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--- classes:2009:fall:phys4101.001:lec_notes_1207 [2009/11/23 22:15] – created yk
+++ classes:2009:fall:phys4101.001:lec_notes_1207 [2009/12/15 23:58] (current) – ely
@@ Line 1: / Line 1: @@
 ===== Dec 07 (Mon)  =====
-** Responsible party: John Galt, Dark Helmet **
+** Responsible party: John Galt, Dark Helmet, Esquire **
 **To go back to the lecture note list, click [[lec_notes]]**\\
@@ Line 8: / Line 8: @@
 **Main class wiki page: [[home]]**
-Please try to include the following
+====Chapter 6: Time Indepent Perturbation Theory====
-  * main points understood, and expand them - what is your understanding of what the points were.
+We do it becuase it is a useful tool
-    * 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.\\
-\\
+Shroedinger's equation is the must fundamental tool for QM.
+We take solutions and eigenstates/eigenvectors to get energy levels
+===Non-Degenerate Case===
+This is the simplest case
+single energy->single equation
+<math> H_0|\psi_n^{(0)}>=E_n^{(0)}|\psi_n^{(0)}> </math>
+<math> <\psi_m^{(0)}|\psi_n^{(0)}>=\delta_{mn} </math>
+<math> H=H_0 +\lambda H' </math> ⇒ <math> H|\psi_n>=E_n|\psi_n> </math>
+The goal is to seek an approximation of this new Hamiltonian expression. Specifically we want...
+<math> E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^{2} E_n^{(2)} </math>
+We define  <math> <\psi_n^{(0)}|\psi_n^{(1)}>=<\psi_n^{(0)}|\psi_n^{(2)}>=0 </math>
+A Fourier expansion can be used to express <math> |\psi_n^{(1)}>=\Sigma C_{mn}|\psi_n^{(0)}> </math> where m≠n
+Plugging this into the new Hamiltion yields
+<math> (H_0+\lambda H')(|\psi_n^{(0)}>+\lambda|\psi_n^{(1)}>)=(E_n^{(0)}+\lambda E_n^{(1)})(|\psi_n^{(0)}>+\lambda|\psi_n^(1)>) </math>
+<math> H_0|\psi_n^{(0)}>+\lambda(H'|\psi_n^{(0)}>+H_0|\psi_n^{(1)}>)=E_n^{(0)}|\psi_n^{(0)}>+\lambda(E_n^{1}|\psi_n^{(0)}>+E_n^{(0)}|\psi_n^{(1)}>) </math>
+<math> H'|\psi_n^{(0)}>+H_0|\psi_n^{(1)}>=E_n^{(1)}|\psi_n^{(0)}>+E_n^{(0)}|\psi_n^{(1)}> </math>
+Now using the Fourier expansion expression
+<math> H'|\psi_n^{(0)}>+\Sigma C_{nm}H_0|\psi_m^{(0)}>=E_n^{(1)}|\psi_n^{(0)}>+E_n^{(0)}\Sigma C_{nm}|\psi_m^{(0)}>  </math>
+<math> H'|\psi_n^{(0)}+\Sigma C_{nm}E_m^{(0)}|\psi_m^{(0)}>=E_n^{(1)}|\psi_n^{(0)}>+E_n^{(0)}\Sigma C_{nm}|\psi_m^{(0)}> </math>
+Using this, one can find an expression for the expectation of the new Hamiltonian as follows
+<math> <\psi_n^{(0)|H'|\psi_n^{(0)}>+\Sigma C_{nm}E_m^{(0)}<\psi_n^{(0)}|\psi_m^{(0)}>=E_n^{(1)}<\psi_n^{(0)}|\psi_n^{(0)}>+E_n^{(0)}\Sigma C_{nm}<\psi_n^{(0)}|\psi_m^{(0)}> </math>
+<math><\psi_n^{(0)|H'|\psi_n^{(0)}>=E_n^{(1)} </math>
+Now one can introduce a new parameter l≠n but l can equal m and show
+<math><\psi_l^{(0)|H'|\psi_n^{(0)}>+\Sigma C_{nm}E_m^{(0)}<\psi_l^{(0)}|\psi_m^{(0)}>=E_n^{(1)}<\psi_l^{(0)}|\psi_n^{(0)}>+E_n^{(0)}\Sigma C_{nm}<\psi_l^{(0)}|\psi_m^{(0)}> </math>
+<math><\psi_l^{(0)|H'|\psi_n^{(0)}>+C_{nl}E_l^{(0)}=C_{nl}E_n^{(0)}  </math>
+⇒<math>C_{nl}=<\psi_l^{(0)|H'|\psi_n^{(0)}>/(E_n^{(0)}-E_l^{(0)}) </math>
+⇒<math>E_n^{(2)}=\Sigma|<\psi_l^{(0)|H'|\psi_n^{(0)}>|^2/(E_n^{(0)}-E_l^{(0)}) </math>
+This was all i had for notes as well-Dark Helmet
 ------------------------------------------

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