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classes:2009:fall:phys4101.001:lec_notes_1012 [2009/10/14 08:16] – x500_spil0049 | classes:2009:fall:phys4101.001:lec_notes_1012 [2009/10/14 08:35] (current) – x500_spil0049 | ||
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We began class by dividing into groups in hope to gain more class participation. | We began class by dividing into groups in hope to gain more class participation. | ||
Then we took a vote and decided to deal with the odd or antisymmetric solutions for a finite square well. | Then we took a vote and decided to deal with the odd or antisymmetric solutions for a finite square well. | ||
- | Therefor, | + | Therefor, |
- | Lets consider the main the main points in setting up solutions for a potential well problem: | + | Lets consider the main the main points in setting up solutions for a potential well problem:\\ |
- | 1st | + | 1st\\ |
- | One must of course know the potential. Where it is located in a two dimensional plane and its amplitude | + | 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 " | ||
+ | 2nd\\ | ||
+ | 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 | ||
The first point to consider is that an odd solution will be antisymmetric across the origin. This is what we focused on today. | The first point to consider is that an odd solution will be antisymmetric across the origin. This is what we focused on today. | ||