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classes:2009:fall:phys4101.001:lec_notes_1204 [2009/12/06 22:25] – x500_porte210classes:2009:fall:phys4101.001:lec_notes_1204 [2009/12/07 12:04] (current) – yk
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 Yuichi put up a projection of the Clebsch Gordon Tables and asked the class to find the coefficients for |j,j_z> = |3/2,1/2>, at which point Joe asked for clarification on how to know when to look at the rows or the columns of the table, and for what information, since they are a little tricky to work with. Yuichi put up a projection of the Clebsch Gordon Tables and asked the class to find the coefficients for |j,j_z> = |3/2,1/2>, at which point Joe asked for clarification on how to know when to look at the rows or the columns of the table, and for what information, since they are a little tricky to work with.
  
-Around this point in the lecture, Yuichi wrote Spin-Orbit Coupling on the board, but the reason why remains a little unclear to me. +Around this point in the lecture, Yuichi wrote Spin-Orbit Coupling on the board, but the reason why remains a little nuclear to me. 
  
-Another valid question was raised by Kirby and Kate when they noticed negative representations of probability coefficients in the table and Yuichi showed us that when the functions are expressed properly in the Bra and Ket notation, they do not result with the negative inside the square root: +Another valid question was raised by Kirby and Kate when they noticed negative representations of probability coefficients in the table and Yuichi showed us that when the functions are expressed properly in the Bra and Ket notation, they do not result with the negative inside the square root:\\ 
-<math>|3/2,1/2> =  sqrt(3/5)|1>|-1/2> + sqrt(-2/5)|0>|1/2></math>// +<math>|3/2,1/2> =  sqrt(3/5)|1>|-1/2> + sqrt(-2/5)|0>|1/2></math>\\ 
-but instead:// + 
-<math>|3/2,1/2> =  sqrt(3/5)|1>|-1/2> - sqrt(2/5)|0>|1/2></math>//+but instead:\\ 
 +<math>|3/2,1/2> =  sqrt(3/5)|1>|-1/2> - sqrt(2/5)|0>|1/2></math>\\
 with the negative outside. with the negative outside.
  
-Yuichi then asked us to calculate the coefficients of the different states if we had used a plus rather than a minus and Nicole obliged.+Yuichi then asked us to calculate the coefficients of the different states if we had used a plus rather than a minus and Nicole obliged by (I'm a little fuzzy on this) decomposing the parts of the function corresponding to l and s.
  
  
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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. 
 + 
 +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>.
  
-(<math>P(Theta Psi)=|PSI|^2</math>)+<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>
  
-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:+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)d(\cos\theta)=[\int P(\theta \psi)d\phi]d(\cos\theta)</math>
  
-(<math>P(Theta Psi)dOmega)=P(Theta Psi)dCos(Theta)dPsi</math>)\\ +When <math>d(\cos\theta)</math> is expanded to be <math>\sin\theta d\theta</math>,  
-(<math>P'(Theta)=INTEGRATE(P(Theta Psi)d(Psi)d(Cos(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//, ... 
-(<math>P'(Theta')dTheta=(P(Theta,Psi))^2Sin(Theta)</math>)\\+
  
  
 ====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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 **To go back to the lecture note list, click [[lec_notes]]**\\ **To go back to the lecture note list, click [[lec_notes]]**\\
 **previous lecture note: [[lec_notes_1202]]**\\ **previous lecture note: [[lec_notes_1202]]**\\
 **next lecture note: [[lec_notes_1207]]**\\ **next lecture note: [[lec_notes_1207]]**\\
classes/2009/fall/phys4101.001/lec_notes_1204.1260159933.txt.gz · Last modified: 2009/12/06 22:25 by x500_porte210

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