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For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

A quantum superposition is a linear combination of quantum state vectors in a complex Hilbert space. In a specified orthonormal basis, a normalized pure state has the displayed expansion, where $a_j$ are probability amplitudes. A nonzero linear combination can always be normalized; an arbitrary sum of normalized vectors need not already have norm one. Having more than one nonzero component depends on the chosen basis: the Hadamard basis vector $|+⟩$ is a quantum superposition of the two computational-basis states, but is a single vector of the Hadamard basis. ^[ref].

The distinction from classical ignorance is quantitative. Consider the normalized qubit quantum stateIts density operator, written in the computational basis, isPreparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$, while discarding the preparation label, instead gives the mixed stateThe diagonal entries agree, so both preparations give the same computational-basis measurement probabilities. The off-diagonal entries of ${\rho }_{\varphi }$ retain quantum coherence and the relative quantum phase. In the Hadamard basis, the Born rule givesThe density matrix representation retains this distinction between a pure state and a mixed state. ^[ref].

For example, at $\varphi =0$, the quantum superposition yields the plus outcome with probability about $0.9330$, whereas the mixed state yields it with probability $0.5$. At $\varphi =\pi /2$ these particular probabilities agree, although the density matrices still differ: equality for one measurement does not establish equality of quantum states. The probability amplitude figure shows the comparison throughout a full cycle of relative quantum phase.

This is why a quantum superposition is useful: quantum gates act linearly on its probability amplitudes, and unitary operators preserve the full pure state information, including relative quantum phases. A Hadamard gate can turn otherwise hidden quantum coherence into different outcome probabilities. The beautiful feature is that the same basis weights can support different predictions when amplitudes recombine. The expansion specifies a preparation and its measurement predictions; it does not assign every unmeasured outcome a simultaneous classical value. See projective measurement for the separate question of the conditional state after a measurement.

Quantum theory supplies a common language for quantum mechanics, measurement in quantum mechanics, and quantum information: prepare a quantum state, transform it, and ask which experimental outcomes occur. In a finite-dimensional idealization, a pure state is a normalized vector of complex numbers. Its entries are probability amplitudes; the Born rule turns their squared magnitudes into probabilities. Between measurements, an isolated system evolves by a unitary operator, preserving its norm and inner products. A quantum circuit makes this evolution explicit as a sequence of quantum gates.

The extra information carried by quantum phases is essential. Relative quantum phases affect how probability amplitudes add when different contributions reach the same output. Their interference can enhance one outcome and cancel another. A global phase, multiplying the entire quantum state by the same complex factor of magnitude one, changes neither these relative phases nor observable predictions. The Hadamard gate below provides a small experiment in which this distinction becomes a deterministic change of output.

Quantum coherence describes this capacity to interfere relative to a specified orthonormal basis. In a density matrix, it appears in off-diagonal entries; a diagonal mixed state retains the populations but lacks that basis coherence. Thus matching the probabilities of one measurement need not match the predictions of a later experiment. Baumgratz, Cramer, and Plenio made this basis-dependent viewpoint precise in their resource framework for quantum coherence. ^[ref].

This makes quantum information an operational question: which distinctions can a preparation, an available unitary operator, and measurement in quantum mechanics reveal? Amplitude amplification illustrates how carefully arranged interference converts a weakly represented answer into a likely observable result. Preserving the relevant quantum coherence throughout the computation is part of the physical task. ^[ref].

For a prepared normalized pure state $|i⟩$ and a final projective measurement onto the normalized state $|f⟩$, unitary time evolution gives the displayed probability, by the Born rule. This depends on the preparation, final measurement, and time-evolution operator; it does not assume a classical Markov chain of transitions.

A qubit, or quantum bit, is a quantum system whose state space is a two-dimensional complex Hilbert space. Choose an orthonormal basis $|0⟩,|1⟩$, called the computational basis. A pure state has a normalized representative $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, where $\alpha ,\beta$ are complex probability amplitudes and $|\alpha {|}^2+|\beta {|}^2=1$. General qubit states, including mixed states, require a density matrix. ^[ref].

