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--- classes:2009:fall:phys4101.001:lec_notes_1214 [2009/12/15 14:12] – ludeman
+++ classes:2009:fall:phys4101.001:lec_notes_1214 [2009/12/17 23:11] (current) – fitch
@@ Line 18: / Line 18: @@
 \\
-====Fine Structure====
+===Perturbation Theory===
-Fine structure is due to two mechanisms: **relativistic correction** and **spin-orbit coupling**. In other words, a very small perturbation (correction) to the Bohr energies. The equation of which is: <math>H_{fs}=\alpha\vec{S}\vec{L}</math>
+What do we need in perturbation theory?  We want to find the energy:
+<math>E=E^{(0)} + E_{r} + (E_{FS}+E_Z)</math> where <math>E^{(0)} = \frac{-13.6 \mathrm{eV}}{n^2}</math>
+===Fine Structure===
+Fine structure is due to two mechanisms: **relativistic correction** and **spin-orbit coupling**. In other words, a very small perturbation (correction) to the Bohr energies.
+The equation of which is: <math>H_{fs}=\alpha\vec{S}\vec{L}</math>, where <math>\alpha = \frac{e^2}{4\pi \epsilon_0}\cdot\frac{1}{m^2 c^2 r^3}</math>, like a [[http://en.wikipedia.org/wiki/Magnetic_dipole–dipole_interaction|dipole-dipole interaction in E&M]].
+By considering the relativistic version of momentum: <math>p=\frac{mv} {\sqrt{1-(\frac v {c})^2}}</math>. We can derive the relativistic equation for kinetic energy: <math>T=\sqrt{m^2c^4+p^2c^2}-mc^2</math>.
+With <math>p<<mc</math> (the non-relativistic limit) we get <math>T=\frac {p^2} {2m}-\frac{p^4} {8m^3c^2}</math>
+This gives us the lowest-order relativistic correction <math>H'_r=H-H^0</math> with <math>H'_r=\frac{-p^4} {8m^3c^2}</math>
+With this we can then move to find <math>E^1_r</math> which is equal to: <math><H'_r></math> which is equal to: <math>-\frac{1} {8m^3c^2} <p^2\psi|p^2\psi></math>
+If <math>p^2\psi=2m(E-V)\psi</math> then the prior equation gives us an <math><\psi|(E-V)^2|\psi></math> to deal with. Following Hermitian Operator rules we get:
+<math><\psi|E^2|\psi>=E^2</math>
+<math><\psi|EV|\psi>=E<V> \appr E<\frac{1} r> \appr \frac{E} {n^2a}</math>
+<math><\psi|V^2|\psi> \appr <\frac{1} {r^2}> \appr \frac{1} {(l+\frac{1} 2)n^3a^2}</math>
+Putting it all together and we get: <math>E^1_r=-\frac{E_n^2} {2mc^2}[\frac{4n} {l+\frac{1} 2} - 3]</math>  Dividing both sides by <math>E_n</math> and we get a relativistic correction of about <math>2</math>x<math>10^{-5}</math>
+===Higher-Order Degeneracy===
+Enter Notes Here
+===Spin-Orbit Coupling===
+Enter Notes Here

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