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--- classes:2009:fall:phys4101.001:lec_notes_1012 [2009/10/14 07:30] – x500_spil0049
+++ classes:2009:fall:phys4101.001:lec_notes_1012 [2009/10/14 08:35] (current) – x500_spil0049
@@ Line 17: / Line 17: @@
   * wonderful tricks which were used in the lecture.
 =====Main Points =====
+We began class by dividing into groups in hope to gain more class participation.
-The main purpose of today's lecture was to solve for the odd solutions to a finite square well.
+Then we took a vote and decided to deal with the odd or antisymmetric solutions for a finite square well.
-A first point to consider is that an odd solution will be antisymmetric across the origin. This is what we focused on today.
+Therefor,the main purpose of today's lecture was to solve for the odd solutions to a finite square well.\\
+Lets consider the main the main points in setting up solutions for a potential well problem:\\
+st\\
+One must, of course, know the potential. Where it is located in a two dimensional plane and its amplitude.\\
+Next we need to know the energy of the system under consideration. Is E < 0 or E > 0.\\
+With this information one can now divide the problem into regions. That is, both regions outside the potential well and the region in between the potential "walls".\\
+nd\\
+We impose boundary conditions.\\
+ψ and dψ must be continuous at the boundarys\\
+Upon analysis of the potential being an even function we can assume that the solutions are either even or odd. This is advantages b/c we only need to impose boundary conditions  on one side of the well.
+The first point to consider is that an odd solution will be antisymmetric across the origin. This is what we focused on today.
 For the finite square well we choose the axis in the middle of the well instead of the edge of the well like in the infinite square well this is standard convention. As will be seen the algebra is easier because of this convention.
@@ Line 36: / Line 46: @@
 k= <math> \frac{sqrt{-2mE}}{\hbar} </math>\\
 And,\\
-k'= <math> \frac{sqrt{(V_0+E)2m}}{\hbar} </math>\\ b/c E<0
+k'= <math> \frac{sqrt{(V_0+E)2m}}{\hbar} </math> b/c E<0
 \\
 Since we know that k and k' are related to E we can say that with the wavefunctions we have 4 unknowns. At this point we must be careful due to the fact that we are only looking for odd solutions. These are:\\
 <math>\psi_1 (-a)=\psi_2 (-a) </math>\\
-⇒<math>=Ae^{k(-a)}=Be^{ik'(-a)}+Ce^{-ik'(-a)}</math>
+⇒<math>Ae^{k(-a)}=Be^{ik'(-a)}+Ce^{-ik'(-a)}</math>
 \\
 <math>\partial_x\psi_1 (-a)=\partial_x\psi_2 (-a) </math>\\
-⇒<math>=kAe^{k(-a)}=ik(Be^{ik'(-a)}-Ce^{-ik'(-a)})</math>
+⇒<math>kAe^{k(-a)}=ik(Be^{ik'(-a)}-Ce^{-ik'(-a)})</math>
 \\
-At this point we impose only odd solutions boundary conditions:
+At this point we impose only odd solutions boundary conditions:\\
-<math>\psi_2 (0)=0 </math>\\
+<math>\psi_2 (0)=0</math>\\ Due to the symmetry of the potential well and the fact that we are looking for antisymmetric solutions.\\
 ⇒B=C
 \\
 And\\
-<math>\psi_{1} (x)=<math>\psi_{3} (-x)
+<math>\psi_{1}=-\psi_{3}</math>\\
-⇒<math>\psi_{1}=-<math>\psi_{3}
+⇒<math>Ae^{kx}=-Ae^{-kx}</math>
+The remainder of this solution will be dealt with in the following lecture.\\
 **incomplete at this time**
 \\

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