Elements of Modern Physics
116
electron. The magnetic field at the nucleus, due to the orbital motion of the
electron can be deduced from Eq. (4.49) as
B =
3
2
0
( / )
4
− πε e
e
r
m c
L
(4.69)
Therefore, the interaction energy due to this field is
〈 V orb 〉 =
2
3
2
0
1/
4
〈
〉
πε e p
ge
r
m m c
I.L
for l ≠ 0
(4.70)
= 0 for l = 0
This is smaller than the fine structure terms by a factor of about m e /m p
~ 1/1000.
For calculating the field due to the intrinsic magnetic moment of the electron,
we note that the vector potential due to a magnetic dipole moment µ is
A =
2
3
0
1
4
µ ×
πε
c
r
r
(4.71)
From this, the magnetic field comes out as
B = ∇ × A
=
2
3
3
0
1
.
( . )
4
µ ∇
− µ ∇
πε
c
r
r
r
r
(4.72)
Therefore, the energy of the nuclear magnetic moment µ N interacting with
this field is
V spin =
2
3
3
0
1
. ( . )
.
4
−
∇ µ µ
− µ
µ
πε
N
N
c
r
r
r
r
(4.73)
With this, the perturbative expression for the interaction energy comes out
to be
〈 V spin 〉 =
2
3
5
0
.
. .
1
3
4
µ µ
µ µ
〈
−
〉
πε
N
N
c
r
r
r r
for l ≠ 0
(4.74)
For l = 0, the angular integration in Eq. (4.74) gives zero for r ≠ 0. For
obtaining the correct value of the contribution from r = 0, Gauss theorem is used
in the expectation value of the expression in Eq. (4.73), to get
〈 V spin 〉 l = 0 =
2
2
0
1
8
( . )
| (0) |
3
4
N
c
π
−
µ µ
ψ
πε
116
electron. The magnetic field at the nucleus, due to the orbital motion of the
electron can be deduced from Eq. (4.49) as
B =
3
2
0
( / )
4
− πε e
e
r
m c
L
(4.69)
Therefore, the interaction energy due to this field is
〈 V orb 〉 =
2
3
2
0
1/
4
〈
〉
πε e p
ge
r
m m c
I.L
for l ≠ 0
(4.70)
= 0 for l = 0
This is smaller than the fine structure terms by a factor of about m e /m p
~ 1/1000.
For calculating the field due to the intrinsic magnetic moment of the electron,
we note that the vector potential due to a magnetic dipole moment µ is
A =
2
3
0
1
4
µ ×
πε
c
r
r
(4.71)
From this, the magnetic field comes out as
B = ∇ × A
=
2
3
3
0
1
.
( . )
4
µ ∇
− µ ∇
πε
c
r
r
r
r
(4.72)
Therefore, the energy of the nuclear magnetic moment µ N interacting with
this field is
V spin =
2
3
3
0
1
. ( . )
.
4
−
∇ µ µ
− µ
µ
πε
N
N
c
r
r
r
r
(4.73)
With this, the perturbative expression for the interaction energy comes out
to be
〈 V spin 〉 =
2
3
5
0
.
. .
1
3
4
µ µ
µ µ
〈
−
〉
πε
N
N
c
r
r
r r
for l ≠ 0
(4.74)
For l = 0, the angular integration in Eq. (4.74) gives zero for r ≠ 0. For
obtaining the correct value of the contribution from r = 0, Gauss theorem is used
in the expectation value of the expression in Eq. (4.73), to get
〈 V spin 〉 l = 0 =
2
2
0
1
8
( . )
| (0) |
3
4
N
c
π
−
µ µ
ψ
πε