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From asking several questions about virtual particles on this website, the most popular consensus (on PSE at least) seems to be that they are nothing more than a convenient way of expressing a perturbative expansion and therefore one should not read too much into them.

However other sources such as here make arguments about virtual particles that say even 'real' particles can be virtual depending on 'what scale' you do a calculation, with a classic example being 'in the process of a photon emitted from Alpha Centauri being absorbed by our eye, the photon is a virtual particle' (presumably since you can describe it by an internal leg of a Feynman diagram).

This doesn't seem to sit right with my current intuition that virtual particles are just artifacts of perturbative expansions. To me this process isn't an interaction between particles in Alpha Centauri and particles in my eye (which can be calculated by decomposing it into virtual particles) but is instead a process where a real photon has been emitted and transported and then absorbed.

How would you respond to this?

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    $\begingroup$ Feynman diagrams are not processes. There are no virtual particles because an e&m scattering is not, "an electron emits a photon, that then gets absorbed by another photon, resulting in momentum transfer" combined with "an electron emits a photon, the photon emits a positron-electron pair, that collides to form a photon, which scatters the secon electron" and so on and so on. Photons are on-shell states in the real electromagnetic field, full stop. $\endgroup$ Mar 31, 2021 at 19:26
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    $\begingroup$ Does this answer your question? Can virtual particles be thought of as off-shell Fourier components of a field? $\endgroup$
    – Rococo
    Mar 31, 2021 at 19:26
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    $\begingroup$ @Rococo it is very helpful although doesn't quite answer my question $\endgroup$
    – Alex Gower
    Mar 31, 2021 at 19:35
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    $\begingroup$ While the other question seems different at first glance, it refers to the same blog post by Strassler and I would write the same answer here - Strassler is playing a bit of a word game and uses "virtual particle" as synonymous with "intermediate state that's not a pure particle" in a manner different from "internal line in a Feynman diagram". Perhaps you can be more specific what about the existing answers to the linked question (in particular Rococo's and mine, which more or less together give "both sides of the coin" here) is lacking in your eyes? $\endgroup$
    – ACuriousMind
    Mar 31, 2021 at 20:09
  • $\begingroup$ @ACuriousMind actually your answer there does apply here. I like your comments about associating virtual particles with inserting identities. You do also stress there that the expansion is of the time evolution operator not the field. What then would you think about @ user1379857’s EM field analogy in their answer? $\endgroup$
    – Alex Gower
    Mar 31, 2021 at 23:41

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Part of me thinks that if the LSZ formula and asymptotic states (see here) were discussed rigorously in introductory QFT books, instead of the sloppy presentation that is usually given (if at all), all this confusion around virtual particles would be greatly reduced.

QFT is just infinite dimensional QM, and in QM, all you have are states and a Hamiltonian. Nothing more. There's nowhere for "virtual" states to hide.

A photon that travels from alpha centauri to your eye is a real photon, end of story. The little wavepacket that the photon that travels freely through space corresponds to a state which is recognizable in the Hilbert space.

Edit:

I invite you to read closely the blog post of Matt Strassler you posted, because I strongly agree with the view he advances.

In reading it just now, it occured to me that, in some sense, the difference between a particle and a virtual particle has nothing to do with quantum mechanics at all, but is even present in classical physics. Take for instance Maxwell E&M. The Coulomb $1/r^2$ field surrounding a charge is what is directly responsible to the attraction of charges over a distance. However, there are no light waves in this solution. Light waves are created when the charge is wiggled. This creates a propagating $1/r$ field. Griffiths in his E&M textbook gives a nice little analogy, saying that the $1/r^2$ Coulomb field is like the flies buzzing around near a garbage truck, but if a group of those flies splits off and flies off in some random direction, that's radiation, A.K.A. "light."

In classical Maxwell E&M, all that exists are the $E$ and $B$ field. Some of those solutions we identify with light waves (little propagating sinusoidal waves) but NOT all solutions! There are no light waves present in the $1/r^2$ Coulomb potential solution, and THIS is the solution mainly responsible for the attraction of charges.

Just the $E$ and $B$ fields are all that exists in classical E&M, the Hilbert space is all that exists in QFT. Sometimes a state in the Hilbert space has a natural interpretation as being made of particle states, like a photon emitted from alpha centauri moving across the cosmos, but not every state is understandable in terms of particles. Roughly speaking, I think the following analogy is a pretty good one:

Particle : Virtual particle = Light wave : Coulomb Field.

(However, don't get toooo excited about this analogy. As ACuriousMind comments, 'Strassler is playing a bit of a word game and uses "virtual particle" as synonymous with "intermediate state that's not a pure particle" in a manner different from "internal line in a Feynman diagram".' I think this is also a fair take, and really gets more into the language of what the heck a virtual particle is anyway.)

Also @Deschele Schilder's answer seems to disagree with yours somewhat, how you would respond to the difference?

