Go to the U of M home page
School of Physics & Astronomy
School of Physics and Astronomy Wiki

User Tools

Trace: • uploading • test_page • disposal
classes:2009:fall:phys4101.001:lec_notes_0925

Differences

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

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 12:22] – x500_moore616classes:2009:fall:phys4101.001:lec_notes_0925 [2009/09/28 23:56] (current) – x500_moore616
Line 23: Line 23:
  
   *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>.
  
  
Line 43: Line 43:
 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>
  
  
Line 55: Line 55:
 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>
  
  
Line 67: Line 67:
  
 <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. 
Line 76: Line 76:
 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 
classes/2009/fall/phys4101.001/lec_notes_0925.1254158575.txt.gz · Last modified: 2009/09/28 12:22 by x500_moore616

Page Tools