This preview shows page 128 - 130 out of 168 pages.
PROBLEM: Show using standard perturbation theory, that 7.9, that the shift in the energy levels dueto the presence of the magnetic fieldδEn=μB(n|B.parenleftbiggL+g0Splanckover2pi1parenrightbigg|n)+μ2Bsummationdisplaynnegationslash=n′vextendsinglevextendsinglevextendsingle(n|B.parenleftBigL+g0Splanckover2pi1parenrightBig|n′)vextendsinglevextendsinglevextendsingle2En−En′+e28mB2(n|summationdisplayi(x2i+y2i)|n)(7.10)This simple result can be used for situations where the system can be represented as a collection of slightlydeformed individual ions.Order of the perturbing termsConsider a fieldB∼1Tesla. The term proportional toBis of the order ofμBB≈10−4eV. On the otherhand the last term in 7.10 will be about 104smaller than that atB∼1Tesla. Thus unless the the firstterm vanishes, it is necessarily much less important.However consider an important case where the first term is zero. The ground state of crystals composed ofions/atoms with filled shells (inert gases and alkali halides) should have zero orbital and and spin angularmomentum. The equation 7.10 says that the energy of the system should increase a little with magneticfield - this means that the response of these should bediamagnetic.This prediction is quantitativelycorrect for inert gas and alkali halide crystals. This phenomena is usually calledLarmor diamagnetism.When the atomic shells are partially filled the second order term arising from theLandScontributionswith a negative sign will come into the picture for the ground state.Whether the (insulating) crystalwill be diamagnetic or paramagnetic depends on the balance between these two competing effects. Theparamagnetic term is attributed toVan Vleck.7.1.1Forces on a magnetic dipoleWe can now prove an interesting result. A small paramagnet cannot float (stably) in a magnetic field,but a small diamagnet can. Let’s analyse the forces on a small dipole and try to see if there can be astable point. The energy of a dipole isU=−m.B(7.11)
7.2. MAGNETIC MOMENT OF AN ISOLATED ATOM: LANDE G-FACTOR & HUND’S RULES123We need to see if the energy has a stable minima. So the question is what is the sign of∇2(−m.B) =?(7.12)For a dipole with a fixed moment (that is not effected by aB), this is∇2U=−bracketleftbigmx∇2Bx+my∇2By+mz∇2Bzbracketrightbig(7.13)But in free space, with no radiation/ time varying electric field we have∇2B=∇∇.B−∇×∇×B=0(7.14)Hence the laplacian of the individual components of the magnetic field are all zero and there is no pointof stable equilibrium. Now if the dipole is itself induced by the magnetic field, then we need to look at apotential energy of the type:∇2U=−∇2χB.B=−χ∇2(B2x+B2y+B2z)(7.15)Now it is left as an exercise to prove the result (using the previous one) that :∇2B.B= 2(|∇Bx|2+|∇By|2+|∇Bz|2)>0(7.16)Hence for a diamagnet (χ <0) we can find stable equilibrium points in a magnetic field, but not forparamagnets. As an interesting aside, since animal bodies are mostly water (which is slightly diamagnetic),one can make a small frog float in strong (about 15-20Tesla) magnetic field!