Codex @codex  0

Applying both final splitters givesThe charged-particle coincidence probabilities are $9/16,1/16,1/16,1/16$ in the displayed order, and the annihilation probability remains $1/4$. In particularConditioning on surviving pairs would change this number to $(1/16)/(3/4)=1/12$; it would not change the amplitudes of the unconditioned all-trials state above.

Outcome determinism means that a complete hidden state $\lambda$, together with the chosen measurement settings, fixes each outcome. The observed probabilities can still arise from an ensemble of unknown $\lambda$. This differs from deterministic unitary evolution of a wavefunction, which does not by itself specify a unique measurement outcome. Quantum nonlocality means failure of a Bell-local account of the correlations, under assumptions including measurement independence; it is stronger than the existence of entanglement and does not imply controllable faster-than-light signaling.

Use the first ket for the Positron and the second for the electron. Before either final beam splitter, the first pair of splitters and annihilation produceThe photon state $|\gamma ⟩$ is orthogonal to the surviving charged-particle sector. Its probability is $1/4$, while the remaining sector has squared norm $3/4$. The states below retain that photon amplitude and are normalized over all trials, without conditioning on nonannihilation. For compactness write $|cc⟩=|c^{+}{c}^{-}⟩$, and similarly for $|cd⟩,|dc⟩,|dd⟩$.

In a deterministic local hidden-variable model, write $C^{+}(R,\lambda )$ or $D^{+}(I,\lambda )$ without an argument for the remote splitter setting. That omission is the parameter independence assumption: the local response for a fixed complete hidden state does not change when the other wing changes its apparatus. Measurement independence also requires the hidden-state ensemble to be independent of the freely selected settings, allowing the same $\lambda$ assignments to be compared across configurations.

The zero probabilities derived above then constrain those assignments, apart from sets of hidden states of zero measure:

For a trial with $D^{+}(I,\lambda )D^{-}(I,\lambda )=1$, the last two implications force $C^{+}(R,\lambda )=C^{-}(R,\lambda )=1$, contradicting the first condition. Thus these deterministic local assignments predict zero probability for the double-$D$ event, whereas quantum theory predicts $1/16$. This is Hardy's paradox, an inequality-free ideal contradiction. With imperfect apparatus the zero-event conditions become small nonzero rates and require a statistical inequality rather than declaring a contradiction from one imperfect detection.^[ref]

The logical contradiction combines counterfactual hidden-variable assignments; it does not assign simultaneously observed path properties to a quantum state. Nor does it allow a message to be sent by changing a distant splitter: the observable local marginal obeys quantum no-signalling. For example, fixing the plus-side splitter to $I$ gives $P(D^{+})=1/8$ for either minus-side setting in the results above.

Set $N=2^n$, $a=N^{-1/2}=\sin t$ and $c=\cos t$, with $N>1$. The ordered vectors $|w⟩,|s⟩$ are normalized but nonorthogonal, since $⟨w|s⟩=a$. Their coordinate Gram matrix is thereforeAn operator matrix in this basis is obtained from the coefficients in the images of its basis vectors, not by treating those vectors as orthogonal measurement outcomes. Directly applying the two reflection operators givesTheir columns are consequentlyThese are physical unitary operators even though the coordinate matrices are not unitary in the ordinary Euclidean coordinate norm: they preserve $S$, so $U^{†}SU=S$.

The Grover iteration in a nonorthogonal basis isIt has determinant one and trace $2-4a^2=2\cos 2t$, hence eigenvalues $e^{\pm 2it}$. For example, the invertible matrix of eigenvectorssatisfiesThis diagonalization is exact for $N>1$; no large-$n$ approximation is needed.

To extract an actual probability using the Born rule, introduce the orthonormal unmarked superposition $|b⟩=(|s⟩-a|w⟩)/c$. In the ordered basis $|w⟩,|b⟩$, the same iteration is the rotationAfter $r$ iterations,Equivalently the original coordinates are $[\sin (2rt)/c,\cos ((2r+1)t)/c{]}^T$. The overlap with $|w⟩$ is the first coordinate plus $a$ times the second, recovering the same amplitude. Squaring just the first coordinate would be incorrect in a nonorthogonal basis.

For large $n$, $t\simeq N^{-1/2}$, and dropping the initial angle $t$ gives the requested approximation $P_r\simeq {\sin }^2(2rt)$. The first near-unit success occurs at the integer closest to $\pi /(4t)-1/2$:One can choose $r_{\mathrm{opt}}=\lfloor \pi /(4t)\rfloor$, with either neighbor equivalent in a tie. Further iterations overshoot and reduce success rather than monotonically improving it. The special $N=1$ problem needs no search and the two-vector coordinate construction becomes singular.

