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Homework Help: SHO potential in QM

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data
    My QM book says that the standard SHO potential is [tex] m \omega^2 x^2/2 [/tex]. Can someone give me an example of a physical situation that is governed by this potential. It seems rather out of nowhere to me...

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 11, 2007 #2


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    Staff Emeritus
    Science Advisor

    Any situation that is governed by Hookes Law: f=-kx, will have a potential V=kx^2/2
  4. Nov 11, 2007 #3
    Yes but when is k = m \omega x? I guess \omega = g works, but
    omega is usually an angle.
  5. Nov 11, 2007 #4


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    Homework Helper
    Gold Member

    You are probably thinking of the oscillator potential energy using the formula:


    Remember that Omega is the angular frequency of the oscillation. Also, remember that for an oscillator obeying Hooke's Law the angular frequency is:

    [tex]\omega = \sqrt{k/m}[/tex]

    If you solve this for k, you should see where the m*omega^2 is coming from in the book's expression for the potential energy.
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