--- title: The Deuteron and the Tensor Force module: The Nuclear Force moduleNumber: 2 lessonNumber: 2 order: 202 summary: > The deuteron is the only bound two-nucleon state: one shallow level at 2.22 MeV, no excited states. A square-well fit fixes a depth near 35 MeV over a 2 fm range, yet the wavefunction leaks so far past the edge that most of the probability lies outside the force. Its spin-1 ground state, magnetic moment close to the sum of the free-nucleon moments, and small but nonzero electric quadrupole moment together force a D-state admixture and a non-central tensor force. topics: [The Nuclear Force] sources: - book: Krane ref: "Ch. 4 — The Force Between Nucleons; §4.1 The Deuteron" - book: Wong ref: "Ch. 3 — Nuclear Force and Two-Nucleon Systems; §3-1, §3-2" - book: Povh, Rith, Scholz, Zetsche ref: "Particles and Nuclei; Ch. 16 The Nuclear Force" draft: false --- The deuteron $^2\mathrm{H}$, one proton bound to one neutron, is the only bound two-nucleon system. There is no bound di-proton and no bound di-neutron, and the deuteron itself has no excited states — every excitation lies in the continuum. This makes it the hydrogen atom of nuclear physics: the single case where the two-body problem is not obscured by the presence of other nucleons, and every measured property maps directly onto a feature of the nucleon-nucleon force.[^kr-deut] The measured ground-state properties are few and precise.[^codata] - **Binding energy** $B = 2.224573\,\mathrm{MeV}$, from the threshold of the photodisintegration $\gamma + d \to p + n$ and from the neutron-capture line $n + p \to d + \gamma$. - **Spin and parity** $J^\pi = 1^+$. - **Isospin** $T = 0$. - **Magnetic dipole moment** $\mu_d = +0.857438\,\mu_N$. - **Electric quadrupole moment** $Q_d = +0.2860\,\mathrm{fm}^2$. Each of these constrains the force, and the last two force a piece of the interaction that no central potential can supply. ## A single shallow bound state Take the interaction to be a central square well of depth $V_0$ and range $R$, $$ V(r) = \begin{cases} -V_0 & r < R, \\ 0 & r > R, \end{cases} $$ and work in the two-body centre-of-mass frame, where the relative coordinate $r$ carries the reduced mass $$ \mu = \frac{m_p m_n}{m_p + m_n} \approx \tfrac12 m_N = 469.46\,\mathrm{MeV}/c^2 . $$ The ground state is $\ell = 0$. Writing the radial function as $u(r) = rR(r)$, the radial Schrödinger equation $$ -\frac{\hbar^2}{2\mu}\frac{\d^2 u}{\d r^2} + V(r)\,u = -B\,u $$ reduces inside and outside the well to $$ \frac{\d^2 u}{\d r^2} + k_1^2\,u = 0 \quad (r < R), \qquad \frac{\d^2 u}{\d r^2} - \kappa^2\,u = 0 \quad (r > R), $$ with $$ k_1 = \frac{\sqrt{2\mu(V_0 - B)}}{\hbar}, \qquad \kappa = \frac{\sqrt{2\mu B}}{\hbar}. $$ Regularity at the origin ($u(0)=0$) and normalizability at infinity select $$ u_{\text{in}}(r) = A\sin(k_1 r), \qquad u_{\text{out}}(r) = C\,e^{-\kappa r}. $$ Matching $u$ and $u'$ at $r = R$ — equivalently matching the logarithmic derivative — gives the transcendental eigenvalue condition $$ k_1 \cot(k_1 R) = -\kappa . $$ $$ % caption: Square-well model of the deuteron. The single bound level sits at % energy -B, only a little above the well bottom -V0; with V0 about 35 MeV and R % about 2.1 fm the well is barely deep enough to bind at all. