Nuclear Structure Models: Advanced Shell Model & Collective Dynamics

1. The Independent Particle Mean Field Approximation

Although the bare nucleon-nucleon interaction contains strong repulsive cores and tensor components, inside a many-nucleon nucleus each nucleon moves predominantly in an average, smooth, spherically symmetric mean-field potential $V_{\text{MF}}(\vec{r})$ created by the collective action of all other $A-1$ nucleons:

$$\hat{H} = \sum_{i=1}^A \left[ -\frac{\hbar^2}{2M}\nabla_i^2 + V_{\text{MF}}(\vec{r}_i) \right] + \hat{V}_{\text{residual}}$$

This independent-particle motion is made possible by the Pauli exclusion principle: when two nucleons collide inside a nucleus, the states into which they could scatter are already occupied by other nucleons, suppressing short-range collisions and imparting nucleons with long mean free paths ($\lambda_{\text{mfp}} \gg R$).

2. The Degenerate Nuclear Fermi Gas Model

As the simplest microscopic mean-field model, consider protons and neutrons as two independent, non-interacting ideal Fermi gases confined within a spherical nuclear volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$ of constant density $\rho_0 \approx 0.16\text{ fm}^{-3}$.

Each nucleon state occupies a phase space volume of $h^3 = (2\pi\hbar)^3$. For $Z$ protons and $N$ neutrons with spin degeneracy $g = 2$:

$$N = 2 \frac{V}{(2\pi)^3} \int_0^{k_{F,n}} 4\pi k^2 dk = \frac{V}{\pi^2} \frac{k_{F,n}^3}{3} \implies k_{F,n} = \left( 3\pi^2 \rho_n \right)^{1/3}$$ $$Z = 2 \frac{V}{(2\pi)^3} \int_0^{k_{F,p}} 4\pi k^2 dk = \frac{V}{\pi^2} \frac{k_{F,p}^3}{3} \implies k_{F,p} = \left( 3\pi^2 \rho_p \right)^{1/3}$$

For a symmetric nucleus with $N = Z = A/2$ and $\rho_n = \rho_p = \rho_0 / 2 \approx 0.08\text{ fm}^{-3}$:

$$k_F = \left( 3\pi^2 \frac{\rho_0}{2} \right)^{1/3} = \left( \frac{3\pi^2}{2} \times 0.16 \right)^{1/3} = (2.3687)^{1/3} \approx 1.33\text{ to }1.36\text{ fm}^{-1}$$

The maximum kinetic energy at absolute zero is the Fermi Energy $E_F$:

$$E_F = \frac{\hbar^2 k_F^2}{2 M} = \frac{(197.3\text{ MeV}\cdot\text{fm})^2 (1.36\text{ fm}^{-1})^2}{2(938.9\text{ MeV})} = \frac{38938 \times 1.8496}{1877.8} \approx 38.35\text{ MeV}$$

Because nucleons are bound by an average separation energy $B/A \approx 8\text{ MeV}$, the total potential well depth $V_0$ must be:

$$V_0 = E_F + B \approx 38\text{ MeV} + 8\text{ MeV} \approx 46\text{ MeV}$$

The average kinetic energy per nucleon is $\langle E_k \rangle = \frac{3}{5} E_F \approx 23\text{ MeV}$.

1. The 3D Isotropic Harmonic Oscillator Potential

An analytically solvable model for the nuclear mean field is the 3D harmonic oscillator:

$$V_{\text{HO}}(r) = -V_0 + \frac{1}{2} M \omega^2 r^2$$

The energy eigenvalues are governed by the principal oscillator quantum number $N_{\text{osc}} = 2(n - 1) + l = 0, 1, 2, 3, \dots$:

$$E_N = \left( N_{\text{osc}} + \frac{3}{2} \right) \hbar\omega$$

where the standard empirical oscillator frequency scales with mass number as $\hbar\omega \approx 41 A^{-1/3}\text{ MeV}$.

The degeneracies of the harmonic oscillator shells are:

• $N_{\text{osc}} = 0$: $1s$ (Degeneracy $2$) $\implies$ Cumulative: **2**
• $N_{\text{osc}} = 1$: $1p$ (Degeneracy $6$) $\implies$ Cumulative: **8**
• $N_{\text{osc}} = 2$: $1d, 2s$ (Degeneracy $10 + 2 = 12$) $\implies$ Cumulative: **20**
• $N_{\text{osc}} = 3$: $1f, 2p$ (Degeneracy $14 + 6 = 20$) $\implies$ Cumulative: **40** (Not 28!)
• $N_{\text{osc}} = 4$: $1g, 2d, 3s$ (Degeneracy $18 + 10 + 2 = 30$) $\implies$ Cumulative: **70** (Not 50!)

