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I'm trying to understand different ways to solve a PIB model.

I'm given a molecule with a tetrahedral geometry and there are two related ground states - electron pair pointing up or down. I'm being asked to draw the n=1 and n=2 states in three separate cases of a finite potential barrier in the middle of the box -> VBarrier = 0, VBarrier = V0, and VBarrier = ∞. So that left of the barrier are the wavefunctions of the up orientation and right is the down orientation.

I know that for the case of VBarrier = ∞, it'll just be the same sin(n pi x/L) wavefunctions on both sides of the barrier. I don't understand how they would change for V=0 and V=V0, and why these barriers would have different heights? I'd be grateful for any help.

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    $\begingroup$ Some things seem very strange about your question, PIB questions never deal with any kind of tetrahedral molecule or spin of any kind. What you are dealing with is a 1 dimensional problem, or in other words a line which doesn't correspond to any molecule. In any case, for the no barrier you have a box of L, rather than 1/2L. for the barrier in the middle you will have tunneling of the electron from one box to the next $\endgroup$
    – Cody Aldaz
    Apr 24, 2020 at 17:53
  • $\begingroup$ Sorry, I may have made the question too confusing by paraphrasing. The actual question deals with ammonia's tetrahedral geometry, which inverts at a certain frequency. So (I think) I'm being asked to model a PIB by drawing and finding wavefunctions for each geometry, which will lead to the wavefunction for the composite system $\endgroup$
    – AmyWitcher
    Apr 24, 2020 at 18:07

2 Answers 2

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for the infinite barrier I'd put an electron on just one side of the box with length L/2, meaning that electron is trapped on one side/face of the ammonia.

For the no barrier part treat the electron as a box with length L meaning it can easily go from one side to the other without barrier (for example maybe the ammonia is at the planar transition state).

For the finite barrier, I'd show tunneling from one side to the second, however, normally for PIB the tunneling region has finite length.

Assuming you have a tetrahedral geometry, according to the Born-Oppenheimer separation of electron and nuclear motion the finite barrier would extend from L/2 to L (or equivalently from 0 to L/2). Therefore the tunneling would be from the middle to the edge.

Image source

enter image description here

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In the case of ammonia the N atom inversion can be considered as a 1D problem as it moves from one side of the H atoms to the other. There is a barrier of about $2000$ cm$^{-1}$ to cross. The figure shows a calculation which approximates that for ammonia.

ammonia potential

You can see that the energy levels are doubled below the potential and this is because of tunnelling from one side to the other, i.e. one side feels the potential from the other and this splits the levels. (The lowest two levels are so close they appear as one in the figure.) The wavefunctions take on 'odd' end 'even' character, i..e are inverted (odd) or reflected about the middle of the potential. Even above the barrier the wavefunction is distorted compared to that at far higher energy where it approximates to a sine wave as seen in a normal particle in a box.

Also notice that as the potential is not infinitely high the wavefunction tunnels into the potential either side. When tunnelling the wavefunction decays exponentially.

(A hartree is $\approx 27.211$ eV, which is $\approx 2.195\cdot 10^5$ cm$^{-1}$ and $a_0$ is the Bohr radius $0.529\cdot 10^{-10}$ m.)

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