Past exam of the quantum information course of the University of Cambridge

Original public paper: Physical PDF pages 835–838. All three questions and all alternatives in question 3 are answered. Entropies use logarithms to base two.

An eigenvalue of an observable in a known corresponding eigenstate is a simple example satisfying the Einstein–Podolsky–Rosen criterion of reality: its value is predictable with certainty. For example, a spin angular momentum component in a prepared spin-up eigenstate has a definite predicted value. For a distant system, Albert Einstein, Boris Podolsky and Nathan Rosen invoked locality: a measurement performed on a sufficiently separated, noninteracting partner should not disturb that system. In the relativistic version here, the measurement events have spacelike separation.

The Pauli matrices obey $[{\sigma }_z,{\sigma }_x]=2i{\sigma }_y$, soTheir nondegenerate eigenstates are incompatible: an $S_z$ eigenstate has equal probabilities for the two $S_x$ outcomes, and conversely. Thus no single quantum state has both components sharply determined. The Robertson uncertainty relation gives $\Delta S_z\Delta S_x\ge \hslash |⟨S_y⟩|/2$, but its right-hand side may be zero; that alone is not a proof that simultaneous zero uncertainties exist. The absence of a common eigenstate supplies the stronger conclusion for this spin-half pair.

For the specified Bell state, Bob obtains $+\hslash /2$ or $-\hslash /2$ with probability $1/2$ when he measures $S_z$. Conditional on his result, Alice obtains the same sign with certainty if she measures $S_z$. Under the EPR locality premise, this certainty supplies an element of reality for her $S_z$; it is a conditional prediction after Bob learns his own random result, not a premeasurement prediction of its sign.

The normalized $S_x$ eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Expanding their tensor products givesConsequently the same Bell state has equal, perfectly correlated $S_x$ outcomes. Bob's two outcomes again each have probability $1/2$, and the same locality-and-certainty argument applies separately to Alice's $S_x$.

The disputed step is to combine predictions conditional on two incompatible choices into two simultaneous, measurement-independent pre-existing values. Quantum theory provides the two conditional ensembles, not a joint sharp assignment of $S_x$ and $S_z$. This is quantum steering. Bob cannot measure both incompatible observables sharply on the same member of the pair. Nor does conditional steering allow signalling: ignoring his result leaves Alice's reduced density matrix equal to $I/2$ for either measurement choice, by the no-communication theorem.

To test the stronger claim of a local hidden-variable theory, Alice needs several correlation settings, not just perfect agreement along the two common axes. For the CHSH inequality, choose observables with outcomes $\pm 1$:In the given Bell state, $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and the crossed correlations vanish. Henceexceeding the local bound $|\mathcal{S}|\le 2$. The bound assumes Bell local causality and measurement independence, with the appropriate treatment of sampling and settings.

Experimental Bell tests support violations rather than that local assignment. The Aspect switching experiment changed analyzer settings during separated photon flights, while leaving sampling and switching limitations. By the time of this examination, the Delft electron-spin Bell test had used electron spins at stations $1.3\mathrm{km}$ apart, efficient spin readout and rapid random basis choices. The combination addressed both detection and locality loopholes, providing evidence against the tested local model class.^[ref]

This does not prove that every conceivable notion of reality is false, or that noncommutation alone rules out every hidden-variable theory. A nonlocal theory can evade the local premise. What Alice can challenge is Bob's claim that locality, measurement-independent simultaneous values and all observed correlations can be retained together. An ordinary measurement disturbance at Bob's lab is not by itself a local explanation of Alice's conditional ensemble.

Use the thin singular value decomposition $a=Ud{V}^{†}$, with $U^{†}U=I_N$, $V$ unitary and $d=\mathrm{diag}(d_k)$ nonnegative. Expanding its entries givesThe columns of $U$ and of $V^{\ast }$ are orthonormal, establishing the Schmidt decomposition. Bob's basis uses the complex-conjugated columns of $V$. Omitting that conjugation would not reproduce the coefficient matrix. The squared Schmidt coefficients are the eigenvalues of $a^{†}a$; equivalently $d^2=V^{†}{a}^{†}aV$ and $d=U^{†}aV$. The two reduced density matrices are $a{a}^{†}$ and $(a^{†}a{)}^T$, which have the same nonzero eigenvalues.

