===== Oct 07 (Wed) delta-function potential well - Scattering =====
** Responsible party: poit0009, Pluto 4ever ** 

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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 ====

===Double Delta Potential===

We discussed the discussion problem from yesterday (Oct 6).  We have the potential:

<math>V(x)= -\alpha[\delta(x-a)+\delta(x+a)] </math>

We can separate <math>\psi(x)</math> into three regions:

<math>\psi_1(x)</math>, where x < -a

<math>\psi_2(x)</math>, where -a < x < a

<math>\psi_3(x)</math> , where x > a

The wave functions end up being:

<math>\psi_1(x)=Ae^{(kx)}</math>

<math>\psi_2(x)=Be^{(kx)}+Ce^{(-kx)}</math>

<math>\psi_3(x)=De^{(-kx)}</math>

with <math>k^2=-\frac{2mE}{\hbar^2}</math>

as opposed to the situation of <math>k^2=\frac{2mE}{\hbar^2}</math>

which is defined by the complex equation <math>\psi(x)=e^{(-ikx)}</math>.

Normally, this would be the case for E>0 thus as k increases so does the energy. Yet, here we are dealing with E<0 so what ultimately happens is that as k gets larger positively the energy gets larger negatively.

We end up getting:

<math>{\small^+_-} e^{-2ak}=2bk-1</math>, where <math>b=\frac{\hbar^2}{2m\alpha}</math>

Here a specifies how quickly the slope for the first solution decays, and b deals with the depth of the potential well.

There may be 2 values for k, depending on the values of a and b.

If b<a, then there are two possibilities for k.  

If b>a, then only one k exists.  The other solution is for a case when <math>\psi(x)</math> is not normalizable.

This is mainly determined by <math>\Delta\psi=\frac{2m\alpha}{\hbar^2}</math> which is a constant and as such there is only one actual angle for which the change in slopes at the barriers can occur. 

As in the discussion, we have two potentials for the finite well at points -a and a. When the energies for both potential are either positive or negative then the transition between the two barriers is stable creating the new slope for the region -a < x < a.

When one potential is positive and the other negative we have two decaying slopes that join at zero. When these two points are far apart then <math>\Delta\psi</math> agrees with the angle between the change of the slopes at the barriers.

However, if those two points are to close together then the decaying slopes in the region of -a < x < a become exceedingly steep. Therefore, the resulting angle is smaller than <math>\Delta\psi</math>, then <math>\Delta\psi</math> will bend the outside equation till the corresponding angle is achieved. This will cause the function to diverge leave only the one true solution to the equation

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