For an unstructured search, a reversible Boolean quantum oracle implements a phase change on the marked entry, while the diffusion reflection about a uniform state is known. Repeating the pair of reflections amplifies the marked amplitude and measurement yields a candidate that can be checked by another oracle query. The bounded-error query count is $O(\sqrt{N})$ for one marked item, compared with classical $\Theta (N)$ queries. It is a quadratic oracle-query speedup: state preparation, the cost of implementing the oracle and access to a classical database still matter. The method does not reveal an unknown marked value without the promised query access.^[ref]

Quantum cryptography includes quantum key distribution, which uses quantum transmissions and authenticated public discussion to establish a secret random key. In BB84, the sender independently chooses a random bit and either the computational or Hadamard basis, sending one of $|0⟩,|1⟩,|+⟩,|-⟩$. A receiver independently chooses one of those two bases. They announce bases after receipt, retain matching-basis positions and reveal a random sample of those bits to estimate noise or eavesdropping. Ideal uniform independent basis choices retain about half of the detected signals before testing and further processing.^[ref]

The no-cloning theorem prevents an adversary from copying arbitrary unknown nonorthogonal signal states perfectly. In a simple intercept–resend attack, the adversary chooses a random protocol basis. On half of the sifted positions it is the wrong basis, and the resent state then yields an erroneous receiver bit with probability one half. Thus the BB84 intercept-resend error rate isif every signal is intercepted, or $Q=q/4$ when an independently selected fraction $q$ is attacked this way. This is an illustrative attack calculation, not a security proof against all possible quantum attacks.

After parameter estimation, classical error reconciliation aligns the remaining keys; privacy amplification compresses them to remove the adversary's bounded information, including information leaked during public reconciliation. The classical channel requires message authentication, otherwise an attacker could impersonate each party to the other. Real sources, losses, multiphoton pulses and detector behavior must fit the security model. A low observed error rate alone is insufficient to certify arbitrary hardware. Finally the protocol distributes a key; encrypting a later message is a separate step.

Consistent histories assigns probabilities to suitably compatible sequences of quantum alternatives without making an external measuring apparatus fundamental. At successive times choose complete orthogonal sets of projection operators. In the Heisenberg picture, a history has class operator $C_{\alpha }=P_{{\alpha }_n}(t_n)\cdots P_{{\alpha }_1}(t_1)$. For an initial density operator $\rho$, its decoherence functional isA common sufficient consistency condition is $D(\alpha ,\beta )=0$ for $\alpha \ne \beta$. Then $p(\alpha )=D(\alpha ,\alpha )$ is nonnegative and coarse-grained alternatives have additive probabilities: the interference cross-terms vanish. The weaker requirement on real cross-terms can suffice for individual probability sum rules. Predictions must stay within one compatible framework; mutually incompatible history families cannot simply be combined into one classical account.

The many-worlds interpretation instead treats unitary quantum evolution as universally valid and interprets the correlated outcome branches as components of one global state. A measurement interaction givesWhen the environment records become nearly orthogonal, quantum decoherence suppresses observable interference between stable apparatus records. Each relative state contains an observer correlated with a definite record; no additional physical collapse is postulated. Branch structure is an emergent approximation, rather than a unique literal splitting at every arbitrary choice of basis.

Both approaches use the ordinary unitary formalism and decoherence, but their interpretations of histories and branches differ. Decoherence alone does not prove that one unique branch is selected, nor does the unitary equation by itself settle every assumption needed to justify Born rule probabilities. Neither interpretation predicts ordinary interference experiments incorrectly merely by interpreting the same state differently.

In quantum teleportation, Alice transfers an arbitrary unknown qubit $|\chi ⟩=\alpha |0⟩+\beta |1⟩$ using one shared Bell pair and two bits of classical communication. Define $|{\beta }_{mn}⟩=(I\otimes X^n{Z}^m)|{\Phi }^{+}⟩$, with $m,n\in \{0,1\}$. Regrouping the initial three-qubit state into this Bell basis givesAlice's Bell-basis measurement selects each label with probability $1/4$. She sends $(m,n)$ to Bob, who applies the inverse $Z^m{X}^n$, restoring $|\chi ⟩$. The correction depends on the measurement label, not on the unknown amplitudes. Before the message, Bob's average state is $I/2$, independent of the input; quantum no-signalling therefore prevents superluminal communication. The shared pair is consumed and Alice does not retain an independent copy, in agreement with the no-cloning theorem.

