I Quantum behaviour of an electron around a positively charged sphere

[hokhani]

TL;DR Summary

Does an electron collide with a positively charged sphere in quantum mechanics?

If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

 

[GlitchedGluon]

Well, I think that is the whole point of Quantum Mechanics. If you could predict quantum phenomena using classical physics, you would not need quantum physics.

 

[fresh_42]

TL;DR Summary: Does an electron collide with a positively charged sphere in quantum mechanics?

So, it seems that the quantum and classics predict different behaviours!

I think this is why De Broglie brought waves into the game.

The idea of the electron as a classical particle doesn't get us anywhere here. De Broglie explained the stability of atoms by interpreting Bohr's quantum condition with the assumption of electron waves. As a wave phenomenon, an electron cannot transfer energy in space only if the wave pattern is temporally fixed in space.

Source: https://www.uni-ulm.de/fileadmin/we...51/Didactics/quantenchemie/html/Schroedi.html

 

[weirdoguy]

the initial state of electron is a linear combination of Hydrogen-like states.

Why? Positively charged sphere is something way different than proton. Real world sphere would consist of milions of atoms itself!

 

[PeroK]

TL;DR Summary: Does an electron collide with a positively charged sphere in quantum mechanics?

If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

How are you going to model a charged sphere in QM? If you replace the hydrogen nucleus with a greater positive charge, then the lowest energy state will have a greater energy and a smaller expected radius. Eventually, for a very large charge, this model will break down.

Alternatively, a sphere of finite macroscopic size cannot be taken as a direct replacement for the hydrogen nucleus.

 

[hutchphd]

There are a host of atomic behaviors that do not comport with classical mechanics. This was seen to be a problem at the time, and most of them were illuminated by quantum mechanics, although for some the solution seemed as daunting as the problem.

 

[JimWhoKnew]

the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent.

In addition to the wise things that were already said above, the quoted claim is simply wrong. A superposition of different energy eigenstates is not necessarily an eigenstate (we've been there before... ). In General, the relative phases between the participating eigenstates will vary over time (Schrödinger Equation). Thus the state evolves.

(Even more so if the initial state contains unbound components).

 

[PeterDonis]

If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states.

What are you basing that claim on? Do you have a reference?

 

[hokhani]

Why? Positively charged sphere is something way different than proton. Real world sphere would consist of milions of atoms itself!

Ok, thanks. but the problem persists for a proton instead.

 

[hokhani]

In addition to the wise things that were already said above, the quoted claim is simply wrong. A superposition of different energy eigenstates is not necessarily an eigenstate (we've been there before... ). In General, the relative phases between the participating eigenstates will vary over time (Schrödinger Equation). Thus the state evolves.

(Even more so if the initial state contains unbound components).

Right, but to analyse the problem we solve Schrodinger equation, considering the initial conditions. Although the relative phases vary, $\langle \hat{r} \rangle$ is always constant. It seems that this problem also persists for all the other QM systems which have stationary states.

 

[hokhani]

What are you basing that claim on? Do you have a reference?

I got to this problem when I was thinking about the electronic transition among atomic states. I am also interested in possibly making correspondence between quantum and classic physics.

 

[PeroK]

Ok, thanks. but the problem persists for a proton instead.

No, it doesn't. QM applies for a hydrogen atom. And classical electromagnetism does not apply.

 

[hokhani]

No, it doesn't. QM applies for a hydrogen atom. And classical electromagnetism does not apply.

I would appreciate if you could please explain what really occurs in this case, with considerations beyond the classical electromagnetism?

 

[JimWhoKnew]

Although the relative phases vary, $\langle \hat{r} \rangle$ is always constant.

Can you please support this claim by showing how you derive it?

 

[PeroK]

I would appreciate if you could please explain what really occurs in this case, with considerations beyond the classical electromagnetism?

I'm not sure what you mean by really occurs. The hydrogen atom satisfies the relevant Schrodinger equation. That gives a sequence of bound energy states. That is QM. You have a bound energy state of a two-particle system. You do not have two classical particles doing some classical thing.

 

[hokhani]

Can you please support this claim by showing how you derive it?

I'm terribly sorry for my serious error. I disregarded the non-commutation of position and Hamiltonian. So, it seems that by time evolution, at least this phase change of the components, results in $\langle r \rangle =0$.
Many Thanks again.

 

[PeterDonis]

I got to this problem when I was thinking about the electronic transition among atomic states.

This...

If we release an electron around a positively charged sphere

...has nothing to do with energy level transitions in atoms. An electron undergoing an energy level transition in an atom is not "released", as if somebody was holding it and then dropped it. The electron starts out in a stationary state, and it emits or absorbs a photon to transition to another stationary state. There is no description of this process in terms of an electron "trajectory" at all. It's not a classical process and can't be analyzed using classical intuitions.

 

[JimWhoKnew]

I'm terribly sorry for my serious error. I disregarded the non-commutation of position and Hamiltonian. So, it seems that by time evolution, at least this phase change of the components, results in $\langle r \rangle =0$.
Many Thanks again.

I don't see any reason to be sorry, let alone "terribly". I don't think this error qualifies as "serious", either (you didn't embed it in an any crucial system, right?)

Ehrenfest theorem says that under certain conditions (consult the Wikipedia page for these), if the wavefunction is initially highly localized around a spatial point $~\vec{r}_0~$ , then $~\langle \vec{R}\rangle~$ and $~\langle \vec{P}\rangle~$ will approximately follow the classical trajectory for some time (ignoring other aspects of electromagnetism such as interaction with radiation, etc.). Note that to construct a highly localized wavefunction, the superposition will include both bound and unbound eigenstates.