Is the electric field of an atom a superposition or mean of electron positions?

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TL;DR Summary
How to understand atomic orbitals from QM interpretations perspective?
We usually think about atomic orbital as wave(function), but it was created from e.g. electron and proton approaching ~10^-10m (or much more for Rydberg atoms), and electron has associated electric field.
This wavefunction also describes probability distribution for finding electron (confirmed experimentally e.g. https://journals.aps.org/prb/abstract/10.1103/PhysRevB.80.165404 ).

So what is electric field of atomic orbital - is it superposition of electric fields over positions of electron in wavefunction, or maybe just their mean?

I think e.g. van der Waals force requires that it is indeed superposition (?) - so how to understand it from QM interpretations perspective?
E.g. in Many Worlds Interpretation should we imagine that each World has electric field for one position of electron?

In this superposition electrons stay or move?
If stay, where does orbital angular momentum come from? If move, why no synchrotron radiation?
 

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  • #2
PeroK
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The electric field is a concept from classical electromagnetism, which shouldn't be mixed up with QM.

The theory that describes the behaviour of electrons and the EM field is Quantum Electrodynamics.
 
  • #3
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Ok, nice evade, so how should we imagine QED Feynman diagrams from QM interpretations perspective?
E.g. in Multiple Worlds Interpretation - does each World have one diagram?
 
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PeroK
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Ok, nice evade, so how should we imagine QED Feynman diagrams from QM interpretations perspective?
E.g. in Multiple Worlds Interpretation - does each World have one diagram?
It's not a matter of evasion, it's a question of avoiding a confusion of ideas. Feynman diagrams are used to evaluate the amplitude of QM interactions. This leads to, for example, the probability distribution of scattering angles. Orthodox QM says that the measured interaction is one of these with the appropriate probability. MWI says that all scattering possibilities happen and the system remains in a superposition (along with the experimental apparatus and lab technicians) of all these scattering possibilities.

The calculation of amplitudes and probabilities is the same in both cases.
 
  • #5
Jarek 31
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This leads to, for example, the probability distribution of scattering angles.
What leaves analogous question especially if the scattered particles are charged: have associated electric fields - is the EM field around a superposition or mean over all such scenarios like scattering angles?

For example, if there is another charged particle flying nearby atom or scattering site, its behavior depends on this electric field ... what can be also expressed in perturbative QED language - seeing e.g. e-p Coulomb interaction thorough photon exchange.

Maybe let's not focus on one of languages, but directly ask about behavior - charged particle flying near atom behaves as in electric field of point electron, or as in field averaged over wavefunction?
 
  • #6
PeroK
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What leaves analogous question especially if the scattered particles are charged: have associated electric fields - is the EM field around a superposition or mean over all such scenarios like scattering angles?

For example, if there is another charged particle flying nearby atom or scattering site, its behavior depends on this electric field ... what can be also expressed in perturbative QED language - seeing e.g. e-p Coulomb interaction thorough photon exchange.

Maybe let's not focus on one of languages, but directly ask about behavior - charged particle flying near atom behaves as in electric field of point electron, or as in field averaged over wavefunction?
Unless you stop mixing up concepts from different phtsical theories we cannot have a meaningful discussion. If you study electric fields using classical electrodynamics then, by definition, there are no quantum effects. In particular, the hydrogen atom is not physically possible
 
  • #7
Jarek 31
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I am not asking about classical EM, but the one used for orbitals: in Schrodinger equation, you evaded going to QED - which allows to imagine this Coulomb interaction through photon exchange (different language).

For QM interpretations perspective maybe let's start with Schrodinger - charge particle nearby behaves as in superposition of electric field of point electron, or maybe as in electric field averaged over wavefunction?
 
  • #8
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If this helps to have some intuition about what is going on, you can think of the classical electromagnetic field as arising from quantum fields. After a measurement is completed.
 
  • #9
vanhees71
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Unless you stop mixing up concepts from different phtsical theories we cannot have a meaningful discussion. If you study electric fields using classical electrodynamics then, by definition, there are no quantum effects. In particular, the hydrogen atom is not physically possible
Well, particularly for atoms and molecules you get very far without quantizing the electromagnetic field, i.e., treating the electromagnetic field due to the charges of the atomic nuclei and electrons as classical. For light atoms, particularly hydrogen, you start from describing the proton by its electrostatic Coulomb interaction of the electron and treat only the electron as quantized.
 
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PeroK
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I am not asking about classical EM, but the one used for orbitals.

For QM interpretations perspective maybe let's start with Schrodinger - charge particle nearby behaves as in superposition of electric field of point electron, or maybe as in electric field averaged over wavefunction?

In the QM model of the atom there are no classical EM fields. In the ground state, for example, the electron has zero angular momentum. Which, using the Coulomb force and classical electrodynamics, is impossible.

Likewise, the interaction of two atoms in a hydrogen molecule is a QM model. I'm not aware of a workable model based on EM fields.

And the behaviour of large objects - such as why a solid is solid - cannot be explained by electric fields. It requires the Pauli exclusion principle, which is inexplicable in terms of EM fields.
 
