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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.

Inside the infinite square well, $V=0$, and the time-independent Schrödinger equation isFor $E>0$, write $k^2=2mE/{\hslash }^2$ and $\psi =A\sin kx+B\cos kx$. The first Dirichlet boundary condition gives $B=0$, and the second requires $kL=n\pi$ for a nonzero state. The case $n=0$ is identically zero; negative $n$ only changes an overall quantum phase. For $E=0$ a linear solution satisfying both walls is zero, and for $E<0$ the hyperbolic solutions likewise cannot satisfy both walls nontrivially.

Wavefunction normalization, ${\int }_0^L|\psi {|}^{2}dx=1$, fixes $|A|=\sqrt{2/L}$. The orthonormal energy eigenstates and energy eigenvalues are thereforeEach state vanishes outside the well. Its stationary time-dependent form is ${\Psi }_n(x,t)={\psi }_n(x)e^{-i{E}_{n}t/\hslash }$. General states are normalized superpositions of these modes, rather than additional single energy eigenstates.

Assume the given wavefunction is zero outside $[0,L]$. Wavefunction normalization requiresAn arbitrary overall quantum phase is unobservable, so choose the positive real representative $N=\sqrt{3}/L^{3/2}$.

The linear ramp state in an infinite square well has a boundary subtlety: the limiting value at $L$ is $\sqrt{3/L}$, rather than zero. The function is still a normalized element of the well's Hilbert space $L^2(0,L)$, so position measurement and energy-projection probabilities are meaningful. It is not in the domain of the Dirichlet Hamiltonian, nor is it a finite-energy box state. Altering the value at the single endpoint does not cure its nonzero boundary trace; a physical finite-energy preparation must smooth the function to zero near the wall. The following probabilities refer to the supplied idealized state.

An energy measurement selects the ground state with probability equal to the squared inner product with ${\psi }_1$. Using $N=\sqrt{3}/L^{3/2}$ and integration by parts,ThusThe full Fourier sine series has $c_n=(-1{)}^{n+1}\sqrt{6}/(n\pi )$, so ${\sum }_n|c_n{|}^2=(6/{\pi }^2){\sum }_n{n}^{-2}=1$, an independent wavefunction normalization check. However $E_n|c_n{|}^2=3{\hslash }^2/(m{L}^2)$ for every $n$, hence ${\sum }_n{E}_n|c_n{|}^2$ diverges. This explicitly verifies the infinite-energy expectation qualification in part (b), without invalidating the individual projection probabilities.

Multiplying the quantum harmonic oscillator ground state by $x^2$ produces a quantum superposition of only its $n=0$ and $n=2$ energy eigenstates. In the conventional phase choice for the creation and annihilation operators, $X=\sqrt{\hslash /(2m\omega )}(b+b^{†})$ and $b|0⟩=0$. ThereforeAfter wavefunction normalization, the energy measurement probabilities are $1/3$ at $\hslash \omega /2$ and $2/3$ at $5\hslash \omega /2$. Rephasing the second basis state can change the displayed relative coefficient's sign but cannot change these probabilities. For example, defining the unscaled creation operator as $a_{+}=P+im\omega X=i\sqrt{2m\hslash \omega }{b}^{†}$ reverses the sign of the $n=2$ basis function generated from $(a_{+}{)}^2|0⟩$.

Quantum mechanics describes a quantum system using a quantum state, with observables represented by self-adjoint operators. For a single particle in one dimension, a pure state can be described by a wavefunction $\psi (x,t)$: its squared modulus gives the probability density for position. The Schrödinger equation predicts how this probability amplitude evolves between interventions; measurement in quantum mechanics connects it to experimental outcomes through the Born rule. A wavefunction therefore predicts a distribution of possible positions rather than a single trajectory as in classical mechanics.

The infinite square well makes three central ideas visible with elementary sine functions. Confinement selects discrete energy eigenvalues; wavefunction normalization converts a shape into quantitative probabilities; and a quantum superposition can have a changing probability density even when each constituent energy eigenstate has a stationary one. This combination explains why quantum mechanics uses both linear algebra and differential equations: eigenfunctions describe the building blocks, while quantum phases determine how they interfere. The same tools help explain the discrete energies of chemical atoms and molecules, although their Quantum Hamiltonians require different potentials and often many particles.