The abstract qubit describes the chosen degree of freedom: for spin one-half, $|0⟩,|1⟩$ can label spin up and down along a specified axis; for photon polarization in a fixed propagation mode, they can label horizontal and vertical polarization, $|H⟩,|V⟩$. These implementations share the same two-dimensional linear algebra, but their preparation and measurement in quantum mechanics use different apparatus. A qubit therefore need not be an entire two-level particle. ^[ref].

The usefulness of a qubit lies in controlling quantum superpositions and their relative quantum phases, then converting them into outcome probabilities with quantum gates and an appropriate measurement. One ideal computational-basis measurement returns one label, zero or one. It does not reveal both probability amplitudes, their relative quantum phase, or all the bits of a larger register's expansion. Quantum state tomography instead estimates a qubit state using many repetitions of the preparation and suitable measurement settings. ^[ref].

An actual preparation and measurement example is Dederick and Beck's 2014 teaching-laboratory experiment. A 405-nm laser pumped two nonlinear crystals, producing signal and idler photons near 810 nm. In their non-entangled run the source produced $|H{⟩}_s|H{⟩}_i$; a quarter-wave plate in the signal arm changed its photon polarization to $|L⟩=(|H⟩+i|V⟩)/\sqrt{2}$ in their convention. The plate supplies a relative quantum phase between polarization components. An idler detection selected the signal trials; signal detectors measured horizontal/vertical, diagonal, and circular polarization bases on separate runs. Before the polarization analyzer, a half-wave plate at $22.5$ degrees selected the diagonal basis, and a quarter-wave plate at $45$ degrees selected the circular basis. Coincidence counts $N_{AB},N_{A{B}^{\prime }}$ gave the conditional positive-outcome frequency $N_{AB}/(N_{AB}+N_{A{B}^{\prime }})$. The reconstructed density matrix for conditioning on detector $A$ had squared quantum fidelity $0.96$ with $|L⟩$ and purity of a density operator $0.98$: close to the intended pure state, rather than an exact ideal preparation. ^[ref].

The model diagram explains why those settings matter. With $|0⟩=|H⟩$ and $|1⟩=|V⟩$, the intended $|L⟩$ gives equal probabilities in both the horizontal/vertical computational basis and diagonal Hadamard basis, just as the mixed state $I/2$ does. In the circular basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, however, the ideal $|L⟩=|+i⟩$ always gives the plus outcome, whereas $I/2$ still gives equal probabilities. These are deductions from the Born rule, not the paper's measured counts. Unequal detector efficiencies and finite samples must be accounted for when interpreting experimental frequencies; the ideal comparison assumes equally efficient detection.

A qubit state specifies measurement probabilities, rather than a pre-existing answer to every possible measurement. For a pure state, normalized vectors differing by a global phase represent the same physical state:Equivalently, the pure state is a ray: the one-dimensional complex linear subspace spanned by any nonzero representative. Dividing a nonzero vector by its norm fixes its length, but leaves the global phase free. The density operator is unchanged because $(e^{i\gamma }|\psi ⟩)(e^{i\gamma }|\psi ⟩{)}^{†}=|\psi ⟩⟨\psi |$. In contrast, multiplying only one nonzero component by $e^{i\varphi }$ changes a relative quantum phase and can change measurement probabilities. ^[ref].

After choosing a global phase, every pure state of a qubit can be writtenAt $\theta =0$ or $\pi$, the parameter $\varphi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative quantum phase. The corresponding unit Bloch vector is $\mathbf{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ on the existing Bloch sphere. Its angles describe a space of states; for a photon polarization qubit they are not the photon's position or direction of travel. ^[ref].

The Born rule in the computational basis gives $P(0)={\cos }^2(\theta /2)$ and $P(1)={\sin }^2(\theta /2)$. These probabilities discard $\varphi$. In the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, adding or subtracting the probability amplitudes instead givesThis last equality follows by substituting the angular representative. In the basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, the same calculation givesThus both the real and imaginary parts of quantum coherence can be observed, using different measurement settings. For example, $(|0⟩+|1⟩)/\sqrt{2}$ and $(|0⟩-|1⟩)/\sqrt{2}$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A global phase could never cause this difference.

