A Can we create a random variable using QED effects?

Quantum Electrodynamics (QED) has some observable effects such as the lamb shift, which is mainly caused by the vacuum polarization and the electron self-energy. These effects contribute to the "smearing" of the electron in an unpredictable manner, other than the uncertainty we already have about the electron's position due to Heisenberg's uncertainty principle.


Now I'm curious, can we produce a random variable, the randomness of which is coming from such QED effects, and not the regular quantum uncertainty (aka even if we could somehow circumvent the uncertainty principle, it would still be unpredictable due to such QED effects)?


I myself was thinking that maybe the thermal noise in a conductor might have some QED-related randomness, because the electrons in metallic bonds are subject to strong nucleus electric fields responsible for the lamb shift.
 

king vitamin

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I'm not sure what exactly your litmus test is for an experiment "using QED effects" is, but I would argue that the NIST random number generator described in this this article (technical paper here) is effectively using quantum mechanics (and ultimately QED) to generate random bits. This is done by using entangled photons in a "Bell-test" type experiment, which may be described by a quantum optics setup that certainly involves QED, but it doesn't require computing nonlinear fluctuations like the Lamb shift or anomalous electron magnetic moment do.
 

Cthugha

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Now I'm curious, can we produce a random variable, the randomness of which is coming from such QED effects, and not the regular quantum uncertainty (aka even if we could somehow circumvent the uncertainty principle, it would still be unpredictable due to such QED effects)?
I am not sure I get what you mean. You are talking about typical experiments in, say, condensed matter physics, not in high-energy physics. This means that you get away quite well with low level descriptions using the Coulomb gauge, which end up giving you a description that is not Lorentz covariant, but in this setting you can live with this quite well. In this description you simply map the light field onto the physics of the harmonic oscillator with the two orthogonal field quadratures exactly out of phase with each other are mapped to position and momentum. Accordingly, you get the same uncertainty relations for the light field which you would get from the "regular" uncertainty principle. And this also has the same consequences, e.g. when you repeatedly measure the field of the vacuum state of the light field, you get a Gaussian distribution.
And of course this has already been used to build random number generators simply by repeatedly measuring the vacuum. I think the record random number generation rate based on this approach is now 17 Gbps. See https://www.nature.com/articles/s41467-018-07585-0 and references therein. The paper should be open access.
 

Vanadium 50

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(aka even if we could somehow circumvent the uncertainty principle, it would still be unpredictable due to such QED effects)
I don't see how this is even a well defined question. If you don't have QM, how do you even formulate QED?
 

DarMM

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Quantum Electrodynamics (QED) has some observable effects such as the lamb shift, which is mainly caused by the vacuum polarization and the electron self-energy. These effects contribute to the "smearing" of the electron in an unpredictable manner, other than the uncertainty we already have about the electron's position due to Heisenberg's uncertainty principle
I think you are talking about things like the Darwin term here when you refer to "smearing". The electron in a Hydrogen atom can be approximated as never oscillating finely enough to probe beneath the Compton length scale (otherwise it would have an amplitude for producing an electron-positron pair) and thus it experiences a "smeared" version of the Coulomb field.

However this isn't really some kind of separate source of randomness. It's just a certain low velocity expansion of the full QED case.

The predictions of Quantum Electrodynamics are simply probabilistic like all quantum theories. In certain regimes it approaches the Quantum Mechanical limit of a quantum particle in a classical potential, but that potential can be made more accurate than the naive Coulomb one by retaining some relativistic corrections which can be intuitively understood as a smearing.
 
@DarMM @Cthugha @Vanadium 50
Let me make what I mean more clear with an example. See this paper by Antony Valentini (https://arxiv.org/pdf/quant-ph/0203049.pdf), Section 4. The paper is in the realm of Bohmian mechanics, which is realistic and deterministic (but nonlocal). In BM, we still cannot circumvent Heisenberg's Uncertainty Principle (HUP), since we have to disturb the wavefunction of the particle to measure it. However, because of the consideration of realism, the position actually "exists".

Valentini suggests that in theory, if one has a "non-equilibrium" (not psi-squared distributed) particle with arbitrarily precisely known position, and that particle is entangled with the particle under measurement, it is possible to do measurements violating HUP and know the position of the measured particle arbitrarily precisely. Of course, this is only on paper and in theory, not in practice. But let's discuss it in principle, as it is.

Now, suppose that the author of the paper has achieved an infinite number of such "non-equilibrium" particles, and has infinite computational power with infinite speed available. He wants to revive Laplace's demon by circumventing HUP by his own method, and predict the future of the universe. But, he has no idea about QED effects, and has not considered them in this paper.

Now my question is clearer. QED phenomena like self-energy and vacuum polarization are known to exist, and they "smear" the positions of particles like electrons in a probabilistic manner. Now, do these effects make random changes to the positions of electrons and other charged particles, frustrating Valentini, who has in principle found a way to measure particle positions with infinite accuracy?!
 

