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classes:2009:fall:phys4101.001:lec_notes_1007 [2009/10/05 19:57] – created ykclasses:2009:fall:phys4101.001:lec_notes_1007 [2009/10/07 18:55] (current) – kuehler
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-===== Oct 05 (Mon) =====+===== Oct 07 (Wed) delta-function potential well - Scattering =====
 ** Responsible party: poit0009, Pluto 4ever **  ** Responsible party: poit0009, Pluto 4ever ** 
  
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 ==== Main Points ==== ==== 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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classes/2009/fall/phys4101.001/lec_notes_1007.1254790663.txt.gz · Last modified: 2009/10/05 19:57 by yk

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