Go to the U of M home page
School of Physics & Astronomy
School of Physics and Astronomy Wiki

User Tools

Trace: • quiz_1_1002 • lec_notes_1026 • lec_notes_1005 • lec_notes_1007 • lec_notes_1106
classes:2009:fall:phys4101.001:lec_notes_1106

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
classes:2009:fall:phys4101.001:lec_notes_1106 [2009/11/06 16:47] – x500_liux0756classes:2009:fall:phys4101.001:lec_notes_1106 [2009/11/07 21:36] (current) – yk
Line 1: Line 1:
-===== Nov 06 (Fri)  =====+===== Nov 06 (Fri) Legendre polynomials, Radial wave equation =====
 ** Responsible party: liux0756, Dagny  **  ** Responsible party: liux0756, Dagny  ** 
  
Line 55: Line 55:
 Define <math> R(r)=\frac{u(r)}{r}</math>, then the differential equation is transformed from <math>R(r)</math> to <math>u(r)</math>: Define <math> R(r)=\frac{u(r)}{r}</math>, then the differential equation is transformed from <math>R(r)</math> to <math>u(r)</math>:
  
 +<math> -\frac{\hbar^2}{2m} \frac{d^2u}{dr^2}+[V+\frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}]u=Eu </math>
 +
 +This equation is similar to 1D Schrodinger equations discussed in Chapter 2.
 +
 +The equation above cannot be solved further before one knows the potential distribution in the system.
 +
 +=== Example ===
 +
 +Since we cannot go much further without specifying potential energy, now consider a 3D infinite square well.
 +
 +When <math>r>a</math>, the potential goes to infinity.
 +
 +When <math>r<a</math>, the potential is 0.
 +
 +The radial differential equation is:
 +
 +<math>\frac{d^2u}{dr^2}-[\frac{l(l+1)}{r^2}-k^2]u=0</math>
 +
 +where <math>k^2=\frac{2mE}{\hbar^2} </math>
 +
 +The solution is Bessel function <math>j_l(kr)</math>
 +
 +Now take a look at the 0th order solution.
 +
 +When <math>l=0</math>, the differential equation is <math>\frac{d^2u}{dr^2}=-k^2u</math>
 +
 +The solution is: <math>u(r)=Ae^{ikr}+Be^{-ikr}=A'sinkr+B'coskr</math>
 +
 +The boundary condition is <math>u(a)=0</math>
 +
 +At r=0, if u is not 0, then <math>R=\frac{u}{r}</math> will go to infinity. So <math>u(0)=0</math> should be satisfied. (However, this is not quite correct because even if the value of wave function is infinite, the wave function may still be able to be normalized: <math>\int_0^a \frac{1}{r^2}dV = \int_0^a \frac{1}{r^2} r^2 sin\theta dr d\theta d\phi=\int_0^a sin\theta dr d\theta d\phi</math> is normalizable - finite)
 +
 +So the <math>B'coskr</math> term should be 0. (In fact this term is allowed)
  
  
classes/2009/fall/phys4101.001/lec_notes_1106.1257547642.txt.gz · Last modified: 2009/11/06 16:47 by x500_liux0756

Page Tools