The deuteron ground state has and

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Nuclear Physics · The nuclear force and nuclear models

The Nucleus Resolved, Ch The deuteron and what two nucleons do to each other

Problem

The deuteron ground state has Jπ=1+J^\pi=1^+ and, from antisymmetry, is the mixture ∣d⟩=1−PD ∣3S1⟩+PD ∣3D1⟩|d\rangle=\sqrt{1-P_D}\,|^3S_1\rangle+\sqrt{P_D}\,|^3D_1\rangle, with PDP_D the DD-state probability. The magnetic-dipole operator is built only from L⃗\vec L and S⃗\vec S and is diagonal in orbital angular momentum, so it has no matrix element between the 3S1^3S_1 and 3D1^3D_1 pieces (which differ by ΔL=2\Delta L=2): the deuteron's moment is a population-weighted average, μd=(1−PD)μS+PDμD,\mu_d=(1-P_D)\mu_S+P_D\mu_D, where μS=μp+μn\mu_S=\mu_p+\mu_n is the moment of the pure 3S1^3S_1 state and the vector model gives the pure 3D1^3D_1 moment as μD=34−12(μp+μn)\mu_D=\tfrac34-\tfrac12(\mu_p+\mu_n), both in units of the nuclear magneton μN\mu_N. Take μp=+2.7928 μN\mu_p=+2.7928\,\mu_N, μn=−1.9130 μN\mu_n=-1.9130\,\mu_N and the measured deuteron moment μd=0.8574 μN\mu_d=0.8574\,\mu_N. ((a)) Evaluate μS\mu_S and μD\mu_D. ((b)) Rearrange the mixture formula for PDP_D and evaluate it. ((c)) The deuteron's quadrupole moment is measured to be Qsp=+0.286 fm2Q_{sp}=+0.286\,\mathrm{fm^2}, not zero. Explain, from the tensor character of the quadrupole operator, why this quantity cannot be written as a simple population average of the same two pure-state values in the way μd\mu_d was, and why it is nonetheless firm evidence for the same L=2L=2 admixture.

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