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Another uncertainty question, this came up in another forum.

As I understand it an electron, for example, is a point-like particle. I take this to mean it exhibits dimensionless properties, but being only point-like, lacks a defined location. Instead of a location it has a wave equation.

I asserted this means that all of the properties (eg. charge and mass), of an electron would factually act in a manner that directly matched the wave equation, whatever it was at that moment. In other words, if the wave equation for the electron was an "electron cloud" in an atom, then the electron's properties would be that also. It would generate an electric field exactly as if it factually was a cloud around the nucleus until such time as it's wave equation changed.

The other poster insisted I was wrong. I am sure I can't explain their view exactly but they said electrons don't become "smeared out" (their term) like that and that the electron was just a point particle, we were just uncertain about where it was.

I just can't see it working that way, I believe that if the electron is in any wave equation state, all of it's properties and fields would have to be in the same wave equation state as well. Of course the location can be confined to a smaller location, different wave equation, but it can never be confined to a point. So as a practical matter, the properties of an point particle must always be spread over some finite area.

Is this wrong, and if so why?

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  • $\begingroup$ Yes, that is the core of my question. My thinking goes like this. Consider an external test charge, it seems that the only way to localize the charge to either A or B is to collapse the wave function and localize the electron also. If the test charge is such that it doesn't localize the electron's charge, then net effect on the charge must be the same as interacting with both probabilities. Kind of analogous to how in the two slit experiment the results must conform with the electron wave having pass through both slits. $\endgroup$ – Whitethorn Mar 7 '19 at 5:45
  • $\begingroup$ @DanYand If I correctly understand this new question (is the word before se second instance of B and "at" instead of "and' ?) the two situations that you describe don't really exists, at least if A and B are point, as. the electron can never be localised at a given point. If A and B are small regions of space, it dos make sense, and the answer to your question Yes the two situations are different. $\endgroup$ – Jhor Mar 7 '19 at 14:52
  • $\begingroup$ @Whitethorn Your new question in the comment is much more explicit. First, you do not need to "localize" the electron, and the interaction with another charge does not induce any collapse. Secondly, quantum wave functions can be added (like any waves) but the probabilities , given by the square modulus of the wavefunction (like electromagnetic energy proportional to the square of electric field).don't simply add, because you have interferences terms : |f_1(x)+f_2(x)|^2 is not equal | f_1(x)|^2+|f_2(x)|^2 but you have to take into account the extra term 2\Re[f_1^*(x) f_2(x)]. $\endgroup$ – Jhor Mar 7 '19 at 15:08
  • $\begingroup$ TY QM is notorious for being difficult and thank you for your answers BTW. I believe QM waves are effectively real objects, an electron may be a "point-like" or more accurately dimensionless particle, but should not be considered a point because it and all of it's properties can never be separated from the wave equations that describe it. So it is factually a "fuzzy smeared out" object as opposed to simply a point with an uncertain location. I do understand, that QM equations are still wave equations and don't behave the same as classical charge density. Is this correct? $\endgroup$ – Whitethorn Mar 12 '19 at 6:11
  • $\begingroup$ What do you mean by your last sentence of the comment. Especially what is the meaning of "behaves" ? As explained in my answer, the charge density distribution actually results from the (selected solution of) wave equation, but is nothing more (or less) than a charge density distrinution, in the usual sense. Of course this density is not rigid, so if you add another charge, the electric potential involved in the wave solution is modified, so the electron cloud and the charge density distribution is modified accordingly. $\endgroup$ – Jhor Mar 13 '19 at 9:07
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YES, allmost all of this is wrong.

1) The fact that the electron is (or more accurately can be in this model considered as) punctual has nothing to do with its wave behaviour.

2) The latter comes from its mass and (typical) velocity, which result in a de Broglie wavelength lamba_dB=h/mv in the range of 0.1 mm, that is to say the typical size of the atom. And thus imply that its position can not be known accurately at this length. This lead to use the wave equation and its probabilistic interpretation.

3) due to the attractive potential, the related wave is not the propagative wave you seem to have in mind, but a localised wave, that must fulfill a specific wave equation, named Schödinger equation. Typically, the most simple solution is psi[r]=A exp[-r/a_0] where a_0 is Bohr's radius, corresponding to the ground state of the atom. Its squares modulus defines the volumic probability density of what is often described as the "electronic cloud" to account of the fuzziness of the electron's position.

4) this density is actually associated to an electric charge density which interacts with the positive nucleus. And the corresponding negatve potential energy is the opposite of the energy to provide to break the bounds between them, that is to say the ionization energy of the atomic ground state.

Finally your last sentence must be reformulated into : if the electron is in any state, (defined as a solution of the wave equation), all of it's properties ( including the electric field that it creates) are described by this solution.

Is it more clear like this ?

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  • $\begingroup$ Sorry it isn't clear, that or you may be saying something similar to me from a different perspective. You mentions charge density around the atom, do you mean that effectively the electron's charge is distributed around the atom? $\endgroup$ – Whitethorn Mar 7 '19 at 5:49
  • $\begingroup$ If it's not clear, you have to state what is the point which remains obscure or better, study some basic quantum mechanics. For the new question, yes, the charge density beeong \rho = e × |\psi|^2, its is a totally distributed around the nucleus, as well as |\psi|^2 is. $\endgroup$ – Jhor Mar 7 '19 at 7:45

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