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Help Request QHO series method from Griffiths

  1. Sep 14, 2010 #1
    I Uwould be grateful if anybody can kindly help me with this particular doubt that cropped when I was working through Introduction to Quantum Mchanics by Griffiths ( Griffiths page 66 )

    My question is how is the following obtained :

    aj = C / (J/2)! ( Please read j as subscript of a )

    I would be grateful to anybody who who can clarify my doubt.

    Thanks

    Relativist
     
  2. jcsd
  3. Sep 14, 2010 #2

    George Jones

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    In my copy of Griffiths, this is on pages 53-54.

    What do you get when use j = 2 in

    [tex]a_{j+2} = \frac{2}{j} a_j?[/tex]

    j = 4? j = 6? j = 8?
     
  4. Sep 14, 2010 #3
    Thanks for your reply. But this leads to

    a10 = a2 / 4! when it should be a10 = a2 / 5! according to aj = C / (j / 2)!. Can you please tell me how this can be approximately coorect when we are considering large values of j. I worked for j = 100 and found a2 / 49 ! when it should be a2 / 50.

    Regards,

    Relativist
     
  5. Sep 14, 2010 #4

    George Jones

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    Okay, how about reworking the argument in Griffiths as follows.

    Consider the series

    [tex]h \left( x \right) = \sum^\infty_{j=0} a_j x^j[/tex]

    with

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

    for large [itex]j[/itex]. Now consider

    [tex]e^{x^2} = \sum^\infty_{n=0} \frac{x^{2n}}{n!} = \sum^\infty_{n=0} c_{2n} x^{2n},[/tex]

    so, for large [itex]j[/itex],

    [tex]c_{j+2} = \frac{2}{j} c_j.[/tex]

    Therefore, asymptotically, [itex]h \left( x \right)[/itex] looks like [tex]e^{x^2}[/tex].
     
  6. Sep 14, 2010 #5
    Thanks very much for the clear explanation. I can see it now.

    Thanks once again for your time.

    Regards,

    Relativist
     
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