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.
  

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 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 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>

Enter Notes Here

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