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.2,0) node[right, black!70] {$r$}; \draw[->, black] (0,-3.6) -- (0,0.9) node[above, black!70] {$V(r)$}; % well \draw[very thick] (0,0) -- (0,-3.2) -- (2.6,-3.2) -- (2.6,0) -- (7.0,0); % bound level just above the bottom \draw[acc, thick, dashed] (0,-0.35) -- (5.4,-0.35); \node[acc, anchor=west, font=\scriptsize] at (5.45,-0.35) {bound level}; % markers \draw[black] (2.6,0.08) -- (2.6,-0.08); \node[black, anchor=south, font=\scriptsize] at (2.6,0.08) {range}; \draw[<->, black] (0.25,-3.2) -- (0.25,-0.35); \node[black, anchor=west, font=\scriptsize] at (0.32,-1.9) {well depth}; \draw[<->, black] (3.1,-0.35) -- (3.1,0.0); \node[black, anchor=west, font=\scriptsize] at (3.2,-0.18) {binding}; \end{tikzpicture} $$ The zero-binding limit $B \to 0$ pushes $\kappa \to 0$, so $\cot(k_1 R) \to 0$ and $k_1 R \to \pi/2$: the shallowest well that binds at all satisfies $$ V_0^{\min} = \frac{\pi^2 \hbar^2}{8\mu R^2} = \frac{\pi^2 (\hbar c)^2}{8\,\mu c^2\,R^2} \approx 23.2\,\mathrm{MeV} \quad (R = 2.1\,\mathrm{fm}). $$ The physical deuteron binds with $B = 2.22\,\mathrm{MeV}$, only slightly above threshold. Solving the transcendental condition with $R = 2.1\,\mathrm{fm}$ gives $k_1 R = 1.88$, hence $$ V_0 = \frac{\hbar^2 k_1^2}{2\mu} + B \approx 33.2 + 2.2 \approx 35\,\mathrm{MeV}. $$ The depth exceeds the minimum-to-bind by only about $50\%$. A well this shallow holds exactly one $\ell = 0$ level and no second bound state, which matches the observed absence of deuteron excited states. The range $R$ and depth $V_0$ are not separately determined by $B$ alone — only their combination is fixed — so the value $V_0 \approx 35\,\mathrm{MeV}$ is tied to the assumed $R \approx 2\,\mathrm{fm}$ of the force. > **Worked example (Why the deuteron is fragile).** A weakly bound level sits > just below the continuum, so a small perturbation unbinds it. The single scale > that governs the exterior is > $$ > \kappa = \frac{\sqrt{2\mu B}}{\hbar} > = \frac{\sqrt{2 (469.46)(2.22)}}{197.3\ \mathrm{MeV{\cdot}fm}} > = 0.232\,\mathrm{fm}^{-1}, > $$ > so the decay length of the wavefunction is $1/\kappa = 4.3\,\mathrm{fm}$, twice > the range of the force. The deuteron is more halo than sphere. ## A wavefunction outside its own well The exterior solution $u_{\text{out}} = C e^{-\kappa r}$ decays on the length $1/\kappa = 4.3\,\mathrm{fm}$, far larger than the range $R \approx 2\,\mathrm{fm}$. The nucleons spend most of their time beyond the reach of the force that binds them. Integrating the matched wavefunction, $$ P_{\text{out}} = \frac{\displaystyle\int_R^\infty u_{\text{out}}^2\,\d r} {\displaystyle\int_0^\infty u^2\,\d r} \approx 0.6 , $$ so roughly $60\%$ of the probability density lies outside the well. The mean separation $\langle r\rangle$ is comparable to $1/\kappa$, and the measured rms matter radius is $1.97\,\mathrm{fm}$ — large for a two-nucleon system. $$ % caption: The deuteron radial wavefunction u(r)=rR(r): a quarter-sine inside the % range, joining smoothly to a slow exponential tail that carries most of the % probability well beyond the edge of the potential (dashed line at the range). \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.4,0) node[right, black!70] {$r$ (fm)}; \draw[->, black] (0,0) -- (0,3.2) node[above, black!70] {$u(r)$}; \foreach \x/\lab in {1.05/1,2.1/2,3.15/3,4.2/4,5.25/5,6.3/6} \node[anchor=north, black, font=\scriptsize] at (\x,-0.02) {\lab}; % interior: rising sine up to the edge at r=2.1 (x=2.205 scaled ~ use x=2.2) \draw[acc, very thick] (0,0) .. controls (0.7,1.3) and (1.5,2.55) .. (2.2,2.8); % exterior: exponential tail from (2.2,2.8) \draw[acc, very thick] (2.2,2.8) .. controls (3.2,2.0) and (4.4,1.05) .. (5.6,0.55) .. controls (6.2,0.38) and (6.8,0.28) .. (7.2,0.22); % range marker \draw[black, dashed] (2.2,0) -- (2.2,2.8); \node[black, anchor=south west, font=\scriptsize] at (2.2,0.05) {range $R$}; % annotate tail \node[acc, anchor=west, font=\scriptsize] at (4.4,1.35) {exponential tail}; \node[black, anchor=south, font=\scriptsize] at (1.1,1.6) {inside}; \end{tikzpicture} $$ The size of $P_{\text{out}}$ is the reason the deuteron's properties depend so weakly on the shape of the well