Vanadium 50

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I do not understand Valentini's paper nor how it is relevant to this, but still do not see how you can get QED out of a theory that starts out with "QM is wrong".

In QM (and QFT), [itex]\hbar[/itex] comes about in the commutator of conjugate variables, e.g.: [itex][P,Q]=i\hbar[/itex]. If you don't have an uncertainty principle, that means everything commutes, so the commutators are all zero and that means you have nowhere to get an [itex]\hbar[/itex] from. I don't see how you can get any QFT to get the right predictions without an [itex]\hbar[/itex] somewhere.
 

Cthugha

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I am not really experienced with Bohmian mechanics, but my impression is as follows. I might be wrong, though: Valentini assumes that the uncertainties we see are a consequence of a kind of equilibration process for these particles in question and there may be parts of the universe, which are still not in equilibrium and may therefore show different distributions or no noise at all. Personally, I think this is not plausible, but whatever.

If we follow this thought, the light field is just a different system, which is in equilibrium in our part of the universe and the vacuum field shows field fluctuations due to the same reason. Accordingly, I would assume that the resulting physics depends on how fast these different systems reach equilibrium. If both the particles and the light field are at equilibrium, you should get standard physics. If the particles are not in equilibrium, but the light field is, you get the HUP violation stuff for the particles, but you should still get QED effects from the vacuum field fluctuations. If both are in non-equilibrium, both systems should show no fluctuations. But to be honest, this sounds like a non-testable and pathological question to me.
 

DarMM

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Now my question is clearer. QED phenomena like self-energy and vacuum polarization are known to exist, and they "smear" the positions of particles like electrons in a probabilistic manner
Here you are thinking of perturbative corrections as active processes, as if the first loop correction "smears" the electron. QED just gives a probability distribution for the locations of localized charges, that's all. It's not correct to think of there being two types of uncertainty: Heisenberg and Self-Energy.

Self-Energy and Vacuum Polarization are just names given to graphs in field theory one uses to remember the integrals used to compute the QED probability distributions. Those distributions then obey various uncertainty principle relations.
 
I am not really experienced with Bohmian mechanics, but my impression is as follows. I might be wrong, though: Valentini assumes that the uncertainties we see are a consequence of a kind of equilibration process for these particles in question and there may be parts of the universe, which are still not in equilibrium and may therefore show different distributions or no noise at all. Personally, I think this is not plausible, but whatever.

If we follow this thought, the light field is just a different system, which is in equilibrium in our part of the universe and the vacuum field shows field fluctuations due to the same reason. Accordingly, I would assume that the resulting physics depends on how fast these different systems reach equilibrium. If both the particles and the light field are at equilibrium, you should get standard physics. If the particles are not in equilibrium, but the light field is, you get the HUP violation stuff for the particles, but you should still get QED effects from the vacuum field fluctuations. If both are in non-equilibrium, both systems should show no fluctuations. But to be honest, this sounds like a non-testable and pathological question to me.
One of the QED uncertainty effects is that we cannot measure the position of the particle with a precision beyond one Compton wavelength, since the photon we would need must so strong that it would create particle-antiparticle pairs. In this case, the measured particle would be indistinguishable from the same particle in the produced pair.

I don't think this effect would vanish even if both the light field and the particles are in non-equilibrium. The non-equilibrium he is talking about just means that the position of the particle doesn't obey the psi-squared distribution, it doesn't have anything to do with their energy or QED effects. You would still need to "measure" that HUP-violating position if you are not the God and don't know it beforehead, and would be limited by the effect I mentioned.
 

Cthugha

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One of the QED uncertainty effects is that we cannot measure the position of the particle with a precision beyond one Compton wavelength, since the photon we would need must so strong that it would create particle-antiparticle pairs. In this case, the measured particle would be indistinguishable from the same particle in the produced pair.

I don't think this effect would vanish even if both the light field and the particles are in non-equilibrium. The non-equilibrium he is talking about just means that the position of the particle doesn't obey the psi-squared distribution, it doesn't have anything to do with their energy or QED effects. You would still need to "measure" that HUP-violating position if you are not the God and don't know it beforehead, and would be limited by the effect I mentioned.
Your reasoning does not make much sense. If (and this is a big if) you "switch off" uncertainty in the sense that you have well defined position and momentum, you need to stick to the scenario. Obviously, you could use any other arbitrarily light particle to measure the position of the particle you are interested in as it is also not subject to uncertainty. Basically, the same should also be true for photons in this scenario, but I would like to avoid the typical discussion about the non-existence of position operators for photons and other issues for relativistic particles.

Anyway, this discussion is rather pointless. If you break the laws of physics, you can break the laws of physics. That is the usual outcome of such topics.
 
Maybe you are right. Valentini's paper says that if we have a non-equilibrium particle with KNOWN exact position with an ARBITRARILY high accuracy, and it's entangled only and only with our particle of interest, it's possible to measure another particle's position with an arbitrary accuracy, violating HUP. His math and theory is true, but the question is "how" the position of the former particle is "known"? It's not available to you until you measure it, where you encounter the limitation I mentioned. Basically the "IF" in the first line is impossible to satisfy, but there is no problem with the theory.

