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.^{[1]}
In quantum physics, a bound state is a quantum state of a particle subject to a potential 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 whose interaction energy exceeds the total energy of each separate particle. One consequence is that, given a potential vanishing at infinity, negativeenergy states must be bound. The energy spectrum of the set of bound states are most commonly discrete, unlike scattering states of 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 "quasibound states".^{[3]} Examples include radionuclides and Rydberg atoms.^{[4]}
In relativistic quantum field theory, a stable bound state of n particles with masses corresponds to a pole in the Smatrix with a centerofmass energy less than . An unstable bound state shows up as a pole with a complex centerofmass energy.
See also: Decomposition of spectrum (functional analysis) § Quantum mechanics 
Let σfinite measure space be a probability space associated with separable complex Hilbert space . Define a oneparameter group of unitary operators , a density operator and an observable on . Let be the induced probability distribution of with respect to . Then the evolution
is bound with respect to if
where .^{[dubious – discuss]}^{[9]}
A quantum particle is in a bound state if at no point in time it is found “too far away" from any finite region . Using a wave function representation, for example, this means
such that
In general, a quantum state is a bound state if and only if it is finitely normalizable for all times .^{[10]} Furthermore, a bound state lies within the pure point part of the spectrum of if and only if it is an eigenstate of .^{[11]}
More informally, "boundedness" results foremost from the choice of domain of definition and characteristics of the state rather than the observable.^{[nb 1]} For a concrete example: let and let be the position operator. Given compactly supported and .
See also: Spectrum (physical sciences) § Continuous versus discrete spectra 
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.^{[12]}^{[13]}
Consider the oneparticle Schrödinger equation. If a state has energy , then the wavefunction ψ satisfies, for some
so that ψ is exponentially suppressed at large x. This behaviour is wellstudied 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.^{[14]} Hence, negative energystates are bound if V vanishes at infinity.
1D bound states can be shown to be non degenerate in energy for wellbehaved wavefunctions that decay to zero at infinities. This need not hold true for wavefunction in higher dimensions. Due to the property of nondegenerate states, one dimensional bound states can always be expressed as real wavefunctions.
Proof 

Consider two energy eigenstates states and with same energy eigenvalue. Then since, the Schrodinger equation, which is expressed as: is satisfied for i = 1 and 2, subtracting the two equations gives: which can be rearranged to give the condition: Since , taking limit of x going to infinity on both sides, the wavefunctions vanish and gives .
Furthermore it can be shown that these wavefunctions can always be represented by a completely real wavefunction. Define real functions and such that . Then, from Schrodinger's equation: we get that, since the terms in the equation are all real values: 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 nondegenerate states in general.

Node theorem states that nth bound wavefunction ordered according to increasing energy has exactly n1 nodes, ie. points where . Due to the form of Schrödinger's time independent equations, it is not possible for a physical wavefunction to have since it corresponds to solution.^{[15]}
A boson with mass m_{χ} mediating a weakly coupled interaction produces an Yukawalike interaction potential,
where , g is the gauge coupling constant, and ƛ_{i} = ℏ/m_{i}c 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 m_{1} and m_{2}, the Bohr radius of the system becomes
and yields the dimensionless number
In order for the first bound state to exist at all, . 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/c^{2}, 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.^{[16]}