Bound state

A bound 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.

In quantum physics, a bound state is a quantum state of a particle subject to a potential energy such that the particle has a tendency to remain localized 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. The energy spectrum of the set of bound states are most commonly discrete, unlike of Free particle, 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".

Examples include radionuclides and .

Thomas F. Gallagher (1994). 9780521385312, Cambridge University Press. . ISBN 9780521385312

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

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Examples

• 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 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 resonance of an electron and a positron. It decays into .
• Any state in the quantum harmonic oscillator is bound, but has positive energy. Note that $\backslash lim\_\{x\backslash to\backslash pm\backslash infty\}\{V\_\{\backslash text\{QHO\}\}(x)\}\; =\; \backslash infty$ , so the below does not apply.
• A atomic nucleus is a bound state of and ().
• The proton itself is a bound state of three (two up quark and one down quark; one color charge, one color charge and one color charge). However, unlike the case of the hydrogen atom, the individual quarks can never be isolated. See confinement.
• The Hubbard model 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.

The JCH Hamiltonian also supports two-polariton bound states when the photon-atom interaction is sufficiently strong.

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Definition Let -finite measure space $(X,\; \backslash mathcal\; A,\; \backslash mu)$ be a probability space associated with Separable space complex number Hilbert space $H$. Define a one-parameter group of unitary operators $(U\_t)\_\{t\backslash in\; \backslash mathbb\{R\}\}$, a density operator $\backslash rho\; =\; \backslash rho(t\_0)$ and an observable $T$ on $H$. Let $\backslash mu(T,\backslash rho)$ be the induced probability distribution of $T$ with respect to $\backslash rho$. Then the evolution

$\backslash rho(t\_0)\backslash mapsto\; U\_t(\backslash rho)(t\_0)\; =\; \backslash rho(t\_0\; +t)$

is bound with respect to $T$ if

$\backslash lim\_\{R\; \backslash rightarrow\; \backslash infty\}\{\backslash sup\_\{t\; \backslash geq\; t\_0\}\{\backslash mu(T,\backslash rho(t))(\backslash mathbb\{R\}\_\{>\; R\})\}\}\; =\; 0$,

where $\backslash mathbb\{R\}\_\{>R\}\; =\; \backslash lbrace\; x\; \backslash in\; \backslash mathbb\{R\}\; \backslash mid\; x\; >\; R\; \backslash rbrace$.

(1980). 9780125850506, Academic Press. ISBN 9780125850506

A quantum particle is in a bound state if at no point in time it is found “too far away" from any finite region $R\backslash subset\; X$. Using a wave function representation, for example, this means

$\backslash begin\{align\}$

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{align} such that

$\backslash int\_X$>

In general, a quantum state is a bound state if and only if it is finitely normalizable for all times $t\backslash in\backslash mathbb\{R\}$ and remains spatially localized. Furthermore, a bound state lies within the pure point part of the spectrum of $T$ if and only if it is an eigenvector of $T$. More informally, "boundedness" results foremost from the choice of domain of definition and characteristics of the state rather than the observable.See Expectation value (quantum mechanics) for an example. For a concrete example: let $H\; :=\; L^2(\backslash mathbb\{R\})$ and let $T$ be the position operator. Given compactly supported $\backslash rho\; =\; \backslash rho(0)\; \backslash in\; H$ and $-1,1\; \backslash subseteq\; \backslash mathrm\{Supp\}(\backslash rho)$. If the state evolution of $\backslash rho$ "moves this wave package to the right", e.g., if $t-1,t+1\; \backslash in\; \backslash mathrm\{Supp\}(\backslash rho(t))$ for all $t\; \backslash geq\; 0$, then $\backslash rho$ is not bound state with respect to position. If $\backslash rho$ does not change in time, i.e., $\backslash rho(t)\; =\; \backslash rho$ for all $t\; \backslash geq\; 0$, then $\backslash rho$ is bound with respect to position. More generally: If the state evolution of $\backslash rho$ "just moves $\backslash rho$ inside a bounded domain", then $\backslash rho$ is bound with respect to position. Properties As finitely normalizable states must lie within the pure point part of the spectrum, bound states must lie within the pure point part. However, as Neumann and Wigner 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 as bound state in the continuum. Position-bound states Consider the one-particle Schrödinger equation. If a state has energy , then the wavefunction satisfies, for some $X\; >\; 0$ $\backslash frac\{\backslash psi^\{\backslash prime\backslash prime\}\}\{\backslash psi\}=\backslash frac\{2m\}\{\backslash hbar^2\}(V(x)-E)\; >\; 0\backslash text\{\; for\; \}x\; >\; X$ so that is exponentially suppressed at large . This behaviour is well-studied for smoothly varying potentials in the WKB 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. Brian C. Hall (2025). 9781461471158, Springer. ISBN 9781461471158 Hence, negative energy-states are bound if $V(x)$ vanishes at infinity. Non-degeneracy in one-dimensional bound states 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. { class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Proof

