- Bound state
In

physics , a**bound state**is a composite of two or more building blocks (particles or bodies) that behaves as a single object. Inquantum mechanics (where the number of particles is conserved), a bound state is a state in theHilbert space that corresponds to two or more particles whoseinteraction energy is negative, and therefore these particles cannot be separated unlessenergy is spent. Theenergy spectrum of a bound state is discrete, unlike the continuous spectrum of isolated particles. (Actually, it is possible to have unstable bound states with a positive interaction energy provided that there is an "energy barrier" that has to be tunnelled through in order to decay. This is true for someradioactive nuclei and for someelectret materials able to carry electric charge for rather long periods.)In general, a

stable bound state is said to exist in a given potential of some dimension if stationary wavefunctions exist (normalized in the range of the potential). The energies of these wavefunctions are negative.In relativistic

quantum field theory , astable bound state of n particles with masses m_{1}, ..., m_{n}shows up as a pole in theS-matrix with a center of mass energy which is less than m_{1}+...+m_{n}. Anunstable bound state (seeresonance ) shows up as a pole with a complex center of mass energy.**Examples*** A

proton and anelectron can move separately; 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. The otherexcited state s are unstable and will decay into bound states with less energy by emitting aphoton .

* A nucleus is a bound state ofproton s andneutron s (nucleon s).

* Apositronium "atom" is an unstable bound state of anelectron and apositron . It decays intophoton s.

* Theproton itself is a bound state of threequark s (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.**In mathematical quantum physics**Let $H$ be a complex separable Hilbert space, $U\; =\; lbrace\; U(t)\; mid\; t\; in\; mathbb\{R\}\; brace$ be a one-parametric group of unitary operators on $H$ and $ho\; =\; ho(t\_0)$ be a statistical operator on $H$. Let $A$ be an observable on $H$ and let $mu(A,\; ho)$ be the induced probability distribution of $A$ with respect to $ho$ on the Borel $sigma$-algebra on $mathbb\{R\}$. Then the evolution of $ho$ induced by $U$ is said to be

**bound**with respect to $A$ if $lim\_\{R\; ightarrow\; infty\}\; sum\_\{t\; geq\; t\_0\}\; mu(A,\; ho(t))(mathbb\{R\}\_\{>\; R\})\; =\; 0$ , where $mathbb\{R\}\_\{>R\}\; =\; lbrace\; x\; in\; mathbb\{R\}\; mid\; x\; >\; R\; brace$.**Example:**Let $H\; =\; L^2(mathbb\{R\})$ and let $A$ be the position observable. Let $ho\; =\; ho(0)\; in\; H$ have compact support and $[-1,1]\; subseteq\; mathrm\{Supp\}(\; ho)$.* If the state evolution of $ho$ "moves this wave package constantly to the right", e.g. if $[t-1,t+1]\; in\; mathrm\{Supp\}(\; ho(t))$ for all $t\; geq\; 0$ , then $ho$ is not a bound state with respect to the position.

* If $ho$ does not change in time, i.e. $ho(t)\; =\; ho$ for all $t\; geq\; 0$, then $ho$ is a bound state with respect to position.

* More generally: If the state evolution of $ho$ "just moves $ho$ inside a bounded domain", then $ho$ is also a bound state with respect to position.

**See also***

Composite field

*Resonance

*Wikimedia Foundation.
2010.*

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