Chapter 1

Nuclear Forces

The Strong Nuclear Force

The strong nuclear force is the fundamental interaction that binds protons and neutrons (collectively called nucleons) together inside atomic nuclei. It is the residual effect of the strong interaction between quarks mediated by gluons in quantum chromodynamics (QCD). At the nuclear scale, this residual force can be described effectively through meson exchange, as first proposed by Hideki Yukawa in 1935.

Key experimental facts about the nuclear force:

- Short range: The force effectively vanishes beyond about 2-3 fm
- Very strong: At short distances (~1 fm), it is about 100 times stronger than the electromagnetic force
- Attractive at medium range: Produces a potential well of depth ~50 MeV
- Repulsive core: Below ~0.5 fm, the force becomes strongly repulsive
- Saturation: Each nucleon interacts only with its nearest neighbors
- Charge independence: The pp, nn, and np forces are nearly identical (after removing Coulomb effects)
- Spin dependence: The force depends on the spin orientation of the nucleons
- Tensor component: A non-central force component exists, similar to the magnetic dipole interaction

The Yukawa Potential

In 1935, Hideki Yukawa proposed that the nuclear force is mediated by the exchange of a massive boson (later identified as the pion). By analogy with the Coulomb potential arising from massless photon exchange, Yukawa derived the potential for massive particle exchange.

Derivation from Klein-Gordon Equation

The static potential from a point source satisfying the Klein-Gordon equation for a massive scalar field gives:

$$(\nabla^2 - \mu^2)\phi = -g\,\delta^3(\mathbf{r})$$

where $\mu = m_\pi c/\hbar$ is the inverse Compton wavelength of the pion. The solution with outgoing boundary conditions is:

$$\phi(r) = -\frac{g}{4\pi} \frac{e^{-\mu r}}{r}$$

The resulting nucleon-nucleon potential is:

$$V_{\text{Yukawa}}(r) = -\frac{g^2}{4\pi} \frac{e^{-\mu r}}{r}$$

The range of the force is determined by the pion mass:

$$R \sim \frac{1}{\mu} = \frac{\hbar}{m_\pi c} \approx \frac{197.3 \text{ MeV}\cdot\text{fm}}{139.6 \text{ MeV}} \approx 1.41 \text{ fm}$$

Comparison with Coulomb Potential

The Coulomb potential corresponds to the $\mu \to 0$ (massless mediator) limit of the Yukawa potential:

$$V_{\text{Coulomb}}(r) = \frac{e^2}{4\pi\epsilon_0} \frac{1}{r} = \frac{1.44 \text{ MeV}\cdot\text{fm}}{r}$$

The exponential suppression $e^{-\mu r}$ in the Yukawa potential ensures that the nuclear force has finite range, in contrast to the infinite-range Coulomb force.

Detailed Force Properties

Charge Independence & Isospin

The strong nuclear force is approximately charge-independent. The pp, nn, and np interactions are nearly equal after electromagnetic effects are removed. This leads to the concept of isospin symmetry:

$$\text{Proton: } |p\rangle = |T=\tfrac{1}{2}, T_3=+\tfrac{1}{2}\rangle$$$$\text{Neutron: } |n\rangle = |T=\tfrac{1}{2}, T_3=-\tfrac{1}{2}\rangle$$

Saturation Property

The binding energy per nucleon is approximately constant (~8 MeV) for $A > 20$, indicating each nucleon interacts only with its neighbors:

$$\frac{B}{A} \approx 8 \text{ MeV (for } A > 20\text{)}$$

If all pairs interacted, we would expect $B \propto A(A-1)/2 \propto A^2$, giving $B/A \propto A$. Saturation implies the nuclear force has limited range.

Tensor Force

The nuclear force has a tensor component, analogous to the interaction between two magnetic dipoles. The tensor operator is:

$$S_{12} = 3(\boldsymbol{\sigma}_1 \cdot \hat{r})(\boldsymbol{\sigma}_2 \cdot \hat{r}) - \boldsymbol{\sigma}_1 \cdot \boldsymbol{\sigma}_2$$

The tensor force is responsible for the quadrupole moment of the deuteron and the mixing of S and D waves in the deuteron ground state.

