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The deuteron has a dominant ${}^3{S}_1$ component and a smaller ${}^3{D}_1$ component coupled by the nuclear tensor force. The electric quadrupole moment rules out a purely isotropic S wave spatial state. A D-state probability depends on the chosen nuclear Hamiltonian and current approximation; it is not an independently model-free observable.

An S wave has $L=0$ and positive parity. The deuteron's $J=1$ therefore requires a spin-one-half triplet state, $S=1$. The two-nucleon isospin description makes this the antisymmetric $T=0$ channel. Its existence proves sufficient attraction to bind in the ${}^3{S}_1$-dominated channel.

A spin one-half pair could also form $S=0$. The absence of a bound ${}^1{S}_0$ deuteron indicates that the singlet interaction does not bind, not that it is everywhere repulsive. Together with triplet binding, this shows that a spin-independent central attraction is insufficient. No radial excited bound level also constrains the strength and range: the attraction supports only a shallow low-energy bound state.

For identical neutrons or protons, two-particle isospin has $T=1$ and is symmetric. At even $L$, spatial exchange is symmetric, so the Pauli exclusion principle requires an antisymmetric spin state, $S=0$. A low-energy $nn$ or $pp$ pair cannot occupy the deuteron's $S=1,T=0$ channel. With approximate charge independence of nuclear forces, their singlet strong interaction corresponds to the unbound neutron-proton singlet; proton-proton electrostatic repulsion further disfavors binding. There is no bound dineutron or diproton in these physical low-energy channels. The inference uses charge independence and singlet-channel behavior; the deuteron's quantum numbers alone do not constrain every imaginable potential.

In a pure $L=0,S=1$ stretched state both spins point along the total angular momentum, with no orbital contribution. The nuclear magnetic moment would beThe measured $0.857{\mu }_N$ differs. More decisively, a pure S wave spatial state is isotropic and has zero electric quadrupole moment. Its nonzero value requires a nonspherical component of the quantum charge distribution.

Positive parity allows even $L$; coupling $S=1$ to $J=1$ permits $L=0$ and $L=2$. The state contains a deuteron D-state admixture:with radial wavefunctions implicit. A nuclear tensor force couples the orbital components. The quadrupole moment involves their interference as well as a D-state contribution, making it sensitive to a small admixture. This is a quantum angular correlation, not two particles permanently fixed in space.

In the nonrelativistic one-body magnetic-moment approximation,The D-state coupling has $⟨S_z⟩=-1/2$ and $⟨L_z⟩=3/2$ at $J_z=1$; the proton supplies approximately half the relative orbital magnetic moment. ThusThe robust conclusion is a predominantly S wave state with a small tensor-induced D wave component. The 4% estimate is model dependent: exchange currents and other corrections also affect the magnetic moment, and a wavefunction component probability is not itself a uniquely measured observable.^[ref]