The Deuteron and the Tensor Force

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.¹

The measured ground-state properties are few and precise.²

• 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 $R$,

and work in the two-body centre-of-mass frame, where the relative coordinate $r$ carries the reduced mass

The ground state is $\ell =0$. Writing the radial function as $u(r)=rR(r)$, the radial Schrödinger equation

reduces inside and outside the well to

with

Regularity at the origin ($u(0)=0$) and normalizability at infinity select

Matching $u$ and $u^{\prime }$ at $r=R$ — equivalently matching the logarithmic derivative — gives the transcendental eigenvalue condition

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

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

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.

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,

so roughly $60\%$ of the probability density lies outside the well. The mean separation $⟨r⟩$ is comparable to $1/\kappa$, and the measured rms matter radius is $1.97\mathrm{fm}$ — large for a two-nucleon system.

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 ${}^3{S}_1$ — 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

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 ${}^1{S}_0$ neutron-proton state; the force in that channel is too weak to bind, so no bound $nn$ or $pp$ system exists.

The magnetic moment: evidence for L = 0

If the deuteron were a pure ${}^3{S}_1$ 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:

The measured value ${\mu }_d=0.857438{\mu }_N$ is close but not equal; the deficit

is small, which already shows the state is $\ell =0$ to good approximation. A pure $\ell =2$ (${}^3{D}_1$) state would give $\mu {(}^3{D}_1)=0.310{\mu }_N$, far from the data. The observed moment lies just below the ${}^3{S}_1$ value, consistent with a few percent of $D$-state mixed in.

The quadrupole moment demands a non-central force

The decisive number is the electric quadrupole moment. A pure ${}^3{S}_1$ state is spherically symmetric: its charge distribution has no preferred axis, so its quadrupole moment vanishes identically. The measured

is small but definitely nonzero and positive. A nonzero $Q$ requires an $\ell \ne 0$ 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.

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 $\ell \ne 0$ component with $J=1$ and even parity is $\ell =2$, the ${}^3{D}_1$ state. The physical ground state is therefore a superposition

Consistency between the two anomalies fixes $P_D$. Expanding the moment to first order in the small admixture,

and inserting $\Delta \mu =0.0224{\mu }_N$ gives

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.³

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 ${}^3{S}_1$ and ${}^3{D}_1$ 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

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 ${}^3{S}_1$ and ${}^3{D}_1$ — the two even-parity, $J=1$ triplet states — and nothing else, producing exactly the required admixture.

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 ${}^3{S}_1$-${}^3{D}_1$ 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, 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.