For a general qubit state, write its density matrix in the computational basis asThese are exactly the Hermitian, unit-trace, positive semidefinite constraints. Positivity means every eigenvalue is nonnegative, not that every matrix entry is nonnegative. The general definition is given in density matrix. ^[ref].

For this two-by-two matrix, the explicit constraint follows from its determinant, $\det \rho =p(1-p)-|c{|}^2$: with trace one, a nonnegative determinant and real eigenvalues rule out a negative eigenvalue. Its eigenvalues areThe pure states have $|c{|}^2=p(1-p)$ and eigenvalues $1,0$; the mixed states have strict inequality and two positive eigenvalues. For a normalized representative, $p=|\alpha {|}^2$ and $c=\alpha {\beta }^{\ast }$. A statistical preparation $\rho ={\sum }_j{w}_j|{\psi }_j⟩⟨{\psi }_j|$, with $w_j\ge 0$ and ${\sum }_j{w}_j=1$, averages probabilities, rather than adding probability amplitudes. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A reduced density matrix of an entangled pair can also be mixed even when the pair has a pure state. ^[ref].

For a concrete check, the pure state $(\sqrt{3}|0⟩+|1⟩)/2$ and the mixed state obtained by preparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$ haveBoth have trace one and the same computational-basis measurement probabilities. Their eigenvalues are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their purity of a density operator $\mathrm{Tr}({\rho }^2)$ are $1$ and $5/8$. Their Hadamard basis plus probabilities are respectively $1/2+\sqrt{3}/4\approx 0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give determinant $-1/16$ and eigenvalue $(2-\sqrt{5})/4<0$: normalized diagonal entries alone do not make a valid qubit state.

The same constraints become especially beautiful in the Bloch ball representation. With the Pauli matrices $\mathbf{\sigma }=(X,Y,Z)$ and the identity operator $I$,The eigenvalues are $(1\pm |\mathbf{r}|)/2$, so the Bloch sphere is precisely the pure state boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the Born rule gives $P_j(\pm )=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\varphi }/2$ and $\mathbf{r}=(\cos \varphi ,\sin \varphi ,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See quantum state tomography for how independent ensembles measured along the three axes estimate these components. ^[ref].

In a well $-a<x<a$, the normalized state proportional to $x$ on $0<x<a$ and zero elsewhere has stationary-state coefficientsThey follow from the orthonormal basis inner products. Their squared magnitudes give energy-measurement probabilities by the Born rule; splitting even and odd indices in the Parseval identity gives convergent integer-coefficient series identities. The expansion is interpreted in the square-integrable state norm.

Ramsey interferometry uses two coherent pulses with a free interval between them: the first creates a quantum superposition, the interval accumulates relative quantum phase, and the second converts that quantum phase into measurement probabilities. Its attraction is that an ordinary population readout can reveal the quantum coherence hidden from an energy measurement during the free interval.

Here is an ideal qubit model with an explicit pulse convention. Let $R_y(\theta )=\exp (-i\theta {\sigma }_y/2)$, prepare $|0⟩$, apply $R_y(\pi /2)$, evolve freely with $U_{\varphi }=\exp (-i\varphi {\sigma }_z/2)$, and apply the inverse pulse $R_y(-\pi /2)$ before a computational-basis measurement. The first pulse gives $|+⟩$. Direct matrix multiplication givesThus the Born rule gives $P(0)={\cos }^2(\varphi /2)$ and $P(1)={\sin }^2(\varphi /2)$. Before the second pulse, the same computational-basis measurement would give $1/2$ for both outcomes for every $\varphi$. At $\varphi =2\pi /3$, the second pulse changes these probabilities to $1/4$ and $3/4$. If the two pulses have the same sign instead, the output labels interchange: the pulse convention is part of the prediction.

In a pulse experiment the phase is measured relative to the oscillator supplying the coherent pulses. In a rotating frame, write the free Hamiltonian operator as $H_{\mathrm{rot}}=\hslash \delta {\sigma }_z/2$ with signed angular frequency difference $\delta$ in the chosen labeling convention. A free interval $T$ then accumulates $\varphi =\delta T$; changing the second pulse's phase changes the analysis direction. This comparison makes Ramsey interferometry useful for measuring frequency differences and diagnosing loss of quantum coherence.