The printed coefficient $0.41231$ slightly rounds a normalized state: each squared norm is $0.9999995361$. Normalize before calculating an entropy. For the two states, the resulting Gram matrices are approximatelyFor a trace-one two-dimensional Hermitian matrix with diagonal entry $u$ and off-diagonal entry $c$, its eigenvalues areSort them decreasingly and take square roots:The Von Neumann entropy is $S=-{\lambda }_0{\log }_2{\lambda }_0-{\lambda }_1{\log }_2{\lambda }_1$. $\Psi$ is more entangled: its Schmidt weights are closer to equal and its reduced-state entropy is larger. Neither state is a product state or maximally entangled on its two-dimensional support.

By Nielsen's pure-state conversion theorem, deterministic LOCC conversion requires the input vector of squared Schmidt coefficients to be majorized by the output vector. Here $0.699941<0.894238$ and both sums are one, so $\lambda (\Psi )\prec \lambda (\Phi )$. Therefore $\Psi \to \Phi$ is possible deterministically; $\Phi \to \Psi$ is not. Entanglement can be discarded by local operations but cannot increase deterministically for these pure inputs. Probabilistic concentration is a separate question and is allowed below.^[ref]

There is a dimensional qualification in the proposed measurement. Write $P=|0_A^{\prime }⟩⟨0_A^{\prime }|+|1_A^{\prime }⟩⟨1_A^{\prime }|$. The two supplied measurement operators obeyThus their effects form a POVM on the occupied support, where $P$ is the identity. Alice's full space is three-dimensional, so these two operators alone do not sum to $I_3$. A full-space completion is $M_3=|2_A^{\prime }⟩⟨2_A^{\prime }|$. It has zero probability on either input, so it changes none of the requested outcomes. Strictly, the POVM consists of the effects $E_r=M_r^{†}{M}_r$; the specified $M_r$ additionally fix a quantum instrument and therefore the conditional states.

In the appropriate Schmidt basis for either input, write $|\psi ⟩=\sqrt{{\lambda }_0}|00⟩+\sqrt{{\lambda }_1}|11⟩$. The two outcomes have probabilitiesand normalized conditional statesThese formulas give both outcomes for each state by substituting its row of the table. For example,The two inputs have different Schmidt bases; the primed measurement basis is adapted separately to the state, as the question specifies, not one common unexamined laboratory basis.

For the $M_1$ outcome to be a maximally entangled state, its two amplitudes must be equal. HenceThe requested numerical results areThus $\Psi$ has the larger success probability, consistent with its greater initial entanglement. The $M_2$ branch cannot be balanced in the allowed angle range for either nonmaximal input: it would require $\tan \theta =\sqrt{{\lambda }_0/{\lambda }_1}>1$.

The success probabilities belong to this particular binary Schmidt filter. They are not the optimum over every possible quantum instrument. The diagonal success operator $\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ instead gives the standard single-copy entanglement concentration probability $2{\lambda }_1$. Postselection can increase entanglement in the retained branch, while the outcome-averaged entanglement cannot increase under LOCC.

A weak measurement couples an observable only weakly to a meter, so the pointer distributions associated with its different eigenvalues overlap strongly. One trial supplies little information and a small disturbance; an ensemble is needed to estimate a small mean shift. It is not an exact determination of a previously hidden eigenvalue with no disturbance.

For pointer position $q$ and conjugate momentum $p$, a short von Neumann measurement interaction has unitary $U=\exp (-igA\otimes p/\hslash )$. An eigenvalue $a$ translates the pointer by $ga$. The weak regime requires those translations to be small relative to the pointer width. Without postselection the first-order mean displacement is $g⟨A⟩$ for an initially uncorrelated, suitably centered meter.

If the system is prepared in $|i⟩$ and postselected in $|f⟩$, the unnormalized meter state isFor a real centered Gaussian pointer with no position–momentum correlation, the conditional position shift is $g\mathrm{Re}{A}_w$; the imaginary part affects the conjugate momentum distribution. The weak value can be complex or have a real part outside the observable's eigenvalue interval, because it is a conditional transition-amplitude ratio, not an eigenvalue.

