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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.

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.

A superconducting weak link is a connection between regions with superconductivity whose coupling is weak enough that a substantial difference of quantum phase can develop across the connection. An insulating tunnel barrier is one realization; a narrow conducting constriction can also couple the regions. The current-phase relation of a superconducting weak link depends on the connection. A sinusoid is the leading tunnel-junction approximation, not a universal law for every superconducting weak link. ^[ref]

At fixed positive bias ratio $s=I_b/I_c$ with $0<s<1$, the Josephson coupling energy and work of the ideal bias source combine intoThis is the effective potential energy for an unwrapped Gauge-invariant Josephson phase; it includes the bias source and decreases by $2\pi s{E}_J$ on advancing one period. A local minimum and the adjacent downhill barrier areTo verify the barrier, solve $U^{\prime }=E_J(\sin \varphi -s)=0$, select $U^{\prime \prime }({\varphi }_m)>0$ and $U^{\prime \prime }({\varphi }_b)<0$, and subtract their potential energies. The small-oscillation angular frequency is fixed by the curvature:Thus the barrier and oscillation frequency both vanish as $s\to 1^{-}$, but with different powers. For $ϵ=1-s\ll 1$, expanding $\arccos (1-ϵ)$ and $\sqrt{2ϵ-{ϵ}^2}$ givesThe first approximation is for a nearly critical bias. At $s=0$ the exact downhill barrier is $2E_J$; at $s\ge 1$ there is no confining local minimum in this model.

For those stipulated parameters, direct evaluation gives $E_J/h=496.684\mathrm{GHz}$, $\Delta U/h=29.7670\mathrm{GHz}$, and ${\omega }_p/(2\pi )=2.89608\mathrm{GHz}$. Hence $\Delta U/(\hslash {\omega }_p)=10.2784$. The nearly critical barrier approximation gives $29.6165\mathrm{GHz}$, about $0.51\%$ below the exact result. These are original circuit calculations, not parameters assigned to the historical samples.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

In the Heisenberg picture, an observable without explicit time dependence hasHere $\hslash$ is the reduced Planck constant. An emitted photon on a transition $n\to m$ with $E_n>E_m$ has positive angular frequency $(E_n-E_m)/\hslash =|{\omega }_{mn}|$. The sign of the array's quantum phase follows its index convention; it is not a negative measured photon frequency. The frequencies involve differences of energy levels, rather than the absolute values $E_n/\hslash$. Adding a constant to $H$ leaves these frequencies unchanged.

Insert the complete orthonormal basis between two linear operators:The sum runs over intermediate states. The second identity follows by cancellation of $E_k$; it makes every term in the product acquire the same time factor $e^{i{\omega }_{mn}t}$. This is a useful reason for matrix multiplication: it composes the two indexed arrays consistently with the addition of transition frequencies. It does not multiply matching entries independently. Swapping $A$ and $B$ generally changes the intermediate-state sum and hence the result.

The particle exchange operator $P_{12}$ swaps the complete particle slots, including spin. Its antisymmetric projector is $A_{12}=(I-P_{12})/2$. For orthonormal one-particle states $u,v$, the normalized multiparticle quantum state isBut attempting the same construction with $v=u$ gives the zero vector:The zero vector cannot be normalized to a quantum state. This is exclusion as cancellation of probability amplitudes, rather than a repulsive force. More generally, for normalized $u,v$ the squared norm of the numerator divided by $\sqrt{2}$ is $1-|⟨u|v⟩{|}^2$; the expression has unit norm only when $u,v$ are orthogonal. An overall quantum phase difference between $u,v$ describes the same one-particle state and also gives zero. Pauli explicitly connects exclusion with a vanishing antisymmetric wavefunction.^[ref]

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

Let each drive region have length $\ell$ and the drive-free gap have length $L$, measured edge to edge. A particle of constant speed $v>0$ experiences $\tau (v)=\ell /v$ and $T(v)=L/v$. For a normalized detected-beam speed probability density function $f(v)$, sum the transition probabilities, not the probability amplitudes of different particles:The displayed $f$ is an ideal effusion flux distribution, with speed scale $u$, uniform transmission and speed-independent detection efficiency. It is not a claim about any particular clock's selected beam. The substitution $x=(v/u{)}^2$ gives $f(v)dv=x{e}^{-x}dx$, proving normalization. Actual apertures, state selection and readout can change the effective weights.

For an original example choose $\ell =0.10\mathrm{m}$, $L=1.00\mathrm{m}$, $u=v_{\ast }=200\mathrm{m}{\mathrm{s}}^{-1}$ and $\Omega =2\pi \times 500\mathrm{rad}{\mathrm{s}}^{-1}$. The reference particle has ${\tau }_{\ast }=0.500\mathrm{ms}$, $T_{\ast }=5.00\mathrm{ms}$ and resonant $\pi /2$ pulses. The instantaneous-pulse width would be $100\mathrm{Hz}$. The exact finite-pulse width for that single speed is about $88.7\mathrm{Hz}$; the chosen distribution also reduces the on-resonance average to approximately $0.738$ because not every particle receives a $\pi /2$ pulse.

Different velocities smear side fringes because $\delta L/v$ varies across the beam. Nevertheless every term is even in $\delta$, so this symmetric ideal-model average has no odd line asymmetry that could displace the central lock point. Reduced contrast or altered width alone is not evidence of a shifted line center. A speed-dependent transition shift or unequal drive quantum phases changes that conclusion; atomic clock frequency shifts explains why these distinctions matter experimentally.

A wavefunction is the position representation of a pure state. In the one-dimensional model used here, $\psi (x,t)$ takes values in the complex numbers, and the Born rule givesThe probability density $\rho$ has units of inverse length; the wavefunction has units of inverse square root of length. A value of $\rho$ at a point is not itself a probability: probabilities come from an integral over a region. For the infinite square well ground state, $\rho (L/2)=2/L$, whereas the probability of the exact single point $L/2$ is zero. The interesting question is how much probability a finite interval contains.

Two wavefunctions differing only by a global phase represent the same pure state. Relative quantum phases between components of a quantum superposition affect matter-wave interference, so retaining the full complex number amplitude is essential. This scalar description covers a particle without active spin degrees of freedom; a general electron wavefunction can have two spin components, with the position probability density obtained by adding the squared modulus of each component.

For two coherent square pulses of equal duration $\tau$, pulse-start separation $T\ge \tau$, resonant Rabi angular frequency $\Omega$, and Rabi detuning $\delta$, the initially lower-state transition probability has the displayed leading weak-excitation form when $|\Omega |\tau \ll 1$. Integrating the first-order probability amplitude over both windows gives a sum of two equal Fourier transforms, with relative quantum phase $\delta T$. The factor four from coherent addition makes the central value $(\Omega \tau {)}^2$. The rotating-wave approximation additionally requires many carrier cycles per pulse and near-resonant weak driving.