A Do QED effects make a huge change to the position of the electrons?

In https://en.wikipedia.org/wiki/Lamb_shift about the lamb shift, it's mentioned that the change in the electron's frequency due to QED effects (vacuum polarization and self-energy correction) is about 1 GHz, which would translate to an energy change of hf = 6.63E-25 J. This is 3E-7 times of the hydrogen electron's kinetic energy (13.6 eV). It might seem small, but given the huge velocity of the electron in an atom (2E6 m/s), the change in the velocity for a factor of 3E-7 in energy would be about 0.3 m/s (dv = dE/mv).

Now, given the very small size of the atom (1E-10 m), even in one nanosecond, a 0.3 m/s uncertainty in the velocity of the electron would make the uncertainty in its position 3 times more than the radius of the atom, which means a complete uncertainty of position inside the atom.

Is this argument correct? So why are QED effects assumed to be small?
 
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given the huge velocity of the electron in an atom
Electrons in atoms don't have definite velocities. They aren't little billiard balls orbiting the nucleus.

Is this argument correct?
No. The position wave function of an electron in an atom does not assign equal probability to all positions. Nor does the position wave function of an electron in an atom change with time--at least if we are talking about electrons in stationary states, which is the assumption that underlies the treatment of phenomena like the Lamb shift. So there is no "increase in position uncertainty" with time as your argument describes.

why are QED effects assumed to be small?
Because they are small in terms of energy, which is what we actually measure.
 
Electrons in atoms don't have definite velocities. They aren't little billiard balls orbiting the nucleus.



No. The position wave function of an electron in an atom does not assign equal probability to all positions. Nor does the position wave function of an electron in an atom change with time--at least if we are talking about electrons in stationary states, which is the assumption that underlies the treatment of phenomena like the Lamb shift. So there is no "increase in position uncertainty" with time as your argument describes.



Because they are small in terms of energy, which is what we actually measure.
Yes, the wavefunction doesn't change that much. But I think IF the electron had a well-defined position (like in Bohmian mechanics), such in velocity would make its position uncertain and unpredictable. I mean, QED effects would make the Bohmian trajectory of the electron dramatically.
 
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the wavefunction doesn't change that much
It doesn't change at all if the electron is in a stationary state--which, as I said, is the usual assumption when analyzing phenomena like the Lamb shift.

IF the electron had a well-defined position (like in Bohmian mechanics)
Bohmian mechanics assigns a "position" to the electron, but that position has nothing to do with the wave function. It's a separate property, and since we never actually measure the positions of electrons in atoms, the Bohmian "position" is irrelevant to the analysis of phenomena like the Lamb shift; all you need is the wave function.

such in velocity would make its position uncertain and unpredictable
No. The Bohmian "position" is not uncertain at all. Bohmian trajectories are perfectly deterministic.

QED effects would make the Bohmian trajectory of the electron dramatically
No, they wouldn't. The Bohmian trajectory of the electron is driven by the wave function.
 
It doesn't change at all if the electron is in a stationary state--which, as I said, is the usual assumption when analyzing phenomena like the Lamb shift.



Bohmian mechanics assigns a "position" to the electron, but that position has nothing to do with the wave function. It's a separate property, and since we never actually measure the positions of electrons in atoms, the Bohmian "position" is irrelevant to the analysis of phenomena like the Lamb shift; all you need is the wave function.



No. The Bohmian "position" is not uncertain at all. Bohmian trajectories are perfectly deterministic.



No, they wouldn't. The Bohmian trajectory of the electron is driven by the wave function.
I get what you say, but Bohmian mechanics does not include QFT. Take a look at https://arxiv.org/pdf/0707.3685v1.pdf . They introduce Bohmian trajectories for fermions and c-number fields for bosons to both keep Bohmian mechanics and explain QFT effects. In this case, the Bohmian picture isn't the complete picture and QFT effects are present and modify Bohmian positions, and what I said can be true.
 
