The Deuteron Problem & Two-Nucleon Bound State

1. The Fundamental Two-Nucleon Bound System

The deuteron (${}^2\text{H}$ or $d$), consisting of one proton and one neutron, is the simplest bound nuclear system. Just as the hydrogen atom serves as the fundamental testing ground for atomic physics and quantum electrodynamics, the deuteron is the foundational benchmark for the microscopic nucleon-nucleon ($N$-$N$) interaction.

The key empirically measured static ground-state properties of the deuteron are:

• Binding Energy ($B$): Measured via high-precision mass spectrometry and the ${}^1\text{H}(n, \gamma){}^2\text{H}$ capture gamma-ray energy ($E_\gamma = 2.224575\text{ MeV}$): $$B = [M({}^1\text{H}) + m_n - M({}^2\text{H})] c^2 = 2.224575 \pm 0.000009\text{ MeV}$$ Compared to typical nuclear binding energies of $\sim 8\text{ MeV/nucleon}$, the deuteron is exceptionally loosely bound ($B/A \approx 1.112\text{ MeV/nucleon}$).
• Total Angular Momentum (Spin $J^\pi$): Experimentally determined to be $J = 1$ with positive parity ($\pi = +1$), written as $J^\pi = 1^+$.
• Isospin ($I$): Because the neutron and proton have $I_3 = -1/2$ and $+1/2$, the total two-nucleon isospin can be $I = 0$ (antisymmetric singlet) or $I = 1$ (symmetric triplet). Because neither the diproton (${}^2\text{He}$, $I_3 = +1$) nor the dineutron (${}^2n$, $I_3 = -1$) exists as a bound state, the bound deuteron must be an isospin singlet ($I = 0, I_3 = 0$).
• Magnetic Dipole Moment ($\mu_d$): Measured via nuclear magnetic resonance (NMR): $$\mu_d = +0.85743823 \pm 0.00000002\text{ }\mu_N$$ This is remarkably close to, but measurably distinct from, the simple algebraic sum of the free proton and neutron magnetic dipole moments: $$\mu_p + \mu_n = +2.792847\text{ }\mu_N - 1.913043\text{ }\mu_N = +0.879804\text{ }\mu_N$$ The small difference $\Delta\mu = \mu_d - (\mu_p + \mu_n) = -0.022366\text{ }\mu_N$ provides direct evidence for orbital angular momentum mixing ($L = 2$).
• Electric Quadrupole Moment ($Q$): $$Q = +0.002859 \pm 0.000030\text{ b} = +0.2859 \pm 0.0030\text{ fm}^2$$ A pure $S$-wave ($L = 0$) state is spherically symmetric and must have $Q \equiv 0$. The non-zero positive quadrupole moment proves that the deuteron is prolate (elongated along its spin axis), demonstrating that the nuclear force contains a non-central tensor component.

1. The Two-Body Schrödinger Equation in Relative Coordinates

Let $\vec{r}_p$ and $\vec{r}_n$ denote the coordinates of the proton and neutron with masses $m_p \approx m_n \equiv M$. Introducing the center-of-mass coordinate $\vec{R} = \frac{1}{2}(\vec{r}_p + \vec{r}_n)$ and relative coordinate $\vec{r} = \vec{r}_p - \vec{r}_n$, the two-body Hamiltonian separates. With reduced mass $\mu = \frac{M\cdot M}{M + M} = \frac{M}{2}$, the relative motion satisfies:

where $E = -B < 0$ is the bound-state energy. Replacing $\frac{\hbar^2}{2\mu} = \frac{\hbar^2}{M}$:

2. The Central Spherical Square Well Approximation

As a first physical approximation, consider a central, spherically symmetric attractive potential well of depth $V_0$ and range $R$:

Assuming a spherically symmetric $S$-state ($L = 0$), the wavefunction is $\psi(\vec{r}) = \frac{u(r)}{r} Y_{00}(\theta, \phi)$, where the radial wavefunction $u(r)$ satisfies:

3. Interior and Exterior Solutions & Boundary Matching

Region I ($r < R$): $V(r) = -V_0$. Defining $k_1^2 \equiv \frac{M(V_0 - B)}{\hbar^2} > 0$:

(the cosine solution is excluded by the boundary condition $u(0) = 0$ to keep $\psi(0)$ finite).

