nonzero weak rotations.

The half-efficiency limit is consistent with the structure of this Elitzur-Vaidman bomb tester. Suppose generally $U|0⟩=a|0⟩+b|1⟩$. A final outcome that certifies a live bomb must be orthogonal to the dud output $U^{\prime }U|0⟩$. Pulling this outcome backwards through $U^{\prime }$ gives the normalized vector orthogonal to $(a,b)$. Its squared overlap with $|0⟩$ is $|b{|}^2$. Thus the safe-certification probability is $|a{|}^2|b{|}^2$, whereas the explosion probability is $|b{|}^2$. The former never exceeds the latter. Summing these mutually exclusive terminal events over repeated, possibly adaptive trials gives $P_{\mathrm{safe}}\le P_{\mathrm{explosion}}$ and hence $P_{\mathrm{safe}}\le 1/2$. This concerns the specified rotate–interact–rotate–measure protocol.

The Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

The resource here is one copy of a two-qubit Werner state, with the physical range $1/2<F\le 1$. The impossibility must include successful conditional branches of local operations and classical communication; average monotonicity alone would not exclude probabilistic enhancement.

For a normalized two-qubit pure state $|\chi ⟩={\sum }_{i,j}{M}_{ij}|ij⟩$, define its concurrence as $C(\chi )=2|\det M|$. Its mixed-state concurrence of two qubits is the convex roof extensionCombining near-optimal decompositions proves convexity. A separable quantum state has zero concurrence, because it has a decomposition into product vectors with zero coefficient determinant.

First establish the concurrence of a two-qubit Werner state. A pure state with squared Schmidt coefficients $\lambda ,1-\lambda$ has $C=2\sqrt{\lambda (1-\lambda )}$. In this Schmidt basis a maximally entangled state has coefficient matrix $V/\sqrt{2}$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$, because $|V_{ii}|\le 1$. Thus the singlet overlap satisfiesAveraging any decomposition of $W_F$ gives $C(W_F)\ge 2F-1$.

For the reverse bound use the orthonormal triplet vectorsEach is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C({\chi }_{j,\pm })=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence

Every fully specified record of an LOCC protocol gives a product Kraus operator $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output asFor invertible $A,B$, the concurrence scaling under invertible local filters isHere is a derivation of that scaling. Each pure coefficient matrix transforms to $AM{B}^{\mathsf{T}}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized concurrence is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_j{p}_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of ${\rho }^{\prime }$, so taking both infima proves equality. If either local filter has rank one, the output is a separable quantum state and the desired bound is immediate.

It remains to show that the filter normalization cannot make this ratio exceed one. Use the Pauli matrices representationWrite the positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, so $a_0\ge |a|$, $b_0\ge |b|$. Taking traces givesFor completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$, and both sides are nonnegative. This proves the single-copy Werner filtering normalization bound, and hence

Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F^{\prime }}$ would therefore obey $2F^{\prime }-1\le 2F-1$, contradicting $F^{\prime }>F$. Thus even heralded single-copy LOCC cannot produce the specified improved Werner state. This proves that single-copy Werner filtering cannot increase concurrence. It does not forbid collective entanglement distillation from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.

Represent any universally quantum device, including its auxiliaries and arbitrary final readout, by a positive operator-valued measure $\{E_i\}$ on the input. Perfect identification would requireSince a positive operator has a positive square root, zero expectation implies $E_i|{\varphi }_j⟩=0$ for $j\ne i$. Also $I-E_i\ge 0$ and its expectation in ${\varphi }_i$ is zero, so $E_i|{\varphi }_i⟩=|{\varphi }_i⟩$. ConsequentlyThis proves that perfect discrimination of pure states requires orthogonality. Any nonzero overlap contradicts perfect identification, independently of whether the device disturbs its input.

Equivalently, a unitary operator coupling the input to a readout apparatus must preserve inner products. Distinct perfectly readable records have zero inner product, so their corresponding input vectors must already be orthogonal. Nonorthogonal candidate states cannot be identified with certainty in one use. Probabilistic unambiguous quantum state discrimination can have an inconclusive outcome, which is excluded by the guaranteed-identification requirement.

Share the Bell state $|{\Phi }_{+}⟩$ between Alice and Bob, and allow their actions to be spacelike separated. The projective measurement postulate gives the remote preparation in conjugate bases identityIf Alice measures in the computational basis, Bob is prepared in $|0⟩$ or $|1⟩$, each with probability one half. If she instead measures in the conjugate basis $U^{\ast }|0⟩,U^{\ast }|1⟩$, Bob is prepared in $U|0⟩$ or $U|1⟩$, again equally likely. The conjugation is essential for a general complex unitary operator.

Interpreting distinct states as distinct physical rays, the two candidate pairs are disjoint. A hypothetical perfect identifier would tell Bob which member he received and therefore which basis Alice chose. Alice could transmit a bit by that basis choice, without sending her outcome or any other message. Bob could read it while spacelike separated from Alice, violating the no-superluminal-signalling requirement of special relativity. This establishes that perfect two-basis discrimination would allow superluminal signalling and rules out the device using the requested physical assumptions.

Ordinary measurement in quantum measurements do not signal: the two equally weighted ensembles have identical reduced density matrixThe contradiction arises from the hypothesized ability to label the ensemble's individual pure members. Global-phase variants of a vector do not count as distinct physical pure states; treating them as distinguishable labels would already introduce an unphysical extra readout.

A guaranteed exact overlap device would also distinguish two remotely prepared ensembles and permit a forbidden signal. Feed it a known reference $|{\psi }_1⟩=|0⟩$ and Bob's unknown state $|{\psi }_2⟩$, obtained from Alice's projective measurement on a shared Bell state. A device promised to work for arbitrary unknown inputs must work on this fixed-reference subset.

Alice chooses either the computational basis or the diagonal basis $|+⟩,|-⟩$. By the same remote preparation in conjugate bases, Bob receives a uniformly random member of the chosen basis. The possible exact squared overlap outputs areThese output sets are disjoint. Bob would infer Alice's basis choice from one run, without her communicating the random measurement result. Spacelike operation would contradict special relativity. Thus exact pure-state overlap readout would allow superluminal signalling, and no guaranteed exact overlap device is compatible with the stated assumptions.

This argument rules out even the phase-invariant magnitude squared. The raw complex inner product has an additional issue: replacing either pure state representative by a global phase changes that complex number without changing the physical input. The no-signalling contradiction therefore remains valid even if some phase convention is supplied. Statistical estimation from many identically prepared copies is not the promised exact single-pair readout.