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.
~p=~p 1 −~p 2
2=~p 1We will add the relativistic correction for both the electron and the positron.
Hrel=−1
8
p^41 +p^42
m^3 c^2=−
1
4
p^4
m^3 c^2=
− 1
32
p^4
μ^3 c^2=
− 1
8 μc^2(
p^2
2 μ) 2
This is just half the correction we had in Hydrogen (withmeessentially replaced byμ).
The spin-orbit correction should be checked also. We hadHSO= 2 mcge 2 S~·~v×~∇φas the interaction
between the spin and the B field producded by the orbital motion. Since~p=μ~v, we have
HSO=
ge
2 mμc^2S~·~p×∇~φfor 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 thesame. We have the same fomula
as in the fine structure section except that we havemμin the denominator. The final formula then
is
HSO=1
2
ge^2
2 mμc^2 r^3L~·
(
S~ 1 +S~ 2
)
=
1
2
e^2
2 μ^2 c^2 r^3~L·
(
S~ 1 +S~ 2
)
again just one-half of the Hydrogen result if we write everything in terms ofμfor the electron spin,
but, we add the interaction with the positron spin.
The calculation of the spin-spin (or hyperfine) term also needs someattention. We had
∆ESS=
2
3
Ze^2 gN
2 meMNc^2S~·~I^4
n^3(
Zαmec
̄h) 3
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 byμ. For positronium, the result is
∆ESS =
2
3
e^22
2 m^2 ec^2~S 1 ·S~ 24
n^3(αμc
̄h) 3
=
2
3
e^28
2 μ^2 c^2S~ 1 ·~S 24
n^3(αμc
̄h) 3
=
32
3
α^4 μc^21
n^3S~ 1 ·S~ 2
̄h^224.3.6 Hyperfine and Zeeman for H, muonium, positronium
We are able to set up the full hyperfine (plus B field) problem in a general way so that different
hydrogen-like systems can be handled. We know that as the massesbecome more equal, the hyperfine
interaction becomes more important.