An example apparatus uses photon polarization as the two-level system, a weak polarization-dependent transverse displacement as the interaction, and an output polarizer for postselection. Repeated position detections estimate the selected beam's displacement. Nearly orthogonal selection can make the conditional shift large, but also reduces the accepted sample and eventually invalidates the first-order expansion. The original weak-value proposal emphasized preselection, a weak coupling and postselection, rather than an anomalous result from an ordinary sharp spin measurement.^[ref]

Weak coupling limits quantum measurement backaction per trial; it does not abolish the information–disturbance tradeoff. Repeating weak observations can accumulate substantial backaction, and discarding rejected trials cannot automatically improve the information available per prepared system. Conditional pointer amplification and a statistically improved parameter estimate are distinct claims.

Quantum teleportation transfers an unknown qubit state using a previously shared Bell pair, local operations and two classical bits. It requires no description of the unknown complex amplitudes and produces no extra copy.

Let the input be $|\chi {⟩}_S=\alpha |0⟩+\beta |1⟩$, and let Alice and Bob share $|{\Phi }^{+}{⟩}_{AB}$. Alice applies a controlled-NOT gate from $S$ to $A$, then a Hadamard gate to $S$. Expanding the circuit gives the Bell-basis teleportation identity in computational-output form:Her projective measurement yields each bit pair $(a,b)$ with probability $1/4$. She sends the two classical bits, and Bob applies $Z^a{X}^b$ to recover $|\chi ⟩$. The order is the inverse of the displayed conditional operator; interchanging these Pauli corrections changes only an irrelevant global phase when both bits are one.

Before receiving the bits, Bob hasThus teleportation respects the no-communication theorem and cannot transmit the usable unknown state faster than the classical message. Alice's input has been measured, and the shared entanglement resource has been consumed, in agreement with the no-cloning theorem. The protocol transfers the state, not the material particle initially carrying it.

A physical implementation must distribute and preserve an entangled pair, implement a Bell-basis measurement and transmit its result to a controllable output system. Imperfect entanglement, lost outcomes and noisy corrections reduce quantum fidelity; a heralded probabilistic experiment must report its success probability separately from conditional fidelity. Ideal deterministic teleportation assumes every Bell outcome can be identified and corrected.

Quantum error correction encodes a logical qubit into a larger Hilbert space, diagnoses a specified error family with an error syndrome, and reverses it without learning the protected logical amplitudes. It must preserve arbitrary superpositions, not just distinguish two classical codewords.

The bit-flip repetition code encodesMeasure the commuting stabilizer generators $Z_1{Z}_2$ and $Z_2{Z}_3$ using ancillary qubits. Their outcomes reveal the parity relations, with the three-qubit bit-flip syndrome tableBoth codewords have the same parity checks, so the syndrome does not reveal $\alpha$ or $\beta$. For example, the $X_2$ error produces $\alpha |010⟩+\beta |101⟩$; applying $X_2$ restores the encoded superposition. A coherent combination of correctable errors is also recoverable after the syndrome is recorded in an ancilla, by linearity.

This three-qubit code corrects one bit flip, not every one-qubit error. A physical $Z_1$ changes the relative logical phase without a nontrivial bit-flip syndrome. General single-qubit noise is a linear combination of $I,X,Y,Z$; protecting arbitrary single-qubit errors therefore needs a code such as the Shor code, the Steane code or the five-qubit code, with the corresponding enlarged syndrome structure.

The general exact-recovery criterion is the Knill--Laflamme condition. For code projector $P$ and error operators $E_r$, there must be constants $C_{rs}$ such thatIt expresses that the syndrome information can depend on the error but not on the logical state. Encoding entangles several physical qubits; it does not create independent copies of an unknown input, so it respects the no-cloning theorem. Fault-tolerant implementations must also prevent noisy syndrome extraction and recovery gates from spreading an initially correctable error into an uncorrectable one.

Original public paper: Physical PDF pages 761–764. All questions and all optional notes are answered. A pure premeasurement state, a nonselective projection, and a state conditional on one observed outcome are distinguished throughout.

