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Questions to understanding the electron


hobz
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As far as I understand, an electron orbiting a proton can be thought of as a standing wave around the proton. The orbits allowed for the electron are dictated by the distances to the center that allows standing waves from its de Broglie wavelength.

 

If this is true, then:

  • how can there be a certain probability for the electron to be outside these very precise distances?
  • how can Heisenberg's uncertainty principle be true as well?

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Your understanding is wrong.

 

1) The term "de Brogile wavelength" should be thought of an anachronism from the time before QM was developed. The modern name would simply be "wavelength" (the addition "de Broglie" merely emphasizes the big surprise that matter can have a wavelength, which of course everyone familiar with QM knows). More importantly, "wavelength" is something that is usually attributed to free particles, not to bound states as this one. In particular, the wavelength of an hydrogen atom would [the way people would usually understand it] be something completely different than a property of the electron circling around the proton.

 

2) What you are probably referring to is something like the possible wave functions for a particle in a 2D plane which is constrained to the circumference of a circle with radius R. There, the wavelength is constrained to be the circumference of the circle divided by some natural number (or formally infinite). But:

2.1) There is no R for which no solution exists. There also is no mechanism which restricts your "wavelength" (term put in quotation marks here because of 2.3). So in principle, every "wavelength" is possible. So in principle, every radius is possible. Only the combinations have to fit.

2.2) The energy eigenstates, i.e. the orbits, are not a single of such fitting combinations, but a superposition. In other words: an orbit does dictate the probability distribution for finding the electron at a certain distance from the proton. But it does not fix the distance to a unique value.

2.3) Hydrogen atoms are usually considered to exist in three spatial dimensions, not 2D. Not much of a conceptual change here (you merely need to replace the discrete wavelength with the discrete spherical harmonics), but you should probably be aware of it.

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In addition to timo's post: the standing wave formulation is a leftover from the Bohr Model, which is incorrect. While quantized angular momentum is a correct concept from the model, the H ground state in the Bohr model has 1 unit of angular momentum, while in reality (i.e. QM), the state has zero orbital angular momentum.

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Thank you both for your insightful answers!

The popular layman literature I tend to get my informations from are not very clear on discerning between the "old" world and the "new" world.

In fact, I did not know that there was such a distinction.

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Me neither, I've never really bothered about atomic models of the past. But since I encounter terms like "Bohr model" or "Rutherford model" rather often on sfn, perhaps Tom can give a very short overview here.

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There are two issues — one is why incorrect models are taught. That's mainly for historical perspective. The model of the atom, relativity and quantum mechanics represent true paradigm shifts, and these don't happen overnight. So one teaches the to show some of the missteps along the way, when one has only a small amount of data to use to build a model. Most science labs people take can't recreate this, because you know what the right answer is. If the experiment has a flaw, you know it because it doesn't fit the equations. You rarely get the chance to do an experiment where there is no equation to go by.

 

The Bohr model was based on a planetary atom with standing wave orbits, and this gives rise to quantized energy. And, as it turned out, the energy values were correct, but the angular momentum values were not. It didn't account for different transition probabilities between states. It also conflicted with the Heisenberg Uncertainty Principle. The behaviors associated with these issues all fall naturally out of the Schrodinger equation solution.

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As far as I understand, an electron orbiting a proton can be thought of as a standing wave around the proton. The orbits allowed for the electron are dictated by the distances to the center that allows standing waves from its de Broglie wavelength.

 

If this is true, then:

  • how can there be a certain probability for the electron to be outside these very precise distances?
  • how can Heisenberg's uncertainty principle be true as well?

 

When you look at electrons as standing waves, they are waves of existence itself. However, an electron being this weird wave of existence seems to pop in an out of existence at seemingly random positions (also called virtual particles), travel along multiple paths at a time, and if the time is brief enough, occupy two different point positions at a time. When you say "orbits", such as a distance of .529x10^-10 meters from the nucleus (also known as the ground state or lowest energy orbital), you're simply describing the most probable location for an electron to be observed as a single coordinate (where in the frequency of a wave function, it's equal to the polar coordinates of where the wave crest is maximum OR minimum since mathematically your using absolute value). The actual places an electron can show up are anywhere in the universe, or another way of putting it is that an electron's wave function extends indefinitely through space, however, electrons spend most of the time in the most probably places since as you get even a little bit further from their most probably location, the probability of finding that electron there is pretty close to nothing.

This is also why the quantum world seems so different from the classical world. The highest probability for a ground state electron to show up is a sphere with just a diameter of .529x10^-10 meters, so its not that the classical world is different, its that the probability for an electron in around in atom in say, a pencil, to appear even one millimeter away from the surface of a pencil is closer to nothing than you could possibly imagine. There there is a chance a pencil could suddenly teleporrt a few feet, but the probability of even one electron in an atom of that pencil doing that is at least 1 in 10^1000 chance of that happening, which is a pretty unimaginably small percentage. There's an equation that could probably tell me more accurately what its probability would be, but I don't remember it.

