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A gallium arsenide/aluminium gallium arsenide semiconductor heterostructure combines lower-gap GaAs with a higher-gap AlGaAs barrier. In a modulation-doped design, AlGaAs donors supply electrons to lower conduction-band states on the GaAs side. An undoped spacer separates the channel from ionized donors and reduces impurity scattering. The interface barrier and band bending quantize motion normal to the interface while allowing in-plane motion: a two-dimensional electron gas.

The normal confining potential is commonly triangular and finite rather than an infinite square well. At suitable density and low temperature, only the lowest normal subband is occupied. modulation doping reduces an important scattering source; it does not eliminate all disorder.

Use negatively biased surface gates to deplete the surrounding electron gas and leave a rectangular central region. Gate voltage controls lateral dimensions and barrier heights. Two constrictions can weakly couple the region to reservoirs, with a plunger gate controlling population. Etched boundaries are possible but may introduce edge disorder. A flat interior and steep edges approximate a box, givingFringing fields, finite barriers and penetration change the ideal result. A 2DEG confines only the normal direction; gates supply the extra lateral confinement. Gate-dependent spectroscopy can test level spacings, while charging measurements separate orbital confinement from Coulomb blockade.

The one-dimensional time-independent Schrödinger equation isFor the infinite square well on $[0,a]$, $V=0$ inside and the wave function is zero outside. The boundary conditions are $\psi (0)=\psi (a)=0$. For positive integer $n$, ${\psi }_n=A\sin (n\pi x/a)$ satisfies them, and ${\psi }_n^{\prime \prime }=-(n\pi /a{)}^2{\psi }_n$. Substitution gives the energy levelsNormalization requires $1={\int }_0^a|{\psi }_n{|}^{2}dx=A^{2}a/2$. The requested real positive constant is therefore

As an estimate, model the confinement by an infinite square well of width $L$ and use the stipulated electron mass. Its energy eigenvalues are $E_j=j^2{\pi }^2{\hslash }^2/(2m{L}^2)$, $j=1,2,\dots$, so the gap controlling thermal occupation of a quantum well isThus $T\ll 130\mathrm{K}$ suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_{B}T)}$. For example $T\lesssim 28\mathrm{K}$ makes this ratio below one percent in this model. There is no sharp cutoff. The ground state energy itself corresponds to about $44\mathrm{K}$, but it is the excitation gap that enters the Boltzmann factor. Finite barrier height and a different effective mass would change the numerical scale; neither is supplied.

For observables with finite variances, the Heisenberg uncertainty principle gives $\Delta A\Delta B\ge |⟨[A,B]⟩|/2$; in particular $\Delta x\Delta p\ge \hslash /2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and Bose-Einstein statistics; half-integer-spin particles have antisymmetric full states and Fermi-Dirac statistics. The latter imply Pauli exclusion from one complete single-particle state, including spin.

Inside an infinite square well, solutions satisfying both zero boundary conditions arewith zero wavefunction outside. For the ground state, direct integration givesThe infinite-well ground-state uncertainty product follows:The ground state is not a minimum-uncertainty Gaussian. Here momentum variance is the usual kinetic-momentum second moment of the confined state; the hard-wall momentum operator has domain subtleties that do not alter this kinetic-energy result.

For spin degeneracy in filling a one-dimensional well, let $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$ and assume the spin levels are degenerate and available. Four spin-zero bosons all occupy the lowest orbital, giving $4E_1$. Four spin-$1/2$ fermions fill two spin states in each of $n=1,2$, giving $10E_1$. Spin-$3/2$ fermions have four spin projections, so all four can occupy $n=1$ in different complete states, giving $4E_1$. These fillings do not impose full spin polarization.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

Write $L=10^{-9}\mathrm{m}$ and use the Heisenberg uncertainty relation separately in the three Cartesian directions. Since a position coordinate is confined to an interval of length $L$, Popoviciu's inequality on variances gives $\Delta x_i\le L/2$. Consequently $\Delta p_i\ge \hslash /L$, and the expectation value of the kinetic energy obeysThus an uncertainty-based lower estimate is $\boxed{K\gtrsim 0.11\mathrm{eV}}$. This confinement energy bound from position support is not sharp: the exact ground-state energy for a cubical infinite square well is $3{\pi }^2{\hslash }^2/(2m_e{L}^2)=1.13\mathrm{eV}$. The Heisenberg uncertainty principle establishes the confinement scale ${\hslash }^2/(m_e{L}^2)$; identifying a box size with an approximate position standard deviation changes its numerical coefficient.

