Commutation of Spin, Angular and Spin-Orbital Momentum | Quantum Science Philippines

Commutation of Spin, Angular and Spin-Orbital Momentum

Marichu T. Miscala

In quantum mechanics, the presence of spin-orbit coupling gives rise to the Hamiltonian that will no longer commute with [eq]\vec{L}[/eq], and [eq]\vec{S}[/eq], so the spin and orbital momenta are not separately conserved.

In order to understand this concept better, a commutation problem for orbital angular momentum [eq]\vec{L}[/eq], spin [eq]\vec{S}[/eq], and spin-orbital momentum [eq]\vec{J}[/eq] is presented here.

• Consider the fundamental commutation relations for angular momentum. The individual components of the spin [eq]\vec{S}[/eq] do not commute with each other. That is,

[eq][S_x,S_y] = i\hbar S_z [/eq];         [eq][S_y,S_z] = i\hbar S_x [/eq];            [eq][S_z,S_x] = i\hbar S_y [/eq];

• The commutation relation for the ‘instrinsic’ angular momentum [eq]\vec{S}[/eq] is much like a ‘carbon copy’ to that of the ‘extrinsic’ angular momentum [eq]\vec{L}[/eq]

[eq][L_x,L_y] = i\hbar L_z [/eq];         [eq][L_y,L_z] = i\hbar L_x [/eq];            [eq][L_z,L_x] = i\hbar L_y [/eq];

But the spin-obit Hamiltonian does commute with [eq]L^2[/eq], [eq]S^2[/eq] and the total angular momentum [eq]\vec{J}[/eq] which is

[eq]\vec{J} = \vec{L} + \vec{S}[/eq]

From the given fundamental commutation relations above, we then now seek the commutators of the following commutation relations:

(a.) [eq][\vec{L}.\vec{S},\vec{L}][/eq]          (b.) [eq][\vec{L}.\vec{S},\vec{S}][/eq]          (c.) [eq][\vec{L}.\vec{S},J][/eq]

(d.) [eq][\vec{L}.\vec{S},L^2][/eq]          (e.) [eq][\vec{L}.\vec{S},S^2][/eq]            (f.)  [eq][\vec{L}.\vec{S},J^2][/eq]

As a hint here, [eq]\vec{L}[/eq] and [eq]\vec{S}[/eq] satisfy the fundamental commutation relations for angular momentum, but [eq]\vec{L}[/eq] and [eq]\vec{S}[/eq] commute with each other.

(a.) [eq][\vec{L} . \vec{S} , L_x] = [L_x S_x + L_y S_y + L_z S_z , L_x] [/eq]

[eq] = S_x [L_x, L_x] + S_y [L_y, L_x] + S_z [L_z, L_x][/eq]
[eq] = S_x (0) + S_y (-i \hbar L_z) + S_z (i \hbar L_y)[/eq]
[eq] = i \hbar (L_y S_z – L_z S_y) = i \hbar (\vec{L} \times \vec{S})_x[/eq]

same goes for the other two-components, so

[eq][\vec{L}.\vec{S}, \vec{L}] = i \hbar (\vec{L} \times \vec {S})[/eq]

(b.) [eq][\vec{L}.\vec{S},\vec{S}][/eq] is identical only with [eq] \vec{L} \Longleftrightarrow \vec{S}[/eq]

[eq][\vec{L}.\vec{S},\vec{S}] = i \hbar (\vec{S} \times \vec{L})[/eq]

(c.) [eq][\vec{L}.\vec{S},\vec{J}] = [\vec{L}.\vec{S},\vec{L}] + [\vec{L}.\vec{S},\vec{S}] = i \hbar (\vec{L} \times \vec{S} + \vec{S} \times \vec{L}) = 0[/eq]

(d.) [eq]L^2[/eq] commutes with all the components of [eq]\vec{L}[/eq] (and [eq]\vec{S}[/eq]), so

[eq][\vec{L} . \vec{S} , L^2] = 0 [/eq]

(e.) Likewise,

[eq][\vec{L} . \vec{S} , S^2] = 0 [/eq]

(f.) [eq][\vec{L} . \vec{S} , J^2] = [\vec{L} . \vec{S} , L^2] + [\vec{L} . \vec{S} , S^2] + 2[\vec{L} . \vec{S} , \vec{L} . \vec{S}][/eq]

where [eq]J^2 = (\vec{L} + \vec{S}) . (\vec{L} + \vec{S}) = L^2 + S^2 + 2 \vec{L} . \vec{S} [/eq]

The first, second and third terms vanish from the results of (d.) and (e.). Thus,

[eq][\vec{L} . \vec{S} , J^2] = 0 [/eq]

This means that the quantities [eq]L^2[/eq], [eq]S^2[/eq] and [eq]\vec{J}[/eq] are conserved. That is, the eigenstates of [eq]L_z[/eq] and [eq]S_z[/eq] are not “good” states to use in perturbation theory, but the eigenstates of [eq]L^2[/eq], [eq]S^2[/eq] and [eq]J_z[/eq] are.

The problem presented here is based on the problem 6.16 from D.J. Griffiths “Introduction to Quantum Mechanics”.

About the author: Marichu T. Miscala is currently taking up her masters in Mindanao State University – Iligan Institute of Technology. She is most interested in pursuing a career in advanced research.