How to Stop an Electron from Falling into a Proton - Comments

[edguy99]

edguy99 submitted a new PF Insights post

https://www.physicsforums.com/insights/quantum-animations-stop-electron-falling-proton/

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https://www.physicsforums.com/insights/quantum-animations-stop-electron-falling-proton/

 

[jerromyjon]

Thanks for the Insights! This is one of the most interesting physics topic for me, the grey area between quantum and classical realms. I've never thought of a covalent bond as an electron trapped in two nuclei simultaneously. It certainly is food for thought!

 

[edguy99]

Thanks for the Insights! This is one of the most interesting physics topic for me, the grey area between quantum and classical realms. I've never thought of a covalent bond as an electron trapped in two nuclei simultaneously. It certainly is food for thought!

It's amazing what images can suggest. I often get a much deeper understanding of a process through animation. It's a bit like an engineer, saying ok, the math is fine, so let's build it and see if it works. Thanks for the comments.

 

[vanhees71]

Well, it's also pretty clear how dangerous images can be. "Though shalt not make images" (other than mathematical ones) is a good advice when it comes to quantum theory.

It's very dangerous to depict electrons in an atom as miniature "billard balls". In the bound states of electrons and an atomic nucleus they are far from having such classical particle-like features. That's why it's one of the "Sins in Physics Didactics" (I should write about as soon as possible in the Insights again, but there's lack of time for that at the moment) to use this particle picture a la the Bohr-Sommerfeld model. Here, it's more appropriate to depict the probability densities as calculated in any quantum-mechanics 1 lecture.

 

[Demystifier]

I've never thought of a covalent bond as an electron trapped in two nuclei simultaneously.

Is there any other way to think of a covalent bond?

 

[jerromyjon]

We could go back to the elementary atom model of colored balls and sticks, then when I started to learn about electricity and magnetism I imagined they had a magnetic bond of some sort. Only within the last year I have seen the probability distribution of electrons in various modes and energy levels. I know there are also "double bonds?" that used to be represented as springs in the ball and stick model. I can't remember if multiple electrons contribute to those... probably because I never assimilated a convincing model.

 

[Greg Bernhardt]

@vanhees71 how would you model an atom via graphic or animation?

 

[vanhees71]

That's a difficult question. The best thing to "depict" atoms is to plot the density distributions of the electrons around the nucleus at rest. For an atom in an energy eigenstate there's no animation, because the distribution is time independent by construction, because the energy eigenstates are the stationary states of the system.

Strictly speaking this holds only for the ground state, because including the quantized em. field you always have sponatneous emission of photons for excited states going into lower states. Here, however, an animation is also very difficult to do, because all we have here is the S-matrix element, telling your the transition probability per unit time for a spontaneous emission of one (or more) photon(s) from an asymptotic state (atom in an excited energy eigenstate in the QED vacuum) to another (atom in another lower-energy eigenstate + 1 Photon (or n photons)). It's not clear, how to define the "transient state" in terms of a process you could follow in time. So to animate this, you'd look at the situation with a time resolution large compared to the typical duration of a transition process, getting spontaneously emitted photons + rearranged (quasi-)stationary states of the atom.

The same holds for the scattering of an electron with the atom, where you can calculate the S-matrix elements for various processes like elastic scattering, Raman scattering, ionization of the atom by kicking out an electron, etc. etc.

 

[jerromyjon]

Couldn't your spontaneous emission just re-energize the adjacent atom in a simulation? i just found this on wikipedia, going to read it:
http://en.wikipedia.org/wiki/Born–Oppenheimer_approximation

 

[Jazzdude]

I'm curious, am I the only one who has severe issues with the physics presented in this insight?

Cheers,

Jazz

 

[edguy99]

@vanhees71, I smiled at "Though shalt not make images", and it got me thinking that the punishment for animations is probably pretty severe.

I agree with "It's very dangerous to depict electrons in an atom as miniature "billard balls"" and tried to qualify things with the paragraph:

Although the model looks interesting, it does not deal with emitted radiation or bond strengths. To properly model bonds, we have to keep track of which electron is bonded to which proton. We will have to track not only the bonds energy level, but also the photons that are emitted or absorbed through the creation and destruction of the bonds.​

Your comment is very much what I have in mind: So to animate this, you'd look at the situation with a time resolution large compared to the typical duration of a transition process, getting spontaneously emitted photons + rearranged (quasi-)stationary states of the atom.

WRT to electron density maps, that also would be great but very hard to do. If you have a look at http://www.ebi.ac.uk/pdbe/emdb/empiar/, and try to download a file. You get an idea of the scope of the issue. Many on the protein molecules are only 1000 or 2000 atoms but yet will take all day to just to download.

Finally, you often see hydrogen orbitals in pictures like this. Is there a reason that there is no distance scale on this?

Thank you for your comments.

 

[vanhees71]

This orbitals are a quite nice way to depict the probability distribution of electrons in the bound states of the hydrogen atom. It's just ##\propto |Y_{lm}(\vartheta,\varphi)|^2\sin \vartheta##. You can put a distance scale, when including the radial part of the wave function. As the typical length scale, of course, you get the Bohr radius.