The Deuteron

The deuteron (d or $^2$H) is the simplest bound nuclear system, consisting of one proton and one neutron. Its properties provide fundamental constraints on the nuclear force:

Binding energy

$B_d = 2.2246$ MeV

Spin-parity

$J^\pi = 1^+$

Isospin

$T = 0$

Quadrupole moment

$Q_d = 0.2860$ fm$^2$

Square Well Model of the Deuteron

As a first approximation, model the nuclear potential as a square well of depth $V_0$and range $R$. The radial Schrodinger equation for the relative motion ($\ell = 0$ S-wave):

$$-\frac{\hbar^2}{2\mu}\frac{d^2 u}{dr^2} + V(r)u(r) = Eu(r)$$

where $\mu = m_N/2$ is the reduced mass and $u(r) = r R(r)$. Inside the well ($r < R$):

$$u(r) = A\sin(kr), \quad k = \sqrt{\frac{2\mu(V_0 - B_d)}{\hbar^2}}$$

Outside the well ($r > R$):

$$u(r) = Ce^{-\kappa r}, \quad \kappa = \sqrt{\frac{2\mu B_d}{\hbar^2}}$$

Matching at $r = R$ gives the transcendental equation:

$$k\cot(kR) = -\kappa$$

For $R = 2.1$ fm and $B_d = 2.2246$ MeV, one finds $V_0 \approx 35$ MeV, confirming the deep, short-range nature of the nuclear potential.

Python Simulation: Nuclear Potentials

Comparison of Yukawa, Coulomb, and Woods-Saxon potentials showing the key features of nuclear interactions: short range, attractive well, and range dependence on mediator mass.

Nuclear Potential Comparison

Python

Compares Yukawa (pion exchange), Coulomb, and Woods-Saxon potentials and shows range dependence

script.py75 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

# ──────────────────────────────────────────────
# Comparison of Nuclear Potentials
# Yukawa vs Coulomb vs Woods-Saxon
# ──────────────────────────────────────────────

r = np.linspace(0.1, 10.0, 500)  # distance in fm

# Physical constants
hbar_c = 197.3  # MeV fm
m_pi = 139.6    # MeV (pion mass)
mu = m_pi / hbar_c  # inverse fm

# 1. Yukawa potential: V(r) = -g^2 * exp(-mu*r) / (4*pi*r)
g2 = 14.0 * hbar_c  # coupling (dimensionless g^2/(4pi) ~ 14)
V_yukawa = -g2 * np.exp(-mu * r) / (4 * np.pi * r)

# 2. Coulomb potential for two protons: V(r) = e^2/(4*pi*eps0*r) = 1.44 MeV fm / r
V_coulomb = 1.44 / r  # MeV

# 3. Woods-Saxon potential: V(r) = -V0 / (1 + exp((r-R)/a))
V0_ws = 50.0   # MeV depth
R_ws = 1.2 * 12**(1.0/3.0)  # fm, for A=12 (Carbon)
a_ws = 0.65    # fm diffuseness
V_ws = -V0_ws / (1.0 + np.exp((r - R_ws) / a_ws))

fig, axes = plt.subplots(1, 2, figsize=(14, 6))
fig.patch.set_facecolor('#0a0a1a')

for ax in axes:
    ax.set_facecolor('#0a0a1a')
    for spine in ax.spines.values():
        spine.set_edgecolor('#334155')

# Left: all three potentials
axes[0].plot(r, V_yukawa, color='#ef4444', linewidth=2.5, label='Yukawa (pion exchange)')
axes[0].plot(r, V_coulomb, color='#3b82f6', linewidth=2.5, label='Coulomb (pp)')
axes[0].plot(r, V_ws, color='#f59e0b', linewidth=2.5, label='Woods-Saxon (A=12)')
axes[0].axhline(y=0, color='#64748b', linewidth=0.5, linestyle='--')
axes[0].set_xlim(0, 8)
axes[0].set_ylim(-60, 30)
axes[0].set_xlabel('r [fm]', color='#94a3b8', fontsize=12)
axes[0].set_ylabel('V(r) [MeV]', color='#94a3b8', fontsize=12)
axes[0].set_title('Nuclear Potential Comparison', color='white', fontsize=14)
axes[0].legend(fontsize=10, facecolor='#1a1a2e', edgecolor='#334155', labelcolor='white')
axes[0].tick_params(colors='#94a3b8')

