PROBLEM Show using standard perturbation theory that 79 that the shift in the

Problem show using standard perturbation theory that

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 due to the presence of the magnetic field δE n = μ B ( n | B . parenleftbigg L + g 0 S planckover2pi1 parenrightbigg | n ) + μ 2 B summationdisplay n negationslash = n vextendsingle vextendsingle vextendsingle ( n | B . parenleftBig L + g 0 S planckover2pi1 parenrightBig | n ) vextendsingle vextendsingle vextendsingle 2 E n E n + e 2 8 m B 2 ( n | summationdisplay i ( x 2 i + y 2 i ) | n ) (7.10) This simple result can be used for situations where the system can be represented as a collection of slightly deformed individual ions. Order of the perturbing terms Consider a field B 1Tesla. The term proportional to B is of the order of μ B B 10 4 eV. On the other hand the last term in 7.10 will be about 10 4 smaller than that at B 1Tesla. Thus unless the the first term vanishes, it is necessarily much less important. However consider an important case where the first term is zero. The ground state of crystals composed of ions/atoms with filled shells (inert gases and alkali halides) should have zero orbital and and spin angular momentum. The equation 7.10 says that the energy of the system should increase a little with magnetic field - this means that the response of these should be diamagnetic . This prediction is quantitatively correct for inert gas and alkali halide crystals. This phenomena is usually called Larmor diamagnetism . When the atomic shells are partially filled the second order term arising from the L and S contributions with a negative sign will come into the picture for the ground state. Whether the (insulating) crystal will be diamagnetic or paramagnetic depends on the balance between these two competing effects. The paramagnetic term is attributed to Van Vleck . 7.1.1 Forces on a magnetic dipole We 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 a stable point. The energy of a dipole is U = m . B (7.11)
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7.2. MAGNETIC MOMENT OF AN ISOLATED ATOM: LANDE G-FACTOR & HUND’S RULES 123 We 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 a B ), this is 2 U = bracketleftbig m x 2 B x + m y 2 B y + m z 2 B z bracketrightbig (7.13) But in free space, with no radiation/ time varying electric field we have 2 B = ∇∇ . B −∇×∇× B = 0 (7.14) Hence the laplacian of the individual components of the magnetic field are all zero and there is no point of stable equilibrium. Now if the dipole is itself induced by the magnetic field, then we need to look at a potential energy of the type: 2 U = −∇ 2 χ B . B = χ 2 ( B 2 x + B 2 y + B 2 z ) (7.15) Now it is left as an exercise to prove the result (using the previous one) that : 2 B.B = 2 ( |∇ B x | 2 + |∇ B y | 2 + |∇ B z | 2 ) > 0 (7.16) Hence for a diamagnet ( χ < 0) we can find stable equilibrium points in a magnetic field, but not for paramagnets. 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!
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