[brabel]

I don’t understand how something that has no clearly defined position like an electron can have a well defined speed. I thought I had understood that at that level, particles are more like clouds, or vibrations in the quantum field, and they had no well defined position until you tried to measure it, causing its cloud to collapse to a smaller region. But if non observed electrons can have a speed that defines the color of a material, that whole understanding seems to be wrong! Where is the error? Are all atoms on a piece of gold being “observed” in the quantum sense?? Even if we just capture the spectrum? Or it’s something else??

[gus_massa]

You are mostly correct.

The idea is that it has not a clearly definite position, but it has a distribution of probability to find it that looks like a "cloud" https://en.wikipedia.org/wiki/Atomic_orbital

In a more abstract sense, has not a clearly definite speed, but it has a distribution of probability to find it in a speed graphic.

The distribution of position and speed are defined by an equation and you must add a relativistic correction to the classic version. For lighter atoms you can just ignore the correction. For heavy atom (like Bismuth in this case) the correction is important.

Informally, the correction is important only when the "average" speed is fast enough to be somewhat close to the speed of light, like 50%c.

The correction changes the energy of the expected distribution of position and speed, and the energy. When an electron jumps from an orbital to another orbital, the difference of energies is related to the color.

> Are all atoms on a piece of gold being “observed” in the quantum sense??

[Ignoring that "observer" is a very misleading word and causes a lot of confusion, but it's the standard one and we are stick with it...]

The observation is only of the energy level of the orbital electron. We know the energy, but we don't know the position or the speed. When you observe some quantum object you don't get magically all the properties, only one of them, in this case the energy. In other experiments you can get only the position, in others only the speed. [And there are a lot of weird cases and technical details.]

[alok-g]

(Newbie here). And then going further, shouldn't there also be acceleration and its distribution? It says classical models could not explain why accelerating electrons were not radiating. If acceleration also shows up in QM, then ... a distribution of radiation?

[gus_massa]

[Sorry for the delay. I really had to watch that soccer game and kids don't have school on Sunday.]

There is an acceleration distribution, but the acceleration operator is strange. I don't remember the details and a quick google search confirms that it's strange.

It's complicated... let's oversimplify some details...

In QM the electron must jump from one orbital to another, and the difference in energy is emitted as radiation as a photon. If the electron jumps from A to B, then B must be empty. So if A is the orbital with less energy then it can't emit. Also if B is full, it can't emit.

For a big enough system, there are plenty of options for B and you get a very good approximation that is the classic rule that says that accelerating electrons emit photons.

There are weird cases, like in a neutron star, there are too many electrons trapped by gravity so all possible B are full, and you have electrons that can't emit.

It you want a tabletop experiment, the keyword is "fermion gas" that are gas of fermions (like electrons) that are very cold and very dense and they have a strange repulsion that is not explained classically. It is caused because there are jumps that are forbidden because the destination is full. (If you heat them or give them more room to be diluted, this strange repulsion almost disappears and you can aproximarte them as a classical gas.)

> a distribution of radiation?

If you measure the radiation far away, you have a distribution of possible colors/energy of the photon, because the electron may choose to jump form A to B1, B2, B3, ... This is like the lines color of gas lamps.

If you measure close enough, you have to draw Feynman diagrams and the photons may have a slightly different value of color/energy. But it's complicated and I'm not sure of the details. I guess it's related to the acceleration distribution, but I'm not sure of the details again.

---

The easy answer is that "acceleration -> radiation" is only a useful approximation when the system is big enough to ignore the quantum effect.

The hard answer is probably that you have to study like 10 years of physics to be sure and explain me the details. :)

[alok-g]

Thanks a lot for sharing. This is insightful for me ... I now know better on how to think and what to look for as I dwell deeper.

[abecedarius]

"High speed" here can be taken in terms like this: the phase of the wave function changes rapidly with position and time. (Changing with position -> a superposition that's heavy on short wavelengths, high momentum; with time -> high frequency, high energy.)

Re "observed all the time": when gold interacts with light, the light's normally of a strength that's a small perturbation on the fields internal to the atom, which is basically why you can treat the atom/light-field system as two weakly coupled quantum systems. It's an "observation" when the light leaves a classical trace such as a current in a CCD.

(I don't expect this to leave you unmystified about QM, but hopefully a bit clearer about it.)

[Sharlin]

The uncertainty principle says that the less well-defined the position, the more well-defined the velocity, and vice versa.