Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 62 (Quantum Foundations)

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.

Let $U_1=e^{-iH(t-t_1)/\hslash }$ and $U_2=e^{-iH(t_2-t)/\hslash }$ describe unitary time evolution. For an ideal projective measurement with the Lüders rule, its unconditioned outcome probability is $p_j=\Vert P_j{U}_1|\psi ⟩{\Vert }^2$. After that outcome, the normalized state is $P_j{U}_1|\psi ⟩/\sqrt{p_j}$. The Born rule probability of successful postselection is then $|⟨{\psi }^{\prime }|U_2{P}_j{U}_1|\psi ⟩{|}^2/p_j$. Multiplying gives the joint probabilityConditional probability therefore gives the Aharonov-Bergmann-Lebowitz rule:The denominator must be positive; otherwise the selected subensemble does not occur. Define the forward-evolved ket $|a⟩=U_1|\psi ⟩$ and backward-evolved ket $|b⟩=U_2^{†}|{\psi }^{\prime }⟩$. The numerator becomes $|⟨b|P_j|a⟩{|}^2$, which is unchanged by interchanging $a,b$. Equivalently, with ${\rho }_a=|a⟩⟨a|$ and ${\rho }_b=|b⟩⟨b|$,This expresses the boundary-state symmetry explicitly. It follows from the ordinary time-asymmetric preparation, Born rule, and state update; it does not posit an additional backward dynamical collapse.

Restore the post-selected vector omitted entirely from the TeX aid by reading the original PDF. Write $D=N^2-N+1$ andIts norm is one because $N+(N-1{)}^2=D$. For the uniform prestate and $H=0$, put $c=1/\sqrt{(N+1)D}$. The individual transition amplitudes areIn experiment $E_i$, the complement amplitude is $c-c=0$. Thus for every $i\le N$,The successful postselection rate in this experiment is $c^2=1/((N+1)D)$.

In the fully resolved experiment $E_0$, the ABL rule squares the individual amplitudes before summing. Their squared sum is $D{c}^2=1/(N+1)$, givingThese probabilities sum to one. For $N=1$ they reduce to a single certain outcome in either quantum measurement. For $N\ge 2$, the N-box pre- and post-selection paradox is that each separate binary question can be answered affirmatively with certainty, although the fully resolved quantum measurement cannot give all those outcomes at once.

There is no inconsistency. In $E_i$, the unresolved complement preserves coherent cancellation between its basis contributions under the Lüders rule. In $E_0$, those alternatives are resolved, so their squared amplitudes add instead. The different projective measurements disturb the state differently and have different postselection success rates. Merely merging the recorded $E_0$ outcomes afterwards does not reproduce $E_i$. This is the context dependence of pre- and post-selected measurements; certainties in mutually alternative experiments do not describe simultaneous measurement-independent properties.

The GRW model takes a normalized wave function in $L^2({\mathbb{R}}^{3N})$. Between jumps it obeys the Schrodinger equation, $i\hslash {\partial }_t\Psi =H\Psi$, with the usual many-particle Hamiltonian operator, for example $H={\sum }_i[-{\hslash }^2{\nabla }_i^2/(2m_i)]+V({\mathbf{x}}_1,\dots ,{\mathbf{x}}_N)$. Each particle has an independent Poisson process of collapse times with rate $1/\tau$. Thus the total jump rate is $N/\tau$, the mean wait is $\tau /N$, and the jumping label is uniform on $\{1,\dots ,N\}$.

At a jump of particle $i$, use the GRW localization operatorIts centre has probability density function $p_i(\mathbf{c})=\Vert L_i(\mathbf{c})\Psi {\Vert }^2$. The post-jump quantum state is $L_i(\mathbf{c})\Psi /\Vert L_i(\mathbf{c})\Psi \Vert$. The normalization is consistent because $\int d^3\mathbf{c}{L}_i(\mathbf{c}{)}^2=I$. The collapse times and labels have these state-independent rates; the centres depend on the wave function through the Born rule-like squared norm.

For precision about the stated spatial event, let $B=Y_{10a}=\{\mathbf{c}:\mathrm{dist}(\mathbf{c},Y)\le 10a\}$. With $H=0$, the GRW collapse-centre event probabilities are exactlyThe last expression is defined only for $p_1(B)>0$. It includes the possibility that the next jump hits the same particle.

Write $f_i^Y(\mathbf{x})=|\varphi (\mathbf{x}-{\mathbf{y}}_i){|}^2$, and similarly $f_i^Z$. Putwith the same definitions for $Z$. The macroscopic separation and negligible packet tails make the two branches essentially orthogonal, so their interference contribution can be neglected. The first-event probability is thereforeThe $Z$ contribution is extremely small: its actual positions are within $r$ of $Z$, whereas centres in $B$ are within $10a$ of $Y$, leaving a distance much larger than $a$ in the Gaussian. Packet-tail errors add to this Gaussian-tail error.

For the next-event calculation, two distinct labels have independent position densities within either product branch, giving factors ${\eta }_i{\eta }_j$. A repeated label instead gives $R_i$, since both localization Gaussians multiply the same position variable. ConsequentlyThese formulas, with $p_{2\mid 1}=p_{12}/p_1$, answer the spatial questions without an extra containment assumption. If the first event is dominated by the $Y$ branch, they simplify toNeglecting the $Z$ term in this conditional expression requires it to be small relative to the retained first-event probability, not merely small in absolute value. The exact operator formulas above cover rare-event exceptions as well.

The intended GRW branch persistence versus geometric containment approximation is obtained if $Y$ and $Z$ contain essentially all the position mass of their corresponding branches. Then positions in $Y$ have a Gaussian-centre margin of $10a$ inside $B$, so ${\eta }_i^Y\simeq R_i^Y\simeq 1$, while ${\eta }_i^Z,R_i^Z\simeq 0$. This givesFor the Gaussian convention used here, the three-dimensional probability of a displacement exceeding $10a$ is $\mathrm{erfc}(10)+20e^{-100}/\sqrt{\pi }$, an extremely small number. A first centre near $Y$ also suppresses the $Z$ branch relative to the $Y$ branch by the very small ratio $|\beta {|}^2{q}_i^Z(\mathbf{c})/(|\alpha {|}^2{q}_i^Y(\mathbf{c}))$ when the latter branch has appreciable weight. All particles are correlated with the same pointer alternative, so one localization selects the macroscopic branch even when the next particle label differs.

The literal PDF, however, only places the packet centres ${\mathbf{y}}_i$ in $Y$; it does not say that the packets themselves fit inside $Y$. Since it permits $r>a$, the two numerical approximations just displayed do not follow for every allowed packet and region. For example, take all ${\mathbf{y}}_i=0$, let $Y$ be a very small ball, and let $|\varphi {|}^2$ be uniform on a ball of radius $r\gg 10a$. Translate the other branch to a similarly small region $Z$ at distance much larger than $100r$. ThenIf these equal packet probabilities are denoted by $\eta$ and their repeated-label integral by $R$, the next probability iswhich approaches $\eta$, not one, for large $N$. Thus packet containment is needed for the stated near-region event to have the usual branch probabilities. The literal general answer is given by the integrals above.

The physical GRW amplification mechanism remains intact: with properly chosen macroscopic branch regions, spontaneous localization selects the alternatives with approximately the squared branch amplitudes and stabilizes the selected record, on a time scale $\tau /N$. Gaussian tails are not exactly eliminated, so the persistence is approximate. The containment qualification concerns how the region is defined, rather than whether the dynamics selects a macroscopically separated branch.