Consider two energy eigenstates states $\backslash Psi\_1$ and $\backslash Psi\_2$ with same energy eigenvalue. Then since, the Schrodinger equation, which is expressed as:$$E\; =\; -\; \backslash frac\; 1\; \{\backslash Psi\_i(x,t)\}\; \backslash frac\{\backslash hbar^2\}\{2m\}\backslash frac\{\backslash partial^2\backslash Psi\_i(x,t)\; \}\{\backslash partial\; x^2\}\; +\; V(x,t)$$is satisfied for i = 1 and 2, subtracting the two equations gives:$$\backslash frac\; 1\; \{\backslash Psi\_1(x,t)\}\; \backslash frac\{\backslash partial^2\backslash Psi\_1(x,t)\; \}\{\backslash partial\; x^2\}\; -\; \backslash frac\; 1\; \{\backslash Psi\_2(x,t)\}\; \backslash frac\{\backslash partial^2\backslash Psi\_2(x,t)\; \}\{\backslash partial\; x^2\}\; =\; 0$$which can be rearranged to give the condition:$$\backslash frac\{\backslash partial\; \}\{\backslash partial\; x\}\; \backslash left(\backslash frac\{\backslash partial\; \backslash Psi\_1\}\{\backslash partial\; x\}\backslash Psi\_2\backslash right)-\backslash frac\{\backslash partial\; \}\{\backslash partial\; x\}\; \backslash left(\backslash frac\{\backslash partial\; \backslash Psi\_2\}\{\backslash partial\; x\}\backslash Psi\_1\backslash right)=0$$Since $\backslash frac\{\backslash partial\; \backslash Psi\_1\}\{\backslash partial\; x\}(x)\backslash Psi\_2(x)-\; \backslash frac\{\backslash partial\; \backslash Psi\_2\}\{\backslash partial\; x\}(x)\backslash Psi\_1(x)=\; C$, taking limit of x going to infinity on both sides, the wavefunctions vanish and gives $C\; =\; 0$. Solving for $\backslash frac\{\backslash partial\; \backslash Psi\_1\}\{\backslash partial\; x\}(x)\backslash Psi\_2(x)\; =\; \backslash frac\{\backslash partial\; \backslash Psi\_2\}\{\backslash partial\; x\}(x)\backslash Psi\_1(x)$, we get: $\backslash Psi\_1(x)\; =\; k\; \backslash Psi\_2(x)$ which proves that the energy eigenfunction of a 1D bound state is unique. Furthermore it can be shown that these wavefunctions can always be represented by a completely real wavefunction. Define real functions $\backslash rho\_1(x)$ and $\backslash rho\_2(x)$ such that $\backslash Psi(x)\; =\; \backslash rho\_1(x)\; +\; i\; \backslash rho\_2(x)$. Then, from Schrodinger's equation:$$\backslash Psi$$ = - \frac{2m(E-V(x))}{\hbar^2}\Psi we get that, since the terms in the equation are all real values:$$\backslash rho\_i$$ = - \frac{2m(E-V(x))}{\hbar^2}\rho_i applies for i = 1 and 2. Thus every 1D bound state can be represented by completely real eigenfunctions. Note that real function representation of wavefunctions from this proof applies for all non-degenerate states in general.

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Node theorem

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

F. A. Berezin (1991). 9780792312185, Dordrecht ; Boston : Kluwer Academic Publishers. . ISBN 9780792312185

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Requirements

A boson with mass gauge boson a weakly coupled interaction produces an Yukawa potential interaction potential,

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

where $\backslash alpha\_\backslash chi=g^2/4\backslash pi$, is the gauge coupling constant, and 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 and , the Bohr radius of the system becomes

$a\_0=\backslash frac$

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