= Solution
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 be
$$
\mu_d^{(S)}=\mu_p+\mu_n=(2.793-1.913)\mu_N=0.880\mu_N.
$$
The measured $0.857\mu_N$ differs. More decisively, a pure <S wave (quantum scattering)> 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>:
$$
|\psi_d\rangle=\sqrt{1-P_D}\,|{}^3S_1\rangle+\sqrt{P_D}\,|{}^3D_1\rangle,
$$
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,
$$
\frac{\mu_d}{\mu_N}=(1-P_D)\mu_s+P_D\left(\frac34-\frac12\mu_s\right),
\qquad \mu_s=\frac{\mu_p+\mu_n}{\mu_N}=0.880.
$$
The D-state coupling has $\langle S_z\rangle=-1/2$ and $\langle L_z\rangle=3/2$ at $J_z=1$; the <proton> supplies approximately half the relative orbital magnetic moment. Thus
$$
P_D\simeq\frac{0.880-0.857}{\frac32(0.880-\frac12)}
=\boxed{0.040\ \text{in this approximation}.}
$$
The robust conclusion is \b[a predominantly <S wave (quantum scattering)> 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.<Modern Theory of Nuclear Forces by Evgeny Epelbaum and others>[§I.F, Observables and not-so observable quantities; §II.B, spin-dependent nuclear interactions]{ref}
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