0
$\begingroup$

Suppose the inverted harmonic oscillator potential $$H=\frac{p^2}{2m}-\frac{1}{2}m\omega^2x^2$$ I'm looking for a form of solution for the case when $E<0$. It's clear that a scattering solution will exist. Suppose I shot a particle from the left-hand side, then How does the wavefunction would look like?

enter image description here

To my knowledge, it should look like this. But I'm not sure in the tunneling region.

$\endgroup$
2

1 Answer 1

1
$\begingroup$

The wave function may roughly look like this.

enter image description here

  • At the classical turning points (marked by ¦) it is $\psi''(x)=0$.
  • In the classical allowed range $\psi(x)$ is oscillating around $0$.
  • In the classically forbidden range $\psi(x)$ is non-oscillating.
$\endgroup$
2
  • $\begingroup$ What about the amplitude? $\endgroup$ Sep 8, 2021 at 18:21
  • $\begingroup$ @YoungKindaichi Amplitude ratio depends on many things (energy $E$, width of forbidden region, boundary conditions). I assumed your scenario: an incident wave from the left. So you get a reflected wave back to the left, and a transmitted wave on the right. Therefore I've drawn the right amplitude smaller than the left amplitude, $\endgroup$ Sep 8, 2021 at 18:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.