Linearity of the detector's unitary time evolution maps the two input branches separately, giving the quantum premeasurement stateIn the ordered tensor product basis $(|\uparrow d_{\uparrow }⟩,|\uparrow d_{\downarrow }⟩,|\downarrow d_{\uparrow }⟩,|\downarrow d_{\downarrow }⟩)$, its pure density matrix isA nonselective projective measurement in the indicated detector basis applies $\rho \mapsto {\sum }_j(I_S\otimes {\Lambda }_j)\rho (I_S\otimes {\Lambda }_j)$, removing the two off-diagonal terms:This is the usual nonselective meaning of von Neumann process 1. It is not equal to the pure projector $|{\Phi }^C⟩⟨{\Phi }^C|$. If “the detector has clicked” means conditioning on an actual up outcome, the normalized postselected state is instead $|\uparrow d_{\uparrow }⟩$, with probability $1/2$. The correlated superposition describes the coherent premeasurement before that selection.

Put $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The detector rotation is real and orthogonal, so its inverse is the same transformation. Substituting it givesThus the requested orthonormal system states areThe pure density operator has not changed: this is only a different Schmidt decomposition of a maximally entangled state. But choosing a nonselective projective measurement in this new detector basis gives a different operation and generally a different state:For an explicit matrix in the original ordered basis, write $q=c^2{s}^2$ and $h=cs(c^2-s^2)/2$. ThenAt $\theta =\pi /2$ the entries $(1,4)$ and $(2,3)$ are $1/4$, whereas they vanish in ${\rho }_{up/down}$. Nevertheless, the system and detector individually each have reduced density matrix $I/2$ for either mixture.

This is the preferred-basis problem: the correlated state alone does not specify which detector observable becomes a stable classical record. Arbitrarily declaring collapse in a Schmidt basis changes the physical prediction. The actual detector interaction and environment must supply the basis selection, rather than a mere choice of notation.

Write $H=g{I}_S\otimes Z_D\otimes X_E$, with $Z_D={\Lambda }_{\uparrow }-{\Lambda }_{\downarrow }$. The orthogonal projection operators obey ${\Lambda }_{\uparrow }{\Lambda }_{\downarrow }=0$ and ${\Lambda }_j^2={\Lambda }_j$. Hence $[Z_D,{\Lambda }_j]=0$, and the remaining tensor factors commute with detector operators. ThereforeThe environment interaction preserves these detector populations, making their basis a candidate pointer basis. It can change their relative coherence without flipping the stored labels.

Since $(Z_D\otimes X_E{)}^2=I$, the solution of the Time-dependent Schrödinger equation isThe two detector eigenvalues generate opposite rotations of the environment. Acting on the initial state givesThese are the indicated $|E_{-}⟩$ and $|E_{+}⟩$. Their conditional environment overlap is $⟨E_{+}|E_{-}⟩=\cos 2A$. The partial trace therefore multiplies the coherence between $|\uparrow d_{\uparrow }⟩$ and $|\downarrow d_{\downarrow }⟩$ by $\cos 2A$. This is an exactly solvable single-qubit environment dephasing model, with fully reversible global evolution.

At $A=\pi /4$, defineThey are normalized and orthogonal. The full state is the pure superposition $(|\uparrow d_{\uparrow }{E}_{\leftarrow }⟩+|\downarrow d_{\downarrow }{E}_{\to }⟩)/\sqrt{2}$. A nonselective von Neumann process 1 selecting the original detector records givesWithout imposing any projection on the full state, tracing out the orthogonal environment states already gives ${\rho }_{SD}={\rho }_{up/down}$ from part (a). These two global descriptions share that reduced state but differ in their global coherence. Quantum decoherence does not make a pure global state literally equal to the corresponding statistical mixture.

Inverting the detector rotation at this time givesFor a representation with one system factor and one environment factor per displayed detector state, each parenthesized system–environment vector would have to be a product state. But the first has coefficient matrix $\mathrm{diag}(c,s)$ in the indicated orthonormal bases; the second has matrix $\mathrm{diag}(s,-c)$. Their matrix ranks are two whenever $cs\ne 0$. Equivalently, their two nonzero Schmidt coefficients are $|c|,|s|$, which no independent basis changes can remove. Thus the proposed factorization is impossible when $\sin \theta \ne 0$.

The exception matters: at $\theta =m\pi$, the detector basis is the original basis up to phases or interchange, and a factorization does exist. In particular, the literal condition $\theta \ne 0$ is insufficient if it includes $\theta =\pi$. The intended obstruction concerns genuinely mixed detector bases.

