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Put ${\omega }_0^2=k/m$, where $k$ is the binding spring constant rather than an optical wavenumber. The steady complex amplitude solution of the forced Lorentz oscillator model isThe acceleration complex amplitude is $-{\omega }^2{\mathbf{x}}_0$. HenceThis is the scattering cross-section of a harmonically bound charge. The displacement and induced electric dipole moment are parallel to the incident linear polarization, and integrating their dipole radiation over all directions gives the total cross section above.

For $\omega \ll {\omega }_0$, binding dominates andrecovering Rayleigh scattering with static electric polarizability $q^2/k$. For $\omega \gg {\omega }_0$, inertia dominates: ${\mathbf{x}}_0\simeq -q{\mathbf{E}}_0/(m{\omega }^2)$, so its second time derivative becomes frequency independent andThis is a classical Thomson scattering regime only while $\hslash \omega \ll m{c}^2$. Exactly at $\omega ={\omega }_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite damping and radiation reaction regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.

Use Fraunhofer diffraction, appropriate to observation in the far field or in the focal plane of a lens, and uniform illumination across the slit. At observation angle $\theta$, points separated by transverse coordinate $u$ acquire a phase difference $ku\sin \theta$, where $k=2\pi /\lambda$ is the wavenumber. The superposition principle for the coherent wave amplitudes givesThe intensity is proportional to the squared modulus of the wave amplitude, soThe limit at $\beta =0$ is one. The noncentral zeros satisfy $w\sin \theta =m\lambda$ for nonzero integers $m$ for which the angle is physically allowed. In the Rayleigh criterion, two mutually incoherent point images are just resolved when the central maximum of one diffraction pattern lies at the first minimum of the other. For single-slit diffraction this givesThe exact statement here takes one image on the slit normal; for nearby paraxial images the angular resolution is $\lambda /w$. The first propagating minimum exists only for $w\ge \lambda$. The numerical factor $1.22$ belongs to a circular aperture and does not apply to this slit.

This is a plane electromagnetic wave with linear polarization. $\mathbf{E}$ is the electric field; $E_0$ its peak amplitude; $x$ the position along propagation; $t$ the time; $k=2\pi /\lambda$ the wavenumber; and $\omega =2\pi f$ the angular frequency. Here $\lambda$ is the wavelength, $f$ the frequency, and $\hat{\mathbf{y}}$ the unit vector in the direction of the electric field. Constant phase $kx-\omega t$ moves along $+x$ with phase velocity $\omega /k$.

Put $\theta =kx-\omega t$. The differential Faraday's law of induction givesThe accompanying oscillating magnetic field, excluding any independent static background, is therefore $\mathbf{B}=(k/\omega )E_0\cos \theta \hat{\mathbf{z}}$. To determine the propagation speed, also use the source-free Ampere-Maxwell law:ThusThe electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation; $\mathbf{E}\times \mathbf{B}$ points along $+x$.

A wave packet is a localized superposition of nearby wavenumbers. Its rapid carrier crests have phase velocity $v_p=\omega /k$; the slowly varying envelope translates at the group velocity. Dispersion means that $\omega (k)$ is not simply a constant times $k$, so different carrier components have different phase velocities. Expand around $k_0$:The integral is an envelope depending on $x-{\omega }^{\prime }(k_0)t$, so $v_g=d\omega /dk$ at the central frequency. Neglected curvature causes packet spreading. For a deep-water gravity wave, $\omega =\sqrt{gk}$, hence $v_p=\sqrt{g/k}$ andFor crest crossings through a dispersive wave packet, an instantaneous group containing six crests has length approximately $L=6\lambda$. A stationary ship takes $L/v_g$ to encounter the group; its crest-arrival frequency is $v_p/\lambda$. ConsequentlyIn the envelope frame the ship moves backwards at $v_g$ while crests move forwards at $v_p-v_g=v_g$. Their relative speed is $2v_g$, explaining why counting a snapshot and counting crossings differ.

To estimate deep-water swell distance from period-dependent arrival times, assume a localized storm emits both periods at the same time and place, propagation is in deep water, and currents and refraction can be neglected. If its distance is $D$, the arrival lag obeysWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $D=15g(86400)/\pi \simeq 4.05\times 10^6\mathrm{m}$, about 4050 km. The corresponding travel times are three and four days. A moving or long-lived storm with frequency-dependent emission times would invalidate the unique distance inference.