The harmonic oscillator correctly predicts the first three magic numbers ($2, 8, 20$), but completely fails to explain the higher magic numbers ($28, 50, 82, 126$).

2. The Realistic Woods-Saxon Potential

In real nuclei, the nuclear density is constant in the interior and drops smoothly to zero at the surface over a skin thickness $t \approx 2.4\text{ fm}$. This is modeled by the Woods-Saxon potential:

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

where $V_0 \approx 50\text{ MeV}$, $R = R_0 A^{1/3} \approx 1.25 A^{1/3}\text{ fm}$, and diffuseness $a \approx 0.65\text{ fm}$.

Because the Woods-Saxon potential is flatter at the center and steeper at the surface than a parabola, it lifts the $l$-degeneracy: states with higher orbital angular momentum $l$ have wavefunctions concentrated closer to the surface and are pulled downward in energy relative to lower-$l$ states. However, even with this flattening, the magic numbers beyond 20 cannot be explained without spin-orbit coupling.

1. The Inverted Nuclear Spin-Orbit Interaction

In 1949, Maria Goeppert Mayer and independently J. Hans D. Jensen (Nobel Prize 1963) solved the mystery of the magic numbers by introducing a powerful, relativistic spin-orbit potential into the nuclear mean field:

$$V(r) = V_{\text{WS}}(r) + V_{so}(r) \vec{l}\cdot\vec{s}$$

where $V_{so}(r) = -V_{so}^{(0)} \frac{1}{r}\frac{dV_{\text{WS}}}{dr}$ is concentrated entirely at the nuclear surface.

Evaluating the operator $\vec{l}\cdot\vec{s}$ from total single-particle angular momentum $\vec{j} = \vec{l} + \vec{s}$:

$$\vec{j}^2 = \vec{l}^2 + \vec{s}^2 + 2\vec{l}\cdot\vec{s} \implies \vec{l}\cdot\vec{s} = \frac{1}{2}\left[ j(j+1) - l(l+1) - s(s+1) \right]$$

For nucleon spin $s = 1/2$, the allowed total angular momenta are $j = l + 1/2$ and $j = l - 1/2$:

$$\langle \vec{l}\cdot\vec{s} \rangle = \begin{cases} +\frac{1}{2} l, & j = l + 1/2 \text{ (spin parallel to orbit)} \\ -\frac{1}{2}(l + 1), & j = l - 1/2 \text{ (spin antiparallel to orbit)} \end{cases}$$

The energy splitting between the two members of the spin-orbit doublet is:

$$\Delta E_{so} = E(j = l - 1/2) - E(j = l + 1/2) = \frac{2l + 1}{2} \langle V_{so} \rangle$$

CRITICAL PHYSICAL DISTINCTION FROM ATOMIC PHYSICS: In atomic physics, spin-orbit coupling is positive ($+\vec{L}\cdot\vec{S}$), placing $j = l - 1/2$ lower in energy. In nuclear physics, the spin-orbit interaction is strongly attractive and negative: the state with $j = l + 1/2$ is pushed dramatically DOWNWARD in energy.

2. The Exact Generation of the Magic Numbers

Because $\Delta E_{so} \propto (2l + 1)$, the splitting grows enormously with orbital angular momentum $l$:

• In the $N_{\text{osc}} = 3$ shell ($1f, 2p$), the $1f_{7/2}$ state ($j = 3 + 1/2$) is pushed downward so strongly that it detaches from the shell and joins the lower shell. With its capacity of $2j + 1 = 8$ nucleons, adding it to $20$ produces the magic number: $$\mathbf{20 + 8 = 28}$$
• In the $N_{\text{osc}} = 4$ shell, the intruder state $1g_{9/2}$ ($2j + 1 = 10$) plunges downward across the major shell gap. Adding it to $40$ yields: $$\mathbf{40 + 10 = 50}$$
• In the $N_{\text{osc}} = 5$ shell, the intruder state $1h_{11/2}$ ($2j + 1 = 12$) plunges downward. Adding it to $70$ yields: $$\mathbf{70 + 12 = 82}$$
• In the $N_{\text{osc}} = 6$ shell, the intruder state $1i_{13/2}$ ($2j + 1 = 14$) plunges downward. Adding it to $112$ yields: $$\mathbf{112 + 14 = 126}$$

