Question

Consider the bound states of a system of two non-identical, nonrelativistic, spin one-half particles interacting via a spin-independent central potential. Focus in particular on the ³P₂ and ¹P₂ levels (³P₂: spin-triplet, L = 1, J = 2; ¹P₂ : spin-singlet, L = 1, J = 1). A tensor force term \(H' = \lambda [3\sigma (1).\hat r\sigma(2). \hat r - a(1).a(2)]\) is added to the Hamiltonian as a perturbation, where \(\hat r\) is a constant, 2 is a unit vector along the line joining the two particles, σ(1) and σ(2) are the Pauli spin operators for particles 1 and 2.

(a) Using the fact that H' commutes with all components of the total angular momentum, show that the perturbed energies are independent of m, the eigenvalues of J_z.

(b) The energy is most easily evaluated for the triplet state when the eigenvalue of J_z takes on its maximum value m = j = 2. Find the perturbation energy \(\Delta E\)(³P₂).

(c) Find \(\Delta E\)(¹P₁).

Answer 1

(a) Use units for which h = 1. As S = 1/2 [σ(l) + σ(2)], the perturbation Hamiltonian H' can be written as

To prove that H' commutes with all components of the total angular momentum J, we show for example [J_z, H'] = 0. AS [S_z, S²] = 0, [L_z, S²] = 0, we have (J_z, S²] = 0. Also, as [S_x, S_y] = ihS_z, etc, L_z = \(-ih \frac\partial{\partial \phi}\), we have

Combining the above results, we have [J_z, H'] = 0. Similarly we can show

[J_x, H'] = [J_y, H'] = 0.

It follows that J₊ = J_x + iJy also commutes with H' J₊ has the property

\(J + |j, m\rangle = a |j,m+1\rangle\).

where a is a constant. Suppose there are two unperturbed states

\( |j, m_1\rangle \) and \( |j, m_2\rangle \) where rn₂ = m₁ + 1,

which are degenerate and whose energies to first order perturbation are E₁ and E₂ respectively. Then

Since the matrix element a \(\ne\) 0, E₁ = E₂, i.e., the perturbation energies are independent of m.

(b) The perturbation energy is

(c) For the state ¹P₁, as S = 0, m_s = 0 and so H’ = 0, we have \(\Delta E\)(¹P₁) = 0.