===== Dec 14 (Mon) ===== ** Responsible party: ** **To go back to the lecture note list, click [[lec_notes]]**\\ **previous lecture note: [[lec_notes_1211]]**\\ **next lecture note: [[lec_notes_1216]]**\\ **Main class wiki page: [[home]]** Please try to include the following * main points understood, and expand them - what is your understanding of what the points were. * expand these points by including many of the details the class discussed. * main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s). * Other classmates can step in and clarify the points, and expand them. * How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture. * wonderful tricks which were used in the lecture.\\ \\ ===Perturbation Theory=== What do we need in perturbation theory? We want to find the energy: E=E^{(0)} + E_{r} + (E_{FS}+E_Z) where E^{(0)} = \frac{-13.6 \mathrm{eV}}{n^2} ===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: H_{fs}=\alpha\vec{S}\vec{L}, where \alpha = \frac{e^2}{4\pi \epsilon_0}\cdot\frac{1}{m^2 c^2 r^3}, like a [[http://en.wikipedia.org/wiki/Magnetic_dipole–dipole_interaction|dipole-dipole interaction in E&M]]. By considering the relativistic version of momentum: p=\frac{mv} {\sqrt{1-(\frac v {c})^2}}. We can derive the relativistic equation for kinetic energy: T=\sqrt{m^2c^4+p^2c^2}-mc^2. With p< (the non-relativistic limit) we get T=\frac {p^2} {2m}-\frac{p^4} {8m^3c^2} This gives us the lowest-order relativistic correction H'_r=H-H^0 with H'_r=\frac{-p^4} {8m^3c^2} With this we can then move to find E^1_r which is equal to: which is equal to: -\frac{1} {8m^3c^2} If p^2\psi=2m(E-V)\psi then the prior equation gives us an <\psi|(E-V)^2|\psi> to deal with. Following Hermitian Operator rules we get: <\psi|E^2|\psi>=E^2 <\psi|EV|\psi>=E \appr E<\frac{1} r> \appr \frac{E} {n^2a} <\psi|V^2|\psi> \appr <\frac{1} {r^2}> \appr \frac{1} {(l+\frac{1} 2)n^3a^2} Putting it all together and we get: E^1_r=-\frac{E_n^2} {2mc^2}[\frac{4n} {l+\frac{1} 2} - 3] Dividing both sides by E_n and we get a relativistic correction of about 2x10^{-5} ===Higher-Order Degeneracy=== Enter Notes Here ===Spin-Orbit Coupling=== Enter Notes Here ------------------------------------------ **To go back to the lecture note list, click [[lec_notes]]**\\ **previous lecture note: [[lec_notes_1211]]**\\ **next lecture note: [[lec_notes_1216]]**\\