# Right: Yukawa vs range
mu_values = [0.5, 0.7, 1.0, 1.5]
colors = ['#ef4444', '#f59e0b', '#22c55e', '#3b82f6']
for i, mu_val in enumerate(mu_values):
    V = -g2 * np.exp(-mu_val * r) / (4 * np.pi * r)
    label = f'mu = {mu_val:.1f} fm' + r'$^{-1}$'
    axes[1].plot(r, V, color=colors[i], linewidth=2, label=f'mu={mu_val:.1f} /fm')

axes[1].axhline(y=0, color='#64748b', linewidth=0.5, linestyle='--')
axes[1].set_xlim(0, 8)
axes[1].set_ylim(-60, 5)
axes[1].set_xlabel('r [fm]', color='#94a3b8', fontsize=12)
axes[1].set_ylabel('V(r) [MeV]', color='#94a3b8', fontsize=12)
axes[1].set_title('Yukawa Potential: Range Dependence', color='white', fontsize=14)
axes[1].legend(fontsize=10, facecolor='#1a1a2e', edgecolor='#334155', labelcolor='white')
axes[1].tick_params(colors='#94a3b8')

plt.tight_layout()
plt.savefig('nuclear_potentials.png', dpi=130, bbox_inches='tight',
            facecolor=fig.get_facecolor())
print("Yukawa range: hbar*c / m_pi =", f"{hbar_c/m_pi:.2f} fm")
print("Woods-Saxon radius (A=12):", f"{R_ws:.2f} fm")
print("Yukawa potential at r=1 fm:", f"{V_yukawa[np.argmin(np.abs(r-1.0))]:.2f} MeV")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran Implementation

Fortran solver for the deuteron binding energy using the square-well potential model. Solves the transcendental matching condition numerically using bisection.

Deuteron Binding Energy Solver

Fortran

Solves for the square-well depth V0 that reproduces the deuteron binding energy for various well radii

deuteron_binding.f9058 lines

program deuteron_binding
  implicit none
  integer, parameter :: dp = selected_real_kind(15)
  real(dp), parameter :: hbar2_over_2m = 20.7_dp  ! hbar^2/(2*mu) in MeV fm^2, mu=m_N/2
  real(dp), parameter :: B_d = 2.2246_dp           ! deuteron binding energy (MeV)
  real(dp), parameter :: pi = 3.14159265358979_dp
  real(dp) :: R, V0, V0_lo, V0_hi, k, kappa, f, f_lo
  integer :: i, iter

kappa = sqrt(B_d / hbar2_over_2m)
  write(*,'(A)') '=== Deuteron Binding Energy Solver ==='
  write(*,'(A)') 'Square-well potential: V(r) = -V0 for r < R, 0 for r > R'
  write(*,'(A,F8.4,A)') 'Target binding energy: ', B_d, ' MeV'
  write(*,'(A,F8.4,A)') 'Decay constant kappa = ', kappa, ' fm^-1'
  write(*,'(A)') ''

! Scan over different well radii
  write(*,'(A)') '  R (fm)    V0 (MeV)    kR       Prob(r>R)'
  write(*,'(A)') '  ------    ---------    -----    ---------'

do i = 1, 6
    R = 1.0_dp + 0.4_dp * i   ! R from 1.4 to 3.4 fm

! Bisection to find V0 such that k*cot(kR) = -kappa
    V0_lo = B_d + 0.1_dp
    V0_hi = 200.0_dp

do iter = 1, 100
      V0 = 0.5_dp * (V0_lo + V0_hi)
      k = sqrt((V0 - B_d) / hbar2_over_2m)
      f = k / tan(k * R) + kappa
      k = sqrt((V0_lo - B_d) / hbar2_over_2m)
      f_lo = k / tan(k * R) + kappa
      if (f * f_lo < 0.0_dp) then
        V0_hi = V0
      else
        V0_lo = V0
      end if
      if (abs(V0_hi - V0_lo) < 1.0d-10) exit
    end do

V0 = 0.5_dp * (V0_lo + V0_hi)
    k = sqrt((V0 - B_d) / hbar2_over_2m)

! Probability of finding nucleons outside the well
    ! P(r>R) = integral of |u|^2 from R to infinity / total
    ! u_out = C*exp(-kappa*r), normalized
    ! P(r>R) = 1 / (1 + (sin(kR))^2 * kappa / (k^2 + kappa^2) * ... )
    ! Simplified: use asymptotic normalization
    write(*,'(F8.2, F12.4, F10.4, F12.4)') R, V0, k*R, exp(-2.0_dp*kappa*R)
  end do

write(*,'(A)') ''
  write(*,'(A)') 'Note: The deuteron is weakly bound (B/nucleon ~ 1.1 MeV)'
  write(*,'(A)') 'Typical nuclear potential depth ~ 35-50 MeV for R ~ 2 fm'
  write(*,'(A)') 'The large spatial extent (small kappa) makes the deuteron'
  write(*,'(A)') 'a diffuse, loosely-bound system.'
end program deuteron_binding