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In this case, the Bohmian picture isn't the complete picture and QFT effects are present and modify Bohmian positions
No, you are misunderstanding these models. These are "Bohmian" models in which there are no particles with positions at all; instead there are fields that are "guided" by the pilot wave (i.e., the wave function or QFT fields that obey the usual quantum equations).

(The paper uses the standard Bohmian pilot wave model with the non-relativistic Schrodinger Equation as an example of the general process it describes for constructing pilot wave models; but all of the other models it describes, including all of the QFT ones, have no particles, only fields.)
 
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Take a look at
All of these models are speculative, and since they all make the same predictions as standard QM/QFT, they are just interpretations, which nobody actually uses for anything.

In this case, the Bohmian picture isn't the complete picture and QFT effects are present and modify Bohmian positions, and what I said can be true.
No, because even if there were Bohmian positions in these models (which there aren't, see post #6), all of the issues I raised in posts #2 and #4 would still apply.
 
Ok, so in this case the usual pilot-wave model with particle positions as beables should be thrown away because it cannot reproduce QFT. Why would one still consider it while it's known it must undergo such a fundamental change (taking out particle positions) in order to be in line with QFT (and its tested predictions)?
 
Yes, the wavefunction doesn't change that much. But I think IF the electron had a well-defined position (like in Bohmian mechanics), such in velocity would make its position uncertain and unpredictable. I mean, QED effects would make the Bohmian trajectory of the electron dramatically.
In the Bohmian model, the electron's position inside an atom is stationary (and well-defined), since the electron is in a bound state. But we know that the electron has a kinetic energy of 13.6eV in the hydrogen atom. Isn't this a contradiction in the Bohmian model?
 
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In the Bohmian model, the electron's position inside an atom is stationary (and well-defined)
No, it isn't. The electron's wave function is stationary. That does not mean its position is stationary.
 
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in this case the usual pilot-wave model with particle positions as beables should be thrown away because it cannot reproduce QFT
Yes.

Why would one still consider it
Who is considering it? It is of historical interest as regards the development of quantum interpretations, but AFAIK nobody is actively promoting non-relativistic Bohmian mechanics as a valid interpretation today.
 
No, it isn't. The electron's wave function is stationary. That does not mean its position is stationary.
But there is no Bohmian trajectory in this case since the wavefunction is not evolving. How come isn't the electron stationary and has kinetic energy? In Bohm's model every motion is guided by the wavefunction.
 
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there is no Bohmian trajectory in this case since the wavefunction is not evolving
The wave function doesn't need to change with time to have a Bohmian trajectory. In the dynamics of the Bohmian particle, the wave function plays the role of a potential (you will see the term "quantum potential" a lot in the literature); a potential that doesn't change with time doesn't stop the particle from having a trajectory.
 
It doesn't change at all if the electron is in a stationary state--which, as I said, is the usual assumption when analyzing phenomena like the Lamb shift.



Bohmian mechanics assigns a "position" to the electron, but that position has nothing to do with the wave function. It's a separate property, and since we never actually measure the positions of electrons in atoms, the Bohmian "position" is irrelevant to the analysis of phenomena like the Lamb shift; all you need is the wave function.



No. The Bohmian "position" is not uncertain at all. Bohmian trajectories are perfectly deterministic.



No, they wouldn't. The Bohmian trajectory of the electron is driven by the wave function.
The Wikipedia page for lamb shift states: "The fluctuation in the electric and magnetic fields associated with the QED vacuum perturbs the electric potential due to the atomic nucleus. This perturbation causes a fluctuation in the position of the electron, which explains the energy shift". so there should be a change in the position of the electron. How does this work? If one is accepting Bohmian positions, it must be them that are fluctuating.


Finally, let me ask my question again. How would one stick to Bohmian mechanics with exact particle positions, while also including QED effects? I think in this case QED effects must be able to change the Bohmian positions or the wavefunction, otherwise how can have any effect?
 
