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 $\vec{L}$ , and $\vec{S}$ , so the spin and orbital momenta are not separately conserved.

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

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

$[S_x,S_y] = i\hbar S_z$ ; $[S_y,S_z] = i\hbar S_x$ ; $[S_z,S_x] = i\hbar S_y$ ;

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

$[L_x,L_y] = i\hbar L_z$ ; $[L_y,L_z] = i\hbar L_x$ ; $[L_z,L_x] = i\hbar L_y$ ;

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

$\vec{J} = \vec{L} + \vec{S}$

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

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

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

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

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

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

same goes for the other two-components, so

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

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

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

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

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

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

(e.) Likewise,

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

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

where $J^2 = (\vec{L} + \vec{S}) . (\vec{L} + \vec{S}) = L^2 + S^2 + 2 \vec{L} . \vec{S}$

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

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

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

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Commutation of Spin, Angular and Spin-Orbital Momentum

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