The infinite walls impose $\psi (0)=\psi (L)=0$, and the interior time-independent Schrödinger equation is ${\psi }^{\prime \prime }+k^2\psi =0$ with $E={\hslash }^2{k}^2/(2m)$. Its general solution is $A\sin kx+B\cos kx$. The first boundary condition gives $B=0$, and the second gives $kL=n\pi$. A nonzero normalizable state needs $n=1,2,\dots$; $n=0$ would give the zero function.

Wavefunction normalization, ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$, gives the infinite square well states and their energy eigenvalues:The wavefunctions vanish outside the well, and their full time dependence is ${\psi }_n(x)e^{-i{E}_{n}t/\hslash }$.

The Born rule gives the required probability by integrating the probability density over the central interval. Put $d=0.1\mathrm{nm}$. For this infinite square well ground state,Substitution of $d/L=0.1$ givesThis exceeds the interval's $20\%$ fraction of the box length because the ground state probability density is largest at the centre; it is not a uniform position distribution.

Equation and interpretation. The supplied equation is the time-independent Schrödinger equation for a nonrelativistic particle with mass $m$ in a prescribed potential energy $V(x)$. Equivalently,It is an eigenvalue problem for the Hamiltonian operator. A stationary energy state has full wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. By the Born rule, $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a small interval $dx$, with wavefunction normalization ${\int }_{-\infty }^{\infty }|\psi {|}^{2}dx=1$.

Infinite barriers. Infinite exterior potential energy excludes any finite-energy probability outside the interval. The infinite square well consequently has $\psi =0$ outside and the hard-wall boundary conditionsThe interior wavefunction must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.

Ground state. Inside the well $V=0$, so ${\psi }^{\prime \prime }+k^2\psi =0$, with $k=\sqrt{2mE}/\hslash$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\dots$. Thus the discrete energy eigenvalues are $E_n=n^2{\pi }^2{\hslash }^2/(2m{a}^2)$. The smallest, $n=1$, is the ground state. Sinceits normalized wavefunction, choosing an irrelevant overall phase to make it positive inside, is

Finite barriers. For a finite step and constant particle mass, both $\psi$ and ${\psi }^{\prime }$ are continuous at $x=0,a$. Wavefunction matching at a finite potential step follows by integrating the Schrödinger equation across a vanishingly small neighborhood of the step: a bounded potential cannot generate a finite jump in ${\psi }^{\prime }$. The bound-state exterior solutions decay rather than vanish identically:Normalizability excludes the growing exterior exponentials. The equivalent interior boundary conditions are ${\psi }^{\prime }(0)=\kappa \psi (0)$ and ${\psi }^{\prime }(a)=-\kappa \psi (a)$; neither boundary value is zero for the nontrivial ground state.

The ground state of a finite square well remains symmetric about $a/2$ and has no nodes. It bends smoothly into exponential tails at both walls, extending over a larger region than the infinite-well state. Its maximum after normalization is lower for the deep well considered here. When $V_w\gg E$, the penetration length $1/\kappa$ is short, so the two shapes are close except near the walls and in the exterior tails.

The finite curve is an illustrative exact matched solution with $V_w=10E_{1,\infty }$; no numerical wall height is specified by the question.

Energy comparison. The finite barriers relax the confinement, allowing a less rapidly varying interior wavefunction and therefore smaller kinetic energy. More explicitly, the even interior solution is proportional to $\cos [k(x-a/2)]$, and matching to the tails gives $k\tan (ka/2)=\kappa$. The lowest root lies at $0<ka<\pi$, while the infinite well has $ka=\pi$. HenceAs $V_w\to \infty$, the tails disappear and the finite-well energy approaches the infinite-well value from below. This comparison uses the same zero of potential energy inside both wells.

The infinite square well levels for width $L$ are $E_n=n^2{h}^2/(8m_e{L}^2)$. Starting in the ground state, the smallest excitation is $1\to 2$, and its electric-dipole matrix element is nonzero. It therefore sets the longest absorption wavelength:The corresponding energy separation is about $12.4\mathrm{eV}$. Elastic scattering of light is a different process from the stipulated absorption into an excited well state.

