This question had me thinking about the related 1D problem of two rectangular potential barriers:

Two rectangular potential barriers.

The potential is zero everywhere but for the two barriers, where it is $V_0$.

As ever I am interested in possible bound states for particles with energy $E<V_0$.

The problem is reminiscent of this model of $\alpha$ decay of $\text{Po-212}$.

Also, if $|b| \to \infty$ then the problem becomes that of the finite potential well where bound states are possible. This suggests that for large $|b|$ bound states may be found.

And for a particle with $E<V_0$ incoming from the left one can see how it would first be part reflected/part transmitted at $x=-b$, then its amplitude reduced between $-b<x<-a$ and then it would undergo symmetric treatment at the right hand potential barrier. This of course would suggest no bound states.

Does anyone know the definitive answer?


THere are no bound states. You've pretty much answered your own question.

Imagine what the lowest energy bound state would look like: an upside-down bell-shaped curve. This will match to exponentials in the forbidden region. But exponentials never go to zero. At $b$ the exponentials have to mate to something: they will mate to sinusoids extending to infinity.

  • $\begingroup$ Does this mean we shouldn't consider an $\alpha$ particle in a Po-212 as a bound particle either? $\endgroup$ – Gert Oct 15 '15 at 14:24
  • $\begingroup$ Strictly speaking, it's not a bound state. But Po-212 will remain Po-212 until it's not. :) $\endgroup$ – garyp Oct 15 '15 at 14:32
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    $\begingroup$ If it were a bound state, it wouldn't decay - it is a metastable state with a lower energy state available by having an alpha 'decide' it is time to be somewhere else (i.e. the wave function outside of the barriers). $\endgroup$ – Jon Custer Oct 15 '15 at 14:49
  • $\begingroup$ @garyp: does it have to be "an upside-down bell-shaped curve"? If $\psi$ is a solution of the SE, then so is so is $-\psi$, no? $\endgroup$ – Gert Oct 15 '15 at 15:56
  • $\begingroup$ $-\psi = \psi e^{i\pi}$, a global phase shift. Perfectly good. I was just focusing on the essentials for the question. $\endgroup$ – garyp Oct 15 '15 at 16:37

According to the book Molecular Electronics, by Cuevas and Scheer, Sec. 4.3, there should be bound states. You can have resonant tunneling in the transmission probability located at the energies of those bound states.

Check for example this. Ok, there is no strictly a bound state, but a quasi-bound state which is in between the barriers bouncing around.


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