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classes:2009:fall:phys4101.001:lec_notes_1111

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classes:2009:fall:phys4101.001:lec_notes_1111 [2009/11/07 21:40] – created ykclasses:2009:fall:phys4101.001:lec_notes_1111 [2009/11/19 10:21] (current) – ludeman
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-===== Nov 11 (Wed)  =====+===== Nov 11 (Wed) Radial wave functions =====
 ** Responsible party: Green Suit, Jake22 **  ** Responsible party: Green Suit, Jake22 ** 
  
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 \\ \\
  
 +====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(r)=Asin(kr)</math>
  
 +And for r>a: <math>u"=k^2u</math> Therefore <math>u(r)=De^{{-\kappa}r}</math>
  
 +We have 3 unknowns(two equations and one normalization). With boundary condition at "a" we get Eq1: <math>u_1(a)=u_2(a)</math> that gives <math>Asin(ka)=De^{{-\kappa}a}</math>
  
 +And Eq2: <math>{\frac d {dr}}u_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>
 +
 +====Radial Wave Function for the Hydrogen Atom====
 +//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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classes/2009/fall/phys4101.001/lec_notes_1111.1257651642.txt.gz · Last modified: 2009/11/07 21:40 by yk

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