The complete sequence of nuclear magic numbers is rigorously explained:

$$\mathbf{2, \quad 8, \quad 20, \quad 28, \quad 50, \quad 82, \quad 126}$$

Nuclei with both $Z$ and $N$ equal to magic numbers (${}^4_2\text{He}_2$, ${}^{16}_8\text{O}_8$, ${}^{40}_{20}\text{Ca}_{20}$, ${}^{48}_{20}\text{Ca}_{28}$, ${}^{208}_{82}\text{Pb}_{126}$) are doubly magic, exhibiting exceptional stability, spherical symmetry, high excitation thresholds, and tiny neutron capture cross sections.

1. Ground-State Spin and Parity for Even-Even and Odd-A Nuclei

The Extreme Single-Particle Shell Model (ESPM) provides robust rules for determining ground-state spins and parities $J^\pi$:

1. Even-Even Nuclei ($Z$ even, $N$ even): All protons pair up in time-reversed orbits with opposite magnetic quantum numbers ($|j, m\rangle$ and $|j, -m\rangle$), and all neutrons pair up identically. The net spin and parity for all even-even ground states without exception is: $$J^\pi = 0^+$$
2. Odd-$A$ Nuclei ($Z$ odd, $N$ even or $Z$ even, $N$ odd): All even nucleons pair off to $J = 0^+$. The total spin and parity of the nucleus are determined entirely by the single unpaired valence nucleon occupying the state $n l_j$: $$J = j_{\text{val}}, \qquad \pi = (-1)^{l_{\text{val}}}$$ Examples:
   • ${}^{17}_8\text{O}_9$: $Z=8$ is magic; $N=9$ has one valence neutron in $1d_{5/2}$ ($l=2$). Predicts $J^\pi = 5/2^+$. (Experiment: $5/2^+$).
   • ${}^{41}_{20}\text{Ca}_{21}$: $Z=20$ is magic; $N=21$ has one valence neutron in $1f_{7/2}$ ($l=3$). Predicts $J^\pi = 7/2^-$. (Experiment: $7/2^-$).
   • ${}^{207}_{82}\text{Pb}_{125}$: $Z=82$ is magic; $N=125$ has a single neutron hole in $3p_{1/2}$ ($l=1$). Predicts $J^\pi = 1/2^-$. (Experiment: $1/2^-$).

2. Odd-Odd Nuclei & Nordheim's Coupling Rules

For nuclei with both $Z$ odd and $N$ odd, the spin arises from the vector coupling of the unpaired proton $(j_p, l_p)$ and unpaired neutron $(j_n, l_n)$:

$$|j_p - j_n| \le J \le j_p + j_n, \qquad \pi = (-1)^{l_p + l_n}$$

The precise value of $J$ is predicted by Nordheim's empirical coupling rules, governed by the Nordheim number $\mathcal{N} \equiv (j_p - l_p) + (j_n - l_n)$:

• Strong Rule ($\mathcal{N} = 0$, intrinsic spins parallel): $$J = |j_p - j_n|$$ Example: ${}^{38}_{17}\text{Cl}_{21}$: Proton hole in $1d_{3/2}$ ($j_p = 3/2, l_p = 2 \implies j_p - l_p = -1/2$). Valence neutron in $1f_{7/2}$ ($j_n = 7/2, l_n = 3 \implies j_n - l_n = +1/2$). Here $\mathcal{N} = -1/2 + 1/2 = 0$. Predicts $J = |3/2 - 7/2| = 2$. Parity $\pi = (-1)^{2+3} = -1$. Predicts $J^\pi = 2^-$. (Experiment: $2^-$).
• Weak Rule ($\mathcal{N} = \pm 1$, intrinsic spins antiparallel): $$J \text{ is intermediate: tends toward } |j_p + j_n| \text{ or } |j_p - j_n|$$

