Does Electron Superposition Eliminate Angular Momentum?

In summary, if the electron is in a superposition in such a way that there is no angular momentum about the nucleus, then classicaly the charge of the electron would be zero from inside the orbit.
  • #1
DeepSeeded
113
1
If the electron is in a superposition in such a way that there is no angular momentum about the nucleus, then classicaly the charge of the electron would be zero from inside the orbit correct?

If the charge was changing anywhere there would be angular momentum.
 
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  • #2
Probably not. If we consider a hydrogen atom, there is no net charge outside, so Gauss's law would give ∫E*n dS = 0 on any surface outside the atom. But the electron does have charge, and ∫E*n dS = -q/e0 for the surface integral around a bare electron, where -q is the charge on an electron. So if there are no field lines outside that atom, and we know an electron is inside, where are its field lines? If we perform the integral outside the atom, and also include the surface just outside the nucleus. we will find ∫E*n dS = -q/e0, so there are field lines between the electron cloud and the nucleus. This is independent of the n and l of the electron state. It is true that in an n=1 (ground) state, there is no orbital angular momentum, so l = 0 and there is no orbital magnetic moment.

∂ ∫ ∏ ∑
 
  • #3
There is charge outside the atom at a specific period of time. There is no net charge because over a period of time the charges cancel themselves out.

If there was never a charge outside the atom then chemical reactions would not happen.

There must be angular momentum if something is moving around a point even it is just a charge difference.
 
  • #4
DeepSeeded said:
There is charge outside the atom at a specific period of time. There is no net charge because over a period of time the charges cancel themselves out.

If there was never a charge outside the atom then chemical reactions would not happen.

There must be angular momentum if something is moving around a point even it is just a charge difference.
If you were say 5 Angstroms outside a neutral hydrogen atom, you will not see any charge unless you ionize it. That requires about 13.6 eV. Water ions are actually H3O+ and (OH)-. To ionize hydrogen or water requires work.

Have you ever watched a pendulum, or a Foucault pendulum? How much angular momentum does it have in the horizontal plane?

For a hydrogen atom without electron spin, look at the Klein Gordon solution for the hydrogen atom. There is no angular momentum in the 1s, 2s, 3s etc. states. With the Dirac solution, the only angular momentum comes from the electron spin, which is not orbital.
 
  • #5
Bob S said:
If you were say 5 Angstroms outside a neutral hydrogen atom, you will not see any charge unless you ionize it.

"And yet it does move", as Galileo said. You still have van der Waals forces, which are certainly electrical in origin. (specifically, an effect of long-range correlation of electronic motion)

Which is all just yet another example of how applying classical electrostatics to explain atomic and molecular phenomena just doesn't work. It leads you wrong more often than not. Which is my immediate reaction to these kinds of questions - "Don't think about it that way!"

Considering electrons as some kind of classical or semi-classical charge density cloud inevitably leads you wrong on the dynamical effects of electronic behavior. Considering them to be moving point-charges inevitably leads you wrong on the delocalized nature of bound electrons. Basically repeating the wave-or-particle dilemma.
 
  • #6
alxm said:
Considering electrons as some kind of classical or semi-classical charge density cloud inevitably leads you wrong on the dynamical effects of electronic behavior. Considering them to be moving point-charges inevitably leads you wrong on the delocalized nature of bound electrons. Basically repeating the wave-or-particle dilemma.

Unless you analyze it in the de Broglie-Bohm interpretation where you have both moving point charges correlating with each other and a separate distinct wave which affects the trajectories of the moving charges and accounts for the 'delocalized nature of bound electrons'. And then everything is crystal clear.

I'm sorry to go on about it, but the de Broglie estate is paying me to propagandize on their late master's behalf.
 

FAQ: Does Electron Superposition Eliminate Angular Momentum?

What is superposition of the electron?

The superposition of an electron refers to the quantum mechanical phenomenon in which an electron can exist in multiple states simultaneously, until it is observed or measured. This is a fundamental concept in quantum mechanics and has been experimentally proven through various experiments.

How is superposition of the electron different from classical physics?

In classical physics, objects are described as having a definite position and momentum at any given time. However, in quantum mechanics, the superposition principle allows for particles to exist in multiple states at once, which is not possible in classical physics.

Can superposition of the electron be observed in everyday life?

No, the effects of superposition are typically only observed at the subatomic level. In everyday life, objects are too large and complex to exhibit quantum behaviors like superposition.

What is the significance of superposition of the electron in technology?

The concept of superposition has played a crucial role in the development of technologies such as transistors, lasers, and MRI machines. These technologies rely on the principles of quantum mechanics, including superposition, to function.

How is superposition of the electron used in quantum computing?

Quantum computers utilize the principle of superposition to perform calculations and store information. By allowing quantum bits (qubits) to exist in multiple states simultaneously, quantum computers have the potential to solve certain problems much faster than classical computers.

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