Nuclear force

The residual strong interaction between nucleons. Its long-distance part involves pion exchange and depends on spin and isospin. It is distinct from directly exchanging gluons between color-singlet nucleons.

Why do protons and neutrons bind into an atomic nucleus, even though protons repel electrically? An attractive nuclear force can lower the total energy below that of separated nucleons, producing nuclear binding energy. The deuteron, containing one proton and one neutron, is the simplest such bound state.

In Quantum chromodynamics, nucleons are composite colour singlets built from quarks and gluons. Their residual nuclear force has a longest-range strong contribution from one-pion exchange, with additional interactions at shorter distances.^[ref]

The approximately bounded binding energy per nucleon and approximately constant interior density of well-bound medium and heavy nuclei. The short effective range and many-body character of the nuclear force prevent each nucleon from binding independently to every other nucleon with an unsuppressed long-range attraction.

Formation of a deuteron from a neutron and a proton with photon emission. For an initial pair at rest, two-body energy and momentum conservation gives $E_{\gamma }=B-B^2/[2(m_p+m_n)]$ in natural units, where $B=m_p+m_n-m_d$. The missing binding energy becomes nuclear recoil.

A two-proton system. Its low-energy spin-singlet strong channel does not give a vacuum bound state, and electrostatic repulsion further opposes binding. This is distinct from proton pairing correlations within an atomic nucleus.

A two-neutron system. At low orbital angular momentum its identical-fermion exchange symmetry forces the spin-singlet, isospin-one channel; there is no bound dineutron in vacuum at physical parameters. This does not forbid correlated neutron pairs inside larger nuclei.

The approximate invariance of the strong nuclear force under isospin rotations. It compares channels with the same spin, orbital and total-isospin quantum numbers; it does not identify the spin-triplet isospin-zero deuteron channel with the spin-singlet isospin-one neutron-neutron channel. Electromagnetic effects and quark mass differences break the approximation.

In 1935, Hideki Yukawa proposed a new massive exchange quantum to explain nucleon binding. His paper began with the inadequacy of an electron–neutrino exchange description of the binding strength. It then required a potential energy falling faster with distance than a Coulomb potential, introduced an exponentially decaying field, and related its inverse range to the exchange particle's rest mass. This was a theoretical proposal, not a detection with an apparatus of his own.^[ref]

The crucial idea is that a finite rest mass supplies a length scale even in empty space. In a scalar model, the static limit of a Klein-Gordon equation produces the modified Helmholtz equation. Its decaying Green's function gives the Yukawa potential. The derivation below isolates this range mechanism; one-pion exchange also carries spin and isospin structure.

The nuclear emulsion track experiment used developed photographic emulsions exposed to cosmic rays and inspected microscopically. Stopping pions gave daughter muon tracks with closely grouped ranges. Comparing penetrating muons with pion-induced nuclear breakup distinguished their interactions; this was not a direct static force measurement. The muon is a lepton, not Yukawa's strongly interacting meson.^[ref]

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

Although the pion has spin zero, one-pion exchange depends on nucleon spin and isospin and includes a nuclear tensor force. Shorter-range and multiple-nucleon interactions also matter. The scalar Yukawa potential isolates a range mechanism rather than specifying the complete nuclear force.^[ref]

The linear operator $V_T(r)S_{12}$ couples nucleon spin orientations to their separation direction $\hat{\mathbf{r}}$. Here ${\mathbf{\sigma }}_i$ are Pauli matrices. A scalar radial potential energy alone omits this directional dependence.

An isotropic pure $L=0$ spatial state has zero electric quadrupole moment. The nonzero deuteron quadrupole moment consequently requires anisotropic structure. With ground-state $J^P=1^{+}$, a nuclear tensor force can mix ${}^3{S}_1$ and ${}^3{D}_1$ waves while preserving total angular momentum and parity. A purely central spin-independent potential cannot produce this mixing. The inference of an orientation-dependent force was already explicit in 1939.^[ref]