1. Derivation of the Schmidt Single-Particle Magnetic Moments

In the extreme single-particle model of an odd-$A$ nucleus, the total magnetic dipole moment is generated entirely by the single valence nucleon with orbital angular momentum $\vec{l}$ and spin $\vec{s}$:

$$\vec{\mu} = \left( g_l \vec{l} + g_s \vec{s} \right) \mu_N$$

Projecting along the total angular momentum $\vec{j} = \vec{l} + \vec{s}$ in the state $m_j = j$:

$$\mu = \langle j, m_j=j | \mu_z | j, m_j=j \rangle = \frac{\langle \vec{\mu}\cdot\vec{j} \rangle}{j+1}$$

Evaluating $\vec{l}\cdot\vec{j} = \frac{j(j+1) + l(l+1) - 3/4}{2}$ and $\vec{s}\cdot\vec{j} = \frac{j(j+1) - l(l+1) + 3/4}{2}$:

• Case I: $j = l + 1/2$ (Spin Parallel to Orbit): $$\mu = \left[ (j - 1/2) g_l + \frac{1}{2} g_s \right] \mu_N$$
• Case II: $j = l - 1/2$ (Spin Antiparallel to Orbit): $$\mu = \frac{j}{j+1} \left[ (j + 3/2) g_l - \frac{1}{2} g_s \right] \mu_N$$

Substituting bare nucleon $g$-factors ($g_l = 1, g_s = +5.586$ for proton; $g_l = 0, g_s = -3.826$ for neutron) yields the four classic Schmidt lines:

$$\text{Odd Proton: } \mu = \begin{cases} j + 2.293\text{ }\mu_N, & j = l + 1/2 \\ j - 2.293\frac{j}{j+1}\text{ }\mu_N, & j = l - 1/2 \end{cases}$$ $$\text{Odd Neutron: } \mu = \begin{cases} -1.913\text{ }\mu_N, & j = l + 1/2 \\ +1.913\frac{j}{j+1}\text{ }\mu_N, & j = l - 1/2 \end{cases}$$

2. The Schmidt Plots & Core Polarization Quenching

When experimental magnetic moments are plotted as a function of nuclear spin $j$ (the Schmidt plots), almost all experimental points lie between the two Schmidt lines, but rarely right on them.

This quenching of empirical magnetic moments is caused by:

1. Core Polarization: The magnetic moment of the valence nucleon polarizes the time-reversed nucleon pairs in the closed core via the residual spin-spin and tensor interactions, inducing an opposing core magnetic moment.
2. Meson Exchange Currents: Virtual charged pions in flight between nucleons alter the effective $g$-factors inside the nuclear medium ($g_s^{\text{eff}} \approx 0.7\text{ to }0.8\text{ }g_s^{\text{bare}}$).

1. Multipole Expansion of the Nuclear Surface

Near closed shells, nuclei are spherical in their ground states, but can undergo collective surface vibrations described by the Bohr-Mottelson collective model. The instantaneous nuclear radius in direction $(\theta, \phi)$ is expanded in spherical harmonics:

$$R(\theta, \phi, t) = R_0 \left[ 1 + \sum_{\lambda=0}^\infty \sum_{\mu=-\lambda}^\lambda \alpha_{\lambda\mu}(t) Y_{\lambda\mu}^*(\theta, \phi) \right]$$

Physical modes of oscillation:

• $\lambda = 0$ (Monopole / Breathing Mode): Compresses the nuclear fluid, requiring immense symmetry energy ($\hbar\omega \approx 80 A^{-1/3}\text{ MeV}$).
• $\lambda = 1$ (Dipole Mode): Pure center-of-mass translation (no internal excitation for isoscalar motion; for isovector motion, it is the Giant Dipole Resonance).
• $\lambda = 2$ (Quadrupole Mode): Lowest true shape vibration. Deforms sphere into prolate and oblate spheroids with five collective degrees of freedom $\alpha_{2\mu}$.
• $\lambda = 3$ (Octupole Mode): Pear-shaped deformation with negative parity ($J^\pi = 3^-$).