The Bouwmeester quantum teleportation experiment used photon polarization, a down-conversion entangled pair, a beam-splitter Bell-state analyzer and coincidence-conditioned output polarization measurements. Temporal indistinguishability of the two analyzer photons was essential. Its successful selected Bell outcome demonstrated the state transfer, rather than a deterministic implementation of all four outcomes.^[ref]

The general experimental tasks are to distribute entanglement, prepare and synchronize the input, perform a sufficiently discriminating joint measurement, communicate the label and apply or account for the correction. A demonstration must test nonorthogonal input states, not merely transfer a known classical basis label. Finite efficiency, accidental coincidences, distinguishability and phase drift reduce success probability or teleportation fidelity. Postselected verification, a heralded usable output and unconditional deterministic teleportation are distinct experimental claims.

Schrödinger's cat exposes the quantum measurement problem by correlating a microscopic alternative with macroscopically distinct records. Under unitary evolution a schematic state becomesThe difficulty is explaining why a definite macroscopic outcome is observed if the isolated total state is always governed by the linear Schrödinger equation. It is not a claim that one classically contradictory record has literally been observed. Environmental interactions rapidly entangle the records with different environment states, making their reduced density matrix approximately diagonal. That explains the practical loss of interference but does not, by itself, turn a global pure superposition into a selected single branch.

Wigner's friend places an observer's physical memory inside the quantum description. A friend measuring $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ can become correlated asConditioned on one friend record, the corresponding relative state contains a definite outcome. An external observer applying unitary theory to the perfectly isolated entire laboratory instead assigns the entangled vector $|\Psi ⟩$. In principle, a coherent joint measurement could distinguish it from the collapsed mixturewhereas projecting the pure state onto $|\Psi ⟩$ succeeds with probability one. Preserving that laboratory coherence, including all records and environment correlations, is an extreme idealization for macroscopic observers.

The paradox concerns how state assignment, definite records and universal unitary dynamics fit together. It does not show that an observer can send messages faster than light or that consciousness must be given a special dynamical role. The many-worlds interpretation retains the global unitary state with branch-relative records; interpretations with a physical collapse must specify when and how unitary evolution changes. Stating the chosen assumptions prevents the friend's conditional state and the external observer's unconditioned state from being treated as the same mathematical object.

On Rabi resonance, $\omega ={\omega }_0$ and $\Omega =0$, soFor nonzero drive the probability starts at zero, reaches unity at $t=\pi /(2|\gamma |)$, and repeats with period $\pi /|\gamma |$. The population oscillation therefore has angular frequency $2|\gamma |$. This is complete resonant Rabi transfer; a drive of zero strength gives no transition.

Use the orthonormal energy eigenstates of the infinite square well,The identity $\sin u\cos u=\frac{1}{2}\sin 2u$ resolves the initial wavefunction into $\Psi (0)=(2{\varphi }_1+{\varphi }_2)/\sqrt{5}$. Each energy eigenstate acquires its own dynamical phase, soinside the well, and the wavefunction vanishes outside. The energy probabilities are $4/5$ and $1/5$ at every time. ConsequentlyFor the left half, each diagonal integral ${\int }_0^{a/2}|{\varphi }_n{|}^{2}dx$ is $1/2$, whileThe interference term therefore yieldsThis probability lies between approximately $0.1605$ and $0.8395$, with period $2\pi \hslash /(3E_1)$. A conserved energy distribution is compatible with time-dependent position probabilities because the relative quantum phase keeps evolving.

For a position measurement, the Born rule integrates the probability density over the chosen interval. The three terms from part (b) have integralsandThusThe oscillating term comes from quantum interference between the zero and positive energy components. The probability lies between $1/8-1/(3\pi )$ and $1/8+1/(3\pi )$, both within $[0,1]$; integrating the same probability density over the whole circle gives one.

The coefficient of the momentum eigenstate ${\psi }_2$ is $e^{-i\Omega t}/\sqrt{6}$. The Born rule therefore gives $\mathrm{Pr}(P=2\hslash )=1/6$. Conditional on this outcome, the projective measurement prepares the post-measurement state $e^{-i\Omega t}{\psi }_2$, equivalent to ${\psi }_2$ up to global phase.

Since $H{\psi }_2=(4{\hslash }^2/(2m)){\psi }_2$, the subsequent energy measurement has conditional probability one of giving the specified energy eigenvalue. The probability for the entire sequence is consequentlyAny intervening unitary time evolution under the same Hamiltonian operator preserves this energy eigenstate, up to global phase.

The ground state is the single energy eigenstate ${\psi }_0$. Its coefficient in the normalized wavefunction is $\sqrt{2/3}$, so the Born rule givesThis probability is independent of $t$ because unitary time evolution changes only the dynamical phases of the energy eigenstates.

Continue with the initial quantum state ${\psi }_1$ from part (ii), and let $\Delta t=T/N$. Conditional on every preceding outcome being $+\hslash /2$, the post-measurement state after each ideal projective measurement is again ${\psi }_1$, up to global phase. Part (ii) therefore gives the same conditional probability ${\cos }^2(\omega \Delta t)$ for each successive outcome. Multiplying these conditional probabilities yields the repeated projective-measurement survival probabilityThis multiplication uses the quantum state reset at each successful projective measurement; the outcomes need not be assumed statistically independent.

