2

I know my question might have problems, but I am curious about it. In quantum field theory, particle-antiparticle pairs continuously pop in and out of existence from vacuum. These particles have a very short lifetime, at the scale of 10^-22 seconds. My question is, taking into account the QED which says that all electromagnetic forces are created by the exchange of virtual photons, can such shortly-living particles exchange any photon in their short lifetime and somehow perturb the electric field around them?

P.S. I know that according to the standard quantum mechanics, "vacuum fluctuations" only occur at the time of observation, and are NOT about dynamics of the system. I am talking in the scope of realistic observer-independent interpretations like Bohmian mechanics. In such interpretations, I think there is no explanation for the fluctuations other than that something is really happening in the real time.

Alex L
  • 1,145
  • 3
    "In quantum field theory, particle-antiparticle pairs continuously pop in and out of existence from vacuum." This simply isn't true -- where did you hear that? It's not even true in Bohmian mechanics, putting aside the fact that nobody has made Bohmian mechanics really work for quantum field theory. – knzhou Jan 06 '19 at 20:52
  • @knzhou As I mentioned, if they claim that their theory is observer-independent, so the fluctuations we see in the vacuum energy as we observe the vacuum, must have occurred in the real time independent of our measurement. And yes, particles popping in and out of existence isn't included in Bohmian mechanics. However, there have been some weird attempts to reconcile BM with QFT, and they have claimed that particles really appear and disappear. I will give you the links in the next comment. – Alex L Jan 06 '19 at 20:57
  • @knzhou https://arxiv.org/abs/quant-ph/0701085 , https://arxiv.org/abs/quant-ph/0208072 – Alex L Jan 06 '19 at 21:00
  • As far as I know, there is neither a rigorous theoretical description nor an experimental verification of the claim of "pairs popping in and out of existence". Any 'explanations' based on this are (extremely) heuristic. Another way of interpreting your question would be to ask if loops in Feynman diagrams have electromagnetic effects, which they surely do as is well known. One is Lamb shift as mentioned by @G. Smith in their answer. Another is the dependence of measured electron charge on probing distance. – Avantgarde Jan 06 '19 at 21:32
  • @Avantgarde I know, but the vacuum energy fluctuates while repeating the measurements. As I mentioned, if someone claims their theory is "observer-independent", then something must have really happened between the measurements which makes them have different results, so number of the (virtual) particles in the vacuum state must have suddenly changed. I myself don't believe this, but I think it's a consequence of what realists claim. – Alex L Jan 06 '19 at 21:43
  • @Avantgarde and let me ask my question in a different way. I asked, IF pairs popped in and out of existence for just 10^-22 seconds, could they have electromagnetic effects according to QED? – Alex L Jan 06 '19 at 21:48
  • @Ali To your last comment - yes, each pair will form a dipole. But I don't know of a resource where such a computation is done quantitatively. – Avantgarde Jan 06 '19 at 22:01

2 Answers2

6

I am going to write a careful answer, but I think it should be noted at the outset that the popular phrase "popped into existence" is entirely without meaning, as far as I can tell. Nothing ever has, or will, or even could "pop into existence". Physics is about cause and effect.

If I have a flat piece of paper and then I shake the paper, then a bump or a crease may appear where there was no such bump before. But it would be odd, I maintain, to say that the bump "popped into existence". The bump is a feature of the shape of the paper, and the paper has been manipulated. Such remarks can be transported quite straightforwardly to particle physics, where the paper is the quantum field and the bump is the type of excitation we call a particle.

Now you have in mind virtual particles, and you suggest that there are virtual particles in the vacuum. So let's consider that.

'Virtual particle' is the name we give to the internal lines in Feynman diagrams. There are plenty of Feynman diagrams having lines leading to loops, including virtual particle-anti-particle loops, and the result of the calculation represented by the diagram then does depend on the presence of these loops. So in this sense these virtual particles have physical effects. But this is putting the cart before the horse! The virtual particle is a part of the physical effect! The physical effect is the interaction of one quantum field with another. The virtual particles are a convenient way to lay out the calculation of that interaction.

All the diagrams I have discussed so far involve external lines: the incoming and outgoing physical entities whose interaction is being calculated. So they are not vacuum diagrams.

One can also draw diagrams containing "vacuum bubbles", i.e. a self-contained set of vertices and lines not connected to anything else. These diagrams represent integrals (as do all Feynman diagrams) and these integrals have strictly no effect on anything at all. They do not influence the outcome of any calculation involving external lines.

Finally, I note your P.S. and that your interest is in trying to figure out how field theory works from a Bohmian point of view. I guess the bottom line is that whatever view of quantum mechanics one takes, one wants ultimately to get accurate predictions of what measuring apparatuses will do. All my comments above are about that very thing: what the prediction is for observable behaviour such as detector clicks.

