| Both sides previous revisionPrevious revisionNext revision | Previous revision |
| classes:2009:fall:phys4101.001:lec_notes_1111 [2009/11/16 15:43] – ludeman | classes:2009:fall:phys4101.001:lec_notes_1111 [2009/11/19 10:21] (current) – ludeman |
|---|
| \\ | \\ |
| |
| ====Final Words on Radial Equation==== | ====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> and for r<a: <math>u"=-k^2u</math> | 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] |
| |
| ------------------------------------------ | ------------------------------------------ |