For a narrow packet, phase velocity is $u_p=\omega /k$ and group velocity is $u_g=d\omega /dk$. Set $b=\sigma /{\rho }_1$. The uncovered surface hasEquality occurs at $k_{\ast }^2=g/b$, also the minimum of $u_p$. Numerically,The long-wave gravity limit has $u_g/u_p\to 1/2$; the short-wave capillary limit has $u_g/u_p\to 3/2$.

A compact packet subsequently spreads because different wavenumbers have different group speeds. In the gravity regime $\omega \simeq \sqrt{gk}$, so smaller $k$ and longer waves move ahead. Individual crests move twice as fast as the envelope, appearing at its rear and disappearing at its front. In the capillary regime $\omega \simeq \sqrt{b}{k}^{3/2}$, shorter waves move ahead; the envelope overtakes its slower individual crests, which drift backward relative to it. Thus both the wavelength ordering within the spread packet and the crest motion relative to its envelope reverse.

For the sheet define $c=a{\rho }_2/{\rho }_1=8.0\times 10^{-5}\mathrm{m}$. The elastic-cover frequency ratio for water waves isAt small $k$, $F=1-ck/2+O(k^2)$: sheet inertia lowers the frequency, while bending first enters at higher order. At sufficiently large $k$, bending dominates the restoring force and sheet inertia dominates the extra mass, giving $F\sim \sqrt{B/(\sigma c)}{k}^{1/2}$. The curve initially dips below one, reaches a minimum, then rises and crosses one. This asymptote describes the supplied ideal dispersion law; arbitrarily large $k$ eventually violates a thin-sheet continuum model.

For a pinned-square surface displacement mode, $k_{nm}=\pi \sqrt{n^2+m^2}/L$ with $n,m\ge 1$. The fundamental has $k_{11}=74.05{\mathrm{m}}^{-1}$. The uncovered frequency is $4.3729\mathrm{Hz}$ and $F(k_{11})=0.997051$, soThis uses a square two-dimensional surface mode, not the one-dimensional $\pi /L$ estimate. A clamped plate or additional hydraulic boundary conditions would require a different mode calculation.

Equation and boundary conditions. The delta potential has strength $V_0$ with units of energy times length, because the Dirac delta function has units of inverse length. The time-independent Schrödinger equation isThe wavefunction is continuous at the origin. A finite jump in $\psi$ would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no corresponding term in this potential. A jump in ${\psi }^{\prime }$ does produce a delta contribution and is allowed. Integrate through a symmetric interval:For continuous finite $\psi$, the right-hand side vanishes as $ϵ\to 0$. ConsequentlyThis derivative jump is the boundary condition associated with the attractive delta potential.

Unique bound state. For $E<0$, write $\kappa =\sqrt{-2mE}/\hslash >0$. Away from the origin, ${\psi }^{\prime \prime }={\kappa }^2\psi$. Normalizability excludes exponentials growing at either infinity, leavingContinuity gives $A=B=C$. The derivative jump then becomes $-2\kappa C=-2m{V}_{0}C/{\hslash }^2$. For a nonzero state,There is exactly one positive value of $\kappa$, so exactly one negative-energy bound state. An odd solution would have $\psi (0)=0$; the allowed decaying exponentials would then both vanish identically. Normalize the even solution usingChoosing the overall phase real and positive gives the bound state in a delta potential:It peaks at the origin, has a downward cusp there, and decays exponentially on both sides. Its value, but not its slope, is continuous.

Scattering amplitudes and fluxes. For $E>0$, put $k=\sqrt{2mE}/\hslash$ and retain $\kappa =m{V}_0/{\hslash }^2$ as a strength parameter. A unit-amplitude wave incident from the left hasContinuity gives $t=1+r$. The jump condition givesSolving,The probability current of a plane wave of amplitude $A$ is $\hslash k|A{|}^2/m$ in its propagation direction. Since the asymptotic wavenumber is the same on both sides, reflection probability and transmission probability are just $|r{|}^2$ and $|t{|}^2$:They satisfy $R+T=1$, expressing conserved probability current in this real potential.

The right-hand plot uses $\eta =\kappa /k=V_0\sqrt{m/(2E)}/\hslash$, proportional to $V_0$ at fixed incident energy. Then $R={\eta }^2/(1+{\eta }^2)$ rises monotonically from zero to one, while $T=1/(1+{\eta }^2)$ falls from one to zero; they are equal at $\eta =1$. Despite the absence of a classical forbidden barrier, an abrupt attractive interaction reflects a quantum wave. Increasing attraction strengthens the derivative mismatch and can approach complete reflection at fixed positive energy. At fixed $V_0$, increasing incident energy instead makes the well increasingly transparent. The negative-energy bound state and the positive-energy scattering states are different stationary solutions, not an absorptive loss channel.