interior: the tail, set entirely by $\kappa$ and hence by $B$, dominates every expectation value. This is the general signature of a **halo state**, seen again in loosely bound light nuclei such as $^{11}\mathrm{Li}$. ## Spin, statistics, and isospin The deuteron ground state has $S = 1$: the proton and neutron spins are parallel, a spin **triplet**. The singlet ($S = 0$) configuration is not bound, a first sign that the nuclear force is spin-dependent and stronger in the triplet channel. In spectroscopic notation the dominant component is $^3S_1$ — total spin $1$, orbital $\ell = 0$, total $J = 1$ — with even parity $\pi = (-1)^\ell = +1$, matching $J^\pi = 1^+$. The two-nucleon wavefunction must be antisymmetric under exchange of the two identical fermions (treating $p$ and $n$ as two isospin states of one nucleon). The total exchange symmetry factors as $$ (-1)^{\ell} \cdot (-1)^{S+1} \cdot (-1)^{T+1} = -1 . $$ For the ground state $\ell = 0$ (symmetric space) and $S = 1$ (symmetric spin), so the isospin part must be antisymmetric: $T = 0$. The deuteron is an isospin **singlet**. The would-be di-neutron and di-proton belong to the $T = 1$ triplet together with the unbound $^1S_0$ neutron-proton state; the force in that channel is too weak to bind, so no bound $nn$ or $pp$ system exists. > **Definition (Tensor operator).** For two nucleons with Pauli spin vectors > $\vec\sigma_1,\vec\sigma_2$ separated along the unit vector $\hat r$, the tensor > operator is > $$ > S_{12} = 3(\vec\sigma_1\cdot\hat r)(\vec\sigma_2\cdot\hat r) - \vec\sigma_1\cdot\vec\sigma_2 . > $$ > It is the unique scalar that is quadratic in each spin, has zero angular > average, and couples the spin orientation to the separation direction. It > vanishes when averaged over directions and acts only in spin-triplet states. ## The magnetic moment: evidence for L = 0 If the deuteron were a pure $^3S_1$ state, its magnetic moment would be the sum of the intrinsic nucleon moments, since $\ell = 0$ contributes no orbital current and the spins are aligned: $$ \mu(^3S_1) = \mu_p + \mu_n = (+2.79285) + (-1.91304) = 0.87981\,\mu_N . $$ The measured value $\mu_d = 0.857438\,\mu_N$ is close but not equal; the deficit $$ \Delta\mu = 0.87981 - 0.85744 = 0.02237\,\mu_N $$ is small, which already shows the state is $\ell = 0$ to good approximation. A pure $\ell = 2$ ($^3D_1$) state would give $\mu(^3D_1) = 0.310\,\mu_N$, far from the data. The observed moment lies just below the $^3S_1$ value, consistent with a few percent of $D$-state mixed in. $$ % caption: Vector picture of the deuteron moment in the dominant L=0 state: the % proton and neutron spins are parallel, so their intrinsic moments add. The % neutron moment points opposite its spin (negative gyromagnetic ratio), leaving a % net moment near the arithmetic sum mu_p + mu_n. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % proton \draw[black, thick] (1.0,-1.4) -- (1.0,1.4); \draw[black, very thick, ->] (1.0,0) -- (1.0,1.2); \node[black, anchor=south, font=\scriptsize] at (1.0,1.25) {proton spin}; \draw[black, thick, ->] (1.35,0) -- (1.35,0.95); \node[black, anchor=west, font=\scriptsize] at (1.42,0.6) {moment}; % neutron \draw[black, thick] (4.2,-1.4) -- (4.2,1.4); \draw[black, very thick, ->] (4.2,0) -- (4.2,1.2); \node[black, anchor=south, font=\scriptsize] at (4.2,1.25) {neutron spin}; \draw[black, thick, ->] (4.55,0) -- (4.55,-0.85); \node[black, anchor=west, font=\scriptsize] at (4.62,-0.55) {moment}; % sum \draw[black, dashed] (5.7,0) -- (7.1,0); \draw[acc, very thick, ->] (6.4,-0.2) -- (6.4,0.95); \node[acc, anchor=south, font=\scriptsize] at (6.4,1.0) {net sum}; \end{tikzpicture} $$ ## The quadrupole moment demands a non-central force The decisive number is