Differences

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

--- classes:2009:fall:phys4101.001:lec_notes_1106 [2009/11/06 16:51] – x500_liux0756
+++ classes: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  **
@@ Line 61: / Line 61: @@
 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)

User Tools

Differences

Page Tools