In his first 1926 paper, Erwin Schrödinger formulated quantization as an eigenvalue problem for a hydrogen atom in nonrelativistic quantum mechanics. On its opening page he compared the emergence of integers with the number of nodes of a standing wave on a vibrating string. The infinite square well below realizes that appealing idea especially transparently: the allowed integers follow from the boundary conditions. This historical analogy concerns the origin of discrete energies; the Born rule supplies the statistical interpretation of the modern wavefunction. ^[ref].

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

For a spherically symmetric potential energy $V(r)$, separation of variables with $\psi =R(r)Y_{lm}(\theta ,\varphi )$ reduces the Schrödinger equation to the displayed radial equation. Writing $u=rR$ instead removes the first derivative:The angular contribution acts as a centrifugal barrier. Boundary behavior and wavefunction normalization must still be imposed; an arbitrary radial function is not an energy eigenfunction.

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.

In the product basis $|m_1,m_2⟩$ for two spin-one particles, total magnetic quantum number zero allows only $|1,-1⟩$, $|0,0⟩$ and $|-1,1⟩$. Applying the total ladder operators to $\alpha |1,-1⟩+\beta |0,0⟩+\gamma |-1,1⟩$ givesA singlet state is annihilated by $S_{+}$ and $S_{-}$, so $\alpha =\gamma =-\beta$. Wavefunction normalization gives the displayed state, up to an overall phase. Here the ket on the left of the title formula has total-spin labels, while its right-hand kets have individual magnetic labels. The state is symmetric under exchanging the particles, although no exchange constraint is needed for distinguishable particles.

Consider the original two-electron example obtained by taking $N=2$ in the Laughlin wavefunction, with no additional polynomial for the centre of mass. Introduce $Z=(z_1+z_2)/2$, relative complex coordinate $w=z_1-z_2$, separation $r=|w|$, and dimensionless separation $u=r/{\ell }_B$. Since $|z_1{|}^2+|z_2{|}^2=2|Z{|}^2+|w{|}^2/2$, the probability density factorizes:Integrating out $Z$ leaves an isotropic relative probability density. Define $D_m(u)$ per unit dimensionless area and $p_m(u)$ per unit dimensionless separation:The physical areal probability density is $D_m(r/{\ell }_B)/{\ell }_B^2$. The probability of a separation between $r$ and $r+dr$ is $p_m(r/{\ell }_B)dr/{\ell }_B$. The extra $u$ comes from the Jacobian determinant in polar coordinates: a thin annulus has area $2\pi rdr$. Omitting that factor changes the question being answered.

Both wavefunction normalization and the peak shift can be checked exactly. The substitution $t=u^2/4$ gives the Gamma function integralFor $m>0$, logarithmic differentiation givesConsequently the areal maximum lies at $u=2\sqrt{m}$, while the most probable separation lies farther out, at $u=\sqrt{4m+2}$. For $m=1$ these are $2$ and $\sqrt{6}\simeq 2.449$; for $m=3$, $\sqrt{12}\simeq 3.464$ and $\sqrt{14}\simeq 3.742$. The same substitution gives $⟨u^2⟩=4(m+1)$, so the root mean square separations are $\sqrt{8}$ and $4$ respectively. The distinction between a probability density maximum and a radial maximum therefore has a directly calculable geometric origin.

This two-electron model explains exchange and pair avoidance. It does not establish a nonzero many-particle spectral gap, an extended quantum Hall plateau, or the topological properties of a fluid in the thermodynamic limit. Those require additional many-electron reasoning and experimental tests.

If a trial quantum state is dominated by two energy eigenstates of distinct energies $E_a<E_b$, then its Rayleigh quotient is approximately their probability-weighted mean. Rearranging that mean gives the displayed relative admixture. Neglected states must make small contributions to the energy numerator as well as to wavefunction normalization; small probabilities alone need not suffice if their energies are large.

Wavefunction normalization makes the total Born rule probability equal to one. In an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\varphi }_n$ is normalized precisely when ${\sum }_n|c_n{|}^2=1$.

For a proposed infinite square well mode $A\sin (n\pi x/L)$, the integral of its squared modulus is $|A{|}^{2}L/2$. Thus wavefunction normalization requires $|A|=\sqrt{2/L}$; choosing $A$ real and positive fixes a convenient global phase. For the equal quantum superposition $({\varphi }_1+{\varphi }_2)/\sqrt{2}$, orthogonality makes the integrated cross term zero, leaving $1/2+1/2=1$. The cross term can nevertheless be nonzero locally: normalization constrains total probability, not the shape of its probability density.