The environment has selected robust detector records: its interaction commutes with the original projectors, and orthogonal environmental records suppress their interference in the reduced density matrix. A rotated detector branch instead leaves the system and environment entangled. This is environment-induced pointer-basis selection, the issue analyzed by W. H. Zurek in his apparatus-environment model.^[ref]

This addresses the basis ambiguity but does not by itself select one realized outcome from the global superposition. Moreover, a two-state environment has recurrences: $\cos 2A$ returns to unit modulus, so the toy model does not establish irreversible macroscopic collapse. A large environment can make records effectively persistent over experimental times, but that requires further dynamical assumptions. The distinction is part of the measurement problem, not an algebraic error cured by a partial trace.

The spin singlet state has zero local means and outcomes $X,Y\in \{+1,-1\}$, so both variances are one. Its Pearson correlation coefficient is therefore $C=\mathbb{E}[XY]$. Summing over the four joint outcomes givesThe equal-sign products contribute $+1$ and the opposite-sign products contribute $-1$. In a finite sample the corresponding relative frequencies estimate these probabilities; the formula is not an assertion that a finite random sample has exactly zero empirical means.

A spin measurement along unit vector $\mathbf{a}$ has outcome projector ${\Pi }_s(\mathbf{a})=(I+s\mathbf{a}\cdot \mathbf{\sigma })/2$, with $s=\pm 1$. The spin singlet state obeys $⟨{\sigma }_i\otimes I⟩=⟨I\otimes {\sigma }_i⟩=0$ and $⟨{\sigma }_i\otimes {\sigma }_j⟩=-{\delta }_{ij}$. Applying the Born rule,Thus $P_{++}=P_{--}=(1-\mathbf{a}\cdot \mathbf{b})/4$ and $P_{+-}=P_{-+}=(1+\mathbf{a}\cdot \mathbf{b})/4$. Substitution into part (a) givesEqual axes give perfect anticorrelation, while orthogonal axes give zero correlation. Each local marginal remains uniform.

The four required dot products areConsequently the specified CHSH inequality combination isThe magnitude reaches the quantum Tsirelson bound and exceeds the local bound two. The negative sign follows this singlet and the stated order of terms; other conventions can reverse it.

Let Eve obtain $u,v\in \{\pm 1\}$. Applying part (b) to her axes gives $p(u,v)=[1-uv{\mathbf{n}}_a\cdot {\mathbf{n}}_b]/4$. Conditional on these rank-one local measurements, the forwarded particles are a product state with individual spin directions $u{\mathbf{n}}_a$ and $v{\mathbf{n}}_b$. HenceConditional independence gives the product of these means. Averaging over Eve's outcomes,The sum is $-{\mathbf{n}}_a\cdot {\mathbf{n}}_b$, soEve has turned the entangled source into an ensemble of product states by an intercept-resend attack. Ignoring her records does not restore the original singlet coherence.

Write ${\mathbf{n}}_a=(\cos \alpha ,\sin \alpha )$ and ${\mathbf{n}}_b=(\cos \beta ,\sin \beta )$. Since ${\mathbf{b}}_1-{\mathbf{b}}_3=\sqrt{2}\hat{\mathbf{x}}$ and ${\mathbf{b}}_1+{\mathbf{b}}_3=\sqrt{2}\hat{\mathbf{y}}$, the CHSH inequality combination becomesThereforeParallel or antiparallel Eve axes attain the lower endpoint; orthogonal axes attain zero. This bound is specific to the two-sided projective attack and these planar settings. It is stronger than the generic local bound $|S|\le 2$ and should not be attributed to every possible eavesdropping strategy.

If Eve varies her axes independently of the users' still-secret random settings, the measured correlations average over her choices. The same linear combination obeys $S=-\sqrt{2}\mathbb{E}[{\cos }^2(\alpha -\beta )]$ and remains in the interval by convexity. Strategies correlated with disclosed measurement settings would violate that independence premise; the protocol must protect it.

This is an entanglement-based quantum key distribution scheme of the type proposed by Artur K. Ekert.^[ref] The measurements generate correlated random data; Alice does not choose a message by choosing their random outcomes.