Let the first ray reflect at the upper surface and the second traverse the film, reflect below, and emerge at a laterally displaced point. Its internal round-trip geometric length is $2d/\cos \theta$, but the external path to a common outgoing wavefront must also be included. The lateral separation is $2d\tan \theta$, whose projection along the external ray is $2d\tan \theta \sin {\theta }_i$. The net optical path difference is thereforeusing Snell's law $\sin {\theta }_i=n\sin \theta$. Multiplying by the vacuum wavenumber givesFor the usual index ordering $1<n<n_2$, including the stated later value $n=1.2$, both reflections are towards higher refractive index and both reverse the electric-field sign. Their relative reflection phase is therefore zero; $\delta$ is also their total relative phase modulo $2\pi$. A different index ordering or polarization-dependent reflection phase would require that additional term.

Let $\rho ={\psi }^{\ast }\psi$ be the probability density. For a real potential energy $V(x)$, the Schrödinger equation and its complex conjugate areDifferentiate $\rho$ and substitute both equations. The potential terms cancel, leavingComparison with the probability continuity equation ${\partial }_t\rho +{\partial }_{x}J=0$ identifiesFor a plane wave of amplitude $A$ and positive wavenumber $k$, $J=\hslash k|A{|}^2/m$, the density times the velocity. Integrating the continuity equation over an interval makes its probability change equal to the incoming current minus the outgoing current.

Zero reflection means $B=0$. Matching the first interface then givesThe specified layer quantum phase is $k_{1}a=\pi /2$, so $e^{i{k}_{1}a}=i$ and $e^{-i{k}_{1}a}=-i$. The second-interface conditions becomeSince $C+D=A$, compatibility requires $\boxed{k_1^2=k_0{k}_2}$. This is a quantum quarter-wave matching layer: the intermediate wavenumber is the geometric mean, and the two interface reflections cancel after their round trip through the layer.

In terms of the physical energies,Rearranging also givesA finite positive energy satisfying the stated ordering requires $V_2/2<V_1<V_2$. Under that condition $E-V_2=(V_2-V_1{)}^2/(2V_1-V_2)>0$, consistent with the assumed propagating transmitted wave. The width must simultaneously satisfy the given quarter-wave quantum phase condition.

Let the aperture occupy the $z=0$ plane, with coordinates $(x,y)$ and opening $\Sigma$. Write ${\psi }_{\Sigma }(x,y)$ for the incident scalar complex amplitude and $h(x,y)$ for its dimensionless amplitude transmission, including any imposed phase. The observation point has transverse coordinates $(X,Y)$, corresponding to the PDF's $(x_0,y_0)$, and axial distance $R$. Also $k=2\pi /\lambda$ is the wavenumber and $i^2=-1$.

The Huygens-Fresnel principle expresses the propagated field, up to common prefactors and an approximately constant obliquity factor, asIn the paraxial approximation, expand the path asThe last term is negligible in the Fraunhofer diffraction limit, while the first two terms and $1/r\simeq 1/R$ are common across the aperture. Factoring them out leavesIt is the two-dimensional Fourier transform of the transmitted aperture field, evaluated at transverse wavevector $(kX/R,kY/R)$. For appreciable observation angles the corresponding exact directional components are used rather than $X/R,Y/R$.

Build a narrow-band wave packet from plane waves with wavenumbers close to $k_0$:The Taylor expansion of the dispersion relation is $\omega (k)\simeq {\omega }_0+{\omega }_0^{\prime }(k-k_0)$, where ${\omega }_0^{\prime }=(d\omega /dk{)}_{k_0}$. Put $\kappa =k-k_0$ to obtainThe carrier's wave phase travels at the phase velocity ${\omega }_0/k_0$, while the envelope has constant shape at constant $x-{\omega }_0^{\prime }t$. Its group velocity is thereforeThe approximation neglects higher derivatives of the dispersion relation. Those derivatives cause the wave packet to spread or distort over sufficiently long propagation times; the result describes the local envelope speed of a sufficiently narrow band.

For the trial displacement $z=\sin (qy)e^{i(kx-\omega t)}$, direct differentiation givesThe two-dimensional wave equation is therefore satisfied ifHere $k$ is the longitudinal wavenumber, $q$ the transverse wavenumber, and $c_m$ the speed of an unconstrained membrane wave. The physical displacement is the real part, multiplied by an arbitrary displacement amplitude. $T$ has units $\mathrm{N}{\mathrm{m}}^{-1}$ and $\sigma$ is surface mass density in $\mathrm{kg}{\mathrm{m}}^{-2}$, so $T/\sigma$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$ as required.

