How does the wavefunction look like for inverted oscillator potential? [duplicate]

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?

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

1 Answer

The wave function may roughly look like this.

• 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.
• What about the amplitude? Sep 8, 2021 at 18:21
• @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, Sep 8, 2021 at 18:30