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

The Woods-Saxon Potential

A more realistic nuclear potential is the Woods-Saxon form, which smoothly interpolates between the interior and exterior of the nucleus:

$$V_{\text{WS}}(r) = -\frac{V_0}{1 + \exp\!\left(\frac{r - R}{a}\right)}$$

where:

- $V_0 \approx 50$ MeV is the potential depth
- $R = r_0 A^{1/3}$ with $r_0 \approx 1.25$ fm is the nuclear radius
- $a \approx 0.65$ fm is the surface diffuseness parameter

The Woods-Saxon potential reproduces the nearly constant nuclear density in the interior and the smooth falloff at the surface observed in electron scattering experiments. Adding a spin-orbit term $V_{\text{so}}(r)\,\mathbf{L}\cdot\mathbf{S}$ is essential for reproducing the nuclear shell structure and magic numbers.

Derivation: Yukawa Potential from Meson Exchange

We derive the Yukawa potential rigorously starting from the Lagrangian density for a massive scalar field coupled to a static nucleon source. The free Klein-Gordon Lagrangian is:

$$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi) - \frac{1}{2}\mu^2 \phi^2 - g\,\phi\,\rho(\mathbf{r})$$

where $\mu = m_\pi c/\hbar$ is the inverse Compton wavelength of the exchanged meson,$g$ is the coupling constant, and $\rho(\mathbf{r})$ is the static nucleon source density. For a point-like nucleon, $\rho(\mathbf{r}) = \delta^3(\mathbf{r})$.

Step 1: Equation of Motion

The Euler-Lagrange equation for the static case ($\partial_t \phi = 0$) gives:

$$\nabla^2 \phi - \mu^2 \phi = g\,\delta^3(\mathbf{r})$$

This is a modified Helmholtz equation. To solve it, we take the Fourier transform. Define$\tilde{\phi}(\mathbf{q}) = \int d^3r\, \phi(\mathbf{r})\, e^{-i\mathbf{q}\cdot\mathbf{r}}$:

$$(-q^2 - \mu^2)\tilde{\phi}(\mathbf{q}) = g$$

$$\tilde{\phi}(\mathbf{q}) = -\frac{g}{q^2 + \mu^2}$$

Step 2: Inverse Fourier Transform

Transforming back to coordinate space:

$$\phi(\mathbf{r}) = -\frac{g}{(2\pi)^3}\int \frac{d^3q}{q^2 + \mu^2}\, e^{i\mathbf{q}\cdot\mathbf{r}}$$

Switching to spherical coordinates with $\mathbf{q}\cdot\mathbf{r} = qr\cos\theta$:

$$\phi(r) = -\frac{g}{(2\pi)^2}\int_0^\infty \frac{q^2\,dq}{q^2 + \mu^2}\cdot\frac{2\sin(qr)}{qr}$$

Evaluating by contour integration in the complex plane (closing in the upper half-plane with a pole at $q = i\mu$):

$$\phi(r) = -\frac{g}{4\pi}\frac{e^{-\mu r}}{r}$$

Step 3: Nucleon-Nucleon Potential

The interaction energy between two static nucleon sources is obtained from the overlap of the field produced by one source with the other source:

$$V(\mathbf{r}) = -g\,\phi(\mathbf{r}) = -\frac{g^2}{4\pi}\frac{e^{-\mu r}}{r}$$

Equivalently, from second-order perturbation theory in quantum field theory, the Born-level amplitude for nucleon-nucleon scattering via single-meson exchange yields the momentum-space potential:

$$\tilde{V}(\mathbf{q}) = -\frac{g^2}{|\mathbf{q}|^2 + \mu^2}$$

whose Fourier transform reproduces the Yukawa result. The propagator $1/(q^2 + \mu^2)$ is precisely the non-relativistic limit of the meson propagator, confirming the connection between force range and mediator mass via the uncertainty principle: $R \sim \hbar/(m_\pi c)$.

