# Can particles popped into existence from the vacuum have electromagnetic effects on other particles?

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.

• "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 6 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. – Ali Lavasani Jan 6 at 20:57
• – Ali Lavasani Jan 6 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 6 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. – Ali Lavasani Jan 6 at 21:43

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.

• 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? – Ali Lavasani Jan 7 at 0:29
• @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 physics.stackexchange.com/questions/53802/…. – Ali Lavasani Jan 7 at 2:32
• @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: en.wikipedia.org/wiki/Vacuum_state – Ali Lavasani Jan 7 at 3:36
• Another page related to the E-T uncertainty: physics.stackexchange.com/questions/259334/… – PM 2Ring Jan 7 at 12:41
• @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 7 at 12:51

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.

• 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). – Ali Lavasani Jan 6 at 21:06
• The virtual pairs in the vacuum do slightly modify how charged particles scatter of each other. – G. Smith Jan 6 at 21:18
• 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 7 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? – Ali Lavasani Jan 24 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 at 1:30