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classes:2009:fall:phys4101.001:lec_notes_0918 [2009/09/20 20:03] ykclasses:2009:fall:phys4101.001:lec_notes_0918 [2009/09/20 20:11] (current) yk
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-<math> \hat{H}\Psi(x)=E\Psi(x),</math>+<math> \hat{H}\psi(x)=E\psi(x),</math>
  
 ==Simple Harmonic Oscillator:== ==Simple Harmonic Oscillator:==
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 <math> (x^2-y^2) = (x+y)(x-y) (1*)</math> <math> (x^2-y^2) = (x+y)(x-y) (1*)</math>
  
-except in the Hamiltonian case, we are looking at something like this+except in the Hamiltonian case, we are looking at something like this (ignoring various factors involving mass, Planck constant, angular frequency, ...
  
 <math> (x^2-\partial^2 x)=(x+\partial x)(x-\partial x).  (2*)  </math> <math> (x^2-\partial^2 x)=(x+\partial x)(x-\partial x).  (2*)  </math>
  
-Only there is one key difference.  The Hamiltonian is an operator and does not behave like a normal variable.  In equation (1*), <math> yx=xy.</math>  But in equation (2*), <math> (x)(\partial x)\not= (\partial x)(x).  </math>.+Only there is one key difference.  The Hamiltonian is an operator and does not //quite// behave like a normal variable.  In equation (1*), <math> yx=xy.</math>  But in equation (2*), <math> (x)(\partial x)\not= (\partial x)(x).  </math>.
  
-To deal with this difference, we introduce the idea of raising and lowering operators <math> (a_-,a_+) </math> They are called this because you can use them to "climb up" (raise) the "ladder of solutions" and vice versa.  We also introduce the canonical commutation relation <math> [x,p}=i\hbar. </math>  Using these, we come up with the equation +To deal with this difference, //we will correct for the difference in the two sides of the inequality above.//  We //also// introduce the idea of raising and lowering operators <math> (a_-,a_+) </math> They are called this because it turns out that you can use them to "climb up" (raise) the "ladder of solutions" and vice versa.  We also introduce the canonical commutation relation <math> [x,p]=i\hbar. </math>  Using these, we come up with the equation 
 <math> [a_-,a_+]=1.  </math>  <math> [a_-,a_+]=1.  </math> 
  
  With these, the Hamiltonian can be written as  With these, the Hamiltonian can be written as
- <math> H=\hbar\omega(a_+,a_- + 1/2) </math>+ <math> H=\hbar\omega(a_+a_- + 1/2) </math>
  
  Which makes the Schrodinger equation   Which makes the Schrodinger equation 
-<math> \hbar\omega(a_-,a_+ (\+/-) (1/2))\psi=E\psi </math>+<math> \hbar\omega(a_\mp a_\pm \pm (1/2))\psi=E\psi </math>
  
-With this, we claim that if <math> \psi </math> satisfies the Schrodinger equation with energy E, then <math> a_+\psi </math> satisfies the Schrodinger equation with energy <math> (E+\hbar\omega).  </math>  We can use this result/assumption to find new solutions for higher and lower energies.  +With this, we claim that if <math> \psi </math> satisfies the Schrodinger equation with energy E, then <math> a_\pm\psi </math> satisfies the Schrodinger equation with energy <math> (E\pm\hbar\omega).  </math>  We can use this result/assumption to find new solutions for higher and lower energies.  
  
 This works well, but a few things must be noted.  Ladder operators can not be used universally.  Also, there is no guarantee that each solution will be normalizable so we define a "lowest rung" to be  This works well, but a few things must be noted.  Ladder operators can not be used universally.  Also, there is no guarantee that each solution will be normalizable so we define a "lowest rung" to be 
classes/2009/fall/phys4101.001/lec_notes_0918.1253495029.txt.gz · Last modified: 2009/09/20 20:03 by yk