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QM, the convergence of the harmonic oscillator function.

  1. Jan 28, 2015 #1
    1. After finding out that the wave function ##\Psi(z) \sim Ae^{\frac{-z^{2}}{2}}## in the limit of plus or minus infinity Griffiths separates the function into two parts ##\Psi(z)=h(z)e^{\frac{-z^{2}}{2}}##

    My question will be about a certain aspect of the function ##h(z)##

    After solving the ODE series method, one finds a certain recursion relationship for the coefficients of h(z) with an even and odd part.

    As Griffiths notes for coefficients ##a## of this function, very far up (k large), ##a_{k+2} \approx \frac{2}{k} a_{k}##

    The next equation suddenly states the following: ##a_{k} \approx \frac{C}{(k/2)!}##

    I don't understand this step. Let me show you why my reasoning leads me to the wrong conclusion:

    Let's start again from what we know: ##a_{k+2} \approx \frac{2}{k} a_{k}##. Applying the same thing for ##a_{k}## and plugging in will give: ##a_{k+2} \approx \frac{1}{k/2} \frac{1}{k/2 - 1} a_{k-2}##

    Keep doing this and I find that ##a_{k+2}=\frac{B}{(k/2)!}## In the book this where I have k+2 stands k and I can't figure out why it's more correct than what I have here.


     
  2. jcsd
  3. Jan 28, 2015 #2

    BvU

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    Hello again,

    I'm pretty convinced you won't be pleased with this, but hear (read) me out... :)

    What matters here isn't the exact value of this ak (or ak+2, for that matter), but the ratio of these coefficients for really big k.

    Anything to do with the precise value can be swept in what Griffiths calls C ("for some constant C" -- anything you don't want to be bothered with: throw it in there. As long as it doesn't keep growing with k you're fine).

    Mind you, you want to be way above j = K/2 before you come even close to aj+2 / aj ##\ \approx\ ## 2/j !

    (I find I'm unconsciously switching to Griffiths' indices j now, sorry..)


    And even after all these years, I find it almost miraculous that from this kind of reasoning one is simply forced to conclude that K can NOT assume arbitrary values, but MUST be exactly equal to 2j + 1 for some j, however big. A hair difference and the power series runs away like ##e^{x^2}##. Awesome !
     
  4. Jan 29, 2015 #3
    You were right, I was hoping for having made a stupid reasoning mistake or having overlooked something. The real reason seems to be less elegant than I would have liked but what can you do. Nothing to do with your answer, it was great, thanks.
     
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