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Approximate the probability of tunneling

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the harmonic scillator potential perturbed by a small cubic term, so that
    [tex]V(x) = \frac{1}{2}m\omega^{2} (x^{2} - \frac{x^{3}}{a})[/tex]
    if a is large compared to the characteristic dimension [itex](\hbar /m\omega)^{1/2}[/itex], the states will all me metastable, since there can be no lowest energy state (as [itex]x\rightarrow\infty[/itex], the energy gets arbitrarily negative). Estimate the probability of tunneling from the ground state to the region on the far right.


    2. Relevant equations
    probability = [tex]|T|^{2} = e^{-2\int^{a}_{0} dx \sqrt{2m(V-E)/\hbar^{2}}}[/tex]



    3. The attempt at a solution
    A rough sketch of what the potential should look like when graphed is attached. However, I cut out the [itex]\frac{1}{2}m\omega^{2}[/itex] part at the front (is that bad?) but anyway that is what [itex]x^{2}-x^{3}/a[/itex] looks like. it crosses the x-axis at x=0 and x=a, and the maximum on the right side is at (2/3)a.

    I have tried several things. Just substituting the potential given in the problem and I'm using E = (1/2)*h-bar*omega as the energy (ground state of the harmonic oscillator), into the equation but I can't solve the integral, even an online automatic integrator doesn't figure it out.
    I have also tried equating the potential to the energy using the E = (1/2)*h-bar*omega and setting that equal to the potential and solving for omega. even though I'm not sure I am allowed to do this, it simplifies things a little bit, but the integral still seems impossible to solve.
    I have also tried to approximate the curve on the right side by a negative parabola, perhaps it would work but I am having some trouble finding a parabola that fits close enough.

    Any help would be very appreciated, or a point in the right direction.
     

    Attached Files:

  2. jcsd
  3. Oct 16, 2009 #2

    gabbagabbahey

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    Gold Member

    That doesn't look right....Shouldn't the probability that a particle tunnels over the barrier into the far right be given by

    [tex]T=\int_{\frac{2a}{3}}^{\infty} |\psi(x)|^2 dx[/tex]

    ?
     
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