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--- classes:2009:fall:phys4101.001:lec_notes_0923 [2009/09/24 22:39] – minor corrections and clarifications yk
+++ classes:2009:fall:phys4101.001:lec_notes_0923 [2009/09/29 19:12] (current) – johnson
@@ Line 1: / Line 1: @@
-===== Sept 23 (Wed) =====
+===== Sept 23 (Wed) stationary state expansion of states =====
 ** Responsible party: joh04684, Aspirin  **
@@ Line 70: / Line 70: @@
     * We may need to do this integral not just for the general case where //m// does not equal 2, but also for where //m// = 2.</del>
   * As we said before, doing the actual integrals is left to the students for an exercise.
+\\
+\\
+== Simple Harmonic Oscillator ==
+  * For a SHO, <math><p^2> = \frac{\hbar\omega m}{2}(2n+1)</math>
+  * Where the kinetic energy is <math><K> = \frac{1}{2m}<p^2></math>
+  * Or <math><K> = \frac{\hbar\omega}{4}(2n+1)</math>, where this is the energy an oscillator has in any state //n//
+    * <math>E_n = (n + \frac{1}{2})\hbar\omega</math> for an oscillator
+  * Here we can see that <math>E_n = \frac{1}{2}K</math> (half of the kinetic energy) so the potential must be the other half, such that
+    * <math><V>_n = \frac{\hbar\omega}{2}(n + \frac{1}{2})</math>
+  * Is this really the case?
+    * <math>V = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2</math>
+    * <math><V> = \frac{1}{2}m\omega^2<x^2>_n</math>
+  * To show that these two are equal, we need to calculate <math><x^2></math>
+    * <math><x^2>_n = \int\psi^*_nx^2\psi_ndx</math>
+      * Here, <math>\psi^*_n</math> are all the <math>\psi_0, \psi_1, \psi_2, etc</math> solutions, but it also might not be some closed-form expression for <math>\psi_n</math>
+    * We can write p and x in terms of the raising and lowering operators:
+      * <math>p = ( )(a_- - a_+)</math>
+      * <math>x = ( )(a_- + a_+)</math>
+== Simple Harmonic Oscillator, Analytical Solution ==
+  * The first step to solving the SHO analytically is to remove dimensions and x and all its constants in terms of one variable:
+    * <math>\xi = ( )x</math>, where the parenthesis contain a combination of <math>m, \hbar, \omega</math>
+    * Next, we need to normalize the energy with hbar
+    * <math>(2n + 1) = K = \frac{2E}{\hbar\omega}</math> or <math> (n + \frac{1}{2}) = k = \frac{E}{\hbar\omega}</math>
+    * Doing this makes the differential equation more simple to work with
+    * <math>\frac{d^2\psi}{d\xi^2} = (\xi^2 - k)\psi</math>
+  * Step two is to consider our variable as it approaches plus or minus infinity (such that x is also approaching these limits)
+    * k is related to energy, so it has to go to large distances to make <math>\xi >> k</math> simplifying things to:
+      *<math>e^{-\frac{\xi^2}{2}}</math>, where we only take the -ve exponent, since as it approaches positive infinity, it is not normalizable.
+  * Step three : Continued on Friday
 \\

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