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Transcript
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 . 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 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). 2