Region II ($r > R$): $V(r) = 0$. Defining $\gamma^2 \equiv \frac{M B}{\hbar^2} > 0$:

(the growing exponential $e^{+\gamma r}$ is rejected for normalizability).

Matching the logarithmic derivatives $\frac{1}{u}\frac{du}{dr}$ at the boundary $r = R$:

Because $B \approx 2.22\text{ MeV} \ll V_0 \approx 35\text{ to }40\text{ MeV}$, the exterior decay parameter $\gamma \approx \sqrt{\frac{M B}{\hbar^2}} \approx 0.232\text{ fm}^{-1}$ is small. In the approximation $\gamma \to 0$ (zero binding energy limit):

For a realistic nuclear force range $R \approx 2.0\text{ fm}$, this yields a well depth $V_0 \approx 35\text{ to }38\text{ MeV}$, demonstrating that the deuteron is a shallowly bound state sitting near the threshold of the potential well.

1. The Deuteron Radial Wavefunction

Normalizing the composite radial wavefunction $\int_0^\infty |u(r)|^2 dr = 1$:

The decay parameter $\gamma$ determines the asymptotic behavior outside the potential:

The characteristic decay length (the "tail" of the deuteron) is:

Because the nuclear potential radius is only $R \approx 1.7\text{ to }2.1\text{ fm}$, the probability of finding the nucleons outside the range of their mutual nuclear interaction is:

Thus, the neutron and proton spend approximately 70% of their time outside the nuclear potential well, illustrating the exceptionally diffuse, halo-like nature of the deuteron.

2. The Non-Existence of Bound Excited States

Can the deuteron have bound excited states?

1. Higher Radial Nodes ($n = 2, S$-wave): A second bound state with $L = 0$ requires $k_1 R > \frac{3\pi}{2}$. This would demand a potential depth: $$V_0 R^2 \ge \frac{9\pi^2 \hbar^2}{4 M} \approx 9 \times 103\text{ MeV}\cdot\text{fm}^2 \approx 927\text{ MeV}\cdot\text{fm}^2$$ which is an order of magnitude deeper than the physical nuclear well ($V_0 R^2 \approx 105\text{ MeV}\cdot\text{fm}^2$).
2. Orbital Excitations ($L \ge 1$, $P$-states): For $L = 1$, the effective potential includes a centrifugal barrier: $$V_{\text{eff}}(r) = V(r) + \frac{\hbar^2 L(L+1)}{M r^2} = V(r) + \frac{2\hbar^2}{M r^2}$$ At $r = R \approx 2\text{ fm}$, $\frac{2\hbar^2}{M R^2} \approx \frac{2 \times 41.47}{4} \approx 20.7\text{ MeV}$. This repulsive barrier pushes the ground state energy far above zero, preventing any bound $P$-state.
3. Singlet Spin State (${}^1S_0, S=0$): In the spin-singlet state, the $N$-$N$ interaction is slightly weaker ($V_0^{(s)} \approx 32\text{ MeV}$ compared to $V_0^{(t)} \approx 38\text{ MeV}$). This depth is insufficient to bind: the singlet state is an unbound virtual state with energy $E_s \approx -0.066\text{ MeV} = -66\text{ keV}$ (a pole on the second Riemann sheet).

Therefore, the deuteron possesses no bound excited states whatsoever—neither radial, orbital, nor spin excitations exist.