The fixed edge at $y=0$ is automatically satisfied by $\sin (qy)$. At $y=a$, the displacement must vanish at every $x,t$, so $\sin (qa)=0$. Thus the allowed transverse normal modes haveThe case $n=0$ gives zero displacement for this sine solution and is not a physical propagating mode. Negative $n$ only reverses the amplitude sign and does not create a distinct transverse mode. This discrete transverse spectrum is a consequence of the Dirichlet boundary conditions, rather than quantization of the longitudinal wavenumber in an indefinitely long guide.

Resistance to bending adds a restoring contribution that grows more rapidly with spatial curvature than the tension contribution. Short-wavelength modes are therefore stiffened especially strongly, raising their frequencies and changing the large-wavenumber behaviour.

For example, a thin plate of flexural rigidity $D_b$ has $\sigma z_{tt}=T{\nabla }^{2}z-D_b{\nabla }^{4}z$. When its edge conditions admit the same sine transverse mode, bending stiffness in a membrane waveguide givesThus the cutoffs increase, and bending-dominated propagation has $\omega \propto k^2$ at large $|k|$, rather than approaching $c_m|k|$. Its phase velocity and group velocity then grow with $|k|$. If “rigidly fixed” means a genuinely clamped plate edge, the slope must also vanish and the mode shapes change; the qualitative stiffening and stronger wave dispersion remain, but simple $q_n=n\pi /a$ need not remain exact.

A Bloch state on the loop has $\psi (x)=u(x)e^{i\varphi x}$, with $u(x+a)=u(x)$. Its circumference is $L=Na$, so $u(x+Na)=u(x)$. Single-valuedness after one complete circuit requires the periodic boundary conditionThereforeThe allowed Bloch wavevectors are spaced by $2\pi /(Na)$. Values of $\varphi$ differing by $2\pi /a$ label the same crystal momentum modulo a reciprocal lattice vector, so there are $N$ inequivalent choices in one Brillouin zone. In this notation $\varphi$ is the Bloch wavenumber, not a dimensionless phase; $\varphi a$ is dimensionless. Spin gives additional states but does not alter this condition.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

Let the middle segment occupy $0<x<l$. Normalize the incident displacement complex amplitude to one, suppress the common time factor $e^{-i\omega t}$, and chooseHere $r$ is the reflected complex amplitude at $0$, $a$ and $b$ are the rightgoing and leftgoing middle-segment complex amplitudes referenced to $0$, and $t$ is the outgoing complex amplitude referenced to $l$. The middle wavenumber is $k_M=\omega /c_M$, andis the propagation phase factor across the lossless middle segment.

At each junction, continuity of displacement and transverse force means continuity of $y$ and $T{y}_x$. Using $T{k}_j=\omega Z_j$ givesThese boundary conditions also include all multiple reflections inside the middle segment: $a$ and $b$ are the total travelling components, not just the first waves to cross a junction.

The expression is a scalar diffraction theory integral for a monochromatic wave. The complex quantity ${\psi }_P$ represents a field complex amplitude, with physical field obtained by taking the real part after multiplication by $e^{-i\omega t}$. The light intensity is proportional to $|{\psi }_P{|}^2$. The coordinates $x,y$ label transverse positions in the aperture plane, $\Sigma$ is its transmitting region, and $h(x,y)$ is the incident field multiplied by the aperture's complex transmission. The wavenumber is $k=2\pi /\lambda$ in the propagation medium, and $R$ is the axial propagation distance if $h$ already includes the incident field.

To derive the Fresnel diffraction, start from Huygens-Fresnel principle contributions with propagation factor $e^{iks}/s$. With the coordinate origin at the projection of $P$ into the aperture plane,The paraxial approximation requires $\rho /R\ll 1$, where ${\rho }^2=x^2+y^2$. Replace the slowly varying $1/s$ and the obliquity factor by their axial values, but retain the quadratic variation of wave phase. Neglect of the next wave phase term requires $k{\rho }^4/(8R^3)\ll 1$ over the relevant aperture. Thus, apart from a common constant,The quadratic wave phase may be of order one or much larger; setting it to zero would require the stronger conditions for Fraunhofer diffraction. The scalar treatment also neglects the detailed vector field at aperture edges and assumes dimensions large enough compared with wavelength for this approximation to be useful.

For a finite-distance axial point source at distance $p$ and an observer at distance $q$, the incident field contributes its own quadratic wave phase. If $h$ instead denotes only the aperture transmission, the combined curvature is described by an effective distanceThese are two equivalent conventions for distributing the incident wave phase between $h$ and the exponential. Both distances must be included in the later calculation.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.