What do you mean by "arbitrary light particle"? All photons are alike, and they need to have a wavelength shorter than the Compton wavelength (and very high energy) to measure the particle's position. This is something fundamental.
 

Cthugha

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What do you mean by "arbitrary light particle"? All photons are alike, and they need to have a wavelength shorter than the Compton wavelength (and very high energy) to measure the particle's position. This is something fundamental.
Well, why should I use a photon if I can just shoot some other light (well, in principle it can even be heavy) massive non-equilibrium particle with perfectly known position and momentum at it?
 
Well, why should I use a photon if I can just shoot some other light (well, in principle it can even be heavy) massive non-equilibrium particle with perfectly known position and momentum at it?
What do you mean by "other light"? What light do we have other than photon?
 

Cthugha

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Huh? What do you mean? I would take a massive particle that has less mass (is lighter) than the particle I want to measure. The problem basically reduces to a classical collision in this scenario.
 
So how can you use the lighter particle to measure the position of the heavier one?
 

Cthugha

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Ehm, just scatter it off it? If you do not mind about changing the state of the other particle significantly, you can even use a heavier particle or one of the same mass. If you already know the position and momentum of the first particle to better accuracy compared to what would be allowed by uncertainty, this is a simple "elastic straight impact on a particle at rest" problem.
 
The problem becomes circular. How do you measure the position of the first particle? Basically you are assuming the violation of the uncertainty to deduce it.
 

Cthugha

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I do not see a problem here. You assumed that such particles do exist. In such a scenario, putting the system into a momentum eigenstate will not randomize the particle position and vice versa, so it is trivial to measure the properties. Alternatively, you can just use the tiniest pinhole you can think of and repeatedly fire your particle of interest at this pinhole. At some point it will pass through. Now you know the position of the particle at this time very precisely. Now just place a second tiny pinhole a huge distance away and repeat the whole scattering process until the particle moves through both tiny pinholes. Now you know the position at two different times very precisely and can also deduce its momentum on its way between the slits. If there is no uncertainty at all, this trajectory will be fully sufficient to calculate everything you want to know (well, unless you toss the particle into a chaotic billard or something like that).
 
I do not see a problem here. You assumed that such particles do exist. In such a scenario, putting the system into a momentum eigenstate will not randomize the particle position and vice versa, so it is trivial to measure the properties. Alternatively, you can just use the tiniest pinhole you can think of and repeatedly fire your particle of interest at this pinhole. At some point it will pass through. Now you know the position of the particle at this time very precisely. Now just place a second tiny pinhole a huge distance away and repeat the whole scattering process until the particle moves through both tiny pinholes. Now you know the position at two different times very precisely and can also deduce its momentum on its way between the slits. If there is no uncertainty at all, this trajectory will be fully sufficient to calculate everything you want to know (well, unless you toss the particle into a chaotic billard or something like that).
You are not wrong, but as I told, the problem is circular. What you mentioned doesn't solve it. How narrow are your pinholes? If they are narrower than the uncertainty limit, you cannot know where they exactly are! You need to assume the violation somewhere.
 
You don't measure it. You control it, by controlling the process that emits it.
Right, but how much you can "control" it depends on the uncertainty limit. "Controlling" is nothing but measuring and correcting (a feedback process).
 

Cthugha

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You are not wrong, but as I told, the problem is circular. What you mentioned doesn't solve it. How narrow are your pinholes? If they are narrower than the uncertainty limit, you cannot know where they exactly are! You need to assume the violation somewhere.
I slowly get the feeling that you do not know the meaning of the uncertainty principle. The uncertainty principle is a statement about variances of statistical ensembles subject to identical preparation. There is nothing that keeps me from backtracking to get to know the position and momentum of some particle up to arbitrary precision, e.g. by performing two position measurements at different times. This is perfectly possible. I just cannot expect to repeat the experiment and get the same result again.

For reference: https://www.physicsforums.com/insights/misconception-of-the-heisenberg-uncertainty-principle/
 
By "uncertainty" I didn't mean HUP, I meant the position measurement limitation due to creation of particle-antiparticle pairs, which occurs while measuring beyond the Compton wavelength.

You CAN violate HUP according to Valentiny, but what I was saying was that you will face another limitation beyond HUP. Of course, Valentiny's method is also only on paper, it's almost impossible to do that in practice, since you need to find a nonequilibrium particle, only entangled with exactly your particle of interest.
 
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how much you can "control" it depends on the uncertainty limit.
Which uncertainty limit? If you mean the HUP, this is not correct; the HUP does not place limits on how accurately you can control the position of a single emitted particle.

If you mean the Compton wavelength limit, what you say is still not correct, because that limit refers to trying to measure a particle's position by hitting it with a photon; it has nothing to do with controlling the particle's position when it is emitted.
 

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