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classes:2009:fall:phys4101.001:lec_notes_1106 [2009/11/06 17:18] – x500_liux0756classes:2009:fall:phys4101.001:lec_notes_1106 [2009/11/07 21:36] (current) – yk
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-===== Nov 06 (Fri)  =====+===== Nov 06 (Fri) Legendre polynomials, Radial wave equation =====
 ** Responsible party: liux0756, Dagny  **  ** Responsible party: liux0756, Dagny  ** 
  
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 The equation above cannot be solved further before one knows the potential distribution in the system. The equation above cannot be solved further before one knows the potential distribution in the system.
  
-Now consider a 3D infinite square well.+=== 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 goes to infinity.
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 The boundary condition is <math>u(a)=0</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)+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) So the <math>B'coskr</math> term should be 0. (In fact this term is allowed)
classes/2009/fall/phys4101.001/lec_notes_1106.1257549495.txt.gz · Last modified: 2009/11/06 17:18 by x500_liux0756

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