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I am not an expert in quantum mechanics. However, after recently having studied wave functions, I was wondering if virtual particles have a wave function.

If not in an accelerating reference frame, can virtual particles appear to be real to an accelerating observer and therefore do they then have a wave function?

5 Answers5

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No, virtual particles do not have wavefunctions, since they do not represent actual states of the quantum theory but merely certain sums over states that appear in the mathematical formulation of perturbation theory. They're just lines in a diagram and do not correspond to an actual physical state, and are an artifact of perturbation theory.

See also this answer of mine for a longer discussion of the nature of virtual particles, and a proposed different meaning of the term which does correspond to an actual physical state and would therefore have a wavefunction.

ACuriousMind
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    The wavefunction of the virtual particles actually exists. It's the cross section of all intermediate particle paths as taken by the virtual particles. Like the wavefunction of a real non-interacting particles is the temporal crossection of all their free paths. – MatterGauge Feb 19 '22 at 23:17
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    @Felicia I'm afraid none of that made any sense to me. If your comment is intended as criticism of my answer, please express it by using standard terminology. – ACuriousMind Feb 19 '22 at 23:28
  • Coming back to that question I asked over 5 years ago, I think there's a way to make intuitive sense of virtual particles in the Schrodinger functional picture of QFT. In this picture, you calculate scattering amplitudes by explicitly keeping track of the intermediate states. When you do this perturbatively, you recover the same series expansion as in the usual Feynman diagrams, but now the terms actually correspond to intermediate field states, which one could call virtual particles. This is all worked out in Hatfield's QFT textbook. – knzhou Feb 20 '22 at 01:33
  • As a simple example, when you have a static, localized point charge, the electromagnetic field does have a state. It's a coherent state whose expectation value is just you'd get in classical field theory from a localized source. And there is a sense in which that field state doesn't obey the dispersion relation for real particles: individual field modes (with nonzero $k$) are excited but do not oscillate in time, $\omega = 0$. – knzhou Feb 20 '22 at 01:36
  • (It's just like what you'd get by applying a static force to a quantum harmonic oscillator. The result is a coherent state, but unlike coherent states for an unforced harmonic oscillator, the vev doesn't oscillate in time.) As a side project, I'm trying to make this all more technically precise, but my point is that the usual statement that virtual particles don't exist is not a statement about reality, and it's not even a statement about QFT. It's only a statement about one particular (though popular) way of setting up calculations in QFT. – knzhou Feb 20 '22 at 01:39
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    @knzhou If you want to write that up with a few more details as an answer either here or on your old question, I'd be interested in reading it. I'm skeptical that this works out, but I haven't seen the computations in the Schrödinger picture you mention, so maybe there's something to it (and then it's a shame that this isn't more widely known!). – ACuriousMind Feb 20 '22 at 01:42
  • Yup, I'll try! The details are definitely thorny, even in the simplest possible cases, which makes it clear why this picture is never used for practical calculations in high energy physics. But it should work in principle, and it has the benefit of being closer to how quantum optics people view QFT. – knzhou Feb 20 '22 at 01:45
  • This answer is incorrect because it neglects the fact that there does exist a wavefunction that includes the interacting particles and the virtual particle representing the interaction between them. – dllahr Feb 20 '22 at 13:10
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    @dllahr Read my linked answer and its comments - the ordinary meaning of "virtual particles" does not correspond to the actual intermediate state during an interaction. There of course is an actual intermediate state, but that's not what the internal lines in a Feynman diagram denote. – ACuriousMind Feb 20 '22 at 13:12
  • @knzhou I also hope you can find the chance to explain this in more detail! – Rococo Feb 20 '22 at 19:45
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Wave functions are the solutions of quantum mechanical equations, $Ψ$ and the complex product $Ψ^*Ψ$ is a real number. In the simple bound state solutions $Ψ^*Ψ$ gives the probability of finding a particle quantum at (x,y,z,t).

As real quantum particle interactions are not only bound states like atoms, but also scatterings and creation of new particles, a new mathematical format developed in order to be able to calculate these many particle reactions. It is quantum field theory (QFT), which is based on the postulates of quantum mechanics, but it is using an expansion in series in order to calculate the values for cross sections and decays.

Virtual particles are a mathematical construct within this series expansion. They are placeholders for the quantum numbers to be correct at the vertices of Feynman diagrams, so they have the name of an exchanged particle, but they are not particles and certainly are not solutions of a wave equation.

Virt

anna v
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Particles don't have wave functions. Only systems have wave functions. So in that sense no. But in a broader sense yes, virtual particles "show up in" wave functions and are closely related to them.

