Spin-orbit coupling for positronium

[Aleolomorfo]

Homework Statement

Considering the atom made of an electron and a positron. The spin-orbit Hamiltonian is:
$$H=\frac{e^2}{4\nu^2c^2r^34\pi\epsilon_0}\vec{L}\cdot\vec{S}$$
with $\vec{L}$ the relative angular momentum, with $\vec{S}$ the total spin and $\mu$ the reduced mass. Finding the levels in which $3d$ states are split.

Homework Equations

The Attempt at a Solution

First I show you the way I solved the problem.
$$\vec{J}=\vec{L}+\vec{S}$$
$$\vec{S}\cdot\vec{L} = \frac{J^2-L^2-S^2}{2}$$
So I can rewrite the Hamiltonian:
$$H=A(J^2-L^2-S^2)$$
with $A$ which is equal to al the constants. The splitting in energy is:
$$\Delta E = A\hbar^2 (J(J+1) - \frac{21}{4})$$
In the last step I put $S=\frac{1}{2}$ and $L=2$. Then I calculate the possible $J$'s:
$$|L-S|\le J \le L+S$$
So there is a double splitting with $J=\frac{3}{2}$ and $J=\frac{5}{2}$.

I think the reasoning is right (isn't it?), but I have a question about the spin. I put $S=\frac{1}{2}$ automatically as in the hydrogen spin-orbit coupling. But why do I consider only the spin of the electron and not the total spin of the system, so the spin of the electron and the spin of the positron?

 

[drvrm]

I think the reasoning is right (isn't it?), but I have a question about the spin. I put S=12S=12S=\frac{1}{2} automatically as in the hydrogen spin-orbit coupling. But why do I consider only the spin of the electron and not the total spin of the system, so the spin of the electron and the spin of the positron?

The metastable electron-positron bound state can exist in different configurations depending on the relative spin states of the positron and the electron.
These are known as para-positronium (p-Ps), with total spin S = 0 and ortho positronium (o-Ps) with S = 1.
though These spin states have very different lifetimes

see- the level diagram may help

1. R. Ley, Appl. Surf. Sci. 194 301(2002)
2. http://pyweb.swan.ac.uk/quamp/quampweb/talks/CASSIDY_QuAMP_2013.pdf