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Linear Harmonic Oscilator - QM

  1. Jan 13, 2007 #1
    Just a quickie:

    A particle is in the first excited Eigenstate of energy E corresponding to the one dimensional potential V(x) = [tex]\frac{Kx^2}{2}[/tex]. Draw the wavefunction of this state, marking where the particles KE is negative.

    Now my question.

    The first excited state will be n=1 correct? The first excited state is not the ground state under a different name is it?

    So if it is n=1, then the wavefunction will look like a sin wave? And the KE will be negative on the left hand side of the sin wave (i.e. where a graph of sin(x) will be negative.)?

    Ta guys.
     
  2. jcsd
  3. Jan 13, 2007 #2

    Dick

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    I think you'll have to do more than just guess. But here's a hint. The first excited state is indeed not the ground state. It has one node in the wavefunction. The ground state has zero.
     
  4. Jan 13, 2007 #3
    Well I've drawn the right form of the wavefunction at least. Whether or not I've correctly identified where the KE is negative I don't know. Back to the text book I go!
     
  5. Jan 13, 2007 #4

    Dick

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    Great idea! Once you have the wavefunctions remember KE is an operator on wavefunctions.
     
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