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Homework Help: Energy eigen value

  1. Feb 23, 2006 #1
    The ground state of a one-dimensional Harmonic oscillator described by the Hamiltonian [itex]H = \frac{p^2}{2m} + \frac{kx^2}{2}[/itex] is of the form, [itex]\psi = Ae^{-ax^2}[/itex]. Determine 'A' and 'a' so that the wavefunction [itex]\psi[/itex] is a normalized eigenstate of the Hamiltonian. What is the energy eigenvalue of the wavefunction?

    Well, I was able to normalize the wavefunction and obtained the value of 'A'.
    [tex]\int_{-\infty}^{\infty}\psi \psi^* dx =1[/tex]

    [tex]A^2\int_{-\infty}^{\infty}e^{-2ax^2}dx =1[/tex]

    [tex]A^2\sqrt{\frac{\pi}{2a}} =1[/tex]

    [tex]A = (\frac{2a}{\pi})^{1/4}[/tex]

    How do I determine 'a'? Any clues to obtain energy eigen value?
     
    Last edited: Feb 24, 2006
  2. jcsd
  3. Feb 24, 2006 #2
    Sorry, I could not correct my errors yesterday. I have rectified the LaTeX typos. Now can someone help me....?
     
  4. Feb 24, 2006 #3

    Hurkyl

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    I would say use the definitions of "eigenstate" and "eigenvalue".
     
  5. Feb 24, 2006 #4
    You mean use the eigenfunction and obtain the eigenvalue?
    [tex]i \hbar \frac{\partial}{\partial t} \psi = \mathcall H \psi[/tex]
     
  6. Feb 24, 2006 #5
    Just apply the harmonic oscillators hamiltonian to the eigenfunction. And I don't think you can determine a but you can set some constraints on it. a just tells you how wide the gaussian is.
     
  7. Feb 24, 2006 #6

    Hurkyl

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    The definitions are that [itex]\psi[/itex] is an eigenfunction of H with eigenvalue [itex]\lambda[/itex] if and only if [itex]H \psi = \lambda \psi[/itex].
     
  8. Feb 24, 2006 #7

    qtp

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    yep you should just be able to operate on the wavefunction with the hamiltonian to obtain the eigenvalues which are the energy values
     
  9. Feb 25, 2006 #8
    Thank you for your time.

    [tex]H = {p^2 \over 2m} + {1\over 2} m \omega^2 x^2[/tex]

    [tex]p = -i \hbar \partial / \partial x[/tex]

    [tex]{-\hbar^2\over 2m}{\partial^2 \psi \over \partial x^2} + {1\over 2} m \omega^2 x^2 \psi = E_n \psi[/tex]

    Looks familiar to me, energy eigen values given by:
    [tex]E_n = \hbar \omega \left(n + {1\over 2}\right)[/tex]
     
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