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, µ qv × r B = 0 , 4π r3 −7 −2 2 where µ0 is the vacuum permeability and is defined as 4π × 10 NC s . 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 qr × p q µ = = L, 2m 2m where L is the angular momentum. The magnitude of the orbital magnetic momentum of an electron with orbital-angular- momentum quantum l is e¯h 1/2 1/2 µ = [l(l + 1)] = µB[l(l + 1)] . 2me Here, µB is a constant called Bohr magneton, and is equal to −19 −34 e¯h (1.6 × 10 C) × (6.626 × 10 J · s/2π) −24 µB = = −31 = 9.274 × 10 J/T, 2me 2 × 9.11 × 10 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 e µB EB = −µ · B = Lz · B = BLz. 2me ¯h Therefore, the Hamiltonian corresponding to an external magnetic field along the z axis is −1 HˆB = µBB¯h Lˆz. 1 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 q1q2 Eelec = , 4π²0r 7 2 −12 2 −1 where ²0 = 10 /4πc = 8.854 × 10 C N 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), q0 q0 E = 1 2 , (1) elec r which was also used in class when discussing Schr¨odinger’sequation. As we discussed in 0 1/2 class, charges in Eq. (1) can be viewed as q = q/(4π²0) . 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 0 1/2 · 0 ¸ 0 e (4π²0) 1/2 µ0 q v × r e 0 EB = −µ · B = Lz · (4π²0) 3 = LzB . 2me 4π r 2mec √ Note that the speed of light c = 1/ ²0µ0 and that the magnetic field is now in cgs-type expression and has units of Gauss (equivalent to 10−4 T). 2.