Download A Supplemental Discussion on the Bohr Magneton

A Supplementary Discussion on Bohr Magneton
Magnetic fields arise from moving electric charges. A charge q with velocity v gives rise to
a magnetic field B, following SI convention in units of Tesla,
B=
µ0 qv × r
,
4π r3
where µ0 is the vacuum permeability and is defined as 4π × 10−7 NC−2 s2 .
Now consider a magnetic dipole moment as a charge q moving in a circle with radius r
with speed v. The current is the charge flow per unit time. Since the circumference of the
circle is 2πr, and the time for one revolution is 2πr/v, one has the current as I = qv/2πr.
The magnitude of the dipole moment is |µ| = I · (area) = (qv/2πr)πr2 = qrp/2m, where p
is the linear momentum. Since the radial vector r is perpendicular to p, we have
µ=
qr × p
q
=
L,
2m
2m
where L is the angular momentum.
The magnitude of the orbital magnetic momentum of an electron with orbital-angularmomentum quantum l is
µ=
eh̄
[l(l + 1)]1/2 = µB [l(l + 1)]1/2 .
2me
Here, µB is a constant called Bohr magneton, and is equal to
µB =
eh̄
(1.6 × 10−19 C) × (6.626 × 10−34 J · s/2π)
=
= 9.274 × 10−24 J/T,
2me
2 × 9.11 × 10−31 kg
where T is magnetic field, Tesla.
Now consider applying an external magnetic field B along the z-axis. The energy of
interaction between this magnetic field and the magnetic dipole moment is
EB = −µ · B =
µB
e
Lz · B =
BLz .
2me
h̄
Therefore, the Hamiltonian corresponding to an external magnetic field along the z axis is
ĤB = µB Bh̄−1 L̂z .
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One may find, however, that the units conventions are different in some physics texts.
For example, in SI units, the electrostatic energy (in units of Joule) is expressed as
Eelec =
q1 q2
,
4π²0 r
where ²0 = 107 /4πc2 = 8.854 × 10−12 C2 N−1 is the vacuum permittivity, where c is the
speed of light. One may find similar expression for electrostatic interaction in cgs unit
convention (erg, 1 erg = 10− 7 Joule),
Eelec =
q10 q20
,
r
(1)
which was also used in class when discussing Schrödinger’s equation. As we discussed in
class, charges in Eq. (1) can be viewed as q 0 = q/(4π²0 )1/2 . This is sometimes referred to as
“statcoulumb.” Current under this unit convention has a similar expression and is referred
to as “statamp.” Now re-write the magnetic field and magnetic moment that was discussed
above in cgs unit, we have
·
¸
e0 (4π²0 )1/2
µ0 q 0 v × r
e0
EB = −µ · B =
Lz · (4π²0 )1/2
=
Lz B 0 .
2me
4π r3
2me c
√
Note that the speed of light c = 1/ ²0 µ0 and that the magnetic field is now in cgs-type
expression and has units of Gauss (equivalent to 10−4 T).
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