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Variational Principle of 3D symmetric harmonic oscillator

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the following trial function:
    [itex] \Psi=e^{-(\alpha)r}[/itex]
    to estimate the ground state energy of the central potential:
    [itex]V(r)=(\frac{1}{2})m(\omega^{2})r^{2}[/itex]

    3. The attempt at a solution

    Normalizing the trial wave function (separating the radial and spherical part):
    [itex]\int(A^{2})e^{-2(\alpha)r}r^{2}=1[/itex]
    where the integral is from 0 to infinity and A is the normalization constant.
    Knowing that the spherical normalization will give 4(pi), then the normalization constant is:
    [itex] A=\sqrt{\frac{\alpha^{3}}{\pi}}[/itex]
    Now finding the hamiltonian:
    [itex] <H>=<V>+<T> [/itex]
    I will start with <V>, since this is where I find my issue:
    [itex]<V>=\int(A)e^{-(\alpha)r}(\frac{1}{2}m\omega^{2}r^{2})Ae^{-(\alpha)r}[/itex]
    where the integral is from 0 to infinity.
    What I obtain for <V> is:
    [itex] <V>=\frac{1}{8\pi}m\omega^{2}[/itex]
    which of course doesn't have units of energy...
    Can someone point out where I went wrong please?
     
  2. jcsd
  3. Nov 9, 2012 #2
    <E>=∫ψ*Hψ dv/∫ψ*ψ dv,is the usual way of getting it.why is it bothering you it clearly has the units of energy.
    Edit: I have not checked your calculation.
     
  4. Nov 9, 2012 #3
    <V> does not have units of energy....
    But thank you for showing me the integration, I did not use dV, which will cause many problems... thanks!
     
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