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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>
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