Question

The deuteron is a bound state of a proton and a neutron of total angular momentum J = 1. It is known to be principally an S (L = 0) state with a small admixture of a D (L = 2) state.

(a) Explain why a P state cannot contribute.

(b) Explain why a G state cannot contribute.

(c) Calculate the magnetic moment of the pure D state n - p system with J = 1.

Assume that the n and p spins are to be coupled to make the total spin S which is then coupled to the orbital angular momentum L to give the total angular momentum J. Express your result in nuclear magnetons. The proton and neutron magnetic moments are 2.79 and -1.91 nuclear magnetons respectively.

Answer 1

(a) The parities of the S and D states are positive, while the parity of the P state is negative. Because of the conservation of parity in strong interaction, a quantum state that is initially an S state cannot have a P state component at any later moment.

(b) The possible spin values for a system composed of a proton and a neutron are 1 and 0. We are given J = L + S and J = 1. If S = 0, L = 1, the system would be in a P state, which must be excluded as we have seen in (a). The allowed values are then S = 1, L = 2, 1, 0. Therefore a G state (L = 4) cannot contribute.

(c) The total spin is S = s_p + s_n. For a pure D state with J = 1, the orbital angular momentum (relative to the center of mass of the n and p) is L = 2 and the total spin must be S = 1. The total magnetic moment arises from the coupling of the magnetic moment of the total spin, µ, with that of the orbital angular momentum, µ_L, where µ_p + µ_n, µ_p, µ_n being the spin magnetic moments of p and n respectively.

The average value of the component of p in the direction of the total spin S is

The motion of the proton relative to the center of mass gives rise to a magnetic moment, while the motion of the neutron does not as it is uncharged. Thus

µ_L = µ_NL_p,

where L_p is the angular momentum of the proton relative to the center of mass. As L_p + L_n = L and we may assume L_p = L_n, we have L_p = L/2 (the center of mass is at the mid-point of the connecting line, taking \(m_p\approx m_n\)). Consequently, µ_L = µ_NL/2.

The total coupled magnetic moment along the direction of J is then

Since J = L + S, S . L = 1/2(J² - L² - S²). With J = 1, L = 2, S = 1 and so J² = 2, L² = 6, S² = 2, we have S . L = -3 and thus L . J = 3, S . J = -1. Hence

Taking the direction of J as the z-axis and letting J_z take the maximum value J_z = 1, we have µ_T = 0.31 µ_N.