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Binding energy is the energy required to separate a bound state into specified free constituents. For an electron bound in a chemical atom, neglecting this cost is an assumption of the free-electron Compton scattering model.

For $V(x)=g[\delta (x-a)+\delta (x+a)]$ with $a>0$ and $\nu =-mg/{\hslash }^2>0$, an even bound state obeys $\kappa =\nu (1+e^{-2a\kappa })$, and an odd bound state obeys $\kappa =\nu (1-e^{-2a\kappa })$, with energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and $\kappa >0$. The even right-hand side is strictly decreasing from $2\nu$, so it meets $\kappa$ once. For the odd equation, $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$ is strictly concave, starts at zero with slope $2a\nu -1$, and tends to negative infinity. It has one positive equation root precisely when $2a\nu >1$. Thus there is one bound state for $-{\hslash }^2/(2ma)\le g<0$, and two for $g<-{\hslash }^2/(2ma)$. For $g\ge 0$ positivity of the Hamiltonian operator excludes negative energy, and square integrability excludes nonnegative-energy exterior solutions. At $g=g_c$ the odd zero-energy solution is bounded but not square integrable: it is a zero-energy scattering resonance, not a second bound state.

In an even, piecewise constant potential, integrate outward from the centre with either even boundary conditions ${\psi }^{\prime }(0)=0$ or odd boundary conditions $\psi (0)=0$. Match the wavefunction and its derivative at every finite step and impose exponential decay outside. The resulting discrete energies alternate in parity, with the fourth bound state odd and containing three nodes of a wavefunction. There is no hard-wall zero at a finite step; the second derivative changes according to ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$.

For a zero-potential interval of width $a$, an infinite wall at its left end, and constant exterior potential $U_0$ to its right, the $j$th bound state branch exists precisely when $U_0>{\mathcal{U}}_j$, for $j=0,1,\dots$. Matching the wavefunction and its derivative gives $z\cot z=-\sqrt{z_0^2-z^2}$ with $z_0=a\sqrt{2m{U}_0}/\hslash$. At equality the new threshold solution is not normalizable, so exactly one bound state corresponds to $U_0\in ({\mathcal{U}}_0,{\mathcal{U}}_1]$.

A bound state of a charm quark and its antiquark.

Compton scattering redistributes photon energy and momentum in a collision with a charged subatomic particle, here an electron. For a free electron initially at rest in the chosen inertial frame, an incident photon of energy $E=hc/\lambda$ leaves with energy $E^{\prime }=hc/{\lambda }^{\prime }$ after deflection through $\theta$. The angle $\theta \in [0,\pi ]$ is between the incident and outgoing photon directions; it is not the electron recoil angle. Four-momentum conservation giveswhere $m_e$ is the electron rest mass and $T_e$ its final kinetic energy. The photon energy loss pays for the electron recoil: neither energy nor momentum disappears.

For an original worked example, stipulate $E=70\mathrm{keV}$, the rounded electron rest energy $m_e{c}^2=511\mathrm{keV}$, and the rounded Compton wavelength ${\lambda }_C=2.426\mathrm{pm}$; $\mathrm{keV}$ means $10^3$ electronvolts. Treat these rounded inputs as the example's constants. They give the following calculated values, with energies rounded to three decimal places:

Since $0\le 1-\cos \theta \le 2$, every outgoing photon has $0<E^{\prime }\le E$ and $T_e\ge 0$. At zero deflection there is no recoil; backscattering transfers the most energy. At fixed $\theta$, the absolute wavelength increase is independent of incident energy, while the fractional energy loss grows with $E/(m_e{c}^2)$. This makes Compton scattering useful for inferring energy transfers from directions in photon detectors.

The free-electron-at-rest assumption matters. An electron in a bound state has binding energy and an initial momentum distribution. Neglecting these is useful when the transferred energy is large compared with the relevant binding energy and the initial momentum is small compared with the resolved momentum transfer. Otherwise the outgoing photon energies can broaden or depart from the stationary-target formula; an initially moving electron can even give energy to the photon.

A material may also scatter X-rays through a coherent elastic scattering channel with an effectively unshifted component. When an intact chemical atom of mass $M$ takes the recoil, the corresponding Compton wavelength $h/(Mc)$ is much smaller than the electron value. The single free-electron curve does not predict that component's weight. More generally, energy versus angle is a kinematic constraint; angular wave intensity requires a differential scattering cross-section, together with target and detector information.

A physical-strip pole of a diagonal S-matrix can come from exchange of an existing particle in the crossed channel. For external masses $m_a,m_b$, use the momentum difference, rather than the sum, to identify the exchanged mass through the displayed invariant. Crossing symmetry turns the pole at $i\alpha$ into a direct pole at $i(\pi -\alpha )$ of the particle-antiparticle channel. For $m_B=2m\cos (u/2)$, the pole of $S_{BA}$ at $iu/2$ has $t=m^2$ and exchanges the existing charge-one particle; it is not a new charge-three bound state.

