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First step analytic solution of Schrodinger equation for the Harmonic Oscillator.

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    I wonder if someone could help me to arrive at equation 2.56 by performing the substitutions. Please see the attachment

    2. Relevant equations

    Please see the attachment for this part. and also for the attempt of a solution.
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2009 #2

    kuruman

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    You don't show your work, so I cannot figure out what you did. This is what I would do. There are two substitutions in the original Schrodinger equation. First replace

    [tex]x^{2}=\frac{\hbar}{m \omega}\xi^{2}[/tex]

    Secondly, you need to replace the second derivative in x with the second derivative in ξ. Use the chain rule

    [tex]\frac{d}{dx}=\frac{d \xi}{dx}\frac{d}{d \xi}=\sqrt{\frac{m \omega}{\hbar}}\frac{d}{d \xi}[/tex]

    Repeat to get the second derivative and plug in. It should come out as advertised.
     
  4. Aug 22, 2009 #3
    First, thanks for replying. Well the main question here is why the next happens:

    [tex]\frac{1}{mw}\frac{d^2\psi}{dx^2}=\frac{d^2\psi}{d\xi^2}[\tex]

    Because if you take the second derivative of xi, you get this:

    [tex]\frac{d\xi}{dx}=\sqrt{\frac{m \omega}{\hbar}}[\tex]

    [tex]\Rightarrow\frac{d^2\xi}{dx^2}=0[\tex]

    Because:

    [tex]\sqrt{\frac{m \omega}{\hbar}}=constant[\tex]

    so you can't get there by taking the second derivative of xi or x.

    Sorry i don't know how to make tex to work but i have attached the reply.
     

    Attached Files:

    Last edited: Aug 22, 2009
  5. Aug 23, 2009 #4

    kuruman

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    I cannot see the pdf yet (pending approval) but the LateX code shows that you think that because

    [tex]
    \frac{d}{dx}=\sqrt{\frac{m \omega}{\hbar}}\frac{d}{d \xi}[/tex]

    you think that the second derivative vanishes because

    [tex]
    \sqrt{\frac{m \omega}{\hbar}}= constant[/tex]

    That is not true. Remember that the derivative operates on a function. Let's put one in (I suggest that you do so until you get used to the algebra of operators) and see what happens. From the top

    [tex]\frac{d \psi}{dx}=\sqrt{\frac{m \omega}{\hbar}}\frac{d \psi}{d \xi}[/tex]

    Then taking the derivative with respect to x once more,

    [tex]\frac{d^{2} \psi}{dx^{2}}=\sqrt{\frac{m \omega}{\hbar}}\frac{d}{dx}(\frac{d \psi}{d \xi})=\sqrt{\frac{m \omega}{\hbar}}\sqrt{\frac{m \omega}{\hbar}}\frac{d}{d \xi}(\frac{d \psi}{d \xi})[/tex]

    Note that when you take the derivative once more, a product rule is implied where only one term in the product is constant. Can you finish it now?
     
  6. Aug 23, 2009 #5
    Yes, i got it now, thanks for the help
     
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