Positronium, the Hydrogen-like bound state of an electron and a positron, has a ``hyperfine'' correction which is as large as the fine structure corrections since the magnetic moment of the positron is the same size as that of the electron. It is also an interesting laboratory for the study of Quantum Physics. The two particles bound together are symmetric in mass and all other properties. Positronium can decay by anihilation into two or more photons.

In analyzing positronium, we must take some care to correctly handle the relativistic correction in the case of a reduced mass much different from the electron mass and to correctly handle the large magnetic moment of the positron.

The zero order energy of positronium states is

\begin{displaymath}\bgroup\color{black}E_n={1\over 2}\alpha^2 \mu c^2{1\over{n^2}}\egroup\end{displaymath}

where the reduced mass is given by \bgroup\color{black}$\mu={m_e\over 2}.$\egroup

The relativistic correction must take account of both the motion of the electron and the positron. We use \bgroup\color{black}$\vec{r}\equiv\vec{r}_1-\vec{r}_2 $\egroup and \bgroup\color{black}$\vec{p}=\mu\dot{\vec{r}} = { m\dot{\vec{r}}_1 - m\dot{\vec{r}}_2\over 2}$\egroup. Since the electron and positron are of equal mass, they are always exactly oposite each other in the center of mass and so the momentum vector we use is easily related to an individual momentum.

\begin{displaymath}\bgroup\color{black} \vec{p}= {\vec{p}_1-\vec{p}_2\over 2} = \vec{p}_1 \egroup\end{displaymath}

We will add the relativistic correction for both the electron and the positron.

\begin{displaymath}\bgroup\color{black}H_{rel}=-{1\over 8}{p^4_1+p^4_2\over{m^3c...
={-1\over{8\mu c^2}}\left({p^2\over{2\mu}}\right)^2\egroup\end{displaymath}

This is just half the correction we had in Hydrogen (with \bgroup\color{black}$m_e$\egroup essentially replaced by \bgroup\color{black}$\mu$\egroup).

The spin-orbit correction should be checked also. We had \bgroup\color{black}$H_{SO}={ge\over 2mc^2}\vec{S}\cdot\vec{v}\times\vec{\nabla}\phi$\egroup as the interaction between the spin and the B field producded by the orbital motion. Since \bgroup\color{black}$\vec{p}=\mu\vec{v}$\egroup, we have

\begin{displaymath}\bgroup\color{black}H_{SO}={ge\over 2m\mu c^2}\vec{S}\cdot\vec{p}\times\vec{\nabla}\phi\egroup\end{displaymath}

for the electron. We just need to add the positron. A little thinking about signs shows that we just at the positron spin. Lets assume the Thomas precession is also the same. We have the same fomula as in the fine structure section except that we have \bgroup\color{black}$m\mu$\egroup in the denominator. The final formula then is

\begin{displaymath}\bgroup\color{black}H_{SO}={1\over 2}{ge^2\over 2m\mu c^2r^3}...
...u^2 c^2r^3}\vec{L}\cdot\left(\vec{S}_1+\vec{S}_2\right) \egroup\end{displaymath}

again just one-half of the Hydrogen result if we write everything in terms of \bgroup\color{black}$\mu$\egroup for the electron spin, but, we add the interaction with the positron spin.

The calculation of the spin-spin (or hyperfine) term also needs some attention. We had

\begin{displaymath}\bgroup\color{black} \Delta E_{SS}={2\over 3}{Ze^2g_N\over 2 ...
...{I}{4\over n^3}
\left({Z\alpha m_ec\over\hbar}\right)^3 \egroup\end{displaymath}

where the masses in the deonominator of the first term come from the magnetic moments and thus are correctly the mass of the particle and the mas in the last term comes from the wavefunction and should be replaced by \bgroup\color{black}$\mu$\egroup. For positronium, the result is

\Delta E_{SS} &=& {2\over 3}{e^2 2\over 2 m_e^2 c^2}\vec{S}_1\...
...alpha^4\mu c^2 {1\over n^3}{\vec{S}_1\cdot\vec{S}_2\over\hbar^2}

Jim Branson 2013-04-22