1. 16.451 Lecture 19: The deuteron 13/11/2003 Basic properties: mass: mc 2 = 1876.124 MeV binding energy: (measured via  -ray energy in n + p  d +  ) RMS radius: 1.963  0.004 fm (from electron scattering) quantum numbers: (lectures 13, 14) magnetic moment:  = + 0.8573  N electric quadrupole moment: Q = + 0.002859  0.00030 bn (  the deuteron is not spherical!....) Basic properties: mass: mc 2 = 1876.124 MeV binding energy: (measured via  -ray energy in n + p  d +  ) RMS radius: 1.963  0.004 fm (from electron scattering) quantum numbers: (lectures 13, 14) magnetic moment:  = + 0.8573  N electric quadrupole moment: Q = + 0.002859  0.00030 bn (  the deuteron is not spherical!....) Important because: deuterium is the lightest nucleus and the only bound N-N state testing ground for state-of-the art models of the N-N interaction. 1

2. Because the deuteron has spin 1, there are 3 form factors to describe elastic scattering: the “charge” (G c ), “electric quadrupole” (G q ) and “magnetic (G M ) form factors. (JLab data) Because the deuteron has spin 1, there are 3 form factors to describe elastic scattering: the “charge” (G c ), “electric quadrupole” (G q ) and “magnetic (G M ) form factors. (JLab data) Electron scattering measurements: 2

3. Interpretation of quantum numbers: Of the possible quantum numbers, L = 0 has the lowest energy, so we expect the ground state to be L = 0, S = 1 (the deuteron has no excited states!) The nonzero electric quadrupole moment suggests an admixture of L = 2 (more later!) introduce Spectroscopic Notation: with naming convention: L = 0 is an S-state, L = 1 is a P-state, L = 2: D-state, etc...  the deuteron configuration is primarily 3

4. Isospin and the N-N system: (recall lecture 13) The total wavefunction for two identical Fermions has to be antisymmetric w.r.to particle exchange: Central force problem: with symmetry (-1) L given by the spherical harmonic functions Spin and Isospin configurations: S = 0 and T = 0 are antisymmetric ; S = 1 and T = 1 are symmetric state can only be T = 0 ! 4

5. Understanding “L” -- the 2-body problem and orbital angular momentum: Classical Mechanics: CM set origin at the center of mass, and let the CM be fixed in space  particles orbit the CM with angular speed  Kinetic energy for each particle and angular momentum L = I  : Total angular momentum: L = L 1 + L 2 with I = I 1 + I 2 Total kinetic energy: K = L 2 /2I 5

6. Equivalent one-body problem: CM Particles orbit the center of mass because of an attractive potential V(r) with no external force, particle kinematics are related in the CM system and the 2-body problem is equivalent to a 1-body problem: Total energy: Moment of inertia can be written: Note: L = total angular momentum of the 2-particle system. 6

7. Quantum mechanics version: H = total energy operator, E = eigenvalue) Angular derivatives define an orbital angular momentum operator: wave function in the relative coordinate has this generic form, where R(r) depends on V(r): Analog of L 2 /2  r 2 in the classical problem! lowest energy state has minimum total L! 7

8. Deuteron Magnetic Moment:  d = 0.857  N In general, the magnetic moment is a quantum-mechanical vector; it must be aligned along the “natural symmetry axis” of the system, given by the total angular momentum: But we don’t know the direction of, only its “length” and z-component as expectation values: in a magnetic field, the energy depends on m J via Strategy: we will define the magnetic moment by its maximal projection on the z-axis, defined by the direction of the magnetic field, with m J = J Use expectation values of operators to calculate the result..... 8

9. Calculation of  : Subtle point: we have to make two successive projections to evaluate the magnetic moment according to our definition, and the spin and orbital contributions enter with different weights.   mJmJ 1. Project onto the direction of J: 2. Project onto the z-axis with m J = J: Next, we need to figure out the operator for 9

10. spin and orbital contributions: We already know the intrinsic magnetic moments of the proton and neutron, so these must correspond to the spin contributions to the magnetic moment operator: For the orbital part, there is a contribution from the proton only, corresponding to a circulating current loop (semiclassical sketch, but the result is correct) p I r For the deuteron, we want to use the magnetic moment operator: 10

11. Details: 1. because m n  m p 2. L = 0 in the “S-state” ( 3 S 1 ) but we will consider also a contribution from the “D-state” ( 3 D 1 ) as an exercise 3. The proton and neutron couple to S = 1, and the deuteron has J = 1 Trick: use and write the operator as: But the proton and neutron spins are aligned, and so the second term has to give zero! 11

12. continued.... So, effectively we can write for the deuteron: Trick for expectation values: 12

13. Comparison to experiment: This is intermediate between the S-state and D-state values: Suppose the wave function of the deuteron is a linear combination of S and D states: Then we can adjust the coefficients to explain the magnetic moment: b 2 = 0.04, or a 4% D-state admixture accounts for the magnetic moment ! 13