# Bound state

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In quantum physics, a bound state is a special quantum state of a particle subject to a potential such that the particle has a tendency to remain localised in one or more regions of space. 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 whose interaction energy exceeds the total energy of each separate particle. One consequence is that, given a potential vanishing at infinity, negative-energy states must be bound. In general, the energy spectrum of the set of bound states is discrete, unlike free 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".[1] Examples include certain radionuclides and electrets.[clarification needed][citation needed]

In relativistic quantum field theory, a stable bound state of n particles with masses ${\displaystyle \{m_{k}\}_{k=1}^{n}}$ corresponds to a pole in the S-matrix with a center-of-mass energy less than ${\displaystyle \textstyle \sum _{k}m_{k}}$. An unstable bound state shows up as a pole with a complex center-of-mass energy.

## Examples

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

## Definition

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 ${\displaystyle \mathbb {R} _{>R}=\lbrace x\in \mathbb {R} \mid x>R\rbrace }$.[dubious ][citation needed]

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.

## Properties

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.[citation needed]

### Position-bound states

Consider the one-particle Schrödinger equation. If a state has energy ${\displaystyle E<\operatorname {max} {\left(\lim _{x\to \infty }{V(x)},\lim _{x\to -\infty }{V(x)}\right)}}$, then the wavefunction ψ satisfies, for some ${\displaystyle X>0}$

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

so that ψ is exponentially suppressed at large x.[dubious ] Hence, negative energy-states are bound if V vanishes at infinity.