A two-proton system. Its low-energy spin-singlet strong channel does not give a vacuum bound state, and electrostatic repulsion further opposes binding. This is distinct from proton pairing correlations within an atomic nucleus.

For the formal one-dimensional operator $H=p^2/(2m)-a|x{|}^{-p}$ with $a>0$ and $p>2$, take a smooth normalized test function supported away from the origin. Its normalized dilation ${\psi }_{\lambda }(x)=\sqrt{\lambda }\psi (\lambda x)$ hasAs $\lambda \to \infty$, this tends to minus infinity. Thus the unregularized expression has fall to the centre for a supercritical inverse-power potential and cannot define a semibounded Hamiltonian operator on a domain containing these test functions.

If a putative bound state has separately finite $T,U$ and admits norm-preserving dilation variations inside its operator domain or energy domain, stationarity gives $2T+pU=0$, hence $E=(1-2/p)T>0$, incompatible with a negative-energy bound state below the zero continuum threshold. For a putative ground state, the second scaling derivative is $2(2-p)T<0$, also incompatible with a minimum. The argument's finite-energy and domain assumptions are essential: it proves absence of a stable ground state and excludes bound states to which this ordinary virial calculation applies. An unqualified assertion about all formal square-integrable solutions at the singular origin needs a specified self-adjoint realization; adding a short-distance cutoff changes the problem.

An electron hole is an unoccupied electron quantum state in an otherwise nearly full energy band, often the valence band. “Filled” refers to occupation of the allowed quantum states, not to filling all space with a continuous fluid of electrons. Each state, including its spin label, can hold at most one electron by the Pauli exclusion principle. The almost full energy band and its few missing occupants are therefore simpler to describe as a background plus a few electron holes. The electron hole is a quasiparticle, not a Positron inserted into the crystal.

Removing an electron of electric charge $-e$ changes the background's electric charge by $+e$. Thus an electron hole carries positive electric charge relative to that background. Promoting an electron into the conduction band creates an electron hole without changing the whole crystal's electric charge: $(-e)+(+e)=0$. An unbound pair still consists of two separate charge carriers; only a bound state of their relative motion is an exciton.

The local cartoon explains why a vacancy can move: when a neighboring electron fills it, another vacancy appears where that electron was. The vacancy has moved opposite to that particular transfer of an electron. In an extended Bloch state, use crystal momentum instead of imagining a permanently empty atomic site. For a completely filled energy band the summed electric current vanishes. Omitting one electron with group velocity ${\mathbf{v}}_v$ removes its contribution $-e{\mathbf{v}}_v$, leaving a contribution $+e{\mathbf{v}}_v$. This fixes both the positive electric charge and the velocity convention of the electron hole.

More precisely, near a parabolic valence band maximum, set its energy $E_v(0)$ as reference and define the electron hole wave vector by ${\mathbf{k}}_h=-{\mathbf{k}}_v$. ThenThe occupied electron branch has negative curvature there, while the electron hole branch has positive effective mass $m_h^{\ast }$. The velocity equality compares the electron hole with the missing Bloch state's group velocity; it does not contradict the opposite vacancy motion in a local transfer cartoon. The sign conventions are listed in Phys 460 Lecture 17.

An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

For the finite square well $V=-V_0$ on $|x|<a$ and $V=0$ outside, with $a,V_0>0$, there is always an even bound state, however small $V_0$ is. Define $\rho =a\sqrt{2m{V}_0}/\hslash$, the dimensionless decay constant $\alpha =a\sqrt{-2mE}/\hslash$, and interior oscillation constant $\beta =a\sqrt{2m(E+V_0)}/\hslash$. Wavefunction matching at a finite potential step gives $\alpha =\beta \tan \beta$ and ${\alpha }^2+{\beta }^2={\rho }^2$.

On $0<\beta <\pi /2$, these equations reduce to $\beta /\cos \beta =\rho$. The left side is continuous, strictly increasing from zero to infinity, with derivative $\sec \beta (1+\beta \tan \beta )>0$. The intermediate value theorem gives exactly one root on this first branch for every $\rho >0$. It has $0<\beta <\rho$, hence $\alpha >0$ and $-V_0<E<0$. The matched cosine interior and exponential tails therefore give a nonzero even square-integrable function with continuous first derivative, which can be normalized. As the well becomes shallow the binding energy approaches zero but remains negative; see shallow finite-square-well energy.

An exciton is a bound state of an electron and an electron hole. Their opposite electric charges attract through the Coulomb potential, producing a neutral quasiparticle with internal relative motion and motion of its center of mass. In a semiconductor, promoting an electron from the valence band into the conduction band leaves an electron hole; the pair can occupy either an exciton bound state or an unbound scattering state. Merely creating the two charge carriers does not establish binding. A neutral exciton can transport excitation energy without transporting net electric charge. Coupling this material excitation strongly to a photon can form a polariton.

