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classes:2009:fall:phys4101.001:lec_notes_0925

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classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 12:16] – x500_moore616classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 23:56] (current) – x500_moore616
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   *Use dimensionless form of DE   *Use dimensionless form of DE
-<math>let \xi = \sqrt{\frac{m\omega}{\hbar}} x </math> +let <math> \xi = \sqrt{\frac{m\omega}{\hbar}} x </math> 
-and <math> K=\frac{2E}{\hbar\omega}.  (C) </math>+and <math> K=\frac{2E}{\hbar\omega}. \ \ \ \ \ \ \ \  (C) </math>
  
 Then we can use the dimensionless form of the Schrodinger Then we can use the dimensionless form of the Schrodinger
 <math> \frac{\partial^2 \psi}{\partial x^2}=(\xi^2-K)\psi </math> <math> \frac{\partial^2 \psi}{\partial x^2}=(\xi^2-K)\psi </math>
  
-We can think of <math> \xi </math> as approximately <math> x </math> and also <math> \psi </math> as approximately <math> e^(\frac{-1}{2} \xi^2)  </math>.+We can think of <math> \xi </math> as approximately <math> x </math> and also <math> \psi </math> as approximately <math> e^{\frac{-1}{2} \xi^2}  </math>.
  
  
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 Differentiate and then Schrodinger's equation becomes Differentiate and then Schrodinger's equation becomes
  
-<math> \frac{\partial^2h(\xi)}{\partial \xi^2}=-2h\xi(\xi)+(K-1)(h(\xi)=0 </math> (A)+<math> \frac{\partial^2h(\xi)}{\partial \xi^2}=-2h\xi(\xi)+(K-1)(h(\xi)=0 \ \ \ \ \ \ \ \ (A) </math>
  
  
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 Differentiate once more: Differentiate once more:
  
-<math> \frac{\partial^2h}{\partial\xi^2}=\sum (j+1)(j+2)a_j+2\xi^j </math>+<math> \frac{\partial^2h}{\partial\xi^2}=\sum (j+1)(j+2)a_{j+2}\xi^j </math>
  
  
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 <math>  <math> 
-a_j+2 = \frac{2j+1-K)a_j}{(j+1)(j+2)} </math>+a_{j+2} = \frac{2j+1-K)a_j}{(j+1)(j+2)} </math>
  
 Now all we need to know is <math> a_0 </math> and <math> a_1 </math> and we can find all a.  Now all we need to know is <math> a_0 </math> and <math> a_1 </math> and we can find all a. 
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 This is good, but not all the solutions that are found are normalizable.  For example, at very large j, the formula is  This is good, but not all the solutions that are found are normalizable.  For example, at very large j, the formula is 
  
-<math> a_(j+2) \approx \frac{2a_j}{j}. </math>+<math> a_{j+2} \approx \frac{2a_j}{j}. </math>
  
 Then the solution is  Then the solution is 
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 I'm not going to include this.  He decided to leave it for the "motivated student."  Basically it relates the Hermite polynomials to the ladder operator <math>a_+</math>. I'm not going to include this.  He decided to leave it for the "motivated student."  Basically it relates the Hermite polynomials to the ladder operator <math>a_+</math>.
  
-== conclusion == +== Conclusion == 
-Okay, hopefully between our two sets of notes you will be okay.  Sorry I took so long to make my contribution.  Peace.  EE.+Okay, hopefully between our two sets of notes you will be okay.  Sorry I took so long to make my contribution.   
 + 
 +The difference between the two methods, as far as I can tell, is that one is simple (ladder operators), but are only convenient for the first few //n// values.  The analytical solution involving Hermite polynomials is a little more complicated, especially in it's derivation, but for large n can save a lot of time because you don't have to find all the previous solutions (smaller-n solutions). 
 + 
 +Peace.  EE.
  
  
classes/2009/fall/phys4101.001/lec_notes_0925.1254158183.txt.gz · Last modified: 2009/09/28 12:16 by x500_moore616

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