Abound state is a composite of two or more fundamental building blocks, such as particles, atoms, or bodies, that behaves as a single object and in which energy is required to split them.^[1]

Inquantum physics, a bound state is aquantum state of aparticle subject to apotential such that the particle has a tendency to remain localized in one or more regions of space.^[2] The potential may be external or it may be the result of the presence of another particle; in the latter case, one can equivalently define a bound state as a state representing two or more particles whoseinteraction energy exceeds the total energy of each separate particle. One consequence is that, given a potentialvanishing at infinity, negative-energy states must be bound. Theenergy spectrum of the set of bound states are most commonly discrete, unlikescattering states offree particles, which have a continuous spectrum.

Although not bound states in the strict sense, metastable states with a net positive interaction energy, but long decay time, are often considered unstable bound states as well and are called "quasi-bound states".^[3] Examples includeradionuclides andRydberg atoms.^[4]

Inrelativisticquantum field theory, a stable bound state ofn particles with masses${\displaystyle \{m_k{\}}_{k=1}^n}$ corresponds to apole in theS-matrix with acenter-of-mass energy less than${\displaystyle \sum _k{m}_k}$. Anunstable bound state shows up as a pole with acomplex center-of-mass energy.

• Aproton and anelectron can move separately; when they do, the total center-of-mass energy is positive, and such a pair of particles can be described as an ionized atom. Once the electron starts to "orbit" the proton, the energy becomes negative, and a bound state – namely thehydrogen atom – is formed. Only the lowest-energy bound state, theground state, is stable. Otherexcited states are unstable and will decay into stable (but not other unstable) bound states with less energy by emitting aphoton.
• Apositronium "atom" is anunstable bound state of anelectron and apositron. It decays intophotons.
• Any state in thequantum harmonic oscillator is bound, but has positive energy. Note that${\displaystyle \underset{x\to \pm \mathrm{\infty }}{lim}{V}_{\text{QHO}}(x)=\mathrm{\infty }}$ , so thebelow does not apply.
• Anucleus is a bound state ofprotons andneutrons (nucleons).
• Theproton itself is a bound state of threequarks (twoup and onedown; onered, onegreen and oneblue). However, unlike the case of the hydrogen atom, the individual quarks can never be isolated. Seeconfinement.
• TheHubbard andJaynes–Cummings–Hubbard (JCH) models support similar bound states. In the Hubbard model, two repulsivebosonicatoms can form a bound pair in anoptical lattice.^[5][6][7] The JCH Hamiltonian also supports two-polariton bound states when the photon-atom interaction is sufficiently strong.^[8]

Letσ-finite measure space${\displaystyle (X,\mathcal{A},\mu )}$ be aprobability space associated withseparablecomplexHilbert space${\displaystyle H}$. Define aone-parameter group of unitary operators${\displaystyle (U_t{)}_{t\in \mathbb{R}}}$, adensity operator${\displaystyle \rho =\rho (t_0)}$ and anobservable${\displaystyle T}$ on${\displaystyle H}$. Let${\displaystyle \mu (T,\rho )}$ be the induced probability distribution of${\displaystyle T}$ with respect to${\displaystyle \rho }$. Then the evolution

${\displaystyle \rho (t_0)\mapsto [U_t(\rho )](t_0)=\rho (t_0+t)}$

isbound with respect to${\displaystyle T}$ if

${\displaystyle \underset{R\to \mathrm{\infty }}{lim}\underset{t\ge t_0}{sup}\mu (T,\rho (t))({\mathbb{R}}_{>R})=0}$,

where${\displaystyle {\mathbb{R}}_{>R}=\{x\in \mathbb{R}\mid x>R\}}$.^{[dubious –discuss][9]}

A quantum particle is in abound state if at no point in time it is found "too far away" from any finite region${\displaystyle R\subset X}$. Using awave function representation, for example, this means^[10]

such that

In general, a quantum state is a bound stateif and only if it is finitelynormalizable for all times${\displaystyle t\in \mathbb{R}}$ and remains spatially localized.^[11] Furthermore, a bound state lies within thepure point part of the spectrum of${\displaystyle T}$if and only if it is aneigenvector of${\displaystyle T}$.^[12]

More informally, "boundedness" results foremost from the choice ofdomain of definition and characteristics of the state rather than the observable.^{[nb 1]} For a concrete example: let${\displaystyle H:=L^2(\mathbb{R})}$ and let${\displaystyle T}$ be theposition operator. Given compactly supported${\displaystyle \rho =\rho (0)\in H}$ and${\displaystyle [-1,1]\subseteq \mathrm{S}\mathrm{u}\mathrm{p}\mathrm{p}(\rho )}$.

