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Raising Operator (Harmonic Oscillator)

  1. Jun 9, 2004 #1
    This is (another!) question I cannot solve
    The ground state wavefunction for the harmonic oscillator can be written as
    [tex]$\chi _0 = \left( {\frac{\alpha }
    {\pi }} \right)^{\frac{1}
    {4}} \exp \left( {\frac{{ - \alpha x^2 }}
    {2}} \right)$

    where [tex]$\alpha = \sqrt {\frac{{mk}}
    {{\hbar ^2 }}} $

    We are then asked to apply the rasing operator:

    where [tex]$y = \sqrt \alpha x$
    First of all, does this mean replace all alphas, and x's with y's? (Is it even possible to get rid of all x's? I end up with a y/x in there), and then apply
    Is that what the operator is doing? I don't see the significane of the answer. If I apply the lowering operator (d/dy + y) I still get 'an' answer, when I figure I should get something to tell me I can't go lower then the groud state (a zero perhaps, or a mathematical impossibility like division by zero)

    Anyway...if anyone can shed some light on this it would eb much appreciated!

    Edit: I cant figure out division on this board! I swear to god I am doing it right...I will post my operator down here...if someone can point out the latex error I'd love to know:
    {{dy}} - y

    Last edited: Jun 9, 2004
  2. jcsd
  3. Jun 9, 2004 #2
    I swear my equations are printing in the wrong order above. For reference:
    Raising operator - d/dy - y
    where y = sqrt(alpha)*x

    Oh and I get
    [tex]-1/2\,{\frac {{e^{-1/2\,\alpha\,{x}^{2}}} \left( -1+4\,\alpha\,{x}^{2}
    \right) }{x\sqrt [4]{\alpha\,\pi }}}

    Is this the first excited state? Pretty sure I would have made an error somewhere there!
  4. Jun 9, 2004 #3
    Dont worry. Completly missed something here. All makes sesnse now. Can I delete this thread somehow?
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