Approximate the probability of tunneling

AntiStrange
Messages
18
Reaction score
1

Homework Statement


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.

Homework Equations


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

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.
 

Attachments

  • graph.jpg
    graph.jpg
    5 KB · Views: 461
AntiStrange said:

Homework Equations


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

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]

?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K