Bound state

An overview of the various families of elementary and composite particles, and the theories describing their interactions

• A proton and an electron 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 the hydrogen atom – is formed. Only the lowest-energy bound state, the ground state, is stable. Other excited states are unstable and will decay into stable (but not other unstable) bound states with less energy by emitting a photon.
• A positronium "atom" is an unstable bound state of an electron and a positron. It decays into photons.
• Any state in the quantum harmonic oscillator is bound, but has positive energy. Note that ${\displaystyle \lim _{x\to \pm \infty }{V_{\text{QHO}}(x)}=\infty }$ , so the below does not apply.
• A nucleus is a bound state of protons and neutrons (nucleons).
• The proton itself is a bound state of three quarks (two up and one down; one red, one green and one blue). However, unlike the case of the hydrogen atom, the individual quarks can never be isolated. See confinement.
• The Hubbard and Jaynes-Cummings-Hubbard (JCH) models support similar bound states. In the Hubbard model, two repulsive bosonic atoms can form a bound pair in an optical lattice.[2][3][4] The JCH Hamiltonian also supports two-polariton bound states when the photon-atom interaction is sufficiently strong.[5]

Let H be a complex separable Hilbert space, ${\displaystyle U=\lbrace U(t)\mid t\in \mathbb {R} \rbrace }$ be a one-parameter group of unitary operators on H and ${\displaystyle \rho =\rho (t_{0})}$ be a statistical operator on H. Let A be an observable on H and ${\displaystyle \mu (A,\rho )}$ be the induced probability distribution of A with respect to ρ on the Borel σ-algebra of ${\displaystyle \mathbb {R} }$. Then the evolution of ρ induced by U is bound with respect to A if ${\displaystyle \lim _{R\rightarrow \infty }{\sup _{t\geq t_{0}}{\mu (A,\rho (t))(\mathbb {R} _{>R})}}=0}$, where ${\displaystyle \mathbb {R} _{>R}=\lbrace x\in \mathbb {R} \mid x>R\rbrace }$.

More informally, a bound state is contained within a bounded portion of the spectrum of A. For a concrete example: let ${\displaystyle H=L^{2}(\mathbb {R} )}$ and let A be position. Given compactly-supported ${\displaystyle \rho =\rho (0)\in H}$ and ${\displaystyle [-1,1]\subseteq \mathrm {Supp} (\rho )}$.

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

Let A have measure-space codomain ${\displaystyle (X;\mu )}$. A quantum particle is in a bound state if it is never found “too far away from any finite region ${\displaystyle R\subseteq X}$,” i.e. using a wavefunction representation,

${\displaystyle {\begin{aligned}0&=\lim _{R\to \infty }{\mathbb {P} ({\text{particle measured inside }}X\setminus R)}\\&=\lim _{R\to \infty }{\int _{X\setminus R}|\psi (x)|^{2}\,d\mu (x)}\end{aligned}}}$

Consequently, ${\displaystyle \int _{X}{|\psi (x)|^{2}\,d\mu (x)}}$ is finite. In other words, a state is a bound state if and only if it is finitely normalizable.

As finitely normalizable states must lie within the discrete part of the spectrum, bound states must lie within the discrete part. However, as Neumann and Wigner pointed out, a bound state can have its energy located in the continuum spectrum.[6] In that case, bound states still are part of the discrete portion of the spectrum, but appear as Dirac masses in the spectral measure.

A boson with mass m_χ mediating a weakly coupled interaction produces an Yukawa-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, and ƛ_i = ℏ/m_ic is the reduced Compton wavelength. A scalar boson produces a universally attractive potential, whereas a vector attracts particles to antiparticles but repels like pairs. For two particles of mass m₁ and m₂, the Bohr radius of the system becomes

${\displaystyle a_{0}={\frac {{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{1}+{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{2}}{\alpha _{\chi }}}}$

and yields the dimensionless number

${\displaystyle D={\frac {{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{\chi }}{a_{0}}}=\alpha _{\chi }{\frac {{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{\chi }}{{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{1}+{\lambda \!\!\!^{{}^{\underline {\ \ }}}}_{2}}}=\alpha _{\chi }{\frac {m_{1}+m_{2}}{m_{\chi }}}}$.

In order for the first bound state to exist at all, ${\displaystyle D\gtrsim 0.8}$. Because the photon is massless, D is infinite for electromagnetism. For the weak interaction, the Z boson's mass is 91.1876±0.0021 GeV/c², which prevents the formation of bound states between most particles, as it is 97.2 times the proton's mass and 178,000 times the electron's mass.

Note however that if the Higgs interaction didn't break electroweak symmetry at the electroweak scale, then the SU(2) weak interaction would become confining.[7]

• Composite field
• Resonance (particle physics)
• Bethe–Salpeter equation
• Cooper pair