Nuclear Force Components in Detail

The general nucleon-nucleon potential can be decomposed into several operator structures. The most general form consistent with translational invariance, rotational invariance, parity conservation, time-reversal invariance, and isospin symmetry is:

$$V_{NN} = V_C(r) + V_\sigma(r)\,\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2 + V_T(r)\,S_{12} + V_{LS}(r)\,\mathbf{L}\!\cdot\!\mathbf{S} + V_{LS2}(r)\,(\mathbf{L}\!\cdot\!\mathbf{S})^2$$

Each term can also carry an isospin dependence $\boldsymbol{\tau}_1\!\cdot\!\boldsymbol{\tau}_2$, doubling the number of independent radial functions. The complete potential has the form:

$$V = \sum_{p=1}^{N_{\rm op}} V_p(r)\, O_p^{(1,2)}$$

1. Central Force $V_C(r)$

The spin- and angle-independent part. Depends only on the internucleon distance. In the singlet (S=0) and triplet (S=1) channels these are different:

$$V_C^{(S=0)}(r) = V_C(r) - 3V_\sigma(r), \quad V_C^{(S=1)}(r) = V_C(r) + V_\sigma(r)$$

The central potential provides the dominant attraction at ~1 fm and is responsible for the bulk of nuclear binding.

2. Spin-Spin Force $V_\sigma(r)\,\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2$

Differentiates between spin-singlet and spin-triplet states. The eigenvalues of$\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2$ are:

$$\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2 = \begin{cases} +1 & S = 1 \text{ (triplet)} \\ -3 & S = 0 \text{ (singlet)} \end{cases}$$

The deuteron exists only in the triplet state ($S=1$), demonstrating that the triplet potential is more attractive than the singlet. The spin-singlet np state is unbound (virtual state with scattering length $a_s = -23.7$ fm).

3. Tensor Force $V_T(r)\,S_{12}$

The tensor operator couples spin and spatial degrees of freedom:

$$S_{12} = 3(\boldsymbol{\sigma}_1\!\cdot\!\hat{r})(\boldsymbol{\sigma}_2\!\cdot\!\hat{r}) - \boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2 = \frac{3}{r^2}(\boldsymbol{\sigma}_1\!\cdot\!\mathbf{r})(\boldsymbol{\sigma}_2\!\cdot\!\mathbf{r}) - \boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2$$

Key properties: $S_{12}$ vanishes for $S=0$ states and mixes $\ell$ values differing by 2. The expectation value depends on $m_S$:

$$\langle S_{12} \rangle_{S=1,m_S=\pm1} = -\langle S_{12} \rangle_{S=1,m_S=0}/2$$

The tensor force mixes $^3S_1$ and $^3D_1$ waves in the deuteron, producing: (a) the deuteron quadrupole moment $Q_d = 0.2860$ fm$^2$, and (b) the D-state probability$P_D \approx 4\text{-}7\%$.

4. Spin-Orbit Force $V_{LS}(r)\,\mathbf{L}\!\cdot\!\mathbf{S}$

The spin-orbit force depends on the relative orbital angular momentum and total spin:

$$\mathbf{L}\!\cdot\!\mathbf{S} = \frac{1}{2}[J(J+1) - L(L+1) - S(S+1)]$$

This two-body spin-orbit force, combined with the Thomas precession term, generates the effective one-body spin-orbit potential in the shell model. It splits each$\ell$-level into $j = \ell \pm 1/2$ doublets and is essential for reproducing the nuclear magic numbers. The splitting is proportional to$(2\ell + 1)$, making it largest for high-$\ell$ orbits.

Derivation: One-Pion Exchange Potential (OPEP)

The pion is a pseudoscalar, isovector meson ($J^\pi = 0^-$, $T=1$). The pion-nucleon interaction Lagrangian in the pseudoscalar coupling scheme is:

$$\mathcal{L}_{\pi NN} = -i\,g_{\pi NN}\,\bar{\psi}\gamma_5\,\boldsymbol{\tau}\cdot\boldsymbol{\phi}_\pi\,\psi$$

In the non-relativistic limit, this reduces to the pseudovector (gradient) coupling:

$$\mathcal{L}_{\pi NN}^{\rm NR} = \frac{f_{\pi NN}}{m_\pi}\,\bar{\psi}\,\boldsymbol{\sigma}\!\cdot\!\nabla\boldsymbol{\phi}_\pi\!\cdot\!\boldsymbol{\tau}\,\psi$$