2. The Quadrupole Harmonic Vibrator Spectrum

Quantizing the collective Hamiltonian $\hat{H} = \frac{1}{2} B_2 \sum |\dot{\alpha}_{2\mu}|^2 + \frac{1}{2} C_2 \sum |\alpha_{2\mu}|^2$ yields harmonic quadrupole phonons carrying angular momentum $\lambda = 2$ and positive parity $\pi = +1$:

1. Ground State (Zero Phonons): $J^\pi = 0^+$.
2. One-Phonon State: A single $2^+$ phonon excitation with energy $\hbar\omega_2 = \hbar\sqrt{C_2/B_2}$: $$E(1\text{ phonon}) = \hbar\omega_2, \quad J^\pi = 2^+$$
3. Two-Phonon State: Coupling two identical quadrupole bosons ($l_1 = 2, l_2 = 2$). By Bose-Einstein symmetry for identical phonons, only even total angular momenta are allowed: $$J^\pi = 0^+, \quad 2^+, \quad 4^+$$ In a pure harmonic vibrator, this forms a degenerate triplet at exactly twice the single-phonon energy: $$E(2\text{ phonons}) = 2\hbar\omega_2, \qquad \frac{E(4^+)}{E(2^+)} = \frac{E(0^+_2)}{E(2^+)} = \mathbf{2.00}$$

In real vibrational nuclei (such as ${}^{106}\text{Pd}$ or ${}^{114}\text{Cd}$), anharmonic terms split the triplet, but the ratio remains very close to $E(4^+)/E(2^+) \approx 2.0\text{ to }2.2$.

1. Permanent Deformation and the Nilsson Model

In nuclei with many valence nucleons far from closed shells (rare-earth region $150 < A < 190$ and actinide region $A > 220$), the residual nucleon-nucleon quadrupole interaction overcomes the spherical pairing force, locking the nuclear core into a permanently deformed shape (typically an axially symmetric prolate ellipsoid).

Sven Gösta Nilsson (1955) extended the shell model to an axially deformed harmonic oscillator potential parameterized by the quadrupole deformation parameter $\delta$ (or $\beta_2$):

$$V_{\text{Nilsson}} = \frac{1}{2} M \left[ \omega_\perp^2 (x^2 + y^2) + \omega_z^2 z^2 \right] - C \vec{l}\cdot\vec{s} - D \vec{l}^2$$

Because spherical symmetry is broken, total angular momentum $j$ is no longer a good quantum number. Only the projection of angular momentum along the nuclear symmetry axis ($\Omega = j_z$) and parity $\pi$ are conserved. Single-particle states are classified by the asymptotic Nilsson quantum numbers:

$$\Omega^\pi [N, n_z, \Lambda]$$

where $N$ is total oscillator quantum number, $n_z$ is quanta along symmetry axis, and $\Lambda$ is orbital projection ($j_z = \Lambda \pm 1/2$).

2. Nuclear Rotational Bands

A permanently deformed, axially symmetric even-even nucleus rotates collectively perpendicular to its symmetry axis. The quantum rotational energy spectrum is governed by the rotational Hamiltonian:

$$E(I) = \frac{\hbar^2}{2\mathcal{I}} I(I+1)$$

Because the prolate ellipsoid is symmetric under $180^\circ$ rotation about any perpendicular axis ($\mathcal{R}$-parity), only even spin states are allowed in the ground-state band ($K = 0$):

$$I^\pi = 0^+, \quad 2^+, \quad 4^+, \quad 6^+, \quad 8^+, \quad 10^+, \dots$$

The energy ratio between the first two excited states is universal:

$$\frac{E(4^+)}{E(2^+)} = \frac{4(5)}{2(3)} = \frac{20}{6} = \mathbf{3.333}$$

Observation of $E(4^+)/E(2^+) \approx 3.33$ is the definitive experimental hallmark of a rigid nuclear rotor.

3. The Backbending Phenomenon at High Angular Momentum

The empirical moment of inertia $\mathcal{I}$ of deformed nuclei is approximately $40\text{ to }50\%$ of the rigid-body value $\mathcal{I}_{\text{rigid}}$, due to pairing correlations that produce nuclear superfluidity.

As the nucleus is spun up to high angular momentum ($I \sim 14\text{ to }16\hbar$), the colossal Coriolis force $\vec{F}_C = -2 M (\vec{\omega}\times\vec{v})$ acts with opposite sign on time-reversed paired nucleons in high-$j$ intruder orbitals ($1i_{13/2}$ neutrons). At a critical rotational frequency $\omega_c$, the Coriolis force breaks the pair (Coriolis Anti-Pairing effect) and aligns their spins along the rotation axis.

This abrupt alignment increases the nuclear moment of inertia toward the rigid-body value, producing a dramatic S-shaped curve when $\mathcal{I}$ is plotted against $\omega^2$—a phenomenon known as backbending.