Edited by steevey
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When you look at electrons as standing waves, they are waves of existence itself. However, an electron being this weird wave of existence seems to pop in an out of existence at seemingly random positions (also called virtual particles), travel along multiple paths at a time, and if the time is brief enough, occupy two different point positions at a time.

 

The electron is not a virtual particle, and superposition will last until there is an interaction that collapses the wave function.

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The electron is not a virtual particle, and superposition will last until there is an interaction that collapses the wave function.

 

But as a single point, an electron can exist in a specific location for a brief period of time, which constitutes for being a virtual particle. With superposition though, an electron is is still a wave after you observe it, however you just can't see its properties as a wave. So, theoretically if the time is brief enough, probably around Planck time, the electron can occupy two single points at the same time.

Edited by steevey
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But as a single point, an electron can exist in a specific location for a brief period of time, which constitutes for being a virtual particle.

That is not what "virtual particle" means. If you insist on the term virtual particle having a meaning other than being a factor in a Feynman diagram, then one could probably say that a virtual particle is one for which there is a mismatch between mass, energy, and momentum.
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Is there any reason given for the presumed fact that electrons can appear anywhere but have a higher probability of appearing in some places more than others? Also, when people are calling it a "standing wave," does that mean they are referring to the totality of what would be called an "orbit" in the Bohr model, except the entire "orbit" is viewed as a single hollow sphere-ish type shape that expands and contracts at a certain rate around the nucleus?

Edited by lemur
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Because you still have boundary conditions that limit the solutions.

 

Classically this is the case; if you have an orbit of a certain energy and maximum angular momentum, for example, you have a circle of radius r and speed v. No other trajectory fulfills those conditions. When you go to the wave regime the classical trajectory goes away, but the notion that certain behavior is physically disallowed does not.

 

 

The standing wave idea is that instead of a circular orbit, you had a sine wave along the trajectory, and you had an integral number of wavelengths so that as you went around the circle to the beginning, it was in phase (if it was not, you would have interference)

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Classically this is the case; if you have an orbit of a certain energy and maximum angular momentum, for example, you have a circle of radius r and speed v. No other trajectory fulfills those conditions. When you go to the wave regime the classical trajectory goes away, but the notion that certain behavior is physically disallowed does not.

This doesn't make sense to me. By trajectory, you mean tracing the path of the circle? Are you calling the orbit an angular momentum of a circle? So a certain amount of energy results in a certain speed at a given radius, i.e. the parameters are fixed relative to each other because of conservation of energy and momentum? Why does the classical trajectory go away in "the wave regime?"

 

The standing wave idea is that instead of a circular orbit, you had a sine wave along the trajectory, and you had an integral number of wavelengths so that as you went around the circle to the beginning, it was in phase (if it was not, you would have interference)

So the electron is modeled as a sine wave, or rather a chain of sine waves that connect in phase around the nucleus (without interference)? Does that mean that the quanta of energy are akin to adding waves to the loop/orbit? Does this also mean that each additional wave causes the radius to increase according to the wavelength of the additional wave? Is this related to the (variable) frequency of the atom? Am I mixing up concepts?

Edited by lemur
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But how much longer? And while its still being observed? That can't be right...

 

If you observe it, it won't be in a superposition.

 

This doesn't make sense to me. By trajectory, you mean tracing the path of the circle? Are you calling the orbit an angular momentum of a circle? So a certain amount of energy results in a certain speed at a given radius, i.e. the parameters are fixed relative to each other because of conservation of energy and momentum? Why does the classical trajectory go away in "the wave regime?"

 

You can't have a classical trajectory for a wave because you don't have a single position to track.

 

 

So the electron is modeled as a sine wave, or rather a chain of sine waves that connect in phase around the nucleus (without interference)? Does that mean that the quanta of energy are akin to adding waves to the loop/orbit? Does this also mean that each additional wave causes the radius to increase according to the wavelength of the additional wave? Is this related to the (variable) frequency of the atom? Am I mixing up concepts?

 

Basically. The wavelength is dictated by the momentum and thus angular momentum, so increasing energy means you have to find a new wavelength — energy and angular momentum are coupled in the Bohr model. But it's wrong, since it turns out that they aren't.

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  • 2 weeks later...

As far as I understand, an electron orbiting a proton can be thought of as a standing wave around the proton. The orbits allowed for the electron are dictated by the distances to the center that allows standing waves from its de Broglie wavelength.

 

If this is true, then:

  • how can there be a certain probability for the electron to be outside these very precise distances?
  • how can Heisenberg's uncertainty principle be true as well?

 

 

There's actually a number of properties that determine where an electron can appear around the nucleus, but I think there's only 4, or 4 quantum numbers to describe a wave particle, such as energy or angular momentum. It's also important to note that no two particles in the same region (such as two electrons around a nucleus) can have the same properties because that would mean that two different particles are occupying the same space.

Edited by steevey
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