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classes:2009:fall:phys4101.001:lec_notes_1204

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classes:2009:fall:phys4101.001:lec_notes_1204 [2009/12/07 01:37] – czhangclasses:2009:fall:phys4101.001:lec_notes_1204 [2009/12/07 12:04] (current) – yk
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 (In this section, the code proved difficult, so fulls words are used to replace simpler notation. (In this section, the code proved difficult, so fulls words are used to replace simpler notation.
  
-<math>|1></math>  and <math> |0> </math> can be replaced by the spherical harmonic functions (<math>Y_{1}^1 (theta psi)</math>) and (<math>Y_{1}^{0}(theta psi)</math>), respectively.+<math>|1></math>  and <math> |0> </math> can be replaced by the spherical harmonic functions (<math>Y_{1}^1 (\theta, \phi)</math>) and (<math>Y_{1}^{0}(\theta, \phi)</math>), respectively.
  
-(<math>P(\theta \psi)=|PSI|^2</math>)+Previously, when the wave function was a function of //x//, <math>P(\theta, \phi)=|\psi|^2</math> represented the probability density as a function of //x//.  We can do the same for wave function as a function of <math>\theta</math> and <math>\phi</math>.  Then the probability for finding a particle in the solid angle is given by <math>P(\theta, \phi)d\Omega)=|\psi(\theta, \psi)|^2d(\cos\theta) d\phi</math>.
  
-And using these, Yuichi said the probability at any angle can be found, and I think he was referring to position probability, but this is still a little nuclear to me, and he continued:+<del>And using these, Yuichi said the probability at any angle can be found, and I think he was referring to position probability, but this is still a little unclear to me, and he continued:</del>
  
-<math>P(\theta \psi)d\Omega)=P(\theta \psi)dcos(\theta)d\psi</math>\\ +If you are only interested in the probability density as a function of <math>\theta</math>, integrating over <math>\theta</math>, and  
-<math>P'(\theta)=\int P(\theta \psi)d(\psi)d(cos(\theta))</math>\\ +<math>P'(\theta)d(\cos\theta)=[\int P(\theta \psi)d\phi]d(\cos\theta)</math> 
-<math>P'(\theta')d\theta=(P(\theta,\psi))^2sin\theta</math>\\+ 
 +When <math>d(\cos\theta)</math> is expanded to be <math>\sin\theta d\theta</math>,  
 +<math>P''\theta d\theta=\int d\phi |\psi(\theta,\phi)|^2\sin\theta d\theta</math> which shows that the probability density absorbs the <math>\sin\theta</math> into itself.  //i.e.// <math>P''\theta =\int d\phi |\psi(\theta,\phi)|^2\sin\theta </math>. This is a bit trickier than the previous case involving Cartesian coordinates, //x//, ... 
  
  
 ====Important Announcement==== ====Important Announcement====
  
-Only material covered up to today, December 4th, will be covered on the third midterm and the final. Chapter 5 is still an interesting chapter to look in to, but its material will appear on neither the third midterm nor the final.+Only material covered up to today, December 4th, will be covered on the third midterm and the final. Chapter 5 is still an interesting chapter to look in to, but its material will appear on neither the last (fourth) midterm nor the final.  **//Yuichi's correction// Materials in Chap 5 through (including) section 5.2.1 could appear in the final exam, though not in quiz 4. **
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classes/2009/fall/phys4101.001/lec_notes_1204.1260171449.txt.gz · Last modified: 2009/12/07 01:37 by czhang

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