the electric quadrupole moment. A pure $^3S_1$ state is spherically symmetric: its charge distribution has no preferred axis, so its quadrupole moment vanishes identically. The measured $$ Q_d = +0.2860\,\mathrm{fm}^2 = +2.86\,\mathrm{mb} $$ is small but definitely nonzero and positive. A nonzero $Q$ requires an $\ell \neq 0$ component in the ground state, so the deuteron cannot be a pure $S$-state. The positive sign means the charge is elongated along the spin axis — a **prolate** (cigar-shaped) deformation rather than an oblate one. $$ % caption: A pure S-state (left) is spherical and carries no quadrupole moment; the % observed positive quadrupole moment means the deuteron is slightly prolate % (right), elongated along the spin axis. The deformation is the visible fingerprint % of the small D-state admixture. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % spherical S-state \draw[very thick] (1.4,0) circle (1.05); \node[black!70, anchor=north, font=\scriptsize] at (1.4,-1.35) {$S$-state (spherical)}; \node[black, anchor=center, font=\scriptsize] at (1.4,0) {$Q = 0$}; % prolate D-admixed \draw[acc, very thick, fill=acc!10] (5.2,0) ellipse (0.8 and 1.35); \draw[black, ->] (5.2,-1.7) -- (5.2,1.7); \node[black, anchor=west, font=\scriptsize] at (5.25,1.5) {spin axis}; \node[black!70, anchor=north, font=\scriptsize] at (5.2,-1.8) {prolate ($Q > 0$)}; \end{tikzpicture} $$ The angular momentum bookkeeping fixes the admixture uniquely. The ground state has $J = 1$ and even parity. An $\ell = 1$ ($P$-wave) admixture is odd-parity and forbidden; the lowest allowed $\ell \neq 0$ component with $J = 1$ and even parity is $\ell = 2$, the $^3D_1$ state. The physical ground state is therefore a superposition $$ \psi_d = \cos\varepsilon\;\psi(^3S_1) + \sin\varepsilon\;\psi(^3D_1), \qquad P_D = \sin^2\varepsilon . $$ $$ % caption: The deuteron ground state as a coherent superposition of a spherical % 3S1 amplitude and a small deformed 3D1 amplitude sharing the same J=1 and even % parity; the tensor force is what couples them. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % S component \draw[very thick] (1.2,0) circle (0.85); \node[black!70, anchor=north, font=\scriptsize] at (1.2,-1.15) {$^3S_1$}; \node[black, anchor=south, font=\scriptsize] at (1.2,0.95) {$96\%$}; % plus \node[black, font=\normalsize] at (2.9,0) {$+$}; % D component (deformed) \draw[acc, very thick, fill=acc!10] (4.6,0) ellipse (0.6 and 1.05); \node[black!70, anchor=north, font=\scriptsize] at (4.6,-1.15) {$^3D_1$}; \node[black, anchor=south, font=\scriptsize] at (4.6,1.15) {$4\%$}; % equals -> deuteron \node[black, font=\normalsize] at (6.1,0) {$=$}; \draw[very thick] (7.6,0) ellipse (0.72 and 0.95); \node[black!70, anchor=north, font=\scriptsize] at (7.6,-1.15) {deuteron}; \end{tikzpicture} $$ Consistency between the two anomalies fixes $P_D$. Expanding the moment to first order in the small admixture, $$ \mu_d = (\mu_p + \mu_n) - \tfrac32\,P_D\!\left[(\mu_p+\mu_n) - \tfrac12\right]\mu_N, $$ and inserting $\Delta\mu = 0.0224\,\mu_N$ gives $$ P_D = \frac{\Delta\mu}{\tfrac32\!\left[(\mu_p+\mu_n)-\tfrac12\right]\mu_N} = \frac{0.0224}{1.5\,(0.380)} \approx 0.04 . $$ A $D$-state probability of about $4\%$ reproduces both the magnetic-moment deficit and, in a full calculation, the size and sign of $Q_d$. The two independent observables agree on the same small number, which is strong evidence for the picture.