In scattering calculations (the usual context where Feynman diagrams and virtual particles are discussed), you treat the lab equipment as classical and located at infinity. In that limit, there is a sharp distinction between real particles, which reach infinity where they have a perfectly well defined energy-momentum that satisfies $E^2-p^2=m^2$, and virtual particles, which don't reach infinity. In real life, the lab equipment is at a finite distance and made of quantum particles, and you can imagine the external lines in the Feynman diagrams of your scattering calculation as internal lines of larger Feynman diagrams that include the preparation and measurement apparatus. In that more realistic picture, there aren't any unambiguously real particles. The distinction between "isolated particles with fairly clear energy-momenta that aren't significantly interacting" and "it's complicated" is one of degree, not kind.

Interacting relativistic quantum field theories are not mathematically well defined, and I'm not sure that anything can be definitively said about the relationship between Feynman diagrams and wave functions in them, but in simpler models you can make a mathematically precise connection, and the equivalents of virtual particles in those models do show up in the wave function. They have to, because the wave-function and virtual-particle descriptions are mathematically equivalent, and there are no unambiguously real particles in the latter.

Many people take the position that there is a classical world that is isolated from the quantum world by the Born rule or something like it, and that wave functions and virtual particles are just calculational tools and have no reality. That's a philosophically coherent position to take. But to say that virtual particles don't show up in the wave function is, as far as I can tell, not a coherent position, because virtual particles and wave functions are the same thing. You don't have to believe either one is real, but they're equally real.

benrg
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    No, this is another myth about virtual particles that doesn't hold up to technical scrutiny. Real particles are not "internal lines in a larger Feynman diagram". Feynman diagrams are representations of terms in a perturbation series, not of physical processes. For instance you can't even define what a "virtual particle" is in non-perturbative language, but the notion of "real particle" certainly needs to exist non-perturbatively. The internal lines do not represent quantum states at all, they are much more different from external lines than just "not reaching infinity". – ACuriousMind Feb 19 '22 at 15:25
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    @ACuriousMind I'd like to see a precise statement of what you're saying in a well-defined toy model. I agree that the internal lines don't represent quantum states – I said that in my first paragraph – but I assume your objection is deeper than that. – benrg Feb 19 '22 at 19:47
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    @ACuriousMind Here's an example of what I don't get in your other answer. You say "virtual particles do not appear because we expand the field. They appear because we expand the interaction part of the time evolution operator". Field vs TEO is a distinction without a difference. The reason it's only the interaction part is you already implicitly summed over the rest, because it's easy and there's no need to organize it with diagrams. It's not because vertices with two edges are fundamentally different from vertices with three, it's a practicality. – benrg Feb 19 '22 at 19:49
  • "Well-defined" is a thorny issue here. The "interaction picture" of QFT that we usually use does not rigorously exist (that's Haag's theorem). 2. If you agree that internal lines are not quantum states, then how can you claim that real particles are internal lines in "large" diagrams? A real particle, as something we can observe, is certainly a quantum state, is it not?
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    @ACuriousMind Re 1, I don't know what to do about that, but I did mention it in the answer. Examples of toy models without the problem are quantum circuits and cellular automata, though of course they lack features of RQFT that you could argue are essential. Newtonian QM may work, I don't know. Re 2, states represent systems, so a particle isn't a state unless the system is trivial. The system during the interaction is complicated but you can still describe it at some intermediate time by a state vector, which in the toy model can be related to "Feynman diagram equivalents". – benrg Feb 19 '22 at 20:57
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    @ACuriousMind A weakness of what I'm saying is there's really no good equivalent of Feynman diagrams in the toy models. I think each Feynman diagram represents a family of paths that are similar enough that you can deal with them as a unit. If you disagree with that then we're getting somewhere. There are no useful groupings in the toy models, so Feynman diagram = path, basically. – benrg Feb 19 '22 at 21:00
  • While the particle may be part of a larger system, it still has a state (though it may be mixed if the state of the total system is entangled). Feynman diagrams are not paths, at least not in the sense of "path" as in "path integral" - the paths in QFT are field configurations on spacetime, not (Feynman) graphs - so I don't know what you mean by that. – ACuriousMind Feb 19 '22 at 21:22
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    @ACuriousMind I think Feynman diagrams correspond to terms in a perturbative expansion of the path integral, and represent "parts of" the path integral, i.e., groups of paths. To define a mixed state you have to identify irrelevant degrees of freedom, and that's always somewhat subjective, unless you believe in objective collapse. A real particle probably won't interact any more, but you don't know for sure until $t=\infty$. That's what I was trying to say in the 2nd and 4th paragraphs. – benrg Feb 19 '22 at 22:13
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    Yeah, that's not the case, they aren't paths - in fact, the standard derivation of Feynman diagrams you'll see in most QFT intro texts happens in a canonical quantization formalism, not the path integral formalism. We're expanding the interaction part of the time evolution operator inside an expectation value - in the path integral formalism, you'd write a path integral to compute this expectation value, but the expansion that gives us the diagrams does not operate in terms of paths. – ACuriousMind Feb 19 '22 at 22:24