===== Oct 12 (Mon) Finite Square Well=====
** Responsible party: spillane, physics4dummies ** 

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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.
=====Main Points =====
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.
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:\\
1st\\ 
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".\\
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  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.
So, the well starts at -a and goes to +a and has a depth of V_0 and E<0. For this well we seperate it into three regions so we can evaluate this problem.
Then for each region we have:

|                ^ Region 1            ^ Region 2            ^ Region 3          ^
^ Range          | <math>x<-a</math>    | <math>-a<x<a</math>  | <math>a<x</math>^
^ Wavefunction   | <math>\psi_1 (x)=Ae^{kx}+A'e^{-kx}</math> | <math>\psi_{2}(x)=Be^{ik'x}+Ce^{-ik'x}</math> | <math>\psi_{3} (x)=De^{-kx}+D'e^{kx}</math> ^
Since E < 0 and we are only looking for odd solutions we can simplify our representation of the the waves listed above.
===1st===
We know that, in region 1 as x-> -∞ <math>\psi_1</math>   must go to 0. A' must be zero or else <math>\psi_1</math> would -> ∞. Cant have it.
Also, for the same logic, in region 3 as x-> ∞, <math>\psi_{2}</math> must go to zero D'=0
===2nd===
Where,\\
k= <math> \frac{sqrt{-2mE}}{\hbar} </math>\\
And,\\
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>\partial_x\psi_1 (-a)=\partial_x\psi_2 (-a) </math>\\
⇒<math>kAe^{k(-a)}=ik(Be^{ik'(-a)}-Ce^{-ik'(-a)})</math>
\\
At this point we impose only odd solutions boundary conditions:\\
<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}=-\psi_{3}</math>\\
⇒<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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