After the measurements, Alice and Bob announce their basis choices over an authenticated public channel. They retain the pairs with equal axes, here $({\mathbf{a}}_2,{\mathbf{b}}_1)$ and $({\mathbf{a}}_3,{\mathbf{b}}_2)$, and Bob flips his outcome signs to match Alice's bits. They sacrifice a random subset for error estimation and use the appropriate other setting combinations to estimate $S$. In the ideal source the same-axis bits match after the flip and $S=-2\sqrt{2}$; the specified intercept-resend attack cannot reproduce that value.

A practical protocol includes information reconciliation, an accounting of information leaked publicly, and privacy amplification. It must abort when its statistical security conditions fail. Authentication is essential: a completely unauthenticated public channel permits a man-in-the-middle attack. A Bell-violation observation alone, without sampling, device and adversary assumptions and a finite-key analysis, is not a proof that Eve knows no key bits.

With a sufficiently uniform secret shared key $k$ of the required length, Alice can send $c=m\oplus k$ publicly using a one-time pad; Bob recovers $m=c\oplus k$. The pad must be secret, as long as the protected message, and never reused. A complete security proof makes the probability or distinguishing advantage of failure small under stated assumptions. The ideal exam correlations diagnose the stated attack; they do not provide an unconditional guarantee against every attack or implementation fault.

A bipartite pure state is entangled when it is not a product state. Its Schmidt decomposition is $|\psi ⟩={\sum }_i\sqrt{{\lambda }_i}|i_A{i}_B⟩$, where ${\lambda }_i\ge 0$ and ${\sum }_i{\lambda }_i=1$. More than one nonzero coefficient means entanglement. The two reduced density matrices have the same nonzero spectrum, and their Von Neumann entropy $-{\sum }_i{\lambda }_i{\log }_2{\lambda }_i$ is the pure-state entanglement entropy. A maximally entangled pair of qubits has one bit of this entropy.

For mixed states, classical correlations are not sufficient: a separable quantum state is a convex mixture ${\sum }_j{p}_j{\rho }_A^{(j)}\otimes {\rho }_B^{(j)}$, and entanglement means that no such decomposition exists. A reduced entropy alone no longer measures mixed-state entanglement. An entanglement witness or a suitable separability criterion can detect it.

Local unitary transformations preserve the Schmidt spectrum. More general local operations and classical communication can discard entanglement but cannot freely create it from a separable input. Nielsen's pure-state conversion theorem gives the exact deterministic criterion: $|\psi ⟩\to |\varphi ⟩$ is possible precisely when ${\lambda }^{\psi }\prec {\lambda }^{\varphi }$. In decreasing order, every partial sum for the input is no larger than the corresponding output sum, with equal totals. The direction matters: a uniform spectrum can be converted to a concentrated one; a product state cannot deterministically become a Bell pair.

A probabilistic local filter can concentrate a partially entangled pair. For $|\psi ⟩=\sqrt{\lambda }|00⟩+\sqrt{1-\lambda }|11⟩$ with $1/2\le \lambda <1$, Alice's successful operator $F=\mathrm{diag}(\sqrt{(1-\lambda )/\lambda },1)$ produces equal amplitudes. The outcome is a Bell pair with probability $2(1-\lambda )$; a complementary operator supplies a complete measurement. Failure is heralded, so there is no deterministic creation of a more entangled pair. This illustrates entanglement concentration.

Many-copy entanglement distillation and entanglement dilution connect noisy or nonuniform resources with standard Bell pairs. Entanglement enables quantum teleportation and other protocols, but classical communication is still required; the no-communication theorem excludes message transmission merely by choosing a distant local measurement.

A qubit has state $\alpha |0⟩+\beta |1⟩$ with $|\alpha {|}^2+|\beta {|}^2=1$. An $n$-qubit register occupies a $2^n$-dimensional Hilbert space and may contain entanglement. This large state space does not mean that a measurement reads out all its complex amplitudes: computational-basis outcomes occur with the Born rule probabilities.

A quantum circuit prepares a register, applies quantum gates, and measures selected outputs, possibly using classical feed-forward. Closed gates are unitary operators and therefore reversible; initialization, measurement and discarding can be treated using a larger system and environment. Single-qubit rotations together with an entangling two-qubit gate give a universal quantum gate set. The CNOT gate acts as $|x,y⟩\mapsto |x,y\oplus x⟩$.