Andrew Steane
  • 58,183
  • Thanks for your great answer. As you pointed out, my question was about certain versions of QM which claim to be realistic and observer-independent. I was talking about the "vacuum energy" or "zero-point energy", not virtual particles in Feynman diagrams. I meant, when we measure the vacuum energy, the expected value is zero, but we have inherent uncertainty and measure some small nonzero value. But interpretations like Bohmian mechanics claim not to be dependent on measurement, so shouldn't something have popped in or out of existence so the difference in the measured value is justified? – Alex L Jan 07 '19 at 00:29
  • @Dan I think the vacuum energy is in fact uncertain, because, according to Heisenberg's uncertainty principle, Δt.ΔE > hbar/2, so in a very short time scale you cannot say that the standard deviation of the vacuum energy is zero, because that would mean ΔE = 0. For large time periods, however, you can say that the vacuum energy is almost certain. In fact, the standard deviation is so small. See https://physics.stackexchange.com/questions/53802/what-is-delta-t-in-the-time-energy-uncertainty-principle. – Alex L Jan 07 '19 at 02:32
  • @Dan No, actually, if you look at the first answer in the stack exchange page I referenced, you'll see that this uncertainty relation is mathematically derived, and has nothing to do with hardware errors and noise. It's something inherent to QM, just as the famous inherent position-momentum uncertainty relationship. Plus, vacuum state is defined to have "minimum" energy, not absolutely zero energy. You can see the Wikipedia page for more explanation: https://en.wikipedia.org/wiki/Vacuum_state – Alex L Jan 07 '19 at 03:36
  • Another page related to the E-T uncertainty: https://physics.stackexchange.com/questions/259334/is-there-an-actual-proof-for-the-energy-time-uncertainty-principle – PM 2Ring Jan 07 '19 at 12:41
  • 1
    @Ali and all; E-T uncertainty is like frequency-time uncertainty. It means that if an interaction (e.g. with a measuring device) lasts only a finite time $T$, then it cannot single out one energy from another better than $\hbar/T$. Energy conservation can only be precisely defined at long times, but whatever happens at short times cannot prevent conservation at long times. – Andrew Steane Jan 07 '19 at 12:51
  • @Andrew I think this is what I meant. Energy is conserved for long times, but not for observations in very short times, and this is why "Vacuum energy fluctuation" means. However, my point is that, IF someone (like Bohmists) claims their theory is measurement-independent, then they have to accept that the vacuum energy is really changing with time, they have no other way to justify the inherent difference between measurement results. – Alex L Jan 07 '19 at 22:30
  • @Dan You maybe confused my point. Vacuum state doesn't vary with time. However, if you "measure" the vacuum energy (even ideally without noise), you will always get a small non-zero value. This is standard QM in which both of us believe. However, some people, including Bohmists are claiming their QM is observer-independent. So I say in their theory something in the real time should have happened between measurements which makes their results differ. So they must accept particles are popping in and out of existence. This is their problem, what you say is right. Look at the P.S. in my question. – Alex L Jan 07 '19 at 22:39
  • @AliLavasani I'm not sure what you mean by "measure the vacuum energy". An inertially moving measuring device in otherwise empty flat spacetime will stay in whatever energy eigenstate it was prepared in. If it was not prepared in an energy eigenstate, it will evolve, but that would not constitute a measurement of anything. – Andrew Steane Jan 07 '19 at 23:25
  • @Dan, I'm not also sure what you mean, I'll quote what I mean from Wikipedia. It has nothing to do with the "measuring device": The presence of virtual particles can be rigorously based upon the non-commutation of the quantized electromagnetic fields. Non-commutation means that although the average values of the fields vanish in a quantum vacuum, their variances do not. The term "vacuum fluctuations" refers to the variance of the field strength in the minimal energy state. Hope it helps. – Alex L Jan 07 '19 at 23:50
  • @Dan You are right, but I think the filed operator is a mathematical thing, you will see the non-zero variance even if you could somehow magically do the measurement not using any physical device made up by molecules. The field strength is a random variable, and this is why it would be different for any measurement, even in identical conditions as the other measurements. – Alex L Jan 08 '19 at 04:56
  • @Dan and yes, in "standard" QFT, the vacuum state is time-independent. But as I told some people are suggesting realistic and deterministic (measurement-independent) interpretations (like Bohmian mechanics), and so they have to make it time-dependent to justify the different measurement results. See arxiv.org/abs/quant-ph/0701085 , arxiv.org/abs/quant-ph/0208072 . They are really doing this! – Alex L Jan 08 '19 at 05:01
  • I think this discussion is very close to the one I put here: https://physics.stackexchange.com/questions/444267/particles-fluctuations-and-the-quantum-vacuum-is-this-right; I hope this is useful – Andrew Steane Jan 08 '19 at 12:49
3

Yes, virtual electron-positron pairs in the quantum vacuum have electromagnetic effects. They modify the standard Lagrangian for electromagnetism into something called the Euler-Heisenberg Lagrangian.

