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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.
  

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>

The steps in solving are similar to that of the 1-D SHO

1. Introduce the dimensionless variable: <math>\rho</math> as <math>\rho</math>→0, <math>u(\rho)</math> ~ <math>\rho^{l+1}</math> and as <math>\rho</math>→<math>\infty</math>, <math>u(\rho)</math> ~ <math>e^{\rho^}</math> The second terms fall off at the boundary condition

2. Introduce a test function <math>v(\rho)</math> to peel of the asymptotic behavior by using the equation: <math>u(\rho)=\rho^{l+1}e^{-\rho}v(\rho)</math>

As <math>\rho</math> → 0, <math>v(\rho)=0</math> and as <math>\rho</math> →<math>\infty</math>, <math>v(\rho)=0</math>

3. Use power series to evaluate <math>v(\rho)=\sum^{\infty}_{j=0} C_j\rho^j</math>

4. Differentiate twice <math>v“(\rho)=\sum j(j+1)C_{j+1}\rho^{j-1}</math>

5. Determine recursion formula <math>C_{j+1}=\frac{2} {(j+1)} C_j</math> Therefore <math>v(\rho)=C_0e^{2\rho}</math>

6. Replace test function gives <math>u(\rho)=C_0\rho^{l+1}e^{\rho}</math>. This requires that <math>C_{(j_{max}+1)}=0</math> and <math>n=j_{max}+l+1</math>

Therefore, <math>\rho_0=2n</math> and the allowed energies of the Hydrogen Atom are in concurrence with Eq. [4.70]

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