For example, suppose the first pulse has prepared a density matrix with equal populations and transverse Bloch vector length $0\le V\le 1$:The same inverse analysis pulse yields $P(0)=(1+V\cos \varphi )/2$. The pure state case $V=1$ gives the full ideal fringe; the mixed state $I/2$ gives the flat result $1/2$. This is a controlled mathematical comparison, not a fit to an experiment. Experimental contrast alone does not identify quantum decoherence: imperfect preparation, pulse errors and readout errors can also reduce it.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

A selective measurement in quantum mechanics conditions on a known outcome. For Kraus operator $A_i$, the Born rule gives $p_i=\mathrm{Tr}(A_i^{†}{A}_i\rho )$ and the conditional density operator is $A_i\rho A_i^{†}/p_i$ when $p_i>0$. Conditioning changes the state and can change the conditional reduced density matrix of a correlated remote system. That conditional state is available as such only to an observer with the outcome information.

Consider an original balanced-readout model, rather than a fit to the experiment. Let $n\in \{0,\dots ,N_{\max }\}$, with a normalized prior probability $p_0(n)\ge 0$ and ${\sum }_n{p}_0(n)=1$. A detected chemical atom has outcome $s\in \{e,g\}$. Define the conditional probabilitiesHere $\chi$ is the accumulated quantum phase per photon, $\varphi$ the controlled analysis quantum phase, and $C$ the fringe contrast. The two probabilities sum to one for every $n$. Actual calibration may also require an offset and asymmetric detection errors; $C$ alone does not specify a quantum instrument.

Assume no photon number change during the record and fresh independently prepared atomic meters. For observed outcomes $s_1,\dots ,s_k$ at recorded settings ${\varphi }_1,\dots ,{\varphi }_k$, the Born rule and Bayes' theorem giveThe posterior probability is defined only for $Z_k>0$. Summing its numerator over $n$ proves normalization. This is also the update of diagonal elements of a field density operator under a measurement in quantum mechanics; it does not assume that an initial quantum superposition was an unknown pre-existing integer. The observed settings are part of the record, especially when selected adaptively.

For an exact finite example take $p_0=(1,1,1,1)/4$, $C=4/5$, and $\chi =\pi /2$. An $e$ outcome at $\varphi =0$ has likelihood $(9,5,1,5)/10$, giving $p_1=(9,5,1,5)/20$. Another $e$ at $\varphi =-\pi /2$ gives $p_2=(9,9,1,1)/20$. A subsequent $g$ at $\varphi =0$ givesThus three atomic records raise the posterior probability of $n=1$ from $1/4$ to $45/68$, while leaving finite uncertainty. No one of these outcomes identifies $n$ with certainty.

The periodic likelihood function imposes an identifiability limit. If $\chi =2\pi /q$ for an integer $q>0$, then $L_s(n+q;\varphi )=L_s(n;\varphi )$ for every setting: the posterior odds between $n$ and $n+q$ never change. Varying $\varphi$ can resolve a cosine reflection ambiguity within a period, but cannot remove this exact period. Changing $\chi$ can supply additional information. A finite cutoff is therefore a physical assumption to justify, not evidence that excluded high photon numbers were absent. At $C=0$ no outcome changes the prior probability; at $0<C<1$ neither binary outcome has an exactly zero likelihood.

The no-loss assumption also has a time limit. In an original zero-temperature loss model, let $\kappa =1/T_c$. A Fock state $|n⟩$ loses its first photon at rate $n\kappa$, so its survival probability is $\exp (-n\kappa t)$ and its mean lifetime is $T_c/n$ for $n>0$. Several records must be acquired within that lifetime to identify a plateau. When losses matter, first predict $p_k^{-}(n)={\sum }_m{T}_{nm}(\Delta t)p_{k-1}(m)$ using a normalized transition matrix, then apply the atomic Bayesian inference update to $p_k^{-}$. Thermal excitation requires upward transitions too. A delayed or broadened inference after a jump need not indicate that an atom caused the jump.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

