Understanding the double square well potential

[hmparticle9]

In my textbook I have the following:

"The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). Does the electron tend to draw the nuclei together, or push them apart?"

Here is a diagram of some stationary states, and I have to explain what happens to the nuclei for each of the stationary states a,b,c,d.

I am not a physicist, I am a mathematician trying his best to understand physics. I don't understand how the stationary states relate to the potential experienced by an electron in a diatomic molecule. As a result of this lack of understanding I do not know how the stationary state affects the nuclei.

 

[anuttarasammyak]

I cannot distinguish d) with b). b) and c) seem bounded states for double well potential located where the peaks of square of wave function are. If they are in the same energy level, Energy of c) > Energy of b)
When the enerygy level is lower than that of single well potential, as usually is, the force is attractive.

 

[hmparticle9]

I still don't understand. So what can we say about a)? and why? The wave function being peaked at the origin means what for the nuclei? I just understand the wave function as the probability density (when squared).

 

[anuttarasammyak]

a) seems like a bound state for single well potential. It is not an energy eigenstate of double well potential.

 

[Steve4Physics]

a) seems like a bound state for single well potential. It is not an energy eigenstate of double well potential.

Agree. In fact both a) and b) look like $\Psi_1$ and $\Psi_2$ for a single finite well:

(diagram from https://www.researchgate.net/figure...lar-finite-deep-potential-well_fig3_379916225)

The OP needs to provide the shape/position of the potential-well(s) for each wavefunction.

 

[256bits]

d) does not look to be a stationary state
b) should be the σ
c) the σ^* , but it doesn't zero cross, so the electrons have some time spent between nuclei

 

[hmparticle9]

This is the full problem.

I am having trouble with part c). I tried (and failed) to and isolate the problem. I am sorry about that. I don't see how the results of part a) and b) relate to part c)

 

[vela]

Part (c) states that "If the nuclei are free to move, they will adopt the configuration of minimum energy." What does (b) tell you about how the energy depends on $b$?

 

[hmparticle9]

b) tells me that the energy is smallest for the ground state in the case that $b=0$. As $b$ increases we can see that the energy for both the ground state and the first excited state tend to similar values.

next hint please! I am still none the wiser

 

[Steve4Physics]

b) tells me that the energy is smallest for the ground state in the case that $b=0$. As $b$ increases we can see that the energy for both the ground state and the first excited state tend to similar values.

next hint please! I am still none the wiser

That's incomplete. A graph would help.

For the ground state, as $b$ increases from zero, does the energy increase or decrease?

For the first excited state, as $b$ increases from zero, does the energy increase or decrease?

Remember that $b$ effectively corresponds to the nuclear separation. Then revisit @vela's Post #8 hint.

 

[hmparticle9]

For the ground state the energy increases. For the first excited state the energy decreases. (As b increases).

I see his hint, it is just not clicking.

 

[hmparticle9]

They will adopt the configuration of minimum energy meaning that when b is small they end up in the ground state....?

 

[Steve4Physics]

For the ground state the energy increases. For the first excited state the energy decreases. (As b increases).

Yes.

They will adopt the configuration of minimum energy meaning that when b is small they end up in the ground state....?

Not quite.

First consider the case where the system is in the ground state and $b \gt 0$. There are two possibilities to consider.

i) The nuclei could move closer (i.e. $b$ gets smaller). In this case the energy decreases.

ii) The nuclei could move further apart (i.e. $b$ gets bigger). In this case the energy increases.

But a system naturally evolves in a way which decreases its energy.

Does that help?

Minor edit.

 

[hmparticle9]

So in the ground state they are going to move together and in the excited state they are moving apart.

"But a system naturally evolves in a way which decreases its energy."

I googled this. Is this something to do with thermodynamics? I have a few chapters to go before I get there.

 

[Steve4Physics]

So in the ground state they are going to move together and in the excited state they are moving apart.

Essentially yes. But remember a force can act even if it produces no movement - provided there is another (balancing force) present. So requiring movement is not a good idea for the answer.

Maybe it would be better to use the language of the (Post #7) question and say something like: in the ground state, the electron tends to draw the nuclei together; in the 1st excited state the electron tends to push the nuclei apart.

"But a system naturally evolves in a way which decreases its energy."

I googled this. Is this something to do with thermodynamics? I have a few chapters to go before I get there.

Not thermodynamics (though some may argue)! Think classical mechanics.

A trivial example: if you drop a stone, it naturally moves in the direction which reduces its potential energy. Note that minus the gradient (w.r.t. height) of the stone's potential energy is the stone's weight - see below.

Suppose an object's potential energy (U) is a function of position. For the simple 1D (say x-axis) case, ithe object experiences a force F where $F = -\frac {dU}{dx}$. More generally we should write $\vec F = -\nabla U$. Do a search for 'force and potential energy gradient' if this is unfamiliar.

EDIT. Have changed this a bit to try to improve my explanation - which I still don't think is very good. Input from others is welcome!

 

[anuttarasammyak]

Not thermodynamics (though some may argue)!

I have an argument though I am not able to go thinking it in detail. Some diatomic molecules dissociate in high temperature. These atoms tend to form a molecule in low temperature. It seems that energy come in / go out of a diatomic molecule system depends on temperature or environment surrounding that system. It may be a usual implicit assumption that the environment has zero energy and it tends to suck energy from the system.

 

[Steve4Physics]

Some diatomic molecules dissociate in high temperature.

They all do!