• If the state evolution of${\displaystyle \rho }$ "moves this wave package to the right", e.g., if${\displaystyle [t-1,t+1]\in \mathrm{S}\mathrm{u}\mathrm{p}\mathrm{p}(\rho (t))}$ for all${\displaystyle t\ge 0}$, then${\displaystyle \rho }$ is not bound state with respect to position.
• If${\displaystyle \rho }$ does not change in time, i.e.,${\displaystyle \rho (t)=\rho }$ for all${\displaystyle t\ge 0}$, then${\displaystyle \rho }$ is bound with respect to position.
• More generally: If the state evolution of${\displaystyle \rho }$ "just moves${\displaystyle \rho }$ inside a bounded domain", then${\displaystyle \rho }$ is bound with respect to position.

As finitely normalizable states must lie within thepure point part of the spectrum, bound states must lie within the pure point part. However, asNeumann andWigner pointed out, it is possible for the energy of a bound state to be located in the continuous part of the spectrum. This phenomenon is referred to asbound state in the continuum.^[13][14]

Consider the one-particle Schrödinger equation. If a state has energy, then the wavefunctionψ satisfies, for some${\displaystyle X>0}$

${\displaystyle \frac{{\psi }^{\mathrm{\prime }\mathrm{\prime }}}{\psi }=\frac{2m}{{\hslash }^2}(V(x)-E)>0\text{ for }x>X}$

so thatψ is exponentially suppressed at largex. This behaviour is well-studied for smoothly varying potentials in theWKB approximation for wavefunction, where an oscillatory behaviour is observed if the right hand side of the equation is negative and growing/decaying behaviour if it is positive.^[15] Hence, negative energy-states are bound if${\displaystyle V(x)}$ vanishes at infinity.

One-dimensional bound states can be shown to be non-degenerate in energy for well-behaved wavefunctions that decay to zero at infinities. This need not hold true for wavefunctions in higher dimensions. Due to the property of non-degenerate states, one-dimensional bound states can always be expressed as real wavefunctions.

Node theorem states that${\displaystyle n\text{th}}$ bound wavefunction ordered according to increasing energy has exactly${\displaystyle n-1}$ nodes, i.e., points${\displaystyle x=a}$ where${\displaystyle \psi (a)=0\ne {\psi }^{\prime }(a)}$. Due to the form of Schrödinger's time independent equations, it is not possible for a physical wavefunction to have${\displaystyle \psi (a)=0={\psi }^{\prime }(a)}$ since it corresponds to${\displaystyle \psi (x)=0}$ solution.^[16]

Aboson with massm_χmediating aweakly coupled interaction produces anYukawa-like interaction potential,

${\displaystyle V(r)=\pm \frac{{\alpha }_{\chi }}{r}{e}^{-\frac{r}{\lambda {\frac{}{}}_{\chi }}}}$,

where${\displaystyle {\alpha }_{\chi }=g^2/4\pi }$,g is the gauge coupling constant, andƛ_i =⁠ℏ/m_ic⁠ is thereduced Compton wavelength. Ascalar boson produces a universally attractive potential, whereas a vector attracts particles to antiparticles but repels like pairs. For two particles of massm₁ andm₂, theBohr radius of the system becomes

${\displaystyle a_0=\frac{{\lambda {}^{{}^{\underset{\_}{}}}}_1+{\lambda {}^{{}^{\underset{\_}{}}}}_2}{{\alpha }_{\chi }}}$

and yields the dimensionless number

${\displaystyle D=\frac{{\lambda {}^{{}^{\underset{\_}{}}}}_{\chi }}{a_0}={\alpha }_{\chi }\frac{{\lambda {}^{{}^{\underset{\_}{}}}}_{\chi }}{{\lambda {}^{{}^{\underset{\_}{}}}}_1+{\lambda {}^{{}^{\underset{\_}{}}}}_2}={\alpha }_{\chi }\frac{m_1+m_2}{m_{\chi }}}$.

In order for the first bound state to exist at all,${\displaystyle D\gtrsim 0.8}$. Because thephoton is massless,D is infinite forelectromagnetism. For theweak interaction, theZ boson's mass is91.1876±0.0021 GeV/c², which prevents the formation of bound states between most particles, as it is97.2 times theproton's mass and178,000 times theelectron's mass.

Note, however, that, if theHiggs interaction did not break electroweak symmetry at theelectroweak scale, then the SU(2)weak interaction would becomeconfining.^[17]

• Bethe–Salpeter equation
• Bound state in the continuum
• Composite field
• Cooper pair
• Exciton
• Resonance (particle physics)
• Levinson's theorem

• Blanchard, Philippe; Brüning, Edward (2015). "Some Applications of the Spectral Representation".Mathematical Methods in Physics: Distributions, Hilbert Space Operators, Variational Methods, and Applications in Quantum Physics (2nd ed.). Switzerland: Springer International Publishing. p. 431.ISBN 978-3-319-14044-5.

• Quantum field theory
• Quantum states

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