One interesting physical effect caused by these vacuum fluctuations is the scattering of light by light. In classical electromagnetism, two light waves pass right through each other without interacting. In QED, the photons scatter off of the charged virtual electrons and positrons in the vacuum.

Another interesting physical effect is that if you create a sufficiently large electric field, the virtual electrons and positrons in the vacuum become real electrons and positrons. So you can create real matter and antimatter from just an intense, constant, uniform electric field.

As far as I know, neither of these effects has yet been observed, but they are predictions of QED, which is extremely well-verified.

Effects of vacuum fluctuations which have been observed include the “Lamb shift” of the energy levels of hydrogen, and the anomalous magnetic dipole moment of the electron. These are subtle effects which are essentially small corrections, as opposed to the first two effects I mentioned, in which vacuum fluctuations allow something totally new to happen.

G. Smith
  • 51,534
  • Well, I think if they can have electrodynamic effects, so they might have a contribution to the thermal noise (by perturbing the trajectories of charged particles chaotically moving around). – Alex L Jan 06 '19 at 21:06
  • The virtual pairs in the vacuum do slightly modify how charged particles scatter of each other. – G. Smith Jan 06 '19 at 21:18
  • 1
    I agree light by light scattering and Lamb shift etc, of course, but if one wishes to speak carefully then the internal dynamics of such processes are strictly not "in the quantum vacuum" as you put it, because other entities are present: photons in one case, the hydrogen atom in another, etc. One is examining the evolution of the fields when they are not in the state commonly called the vacuum state. Therefore in these processes we have not virtual pairs in the vacuum, but virtual pairs in the non-vacuum. I think it is valuable to maintain this distinction, in the interests of clarity. – Andrew Steane Jan 07 '19 at 13:19
  • I am interested in knowing something about QED. In QED, electromagnetic interactions are done by the exchange of virtual photons. Does this mean that the created pairs need some time in order to be able to exchange photons and have an effective electromagnetic effect? In this case, I was doubting maybe they cannot exchange photons in their very short lifetime and can't have EM effect. So do they still modify the EM field? Do particles at all need "time" to exchange a virtual photon in QED? – Alex L Jan 24 '19 at 19:48
  • Virtual electrons and positrons in the vacuum can exchange virtual photons, but it is not necessary for them to do so in order to be able to scatter real photons. (The Feynman diagrams in which they exchange virtual photons simply represent higher-order corrections compared with the diagrams in which thry don’t.) Also, with virtual particles, there is no concept of “how long” it takes them to do somethng. – G. Smith Jan 25 '19 at 01:30
  • Btw I am interested to know "how much" these virtual electron-positron pairs change the Euler-Heisenberg Lagrangian and if their impact on the charged particles can be somehow quantified. I know this question might be vague, but, IF possible, I'd like to have an approximation, for example like virtual particles can change the electromagnetic force on electrons by %1, or something like this. I just want to know how large their effect is, and how much they make particles' motion more unpredictable. – Alex L Feb 01 '19 at 00:18
  • @AliLavasani The virtual electron-positron pairs change the Coulomb potential to the Uehling potential (https://en.wikipedia.org/wiki/Uehling_potential). Using the short-distance approximation, I find that the correction to the potential at, say, 0.1 Compton wavelength away from the electron is about 1/100 of 1%. You have to get extremely close to the electron -- 10^-36 Compton wavelength away -- for the correction to become 1%. You can play with these formulas if you really care about the force more than the potential. – G. Smith Feb 01 '19 at 02:00
  • I can't help you with "how much they make particles' motion more unpredictable" because I don't really know what that means. – G. Smith Feb 01 '19 at 02:00
  • Thanks. Could you please let me know why the correction to the potential gets larger as the distance from the electron decreases? and what does this mean, does it mean that the potential gets more uncertain as we get closer to the electron? – Alex L Feb 01 '19 at 06:50
  • The strong electric field near the electron excites more virtual electron-positron pairs. (https://en.wikipedia.org/wiki/Vacuum_polarization) I don’t see anything “uncertain” about the potential. The correction has a well-defined formula. – G. Smith Feb 01 '19 at 07:00
  • Yes you are right. By "uncertain" I meant the instantaneous changes in the potential (and so the force applied to other particles around) which should be somewhat random due to the random (statistical) nature of particle creation and annihilation. – Alex L Feb 01 '19 at 07:48