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  • #11
vanhees71
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In the QM model you use the semiclassical approximation, i.e., classical em. fields. What's quantized is only the electron. The reason, why there is no problem with radiation loss which you inevitably have in purely classical model as the Bohr-Sommerfeld atom, where you just ad hoc assume that on the "allowed orbits" the accelerated electron doesn't radiate, is that there are stationary solutions (i.e., energy eigensolutions) of the quantum theory of the electron bound in the Coulomb field of the proton (with the Coulomb field treated as classical).

In this context of the hydrogen atom (and also more general atoms) you need em-field quantization for (a) spontaneous emission, which cannot described consistently within the semiclassical approximation and (b) radiation corrections describing the Lamb shift and the anomalous magnetic moment of the electron.
 
  • #13
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In the QM model you use the semiclassical approximation, i.e., classical em. fields.

You use that approximation for the interaction of the atom with external fields. That is not what the OP is asking about. The OP is asking about the "electric field" due to the atom itself (more precisely, of an electron in some orbital of the atom). You cannot describe that with classical EM fields.
 
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  • #14
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you just ad hoc assume that on the "allowed orbits" the accelerated electron doesn't radiate

Which, of course, means that you are not using classical EM to describe the atom itself.
 
  • #15
Jarek 31
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You cannot describe that with classical EM fields.
Why cannot we imagine electric field of atomic orbital as superposition over wavefunction of classical electric fields for all electron positions?
Something like
mean(-field): ##\vec{E}(y)= \int_x |\psi(x)|^2\ \vec{E}(x,y) dx##
vs superposition: ##|\vec{E}(y)\rangle= \int_x |\psi(x)|^2\ |\vec{E}(x,y)\rangle dx##
for ##\vec{E}(x,y)## classical electric field in ##y## for electron in ##x##?
 
  • #16
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Why cannot we imagine electric field of atomic orbital as superposition over wavefunction of classical electric fields for all electron positions?

Because it doesn't work; you make wrong predictions.
 
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  • #17
vanhees71
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You use that approximation for the interaction of the atom with external fields. That is not what the OP is asking about. The OP is asking about the "electric field" due to the atom itself (more precisely, of an electron in some orbital of the atom). You cannot describe that with classical EM fields.
What I meant is that in the calculation of the bound states in the usual treatment of QM the em. field is not quantized. You just use the Hamiltonian (in Heaviside Lorentz units) for the relative motion
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2 \mu} -\frac{e^2}{4 \pi \hat{r}},$$
where ##\mu=m_{\text{e}} m_{\text{p}}/(m_{\text{e}}+m_{\text{p}})## is the reduced mass.

Concerning the em. field of the atom as a whole, it depends on what you want to do with it. Probably I misunderstood the question.
 
  • #18
vanhees71
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Which, of course, means that you are not using classical EM to describe the atom itself.
In the era of old quantum theory nobody had an idea that the em. field has to be quantized at some point.

The history is somewhat funny: The first hint that there's something to be done to the em. field concerning "quantization", I'd date with Einstein "kinetic aproach" to the black-body radiation formula, where he had to introduce spontaneous emission in addition to induced emission and absorption, which latter have classical analogues in 1917. Here of course you don't need the full formalism of what we know call quantum field theory, which of course was not known at this time.

Then in 1925 after Heisenberg's heuristic idea of his famous Helgoland paper (the "quantum reinterpretation" of classical quantities) first Born and Jordan and then Born, Jordan, and Heisenberg worked out what's now known as the matrix formalism of modern quantum theory. In the latter paper, Jordan also quantized the em. field, which is kind of logical, because the em. field is part of the dynamics of charges and fields in classical electron theory too. Most physicists at that stage of the development found this, however, to be too much, i.e., they accepted the quantization of the "mechanical part", i.e., the description of the point-particle motion of electrons in terms of the new abstract quantum formalism, but didn't see the necessity for field quantization.

This had to be reinvented by Dirac a few years later within his formulation of QT, the representation free "transformation-theoretical formulation" in terms of creation and annihilation operators, which is more or less the canonical operator formalism we also use today to formulate QFT.

One should realize that one gets pretty far with the semiclassical approach, i.e., qunatized "particles" and classical "fields". The tree-level results for electron scattering and the Compton effect as well as the photoelectric effect can all be understood in the semiclassical way.

The most simple phenomena which make field quantization unavoidable is spontaneous emission and, of course, the "indivisibility of photons" as well as "quantum beats".
 
  • #19
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What I meant is that in the calculation of the bound states in the usual treatment of QM the em. field is not quantized.

More precisely, the Coulomb potential of the nucleus is not quantized; it is just a function ##V(r)## in the Hamiltonian.

But that Coulomb potential is not "the electric field of the atom" or "the electric field of the electron in an orbital".

In the era of old quantum theory nobody had an idea that the em. field has to be quantized at some point.

All of this is interesting history, but it's irrelevant to the question the OP is asking in this thread.
 

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