One must remember exactly when the LSZ formula applies, because that's when all this scattering and Feynman diagram stuff comes into play. LSZ is relevant when you have widely separated wave packets which come to overlap significantly over some spatial region called the 'interaction zone' (see the answer and picture here). Now, when alpha centauri releases a photon, a widely separated wave packet leaving the star is exactly what you get. In other words, the photon leaving alpha centauri is already the `external leg' of the process that created it.

Also, philosophically, why should the electron in your eye be any more real than the photon that excited it? Who's to say that electrons being excited are "real" measurement, and the photons are just virtual? Couldn't you say that the only way you could ever detect electrons is by the emission and absorption of photons? That seems to me to produce a rather vile game of chicken and egg.

Would I be right in saying if you were wanting to calculate the interaction between an electron in Alpha Centauri and an electron in your eye you would have 'virtual photons' in the QED calculation, but that this is calculating a very separate process to the real photon absorption process in my question?

That's a good question, and I don't actually think you would be correct in saying that. For one, the electron in your eye is bound in an atom, in its own orbital, and is not a free electron. So really the correct Feynman diagram would probably involve an atom absorbing a photon and raising into an excited state. There's also a wide range of stuff floating out in the cosmos which you have to account for, so I would also imagine that decoherence effects would play a big effect on the photon on its journey.

To my mind, the most technically correct treatment would be to have the photon emitted as the external leg of some emission process on alpha centauri, then let it travel as an honest to god particle, and then draw a separate Feynman diagram for the absorption process by the atom. I think such a treatment has a stronger basis in the actual LSZ formula, which is where all of this particle scattering stuff is coming from when you get down to it.

It is crazy how I seem to see these sort of ideas spread by people with quite a large physics educational background through.

I think the fact that physicists with a high level of education disagree strongly on matter of interpretation and language becomes less crazy the more highly educated physicists you meet.

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  • $\begingroup$ It is crazy how I seem to see these sort of ideas spread by people with quite a large physics educational background through. Would I be right in saying if you were wanting to calculate the interaction between an electron in Alpha Centauri and an electron in your eye you would have 'virtual photons' in the QED calculation, but that this is calculating a very separate process to the real photon absorption process in my question? $\endgroup$
    – Alex Gower
    Mar 31, 2021 at 19:26
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    $\begingroup$ Please see my latest edit. $\endgroup$ Mar 31, 2021 at 21:22
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    $\begingroup$ ... decomposed into on shell waves. I am saying that the Coulomb field and waves are different phenomena entirely which both can arise from an underlying quantum field. I also noted that this is the case in classical E&M as well. So I'm not saying that the field can be expanded as the cited question posits. It has nothing to do with 'decomposing the field.' $\endgroup$ Apr 1, 2021 at 0:30
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    $\begingroup$ I don't necessarily want to state that what I said here is 100% the truth and what everyone else has said is 100% wrong. I have tried to explain my reasoning so that informed readers can decide for themselves if what I have said makes any sense. However, I do indeed disagree with the sentiment that the photons we see are virtual, but "very nearly" on shell, for the reasons I have written here. I think such photons are, within the LSZ framework, very easily identifiable with actual particle states. $\endgroup$ Apr 1, 2021 at 4:28
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    $\begingroup$ Ah, sorry, I think I see what you meant now. Using virtual particles to instead mean the 'near-field' disturbances you aren't doing any expansion or decomposing anything, you're just highlighting that the field can take forms that aren't just asymptoptic 'far-field' states. This is a completely different phenomenon to using the term 'virtual particles' to describe an internal leg in a peturbative expansion, but is a physical phenomenon which could make sense to be assigned that term anyway $\endgroup$
    – Alex Gower
    Apr 1, 2021 at 14:01
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I read the same remark in the book "Introduction to elementary particles" by David Griffiths. And in principle, he is right. All photons that find themselves between an emission event and an absorption event are virtual. A photon that is emitted on Alpa Centauri and absorbed in your eye is virtual. You can't observe it because if you observe it, then it isn't able anymore to reach your eye. In this sense, it's virtual. The photon is very near to real though, because it can also be considered as one leg of a Feynman diagram. The legs of these diagrams are considered real particles. And so the photon. You can also consider the interaction between an electron on Alpha Centauri and an electron in your eye as a reaction between two electrons mediated by a photon. The photon is in this case a virtual particle and the two electrons are real. The photon is very near to a real particle. It lasts for a long time, which means that it is close to its mass shell. Likewise, you can call the electrons virtual too, i.e. if they will be annihilated when they meet a positron.

In the context of quantum field theory, the answer is much more complicated (you can't even talk about particles), but I think the picture suffices in this context.

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Photons are described by propagating waves and obey the relativistic energy momentum relation. Virtual photons are just fourier components of the nonpropagating or near field.

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  • $\begingroup$ Wow very succint. I think I agree with half though. I agree that the virtual photons are linked to the nonpropagating or near field but the answers in (physics.stackexchange.com/questions/626172/…) seem to specifically say that virtual photons are not simply a fourier decomposition of the field since they are terms in the time evolution operator expansion not the field. What would you say? $\endgroup$
    – Alex Gower
    Apr 1, 2021 at 17:56

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