[^wong-deut] ## The tensor force An $\ell = 2$ admixture cannot arise from any central potential $V(r)$, because $V(r)$ commutes with $\vec L^2$ and cannot mix states of different $\ell$. What mixes $^3S_1$ and $^3D_1$ is a force that depends on the orientation of the nucleon spins relative to the line joining them — a **tensor force**. The two-nucleon potential in the triplet channel has the form $$ V(r) = V_C(r) + V_T(r)\,S_{12} + V_{LS}(r)\,\vec L\cdot\vec S, $$ where $V_C$ is central, $V_{LS}$ is the spin-orbit term, and $V_T(r)\,S_{12}$ is the tensor term built from the operator $S_{12}$ above. Because $S_{12}$ depends on $\hat r$, it does not commute with $\vec L^2$; it does commute with total $\vec J^2$ and with parity. It therefore connects $^3S_1$ and $^3D_1$ — the two even-parity, $J = 1$ triplet states — and nothing else, producing exactly the required admixture. $$ % caption: The tensor force is orientation-dependent. Two aligned nucleon spins % attract more strongly when the spins point along the line joining them (prolate, % lower energy) than when they point across it (higher energy). This orientation % dependence is what a central potential cannot supply. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % left: spins along the separation axis \fill[acc!18, draw=acc, very thick] (0.6,0) circle (0.4); \fill[acc!18, draw=acc, very thick] (2.6,0) circle (0.4); \draw[acc, very thick, ->] (0.6,-0.55) -- (0.6,0.75); \draw[acc, very thick, ->] (2.6,-0.55) -- (2.6,0.75); \draw[black, dashed] (0.6,0) -- (2.6,0); \node[black, anchor=north, font=\scriptsize] at (1.6,-0.7) {spins along axis}; \node[black!70, anchor=south, font=\scriptsize] at (1.6,0.9) {stronger binding}; % right: spins across the separation axis \draw[very thick] (5.2,0) circle (0.4); \draw[very thick] (7.2,0) circle (0.4); \draw[black, very thick, ->] (4.75,0) -- (5.65,0); \draw[black, very thick, ->] (6.75,0) -- (7.65,0); \draw[black, dashed] (5.2,-0.7) -- (5.2,0.7); \draw[black, dashed] (7.2,-0.7) -- (7.2,0.7); \node[black, anchor=north, font=\scriptsize] at (6.2,-0.7) {spins across axis}; \node[black!70, anchor=south, font=\scriptsize] at (6.2,0.9) {weaker binding}; \end{tikzpicture} $$ The tensor force is the reason the deuteron is prolate: aligning the spins along their separation lowers the energy, so the charge stretches along the spin axis. It is also indispensable to binding — without the $^3S_1$-$^3D_1$ coupling the central force alone barely holds the deuteron, and the tensor term supplies a sizeable fraction of the binding energy. The same operator reappears in every realistic nucleon-nucleon potential and, through the [nucleon-nucleon scattering phase shifts](/nuclear-physics/nuclear-force-deuteron/nucleon-nucleon-scattering), in the mixing of coupled partial waves. The deuteron thus reads off three structural facts of the nuclear force at once: it is spin-dependent (triplet binds, singlet does not), it is barely strong enough to bind (the state is a halo), and it is non-central (a tensor term generates the quadrupole moment). The next lesson turns to scattering, where the same force is probed above threshold and its spin and charge dependence are measured directly. | Property | Value | What it fixes | | --- | --- | --- | | Binding energy $B$ | $2.224573\,\mathrm{MeV}$ | well depth $\times$ range$^2$; single bound level | | $J^\pi$ | $1^+$ | triplet $^3S_1$ ground state, even parity | | Isospin $T$ | $0$ | no bound $nn$ or $pp$ | | $\mu_d$ | $+0.857438\,\mu_N$ | $\ell = 0$ dominance; $P_D \approx 4\%$ | | $Q_d$ | $+0.2860\,\mathrm{fm}^2$ | non-central tensor force; prolate shape | [^kr-deut]: Krane, "Introductory Nuclear Physics," §4.1. [^codata]: Deuteron mass, magnetic moment, and quadrupole moment from CODATA and the NIST fundamental-constants compilation, [physics.nist.gov/cuu/Constants](https://physics.nist.gov/cuu/Constants/); binding energy from the AME atomic-mass evaluation via NNDC, [nndc.bnl.gov](https://www.nndc.bnl.gov/). [^wong-deut]: Wong, "Introductory Nuclear Physics," 2nd ed., §3-1 and §3-2.