The one-dimensional time-independent Schrödinger equation isFor the infinite square well on $[0,a]$, $V=0$ inside and the wave function is zero outside. The boundary conditions are $\psi (0)=\psi (a)=0$. For positive integer $n$, ${\psi }_n=A\sin (n\pi x/a)$ satisfies them, and ${\psi }_n^{\prime \prime }=-(n\pi /a{)}^2{\psi }_n$. Substitution gives the energy levelsNormalization requires $1={\int }_0^a|{\psi }_n{|}^{2}dx=A^{2}a/2$. The requested real positive constant is therefore

In the position representation the Time-dependent Schrödinger equation isFor a normalized state define probability density and probability current byMultiplying the equation by $\overline{\psi }$ and its complex conjugate by $\psi$, then subtracting, cancels the real potential. The resulting identity isThus $\boxed{{\partial }_t\rho +{\partial }_{x}j=0}$. Integrating over an interval gives $d{\int }_b^c\rho dx/dt=j(b)-j(c)$: probability changes only through flux across the endpoints. This is local conservation of quantum probability.

Continuity with the zero exterior imposes zero endpoint values on the infinite square well. Solving the stationary equation with these boundary conditions giveswith zero extension outside. The integral of ${\sin }^2(n\pi x/a)$ is $a/2$, which fixes the normalization. The initial coefficients are $3/5$ and $4/5$, whose squared moduli sum to one. Each energy component acquires its stationary phase:At the midpoint, ${\Psi }_1=\sqrt{2/a}$, ${\Psi }_1^{\prime }=0$, ${\Psi }_2=0$, ${\Psi }_2^{\prime }=-(2\pi /a)\sqrt{2/a}$. HenceTaking its imaginary part proves

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.

Inside the infinite square well, the time-independent Schrödinger equation and Dirichlet boundary conditions areand the wavefunction vanishes outside the interval. For a nonzero state, integration by parts givesThe strict inequality follows because a function with zero derivative and zero endpoint values is zero. Put $k=\sqrt{2mE}/\hslash$. The general solution is $A\sin kx+B\cos kx$; the first boundary condition gives $B=0$, and the second gives $ka=n\pi$, $n=1,2,\dots$. Since ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, the normalized energy eigenstates and eigenvalues areThe normalization fixes the magnitude of the coefficient; its arbitrary global phase has been chosen real and positive.

Inside the infinite square well, the Schrödinger equation and Dirichlet boundary conditions areThe wavefunction vanishes outside the box. These walls constrain its values, not its first derivatives: requiring both to vanish at an endpoint would eliminate all the nonzero stationary solutions.

For a nonzero separated solution, separation of variables givesThe last equation is the time-independent Schrödinger equation. Its energy eigenvalue is real. Multiplication by ${\psi }^{\ast }$ and integration by parts, with the endpoint values zero, givesEquality would force $\psi$ to be constant and hence zero, so a nonzero energy eigenstate has $E>0$. Writing $k=\sqrt{2mE}/\hslash$, the general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$; the right gives $\sin ka=0$. Negative integers reproduce the same functions up to a constant, and $n=0$ gives the zero function, leaving $k=n\pi /a$ for $n=1,2,\dots$.

Finally ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. Thus wavefunction normalization with positive coefficient givesAn arbitrary constant global phase has no effect on the quantum state.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.

The apparent interior identity $[H,P]=0$ cannot justify conservation here: the momentum operator does not preserve the Dirichlet boundary conditions of the Hamiltonian operator's operator domain. For sufficiently regular states, retaining the boundary terms in Ehrenfest theorem instead gives the boundary force in an infinite square well,In the strictly confined interval, $-i\hslash {\partial }_x$ with both endpoint values zero is symmetric but not itself a self-adjoint operator. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior commutator misses the wall force.

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.

For the Dirichlet boundary conditions on $(-a,a)$, take the two-dimensional even trial space spanned by $(a^2-x^2)$ and $x^2(a^2-x^2)$. Using the kinetic-energy quadratic form gives, for $b=\lambda a^2$,The minimum occurs at $b=(-49+4\sqrt{133})/13$ and equals $7-\sqrt{133}/2\simeq 1.2337187027$. It is slightly above the exact infinite square well coefficient ${\pi }^2/8\simeq 1.2337005501$. The zero extension of a trial function has derivative jumps at the walls, so the appropriate admissibility statement uses the Dirichlet kinetic-energy form; it need not belong to the unrestricted full-line second-derivative operator domain.

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].