Prepare the $J_1=+\hslash$ spin coherent state and evolve with Hamiltonian operator $H=\omega J_3$ for time $t$. In the descending-$J_3$ basis of the spin-one angular momentum matrices, the initial vector is $(1,\sqrt{2},1{)}^{\mathsf{T}}/2$ and the evolved vector is $(e^{-i\theta },\sqrt{2},e^{i\theta }{)}^{\mathsf{T}}/2$, where $\theta =\omega t$. The $J_1$ orthonormal basis has vectors $(1,\sqrt{2},1{)}^{\mathsf{T}}/2$, $(1,0,-1{)}^{\mathsf{T}}/\sqrt{2}$ and $(1,-\sqrt{2},1{)}^{\mathsf{T}}/2$. Taking their inner products with the evolved vector gives probability amplitudes $(1+\cos \theta )/2$, $-i\sin \theta /\sqrt{2}$ and $(\cos \theta -1)/2$. Their squared moduli give the Born rule probabilities in the title. At integer half-turns $\theta =\pi k$, the outcome is deterministically $(-1{)}^k\hslash$.

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.

The ideal projective measurement of a spin-one-half component along a unit axis $\mathbf{n}$. The two outcomes have Born rule probabilities $⟨\psi |P_{\pm }|\psi ⟩$. For an input with Bloch vector $\mathbf{r}$, they are $(1\pm \mathbf{n}\cdot \mathbf{r})/2$. This model distinguishes a mathematical spin measurement from engineering a literal free-electron beam splitter.

The sudden approximation treats a rapid, finite change of a Hamiltonian operator as leaving the quantum state approximately unchanged during the switch. The energy measurement can nevertheless change immediately: the same quantum state has different probability amplitudes in the new eigenbasis. Compute their inner products with the new energy eigenstates, then apply the Born rule. The two-level sudden quench below makes this distinction quantitative without assuming the change is small.

An ideal finite step has exact quantum state continuity at a finite Hamiltonian step; a finite-duration implementation has a small evolution error under the bounded-model conditions below. An impulsive quantum kick is different: its Hamiltonian operator grows as its duration shrinks, retaining a finite integrated action and a finite change of quantum state. A sudden approximation therefore concerns short evolution during switching, not conservation of energy during externally driven change. The subsequent unitary time evolution can be substantial.^[ref]

Let orthonormal energy eigenvectors be ${\Psi }_1,{\Psi }_2$, and let a measured observable have eigenvectors $v_{+}=\cos \alpha {\Psi }_1+\sin \alpha {\Psi }_2$ and $v_{-}=-\sin \alpha {\Psi }_1+\cos \alpha {\Psi }_2$. If the initial quantum state is $v_{-}$, unitary time evolution multiplies each energy component by $e^{-it{E}_j/\hslash }$. Its amplitude in $v_{+}$ is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. The Born rule gives the displayed probability. The energy gap fixes the oscillation frequency, while the angle between the measurement and energy bases fixes its amplitude. Energy degeneracy or coinciding bases makes the transition probability vanish.

For two orthonormal energy eigenstates ${\psi }_1,{\psi }_2$, let an observable have distinct nondegenerate eigenstates ${\varphi }_{\pm }=({\psi }_1\pm {\psi }_2)/\sqrt{2}$. Starting in ${\varphi }_{-}$, unitary time evolution gives $(e^{-i{E}_{1}t/\hslash }{\psi }_1-e^{-i{E}_{2}t/\hslash }{\psi }_2)/\sqrt{2}$. Its probability amplitude along ${\varphi }_{+}$ is $(e^{-i{E}_{1}t/\hslash }-e^{-i{E}_{2}t/\hslash })/2$. The Born rule therefore gives the stated transition probability. The energy difference determines the oscillation frequency, while a common energy shift has no effect.

A normalized quantum state $\chi =\sqrt{q}{\psi }_1+e^{i\eta }\sqrt{1-q}{\psi }_2$ in two orthonormal energy eigenstates returns to its original one-dimensional measurement subspace with probabilityThis follows by applying unitary time evolution to each energy eigenstate, then the Born rule. The relative phase $\eta$ cancels from the return amplitude. Degenerate energies make the probability identically one.