For example, a Hadamard gate on the first qubit of $|00⟩$ followed by a CNOT creates $(|00⟩+|11⟩)/\sqrt{2}$. A Hadamard produces amplitudes with relative signs, so later gates can make unwanted alternatives interfere destructively and useful alternatives reinforce. Quantum algorithms organize this interference to improve a specified computational task; superposition alone does not supply an arbitrary speedup or permit reading every branch.

A finite discrete universal set such as Clifford gates together with a non-Clifford phase gate approximates general unitaries to chosen precision. Runtime includes the number of gates, repetitions and classical processing. Measurement outcomes may require multiple runs to estimate a probability or amplify success.

Quantum decoherence and imperfect gates limit an implementation. Quantum error correction encodes logical information into several physical qubits and measures error syndromes without directly learning the encoded amplitudes. Under appropriate noise and control assumptions, fault tolerance permits reliable longer computations; simply maintaining an isolated ideal state is not a realistic account of a large processor.

In de Broglie-Bohm theory, the system has both a wavefunction and an actual configuration of particle positions. The wavefunction obeys the Time-dependent Schrödinger equation; the configuration follows the guidance equation. For spinless particles without electromagnetic vector potentials, write $\Psi =R{e}^{iS/\hslash }$ using a dimensionful action $S$. ThenThe actual configuration selects a trajectory, while the guiding wave generally retains all branches. A measurement correlates system and apparatus configurations; the actual apparatus position lies in one recorded branch. Effective collapse follows when wave packets have separated sufficiently in configuration space, although the full wave continues its Schrödinger evolution.

Substituting the polar form into that equation yields the probability continuity equation and a modified Hamilton-Jacobi equation,The amplitude-dependent term is the quantum potential. Taking gradients gives a second-order force description, but initial velocities remain constrained by the guidance law. Nodes require care because the displayed quotient is singular there.

If the initial configuration distribution is $|\Psi {|}^2$, the same continuity equation preserves that distribution: quantum equilibrium equivariance. It then reproduces ordinary nonrelativistic quantum measurement statistics within the model's applicability. Equivariance preserves a distribution; it is not by itself a proof that every initial distribution equals $|\Psi {|}^2$.

For entangled particles the guiding wave lives on their joint configuration space, so one particle's velocity can depend on the distant actual configuration. This explicit nonlocality avoids the locality premise of a Bell inequality. In quantum equilibrium it does not permit controllable faster-than-light signalling. Spin extensions use a spinor wavefunction and its probability current; they need not assign a classical little spinning vector with simultaneous definite values for every spin component.

David Bohm developed this account in his two 1952 papers. The first sets out the dynamics and the second treats measurement and the relation to ordinary predictions.^[ref][ref] The theory supplies additional dynamical variables and a particular nonlocal account of outcomes, rather than deriving a unique interpretation from the standard wavefunction alone.

Original public paper: Physical PDF pages 684–686. All questions and optional notes are covered. The equal-length period-register assumption is distinguished from the general normalized order-finding state.

Nielsen's pure-state conversion theorem concerns exact deterministic transformations of bipartite pure states by local operations and classical communication. Let $\lambda (\psi )$ be the decreasing vector of squared Schmidt coefficients, equivalently the eigenvalues of either party's reduced density matrix, padded with zeros when necessary. ThenThe majorization relation means ${\sum }_{j=1}^k{\lambda }_j(\psi )\le {\sum }_{j=1}^k{\lambda }_j(\chi )$ for every $k$, with equality for the total sum. The direction allows a maximally entangled state to become a product state, not the reverse. Local-unitary invariance of reduced-state spectra alone would require equality of all the weights; measurements and communicated outcomes permit this larger class of conversions. A comparison of entanglement entropies alone is not a sufficient test in higher dimension.^[ref]

For the supplied states, the weight vectors and cumulative sums areAt $k=2$, $0.8>0.75$, violating the required inequality. The desired conversion is impossible with certainty using LOCC alone. The reverse conversion also fails, already at $k=1$, so these states are incomparable under this deterministic ordering.

