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classes:2009:fall:phys4101.001:lec_notes_1111

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Nov 11 (Wed) Radial wave functions

Responsible party: Green Suit, Jake22

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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.


Final Words on the Radial Equation

For bound state, E<0 with l=0, equation [4.37] reduces from 3-D to 1-D with <math>=-k^2=\frac{2m(E+V_0)} \hbar^2</math>

Then for r<a: <math>u“=-k^2u</math> Therefore <math>u®=Asin(kr)</math>

And for r>a: <math>u”=k^2u</math> Therefore <math>u®=De^math_frac_d_dru_1(a)={\frac d {dr}}u_2(a)</math> that gives <math>{\kappa}Acos(ka)=-{\kappa}De^{{-\kappa}a}</math>

Divide Eq1/Eq2 = <math>{\kappa}cot(ka)=-\kappa</math>. Setting <math>cot(ka)=\frac{-\kappa} k</math> we get <math>sqrt{(\frac{z_0} z)^2-1}</math>

The key point here is that tan(z) doesn't exist in 3-D. In descriptive terms, if the well is to shallow there will not be a solution. There needs to be a wide enough and deep enough well for a bound state.

Also there are no allowed energies at n=0. Recall that for cot(z) the lowest allowed energies are <math>\frac{\pi} 2<z<\pi</math>


To go back to the lecture note list, click lec_notes
previous lecture note: lec_notes_1109
next lecture note: Quiz_3_1113
next lecture note: lec_notes_1116

classes/2009/fall/phys4101.001/lec_notes_1111.1258410016.txt.gz · Last modified: 2009/11/16 16:20 by ludeman