130_notes.dvi

The top part is already diagonal so we only need to work in bottom right 2 by 2 matrix, solving the
eigenvalue problem.

(

A B

B − 3 A

)(

a

b

)

=E

(

a

b

)

where

A≡A 4

B≡μBB

Setting the determinant equal to zero, we get

(A−E)(− 3 A−E)−B^2 = 0.

E^2 + 2AE− 3 A^2 −B^2 = 0

E=

− 2 A±

√

4 A^2 + 4(3A^2 +B^2 )

2

=−A±

√

A^2 + (3A^2 +B^2 )

=−A±

√

4 A^2 +B^2

The eigenvalues for themf= 0 states, which mix differently as a function of the field strength, are

E=−

A

4

±

√(

A

2

) 2

+ (μBB)^2.

The eigenvalues for the other two states which remain eigenstatesindependent of the field strength
are
A
4

+μBB

and
A
4
−μBB.

24.3.5 Positronium

Positronium, the Hydrogen-like bound state of an electron and a positron, has a “hyperfine” correc-
tion 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 Quan-
tum 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

En=

1

2

α^2 μc^2

1

n^2

where the reduced mass is given byμ=m 2 e.

The relativistic correction must take account of both the motion ofthe electron and the positron.

We use~r≡~r 1 −~r 2 and~p=μ~r ̇=m
~r ̇ 1 −m~r ̇ 2
2. Since the electron and positron are of equal mass, they