The striking optical consequence is that an exciton optical transition can absorb a photon with less energy than the threshold for creating unbound electrons and electron holes. The difference is the exciton binding energy, not an extra one-electron energy band inside the band gap. The Wannier-Mott exciton supplies a useful large-radius model; the Frenkel exciton describes the opposite, tightly localized limit.^[ref]

For the ideal Wannier-Mott exciton in energy bands whose edges occur at the same crystal momentum, the pair excitation energy, measured from the unexcited crystal, isHere $\hslash \mathbf{K}$ is the pair's total crystal momentum; the last term is center of mass kinetic energy. For a permitted photon absorption transition with negligible recoil energy, $\hslash \omega \simeq E_g-E_{B,n}$. An electric-dipole selection rule and a nonzero transition matrix element are still needed: a bound state need not produce an observable spectral line. If the relevant energy band edges have different values of crystal momentum, a phonon can also be needed to conserve crystal momentum; subtracting a binding scale from a band gap alone does not specify the optical threshold.

For example, the lowest internal bound state in this hypothetical model costs $1.489115\mathrm{eV}$ to create optically, yet needs only $0.010885\mathrm{eV}$ more to dissociate. The large creation energy and the small exciton binding energy answer different physical questions. An electron-hole recombination event can instead release excitation energy by photon emission or to other degrees of freedom; an emitted photon need not reproduce the original absorption energy if relaxation intervened.

For $V=-A{e}^{-\mu r}/r$, the normalized exponential with scale $\alpha$ gives $E(\alpha )={\hslash }^2{\alpha }^2/(2m)-4A{\alpha }^3/(\mu +2\alpha {)}^2$. Choosing $\alpha =\mu /2$ proves binding if $\mu <Am/{\hslash }^2$. Failure to obtain a negative trial expectation outside this range does not prove absence of a bound state. At equality a small positive admixture of the scale-$\mu$ exponential has a negative cross matrix element and proves binding too.

For a symmetric one-dimensional finite square well, nondegenerate bound states alternate even and odd parity, starting with an even ground state. The $n$th state counted from zero has $n$ interior nodes. Outside the well, it decays with inverse length $\sqrt{2m(U-E_n)}/\hslash$. Matching both the wavefunction and its first derivative at a finite step distinguishes it from the infinite-well approximation.

Consider an original one-dimensional effective mass model with $V(x)=0$ for $|x|<a$ and $V(x)=V_0>0$ outside. Take the same positive effective mass $m^{\ast }$ everywhere, an abrupt interface, and no interactions. The wavefunction represents a transverse envelope; free motion parallel to the layer is outside this calculation. A double heterostructure laser need not be thin enough for this quantization to dominate its operation.

Set $s=x/a$, $ϵ={\hslash }^2/(2m^{\ast }{a}^2)$, $z_0^2=V_0/ϵ$, $z^2=E/ϵ$ and $w=\sqrt{z_0^2-z^2}$. The time-independent Schrödinger equation becomesHere $u(s)$ denotes the dimensionless potential; a bound state has $0<z<z_0$. Matching the wavefunction and its derivative at $s=1$ givesFor the even bound state, the interior solution is $A\cos (zs)$ and the exterior solution is $A\cos z{e}^{-w(|s|-1)}$. For the odd bound state, replace $\cos (zs)$ by $\sin (zs)$ and $\cos z$ by $\mathrm{sgn}(s)\sin z$. The physical wavefunction is $\psi (x)=\varphi (x/a)/\sqrt{a}$.

Wavefunction normalization includes both exterior tails, not just the active layer. Direct integration givesThe first two terms are the interior integral; the last is the total exterior integral. Thus the interior probability is $P_{\mathrm{in}}=A^2[1\pm \sin (2z)/(2z)]$, with the upper sign for even parity.

Choose $z_0=2$. There is one even root in $(0,\pi /2)$ and one odd root in $(\pi /2,2)$; higher branches start beyond $z_0$. Each left-hand side increases on its stated interval while $\sqrt{4-z^2}$ decreases, proving uniqueness there. Numerically,These are calculated values of the chosen finite square well. The excited bound state lies closer to the barrier and has a longer exterior tail: being bound does not mean having unit probability inside the well.

This model demonstrates electronic confinement, not optical refractive index confinement or a complete laser. Unequal effective masses, band bending, interface disorder and interactions require a more detailed model. In particular, continuity of the ordinary derivative above uses the chosen constant effective mass assumption.

For a symmetric attractive well of width $a$ and depth $V_0$, the first odd bound state obeys $-k\cot (ka/2)=q>0$, where $q$ is the inverse tail length. Its lowest branch has $\pi <ka<2\pi$. Since $k^2<2m{V}_0/{\hslash }^2$, existence requires $V_0>V_{\mathrm{odd}}$; it is also sufficient by continuity of the matching equation on that branch. At equality the exterior solution has zero decay rate and is not square-integrable. Thus a well at or below this threshold has only its even ground state.

Radiation from unbound charged particles during a scattering encounter, notably electrons accelerated by ions in a plasma. Both the initial and final electron states are free, distinguishing this process from emission involving an atomic bound state.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.