1. The Classical and Quantum Quadrupole Moment

The electric quadrupole moment measures the departure of the nuclear charge distribution from spherical symmetry. Classically, for a charge distribution $\rho_c(\vec{r})$:

In quantum mechanics, the quadrupole operator for a system of nucleons with coordinates $\vec{r}_i$ and charges $e_i$ is:

For the deuteron, choosing the center of mass as the origin, the proton coordinate is $\vec{r}_p = +\frac{1}{2}\vec{r}$ and the neutron coordinate is $\vec{r}_n = -\frac{1}{2}\vec{r}$. Since the neutron has zero electric charge ($e_n = 0$) and the proton has charge $e_p = e$:

The observable quadrupole moment $Q$ is defined as the expectation value in the substate of maximum magnetic projection ($M_J = J = 1$):

2. Impossibility of Quadrupole Moment in a Pure S-Wave

If the deuteron ground state were a pure $S$-wave ($L = 0$), its spatial wavefunction would be spherically symmetric ($\psi \propto Y_{00}$). Then:

By parity conservation and Wigner-Eckart selection rules, $\langle L=0 | Y_{20} | L=0 \rangle \equiv 0$.

However, high-precision radio-frequency molecular beam spectroscopy (Kellogg, Rabi, Ramsey, Zacharias, 1939) established conclusively that the deuteron has a non-zero quadrupole moment:

The positive sign ($Q > 0$) demonstrates that the deuteron charge distribution is a prolate spheroid (cigar-shaped, elongated along the spin vector $\vec{J}$) rather than oblate ($Q < 0$) or spherical ($Q = 0$). This non-spherical deformation provides irrefutable proof that the nuclear force is non-central.

1. The Phenomenological Tensor Operator

To construct a non-central interaction that satisfies parity conservation ($\pi = +1$), time-reversal invariance, rotational invariance, and isospin symmetry, the interaction must couple the nucleon spin operators $\vec{\sigma}_1, \vec{\sigma}_2$ to the relative spatial coordinate unit vector $\hat{r} = \vec{r}/r$. The unique scalar operator formed from the rank-2 spin tensor and rank-2 spatial coordinate tensor is the tensor operator:

Key mathematical properties of $S_{12}$:

• Angle Averaging: $\int S_{12} d\Omega = 0$. Averaged over all space directions, the tensor force vanishes; it produces no contribution to pure spherically symmetric $S$-states.
• Singlet Annihilation: For a singlet spin state ($S = 0$), $\vec{\sigma}_1 + \vec{\sigma}_2 = 0 \implies \vec{\sigma}_1\cdot\vec{\sigma}_2 = -3$. It is readily proven that $S_{12}|S=0\rangle \equiv 0$. The tensor force acts only in spin-triplet ($S = 1$) states.
• Orbital Coupling: $S_{12}$ does not commute with orbital angular momentum $\vec{L}^2$ or spin $\vec{S}^2$, but commutes with total angular momentum $\vec{J} = \vec{L} + \vec{S}$. It can mix states with $\Delta L = 0, \pm 2$.

2. The Admixed Wavefunction: ³S₁ and ³D₁

With $J^\pi = 1^+$, the allowed quantum states combining $L$ and $S = 1$ with positive parity ($\pi = (-1)^L = +1$) are:

• $L = 0 \implies {}^3S_1$ (Principal component)
• $L = 2 \implies {}^3D_1$ (Admixed component)

(States with $L = 1$, such as ${}^3P_1$ or ${}^1P_1$, are forbidden because they have odd parity $\pi = (-1)^1 = -1$).

Therefore, the general ground-state wavefunction of the deuteron is an exact quantum superposition:

where $u(r)$ is the radial $S$-wave function and $w(r)$ is the radial $D$-wave function, normalized such that:

Empirical analysis of the quadrupole moment and magnetic moment yields a $D$-state probability of:

The asymptotic $D/S$ ratio is denoted $\eta \equiv \lim_{r\to\infty} \frac{w(r)}{u(r)} \approx 0.0256 \pm 0.0004$.

1. Derivation of the Deuteron Magnetic Moment

The magnetic moment operator of the two-nucleon system is given by the sum of orbital and spin contributions:

Since $\vec{l}_p = \vec{l}_n = \frac{1}{2}\vec{L}$, $g_l^{(p)} = 1$, and $g_l^{(n)} = 0$, the orbital contribution is $\vec{\mu}_l = \frac{1}{2}\vec{L}\mu_N$. The spin contribution is $\vec{\mu}_s = (\mu_p \vec{\sigma}_p + \mu_n \vec{\sigma}_n)\mu_N = (\mu_p + \mu_n)\vec{S}\mu_N + (\mu_p - \mu_n)(\vec{\sigma}_p - \vec{\sigma}_n)\frac{\mu_N}{2}$.

