Feb 22, 2001

Notes for Lecture Notes for Lecture #17

Derivation of the hyperfine interaction

Magnetic dipole field

These notes are very similar to the notes for Lecture #13 on the electric dipole field.

The magnetic dipole moment is defined by

m = 1
2
 ó
õ 		d3r¢ r¢ ×J(r¢),
(1)
with the corresponding potential
A(r) = m0
4p
m × ^
r
 
r2
 ,
(2)
and magnetostatic field
B(r) = m0
4p
 ì
ï
í
ï
î
3 ^
r
 
 (m · ^
r
 
 ) - m
r3
 + 8p
3
 m d3(r) ü
ï
ý
ï
þ 		.
(3)

The first terms come form evaluating Ñ×A in Eq. 2. The last term of the field expression follows from the following derivation. We note that Eq. (3) is poorly defined as r® 0, and consider the value of a small integral of B(r) about zero. (For this purpose, we are supposing that the dipole m is located at r = 0.) In this case we will approximate

B(r » 0) » æ
è 		ó
õ

sphere 
 B(r) d3r ö
ø 		d3(r).
(4)

First we note that

ó
õ

r £ R 
 B(r) d3r = R2 ó
õ

r = R 
 ^
r
 
 ×A(r)   dW.
(5)

This result follows from the divergence theorm:

ó
õ

vol 
 Ñ·V d3r = ó
õ

surface 
 V · dA.
(6)
In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate of Ñ×A since Ñ×A = [^(x)] ( [^(x)] ·(Ñ× A) ) + [^(y)] ( [^(y)] ·(Ñ×A) ) + [^(z)] ( [^(z)] ·(Ñ×A) ). Note that [^(x)] ·(Ñ×A) = -Ñ·([^(x)] ×A) and that we can use the Divergence theorem with V º [^(x)] ×A(r) for the x- component for example:
ó
õ

vol 
 Ñ·( ^
x
 
 ×A) d3r = ó
õ

surface 
 ( ^
x
 
 ×A) · ^
r
 
 dA = ó
õ

surface 
 (A × ^
r
 
 ) · ^
x
 
 dA.
(7)
Therefore,
ó
õ

r £ R 
 (Ñ×A) d3r = - ó
õ

r = R 
 (A × ^
r
 
 ) ·( ^
x
 
 ^
x
 
 + ^
y
 
 ^
y
 
 + ^
z
 
 ^
z
 
 ) dA = R2 ó
õ

r = R 
 ( ^
r
 
 ×A) dW
(8)
which is identical to Eq. (5). Now, expressing the vector potential in terms of the current density:
A(r) = m0
4 p
 ó
õ 		d3r J(r¢)
|r - r¢|
 ,
(9)
we can use the identity,
ó
õ 		dW
^
r
 
|r - r¢|
 = 4 p
3
 r <
r > 2
    ^
r¢
 
 .
(10)

Therefore,

R2 ó
õ

r = R 
 ( ^
r
 
 ×A) dW = 4 pR2
3
 ó
õ 		d3r¢    r <
r > 2
    ^
r¢
 
 ×J(r¢).
(11)
If the sphere R contains the entire current distribution, then r > = R and r < = r¢ so that (11) becomes
R2 ó
õ

r = R 
 ( ^
r
 
 ×A) dW = 4 p
3
 ó
õ 		d3r¢   r¢ ×J(r¢) º 8 p
3
 m.
(12)

Magnetic field due to electrons in the vicinity of a nucleus

According to the Biot-Savart law (or the curl of Eq. 9), the magnetic field produced by a current density J(r¢) is given by:

B(r) = m0
4 p
 ó
õ 		d3r¢ J(r¢) ×(r - r¢)
|r - r¢|3
(13)
In this case, we assume that the current density is due to an electron in a bound atomic state with quantum numbers |n l ml ñ, as described by a wavefunction ynlml(r), where the azimuthal quantum number ml is associated with a factor of the form ei ml f. For such a wavefunction the quantum mechanical current density operator can be evaluated:
J(r¢) = -e (h/2p)
2mi
 (ynlml*Ѣ ynlml-ynlml Ѣynlml* ).
(14)
Since the only complex part of this wavefunction is associated with the azimuthal quantum number, this can be written:
J(r¢) = -e (h/2p)
2mi r¢sinq¢
 æ
ç
è 		ynlml*
f¢
 ynlml-ynlml
f¢
 ynlml* ö
÷
ø 		^
f¢
 
 =
-e (h/2p) ml ^
f¢
 
m r¢ sinq¢
 |ynlml|2.
(15)

We need to use this current density in the Biot-Savart law and evaluate the field at the nucleus (r = 0). The vector cross product in the numerator can be evaluated in spherical polar coordinates as:

^
f¢
 
 ×(-r¢) = r¢ æ
è 		- ^
x
 
 cosq¢cosf¢ - ^
y
 
 cosq¢sinf¢ + ^
z
 
 sinq¢ ö
ø
(16)
Thus the magnetic field evaluated at the nucleus is given by the integral:
B(0) = - m0 e (h/2p) ml
4 pm
 ó
õ 		d3r¢|ynlml|2
r¢ æ
è 		- ^
x
 
 cosq¢cosf¢ - ^
y
 
 cosq¢sinf¢ + ^
z
 
 sinq¢ ö
ø
r¢ sinq¢  r¢3
 .
(17)
In evaluating the integration over the azimuthal variable f¢, the [^(x)] and [^(y)] components vanish leaving the simple result:
B(0) = -
m0 e (h/2p) ml ^
z
 
4 pm
 ó
õ 		d3r¢|ynlml|2 1
r¢3
 º - m0 e
4 pm
 Lz ^
z
 
á
 1
r¢3
ñ
 .
(18)

File translated from TEX by TTH, version 2.20.
On 22 Feb 2001, 10:20.