Taking a tensor product with the catalyst multiplies every original weight by each catalyst weight. Regrouping the two local systems on Alice's side and the two on Bob's side leaves a Schmidt decomposition across the Alice–Bob partition. After sorting, the resulting weight vectors areThe actual nonnegative Schmidt amplitudes are $\sqrt{x_j}$ and $\sqrt{y_j}$; the displayed numbers are their squares. The cumulative sums areEvery component obeys $X_k\le Y_k$, so $x\prec y$. HenceThe final catalyst is exactly its initial state and is uncorrelated with the desired output. It can therefore be returned and reused. This is entanglement catalysis; it is stronger than merely retaining the same catalyst marginal while leaving correlations with the output.^[ref]

For the maximally entangled two-level auxiliary state, the sorted input weights are $(0.2,0.2,0.2,0.2,0.05,0.05,0.05,0.05)$, while the output weights are $(0.25,0.25,0.125,0.125,0.125,0.125,0,0)$. Their first four sums are $0.8$ and $0.75$. This maximally entangled auxiliary state cannot catalyze the conversion while being returned unchanged.

Nevertheless, $(0.5,0.5)\prec (0.6,0.4)$, so it can first be converted deterministically into the nonmaximal catalyst. The desired conversion can then proceed, after which Alice and Bob can return $|\varphi ⟩$, rather than the borrowed maximally entangled state. Turning that returned pair alone back into a maximally entangled pair would increase its entanglement entropy and is forbidden by LOCC. No joint restoration procedure can both keep the exact desired $|{\psi }_2⟩$ and return the original maximally entangled auxiliary state deterministically: its composition with the preceding steps would contradict the failed four-weight inequality above.

The input size of an integer factorization problem is its number of bits $n\simeq {\log }_{2}N$. Trial division up to $\sqrt{N}$ is exponential in $n$; advanced classical algorithms do substantially better, but no general polynomial-time classical factoring algorithm is known. The heuristic general-purpose general number field sieve has subexponential rather than polynomial scaling, roughly $\exp (O(n^{1/3}(\log n{)}^{2/3}))$. This describes known algorithmic performance, not a proof that efficient classical factoring is impossible or that every large integer is difficult.

Choose $a$ with $1<a<N$. A nontrivial greatest common divisor $\gcd (a,N)$ already supplies a factor; otherwise find the multiplicative order $r$ of $a$ modulo $N$. If $r$ is even and $a^{r/2}\not\equiv -1(\mathrm{mod}N)$, thenMinimality of the order also excludes $a^{r/2}\equiv 1(\mathrm{mod}N)$, so factor extraction from an even modular order gives nontrivial factors through $\gcd (a^{r/2}\pm 1,N)$. Unhelpful choices of $a$ or odd orders require another trial. For $N=15$, $a=2$ has successive residues $1,2,4,8$ and order four, givingComputing the powers is easy by repeated squaring; discovering a large order is the step accelerated by Shor's algorithm.^[ref]

Put $Q=2^L$ and suppose $Q\ge r$. For each residue class $s=0,\dots ,r-1$, letThese are normalized uniform superpositions of the arguments with the same modular-exponentiation value. Their amplitudes repeat at separation $r$ within the finite register window. Exact cyclic periodicity requires $r\mid Q$. The correctly normalized entangled register state isThe equal prefactor $1/\sqrt{r}$ in the displayed algorithm is exact when $r\mid Q$, for then every $n_s=Q/r$. It is not generally exact for an arbitrary unknown order. A physical implementation uses quantum modular exponentiation as a reversible map; its specified action on second-register input zero is extended to a unitary operator on the whole register space.

For the illustrative $N=15$, $a=2$ and $Q=16$, the four first-register states areApplying the positive-sign quantum Fourier transform to the general state givesThe bracket is a finite geometric Fourier amplitude. If $r\mid Q$, it vanishes unless $k=\ell Q/r$ for $\ell =0,\dots ,r-1$, when it equals $Q/r$. HenceThe second-register states are orthogonal, so their different phases do not change the marginal first-register probabilities. Measurement givesIn the $Q=16$ example the possible outcomes are $0,4,8,12$. A nonzero observed ratio $k/Q$ estimates $\ell /r$, not a prime factor directly. Reducing it gives denominator $r/\gcd (\ell ,r)$; repeating avoids uninformative outcomes or recovers missing order factors. For example, $k=4$ or $12$ gives denominator four, whereas $k=8$ alone gives only two.