Evaluating the expectation value in the admixed state $\psi = a_S \psi({}^3S_1) + a_D \psi({}^3D_1)$:

Substituting $\mu_p + \mu_n = 0.879804\text{ }\mu_N$:

Equating this to the measured value $\mu_d = 0.857438\text{ }\mu_N$:

Relativistic corrections and meson exchange currents (MEC) slightly shift this value, establishing $P_D \approx 4\% \text{ to } 6\%$.

2. The Root-Mean-Square Radius of the Deuteron

The charge radius of the deuteron is measured via high-energy elastic electron-deuteron scattering ($e + d \to e + d$) and precision laser spectroscopy of muonic deuterium:

The matter root-mean-square radius $R_{\text{matter}}$ (separation between proton and neutron centers of mass) is:

This is substantially larger than heavy atomic nuclei ($R_{\text{Pb}} \approx 1.2 \times 208^{1/3} \approx 7.1\text{ fm}$, but nucleon-nucleon average spacing is $\sim 1.8\text{ fm}$), demonstrating once again the extended spatial extent of the deuteron.

1. Kinematics and Threshold Energy

Photodisintegration is the electromagnetic breakup of the deuteron by an incident photon:

Let $E_\gamma$ be the laboratory photon energy. Conservation of energy and linear momentum:

Accounting for the small nuclear recoil energy ($E_{\text{recoil}} = \frac{E_\gamma^2}{2 M_d c^2} \approx 1.3\text{ keV}$), the threshold photon energy in the laboratory frame is:

For $E_\gamma > B$, the excess energy $E_\gamma - B$ is partitioned equally into center-of-mass kinetic energy of the escaping proton and neutron ($E_{\text{rel}} = E_\gamma - B$).

2. Electric Dipole (E1) vs Magnetic Dipole (M1) Cross Sections

The transition operator can act via electric dipole ($E1$) or magnetic dipole ($M1$) mechanisms:

1. Magnetic Dipole Disintegration ($M1$): Dominates near the threshold ($E_\gamma - B \le 100\text{ keV}$). The incident gamma ray couples to the spin magnetic moments, flipping the triplet deuteron (${}^3S_1$) into the unbound singlet continuum (${}^1S_0$). The $M1$ cross section is isotropic ($\frac{d\sigma_{M1}}{d\Omega} \propto \text{const}$) and exhibits a sharp threshold peak: $$\sigma_{M1} \propto \frac{\sqrt{E_\gamma - B}}{E_\gamma}$$
2. Electric Dipole Disintegration ($E1$): Dominates at energies $E_\gamma \ge 3\text{ MeV}$. The electric field of the photon couples to the proton charge displacement relative to the center of mass. This induces an electric dipole transition from the ${}^3S_1$ ground state to the ${}^3P$ continuum states ($L = 1$). The differential cross section has the characteristic $\sin^2\theta$ dipolar angular distribution: $$\frac{d\sigma_{E1}}{d\Omega} = \frac{3}{8\pi} \sigma_{E1} \sin^2\theta$$ The total $E1$ cross section as a function of relative wavevector $k = \sqrt{M(E_\gamma - B)}/\hbar$ is given by Bethe and Peierls: $$\sigma_{E1}(E_\gamma) = \frac{8\pi}{3}\left( \frac{e^2}{\hbar c} \\right) \frac{\hbar^2}{M} \frac{\gamma k^3}{(\gamma^2 + k^2)^3} = \frac{8\pi \alpha}{3 M} \frac{\sqrt{B}(E_\gamma - B)^{3/2}}{E_\gamma^3}$$ This cross section vanishes at threshold ($E_\gamma = B$), rises to a pronounced peak at $E_\gamma = 2 B \approx 4.45\text{ MeV}$ with $\sigma_{\max} \approx 2.5\text{ mb}$, and then decays asymptotically as $E_\gamma^{-3/2}$.

The total photodisintegration differential cross section is parameterized as:

where $a$ is the $M1$ contribution and $b$ is the $E1$ contribution.