Born Amplitude and OPEP Result

The one-pion exchange contribution to the NN scattering amplitude in Born approximation gives the momentum-space potential:

$$\tilde{V}_{\rm OPEP}(\mathbf{q}) = -\frac{f_{\pi NN}^2}{m_\pi^2}\,\frac{(\boldsymbol{\sigma}_1\!\cdot\!\mathbf{q})(\boldsymbol{\sigma}_2\!\cdot\!\mathbf{q})}{\mathbf{q}^2 + m_\pi^2}\,(\boldsymbol{\tau}_1\!\cdot\!\boldsymbol{\tau}_2)$$

Fourier transforming to coordinate space and using the identity$(\boldsymbol{\sigma}_1\!\cdot\!\nabla)(\boldsymbol{\sigma}_2\!\cdot\!\nabla)(e^{-\mu r}/r)$, we obtain:

$$V_{\rm OPEP}(r) = \frac{f_{\pi NN}^2}{m_\pi^2}\,(\boldsymbol{\tau}_1\!\cdot\!\boldsymbol{\tau}_2)\left[\frac{m_\pi^2}{3}\,(\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2)\,\frac{e^{-\mu r}}{\mu r} + S_{12}\left(1 + \frac{3}{\mu r} + \frac{3}{(\mu r)^2}\right)\frac{e^{-\mu r}}{\mu r}\right]$$

where $\mu = m_\pi / \hbar c$. Using the empirical coupling constant$f_{\pi NN}^2/(4\pi) \approx 0.08$ (or equivalently $g_{\pi NN}^2/(4\pi) \approx 14$):

$$V_{\rm OPEP}(r) = \frac{m_\pi c^2}{3}\frac{f^2}{4\pi}\,(\boldsymbol{\tau}_1\!\cdot\!\boldsymbol{\tau}_2)\left[(\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2) + S_{12}\left(1 + \frac{3}{\mu r} + \frac{3}{(\mu r)^2}\right)\right]\frac{e^{-\mu r}}{\mu r}$$

The OPEP has been verified to dominate the long-range ($r > 2$ fm) part of the NN interaction and correctly accounts for the deuteron quadrupole moment and D-state admixture.

Hard Core and Short-Range Repulsion

Nucleon-nucleon scattering data at high energies reveals a strongly repulsive core at short distances ($r \lesssim 0.5$ fm). This repulsive core has several important consequences:

Evidence for the Repulsive Core

- S-wave phase shifts: The $^1S_0$ phase shift passes through zero and becomes negative at $E_{\rm lab} \approx 250$ MeV, indicating the transition from attractive to repulsive interaction.
- Nuclear saturation: Without repulsion, nuclear matter would collapse. The equilibrium density $\rho_0 \approx 0.16$ fm$^{-3}$ results from the balance of attraction and short-range repulsion.
- High-momentum components: The presence of short-range correlations depletes single-particle occupancies below the Fermi surface by ~15-20%.

Physical Origin

In meson-exchange models, the short-range repulsion arises from the exchange of heavy vector mesons, particularly the $\omega$ meson ($m_\omega = 783$ MeV):

$$V_\omega(r) = +g_\omega^2\frac{e^{-m_\omega r/\hbar c}}{4\pi r} \quad (\text{repulsive, vector exchange})$$

The range hierarchy of the nuclear force is:

- Long range ($r > 2$ fm): one-pion exchange ($m_\pi \approx 140$ MeV)
- Medium range ($1 < r < 2$ fm): two-pion exchange and $\sigma$ meson ($m_\sigma \approx 500$ MeV), dominant attraction
- Short range ($r < 1$ fm): $\omega$ and $\rho$ meson exchange ($m \approx 780$ MeV), repulsive core

QCD Perspective

At the fundamental level, the short-range repulsion arises from quark-gluon dynamics. When nucleons overlap significantly, the Pauli exclusion principle among quarks (each nucleon contains 3 quarks occupying color-flavor-spin states) and the one-gluon exchange interaction generate a repulsive force. Lattice QCD calculations confirm the qualitative features of the NN potential including the repulsive core, though quantitative precision remains challenging.

Charge Independence: Experimental Evidence

The charge independence of the nuclear force is one of its most fundamental properties. It states that the strong interaction is the same in all three NN channels (pp, nn, np) when the system is in the same quantum state. This is a consequence of isospin symmetry.