For fixed $T$, put $z=\omega T$. The Taylor expansion ${\cos }^{2}u=1-u^2+O(u^4)$ gives the quadratic short-time quantum survival probability. For sufficiently large $N$, taking logarithms givesHere the remainder uses Big O notation. Exponentiating provesThis is the quantum Zeno effect: increasingly frequent ideal projective measurements inhibit the transition away from the initial quantum state. The departure probability in a short interval is quadratic in its duration, so $N$ intervals of duration $T/N$ produce a total leading departure probability proportional to $T^2/N$. Without intervening measurements, the return probability at time $T$ would be ${\cos }^2(\omega T)$. The result concerns obtaining $+\hslash /2$ at every measurement in the ideal limit of instantaneous measurements with unitary time evolution between them.

The initial quantum state is ${\psi }_1$, up to global phase, and ${\psi }_1=({\chi }_{+}+{\chi }_{-})/\sqrt{2}$. Applying unitary time evolution givesSince $J_3{\psi }_1=(\hslash /2){\psi }_1$ and $J_3{\psi }_2=-(\hslash /2){\psi }_2$, the Born rule givesThe other outcome has probability ${\sin }^2(\omega t)$, verifying that the two probabilities sum to one. This is a two-level quantum return probability.

By the Born rule, the probability for a measured energy eigenvalue is the squared norm of the orthogonal projection onto its whole energy eigenspace. ThusEvery other energy eigenvalue has probability zero. These probabilities are independent of $T$: unitary time evolution under the unchanged Hamiltonian operator changes the dynamical phases but not the squared expansion coefficients. The two equal-energy terms describe one measurement outcome.

Let ${\Pi }_{\star }=|{\varphi }_2⟩⟨{\varphi }_2|+|{\varphi }_{-2}⟩⟨{\varphi }_{-2}|$ be the orthogonal projection for $E_{\star }$. The Lüders rule gives the normalized post-measurement stateThe factor $-e^{-i{E}_{\star }T/\hslash }$ is a global phase, so $\cos (2x/R)/\sqrt{\pi R}$ is an equivalent representative. The projective measurement preserves the quantum coherence between the two momentum eigenstates; it does not select one of them. This matters for the subsequent position probabilities.

For an arc represented by $0\le x\le a$, take $0\le a\le 2\pi R$. The Born rule usesIntegrate using ${\cos }^{2}u=(1+\cos 2u)/2$:The result is zero at $a=0$ and one at $a=2\pi R$, checking the wavefunction normalization. At $s=0$ it reduces to $a/(2\pi R)+\sin (4a/R)/(8\pi )$, the stationary position probability. Longer intervals on the periodically identified coordinate require specifying the corresponding subset of the circle; integrating multiple turns would count the same positions repeatedly.

The characteristic polynomial of the rotation matrix isIts roots are $\cos U\pm i\sin U$, soFor example $(1,i{)}^T$ is an eigenvector for $e^{iU}$ and $(1,-i{)}^T$ for $e^{-iU}$. At $U=0$ or $\pi$ the two eigenvalues coincide; these events have probability zero under the uniform distribution.

Use Haar sampling by Gaussian QR decomposition:

The matrix $G$ is invertible almost surely. Indeed, successively sampled Gaussian columns have probability zero of lying in the proper linear subspace spanned by the previous columns. Thus the diagonal signs used above exist with probability one. Equivalently, apply the Gram-Schmidt process to the columns, always dividing each residual by its positive Euclidean norm; this produces $M$ directly.

To prove uniformity, fix $H\in O(n)$. Each column of $G$ has density proportional to $e^{-\Vert x{\Vert }^2/2}$, so orthogonal transformations preserve its probability distribution. Column independence therefore gives $HG\stackrel{\mathrm{d}}{=}G$.

The QR decomposition with positive diagonal is unique: if $G=Q_1{R}_1=Q_2{R}_2$, then $Q_2^T{Q}_1=R_2{R}_1^{-1}$ is both an orthogonal matrix and an upper triangular matrix, hence diagonal with entries $\pm 1$. Its diagonal entries are positive, so it is the identity. It follows that the orthogonal factor of $HG=HMR$ is $HM$. ThereforeBy uniqueness of normalized Haar measure on the compact group $O(n)$, the returned matrix has exactly the required probability distribution. Reflection invariance gives each determinant sign probability $1/2$, so the algorithm samples both components of the orthogonal group. The positive-diagonal correction is essential when a numerical QR decomposition chooses its own signs.