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Variational Method - Gaussian Trial Wavefucntion

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the variational method with a gaussian trial wavefunction ψ(x) = A[itex]e^{\frac{-a^{2}x^{2}}{2}}[/itex] to prove that in 1 dimension an attractive potential of the form shown, no matter how shallow, always has at least 1 bound state.

    *Figure is of a potential V(x) that has a minimum on the negative y axis, and tends to 0 as x -> ∞ and -∞*

    THERE ARE NO BOUNDARIES SHOWN ON THE DIAGRAM, THIS IS A GENERALISED POTENTIAL

    HINT: prove that it is always possible to make the trial energy negative.

    2. Relevant equations

    [itex]E_{0}[/itex] [itex]\geq[/itex] [itex]\frac{\left\langle ψ | \hat{H} | ψ \right\rangle}{\left\langle ψ | ψ \right\rangle}[/itex]


    3. The attempt at a solution
    We don't quite know that Hamilitonian, but we do know that it has a kinetic energy part. So I've written the Hamiltonian as:

    [itex]\hat{H}[/itex] = [itex]\frac{-h^{2}}{2m}\frac{d^{2}}{dx^{2}} + V(x)[/itex] Since the potential is unknown at this point.

    State should end up negative but larger than or equal to the bound state due to the above condition.

    I tried to normal the scalar [itex]\left\langleψ|ψ\right\rangle[/itex] to start things off so I could find the constant, but I have no clue on how to integrate this stupid gaussian function. I used wolfram alpha and I basically get A = [itex]\sqrt{\frac{2a}{\sqrt{\pi}}}[/itex]

    So I'm stuck at the first hurdle. I need to find the scalar product and then superimpose the hamiltonian and integrate to see if I meet the condition.

    Otherwise, what do I need to do?
     
  2. jcsd
  3. Nov 4, 2011 #2
    Any ideas guys? It's the gaussian that's irritating me. How do I integrate it?
     
  4. Nov 9, 2011 #3
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