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[Q]Question about harmonic oscilator

  1. Nov 18, 2008 #1
    Hi, Finally! I reached harmonic oscilator! Congratulation!

    Most of all QM text book introduced this formula :

    Time independent energy eigenstate equation is

    [tex] ( - \frac{\hbar^2}{2m} \frac{\partial}{\partial x) + \frac{Kx^2}{2} )\varphi = E\varphi [/tex]

    (1)[itex] \varphi_{xx} = -k^2 \varphi [/itex]

    [itex] \frac{\hbar^2k^2(x)}{2m} = E - \frac{K}{2}x^2 > 0 [/itex]

    We focused classically forbidden domain [itex] x^2 > x_{o}^2, E < \frac{Kx^2}{2} [/itex]

    In this case, kinetic energy is negative, so [itex] \varphi_{xx} = k'^2 \varphi [/itex] [itex] \frac{\hbar^2k'^2}{2m} = \frac{K}{2}x^2 - E > 0 [/itex]

    For asymptotic domain, [itex] Kx^2/2 >> E [/itex]

    (2) [itex] \varphi_{xx} = \frac{mK}{\hbar^2}\varphi = \beta^4x^2\varphi [/itex] where subscript means 2nd differential, [itex] \beta^2 = \frac{mw_{o}}{\hbar} [/itex]

    We let (3) [itex] \epsilon = \beta x[/itex]

    (2) appears as (4) [itex] \varphi_{\epsilon\epsilon} = \epsilon^2 \varphi [/itex]

    If [itex] \epsilon >>1 [/itex] then (2) is approximated to

    (5) [itex] \varphi \approx Aexp(\pm\frac{\epsilon^2}{2}) = Aexp(\pm\frac{(\beta x)^2}{2}) [/itex]

    I have a question. Liboff said (2) become (4) by introducing (3). But If (3) is right, I thought (4) should be [itex] \varphi_{\epsilon\epsilon} = \beta^2\epsilon^2\varphi [/itex]. Is it right?

    And I don't know how to derive (5) from (4). Please lead me.
  2. jcsd
  3. Nov 18, 2008 #2


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    I don't have Liboff, but since you make substitution of variables from x to beta*epsilon, the second derivative on phi w.r.t to x will change..

    [tex]\frac{d^2\phi}{dx^2} \rightarrow \beta ^2 \frac{d^2\phi}{d\epsilon^2}[/tex] (chain rule of calculus)

    so it should be: [tex] \phi_{\epsilon\epsilon} = \epsilon^2 \phi [/tex]
  4. Nov 18, 2008 #3
    Thank you for your help! I'm very pleased with you. But Could you tell me how to apply chain rule of calculus on that formula? I just subsitute [itex] x = \beta\epsilon [/itex] into the x of dominator to get your formula. is it right procedure?

    What is more, Can you give me a answer of second question?
  5. Nov 18, 2008 #4


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    [tex]\frac{d\phi}{dx} = \frac{d\phi}{d\epsilon}\frac{d\epsilon}{dx}[/tex]

    what the second question is about is that that "far" away from the potential, (classical allowed region) wave function must go down as an exponential.

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