The wave function doesn't need to change with time to have a Bohmian trajectory. In the dynamics of the Bohmian particle, the wave function plays the role of a potential (you will see the term "quantum potential" a lot in the literature); a potential that doesn't change with time doesn't stop the particle from having a trajectory.
Look at this: http://www.bohmian-mechanics.net/whatisbm_pictures_hydrogen.html

It has clearly depicted Bohmian positions for |1s> and |2s> rest states as stationary.
 
Finally, let me ask my question again. How would one stick to Bohmian mechanics with exact particle positions, while also including QED effects? I think in this case QED effects must be able to change the Bohmian positions or the wavefunction, otherwise how can have any effect?
This is kind of like asking how to drive a car to the moon. BM is not sufficiently developed to handle QED.

But that aside, Bohmian electrons are taken to be, across all H atoms in the universe, distributed in accordance with the equilibrium assumption, which is how the Born rule is satisfied. It doesn't matter whether the orbitals are slightly skewed by the Lamb shift or not. The only difference is the equilibrium condition would have a slightly different target distribution.
 
This is kind of like asking how to drive a car to the moon. BM is not sufficiently developed to handle QED.

But that aside, Bohmian electrons are taken to be, across all H atoms in the universe, distributed in accordance with the equilibrium assumption, which is how the Born rule is satisfied. It doesn't matter whether the orbitals are slightly skewed by the Lamb shift or not. The only difference is the equilibrium condition would have a slightly different target distribution.
The lamb shift and QED effects must change the position of the electron, otherwise how can the position distribution change?

This was my question, that whether or not QED effects modify Bohmian trajectories. They change the electron's energy (velocity), so they should also change the position. If the Bohmian position exists, so it's the Bohmian position that changes. This is what I meant.
 
The Lamb shift changes the energy levels/wavefunction/orbital. The Bohmian particle position is a hidden variable layered on top of this.

The d/t velocity of the Bohmian particle literally has nothing to do with the energy of the orbital, and has no meaning in measurements or role in any physical process. Your own example showed this, where the 1s orbital had motionless Bohmian particles even though 1s has non-zero energy.

Like I said, the Lamb shift would ultimately be factored in as a skew in the aggregate distribution of electrons across all atoms.
 
The Lamb shift changes the energy levels/wavefunction/orbital. The Bohmian particle position is a hidden variable layered on top of this.

The d/t velocity of the Bohmian particle literally has nothing to do with the energy of the orbital, and has no meaning in measurements or role in any physical process. Your own example showed this, where the 1s orbital had motionless Bohmian particles even though 1s has non-zero energy.

Like I said, the Lamb shift would ultimately be factored in as a skew in the aggregate distribution of electrons across all atoms.
Thanks. But my problem isn't resolved. In Bohmian mechanics the configuration of the universe is defined by particle positions. The lamb shift must change that if it's real, and we know it is. If it doesn't change the position of the particle, the position distribution must not change.

Remember that Bohmian mechanics says the position is real, it's not determined upon observation, according to the squared amplitude of wavefunction. It has existed before, and has had a certain value.

We that lamb shift changes the d/t velocity (since it changes the energy of the electron), so it also changes the position. I don't get what you guys say that it doesn't.
 
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The Wikipedia page for lamb shift
This is an "A" level thread. You should not be citing Wikipedia. "A" level means graduate level, which means you should already be very familiar with textbooks and peer-reviewed papers on the topic. Those are what you should be giving as references.

Look at this
Not a valid source. Have you actually read any textbooks or peer-reviewed papers on this topic? If you have, cite them. If you haven't, you should not even be trying to understand an "A" level topic like this until you have mastered the basics, and you certainly shouldn't be posting about it.
 
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In Bohmian mechanics the configuration of the universe is defined by particle positions
No, it's defined by particle positions plus the wave function. The wave function is real in Bohmian mechanics.

How would one stick to Bohmian mechanics with exact particle positions, while also including QED effects?
Since nobody has developed a consistent relativistic version of Bohmian mechanics, the short answer is, you can't.

Thread closed.
 

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