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Minor mathematical query from Griffiths [QM]

  1. May 27, 2006 #1
    Hello all,

    I am trying to understand the analytic method for the solution of the (quantum) harmonic oscillator equation.

    [tex]-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^{2}x^{2}\psi = E\psi[/tex]

    Let

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

    [tex]K = \frac{2E}{\hbar\omega}[/tex]

    We consider

    [tex]\psi(\zeta) = h(\zeta)e^{-\zeta^{2}/2}[/tex]

    The equation transforms to,

    [tex]\frac{d^2 h}{d\zeta^2} - 2\zeta\frac{dh}{d\zeta} + (K-1)h = 0[/tex]

    Assuming the power series solution,

    [tex]h(\zeta) = \sum_{j=0}^{\infty}a_{j}\zeta^{j}[/tex]

    we get the recursion formula,

    [tex]a_{j+2} = \frac{2j+1-K}{(j+1)(j+2)}a_{j}[/tex]

    For large j,

    [tex]a_{j+2} \cong \frac{2}{j}a_{j}[/tex]

    So,

    [tex]a_{j+2} \cong \frac{2}{j}\frac{2}{(j-1)}\frac{2}{(j-2)}\ldots\frac{2}{1}a_{1}[/tex]

    which simplifes to

    [tex]a_{j+2} \cong \frac{2^j}{j!}a_{1}[/tex]

    But the solution given is

    [tex]a_{j} = \frac{C}{(j/2)!} = C \frac{1}{(j/2)}}\frac{1}{((j/2)-1)}\ldots\ldots[/tex]

    I think I am missing something here. What am I doing wrong? Also, I know that the [itex](j/2)![/itex] is just notation but what if the "large j" include some possibly odd values? In that case it wouldn't even be defined. Thanks for your help.

    EDIT: 800th post...yay
     
    Last edited: May 27, 2006
  2. jcsd
  3. May 27, 2006 #2

    George Jones

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    Because the index of [itex]a[/itex] is [itex]j + 2[/itex], I think

    should, for odd [itex]j[/itex], actually be something like

    [tex]
    a_{j+2} \cong \frac{2}{j} \frac{2}{ \left(j - 2 \right)} \frac{2}{\left( j - 4 \right)} \ldots \frac{2}{1} a_{1} \cong \frac{1}{j/2 \left( j/2 - 1 \right) \left( j/2 - 2 \right) \ldots 1/2} a_{1}.
    [/tex]

    For real numbers, the Gamma function generalizes the factorial function.

    Regards,
    George
     
    Last edited: May 27, 2006
  4. May 27, 2006 #3

    Thanks again George! I'll look this up.
     
  5. May 27, 2006 #4

    Gokul43201

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    Your approximation fails with all the terms near the end of the product (ie: j not large). And these terms are significantly larger than terms in the beginning, so the error gets quite big. Or am I messing up ?
     
  6. May 27, 2006 #5
    Yes I think you're right...I thought about it that way (interestingly this is obvious if you expand the "factorials"). But Griffiths has used the approximation anyway.
     
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