Scattering Length Evidence

The $^1S_0$ scattering lengths (after removing electromagnetic effects) are:

pp ($^1S_0$)

$a_{pp} = -17.3 \pm 0.4$ fm

nn ($^1S_0$)

$a_{nn} = -18.9 \pm 0.4$ fm

np ($^1S_0$)

$a_{np} = -23.7 \pm 0.02$ fm

The near equality of $a_{pp}$ and $a_{nn}$ demonstrates charge symmetry (invariance under interchange of protons and neutrons). The slight difference of $a_{np}$ from $a_{pp}$and $a_{nn}$ reflects charge independence breaking, which is small (~few percent).

Mirror Nuclei

Mirror nuclei are pairs that differ by the interchange of protons and neutrons (e.g., $^3$H and $^3$He, or $^{13}$C and $^{13}$N). Their energy level schemes are nearly identical after correcting for the Coulomb energy difference:

$$\Delta E_{\rm Coulomb} = \frac{3}{5}\frac{e^2}{4\pi\epsilon_0 R}\left[Z_>(Z_>-1) - Z_<(Z_<-1)\right]$$

The agreement between predicted and observed energy differences to within ~100 keV provides strong evidence for the charge independence of nuclear forces.

Isobaric Analog States

States in different isobaric nuclei that are related by isospin rotations are called isobaric analog states (IAS). For a nucleus with isospin $T$ and projection $T_3$, the analog state in a neighboring isobar has the same $T$ but different $T_3$:

$$|T, T_3 - 1\rangle = \frac{1}{\sqrt{T(T+1) - T_3(T_3-1)}}\,T_-\,|T, T_3\rangle$$

The excitation energies of IAS are predictable from the Coulomb displacement energy and provide precision tests of charge independence.

Modern NN Potentials

Modern high-precision nucleon-nucleon potentials reproduce the world database of NN scattering data (several thousand data points below 350 MeV) with $\chi^2/\text{datum} \approx 1$.

Argonne v18 (AV18)

The Argonne v18 potential uses 18 operator components acting on spin, isospin, and spatial coordinates:

$$V_{NN} = \sum_{p=1}^{18} V_p(r)\, O_p^{(1,2)}$$

The operators include: $1$, $\boldsymbol{\tau}_1\!\cdot\!\boldsymbol{\tau}_2$,$\boldsymbol{\sigma}_1\!\cdot\!\boldsymbol{\sigma}_2$, $(\boldsymbol{\sigma}\!\cdot\!\boldsymbol{\tau})$,$S_{12}$, $\mathbf{L}\!\cdot\!\mathbf{S}$, $\mathbf{L}^2$, $(\mathbf{L}\!\cdot\!\mathbf{S})^2$, and charge-dependent/charge-asymmetric terms.

Each radial function $V_p(r)$ is parameterized with one-pion exchange at long range, intermediate and short-range phenomenological terms from Woods-Saxon-like forms, and an electromagnetic component. Total of 40 adjustable parameters. Achieves $\chi^2/\text{datum} = 1.09$ for 4301 pp + np data below 350 MeV.

CD-Bonn Potential

The CD-Bonn potential is a charge-dependent one-boson-exchange potential that includes the exchange of $\pi$, $\rho$, $\omega$, and $\sigma$ mesons. It is defined in momentum space and uses covariant Feynman amplitudes:

$$V_{\rm CD-Bonn}(\mathbf{p}',\mathbf{p}) = \sum_{\alpha = \pi,\rho,\omega,\sigma} V_\alpha(\mathbf{p}',\mathbf{p})$$

The CD-Bonn potential has a softer (non-local) core compared to AV18, which makes it more suitable for many-body calculations. It achieves$\chi^2/\text{datum} = 1.01$ and predicts the deuteron D-state probability $P_D = 4.85\%$.

Chiral Effective Field Theory

Modern approaches derive the nuclear force systematically from the symmetries of QCD using chiral perturbation theory. The chiral expansion organizes contributions by powers of $(Q/\Lambda_\chi)$ where $Q$ is a typical momentum and$\Lambda_\chi \approx 1$ GeV:

$$V = V^{(0)}_{\rm 2N} + V^{(2)}_{\rm 2N} + V^{(3)}_{\rm 2N+3N} + V^{(4)}_{\rm 2N+3N} + \cdots$$

A key advantage is the systematic inclusion of three-nucleon forces (3NF) that arise naturally at next-to-next-to-leading order (N2LO). These 3NFs are essential for reproducing the binding energies of light nuclei ($A \leq 12$) and the saturation properties of nuclear matter.

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