Are virtual particles really there?

[kexue]

Quantum mechanics: Myths and facts
H. Nikolic
(Submitted on 21 Sep 2006 (v1), last revised 16 Apr 2007 (this version, v2))

Fair enough. But I have provide quotes, too, that give different opinions.

I thought about posting more quotes, but then, what for?

 

[tom.stoer]

I can not judge this. But it is written down in Michele Maggiore A Modern Introduction to Quantum Field Theory, page 219.

I do not have access to this book; are you sure that he means the quantization itself or only the way it used (simplified). I do not see which step in the canonical quantization uses something that restricts this approach to the perturbative regime.

 

[nismaratwork]

I do not have access to this book; are you sure that he means the quantization itself or only the way it used (simplified). I do not see which step in the canonical quantization uses something that restricts this approach to the perturbative regime.

I do, and it in no way supports this approach... in fact it's the textbook (literally) approach all the way. Page 219 is just the first page on the chapter of "Path Integral Quantization", and the first topic is, "Path Integral Formulation of Quantum Mechanics."... Again, basic, and again, showing a lack of understanding around that material.

edit: Here's the table of contents for Maggiore's book: http://elib.tu-darmstadt.de/tocs/126170703.pdf

 

[kexue]

Even for a free field theory, which doesn't involve any virtual particles whatsoever, we need to integrate over all possible paths.

A free quantum field does not involve any virtual particle whatsoever?

And in the path integral we have classical fields.

So I can not follow.

 

[nismaratwork]

A free quantum field does not involve any virtual particle whatsoever?

And in the path integral we have classical fields.

So I can not follow.

Just ask some specific questions, and people here will answer them. No one can read your mind, so if you want to follow the logic, ask some specific questions. When asked one in return, answer it to the best of your abilities... you may find this improves the pace and quality of this thread.

 

[kexue]

As tom and others keep pointing out, for page after page... you continually make statements as though they're fact, when they are blatantly wrong. When asked simply to support your view with a formula, you evade. So, will you write it out, or not? You're clearly not some hapless newcomer to QM, so it seems odd that you make these sweeping generalizations, share a number of emails, but you won't write out an equation to support your point when politely asked by tom.stoer? I don't think you want mentors going over your posts kexue, you're no exactly being the most helpful conversationalist.

Nismaratwork, I have not heard one substantial input from you in this thread.

Except for disparaging coments about opinions that do not agree with yours, even those from Nobel Prize winners or respected textbooks.

I was asked about my take on virtual particles, here is mine, (again).

Virtual particle transcend perturbation theory. They allows us to have one coherent quantum field theoretical picture of nature. Virtual photons, are not less real, not less an mathematical idea than the electromagnetic field. Can we really see an electromagnetic field? No. We can just feel its effects on how it changes charges.

QFT says that a sea of virtual photons, which are the excitations of an quantized electromagnetic field transmits momentum between two charges. But compared to the picture of an electomagnetic field moving the charges, this picture comes with the huge benefit in that it gives us one picture, a picture that describes field and particle behaviour. That is because when we more and more shake one of the two charges, we get more and more 'less off-shell photons', we turn increasingly "virtual" into "real" photons, we can detect more and more clicks in our measurement apparatus. Very few for radio waves, many more for waves with higher frequencies. These less off-shell photons can travel much farther until they get absorbed, they don't fall off with 1/r^2 as the "virtual" photons, the "more off-shell photons" in the Coulomb field.

Only if a photon lives forever, moves forever, it would be on-shell. Every photon that gets created and absorbed is not.

This is one beautiful picture of how nature works, and that is the picture of quantum field theory. Amen


Where do you disagree?

 

[tom.stoer]

Where do you disagree?

That's not the point. Strictly speaking this is not quantum field theory but paraphrasing quantum field theory. In order to get a clear understanding I would like to ask you again to write down an equation valid non-perturbatively which contains mathematical symbol representing your "virtual particles".

 

[tom.stoer]

A free quantum field does not involve any virtual particle whatsoever?

That's why we insist in writing down a non-perturbative definition of virtual particles.

In perturbative quantization schemes you make an expansion of amplitudes in terms of the coupling constant. Simply speaking a virtual particle is an internal line of a Feynman diagram drawn between two vertices. But in a free theory there are no vertices (b/c each vertex comes with a coupling constant which is zero in a free theory).

If you look at the path integral of a free qm particle in one dim. you will see that there is no coupling / no interaction / no potential, but nevertheless you sum over all paths, not only over one single classical path (which is a straight line).

 

[nismaratwork]

Nismaratwork, I have not heard one substantial input from you in this thread.

Except for disparaging coments about opinions that do not agree with yours, even those from Nobel Prize winners or respected textbooks.

I was asked about my take on virtual particles, here is mine, (again).

Virtual particle transcend perturbation theory. They allows us to have one coherent quantum field theoretical picture of nature. Virtual photons, are not less real, not less an mathematical idea than the electromagnetic field. Can we really see an electromagnetic field? No. We can just feel its effects on how it changes charges.

QFT says that a sea of virtual photons, which are the excitations of an quantized electromagnetic field transmits momentum between two charges. But compared to the picture of an electomagnetic field moving the charges, this picture comes with the huge benefit in that it gives us one picture, a picture that describes field and particle behaviour. That is because when we more and more shake one of the two charges, we get more and more 'less off-shell photons', we turn increasingly "virtual" into "real" photons, we can detect more and more clicks in our measurement apparatus. Very few for radio waves, many more for waves with higher frequencies. These less off-shell photons can travel much farther until they get absorbed, they don't fall off with 1/r^2 as the "virtual" photons, the "more off-shell photons" in the Coulomb field.

Only if a photon lives forever, moves forever, it would be on-shell. Every photon that gets created and absorbed is not.

This is one beautiful picture of how nature works, and that is the picture of quantum field theory. Amen


Where do you disagree?

I disagree the moment you believe that a picture (I'd say paradigm), however useful in this case, makes a particle have a physical reality. I appreciate that you have a singular and unwavering belief in some kind of finality to be found in QFTs, but that is not a view that I believe many share. It's nice that your picture is beautiful, and it's even better that it's so fantastically successful when it comes to predicting nature, but that doesn't make a virtual photon real.

Using your logic I should discard every theory for the next which is more beautiful and complete, even if (unlike SR, GR, QM) there is no experimental evidence or observational data to support it. In fact, the logical step in your reasoning is String Theory, which is far more complete and lovely.

Anyway, the entire issue of absorption and emission isn't settled, but the odds that nature will end up imitating our mathematical artifacts to do so seems silly.

 

[tom.stoer]

Just to give you an impression how a non-perturbative formulation of QCD looks like: http://physik.uni-graz.at/itp/oberw/oberw08/Vortraege/reinhardt_oberwoelz08.pdf

 

[nismaratwork]

That's not the point. Strictly speaking this is not quantum field theory but paraphrasing quantum field theory. In order to get a clear understanding I would like to ask you again to write down an equation valid non-perturbatively which contains mathematical symbol representing your "virtual particles".

Kexue: Can you do what is in bold text above? Yes or No... simple answer? This is what... the sixth time you've been asked for this?

 

[kexue]

That's not the point. Strictly speaking this is not quantum field theory but paraphrasing quantum field theory. In order to get a clear understanding I would like to ask you again to write down an equation valid non-perturbatively which contains mathematical symbol representing your "virtual particles".

No Tom, this is exactly the point! This is the basic idea of QFT. It demands virtual particles to work, they are essential to quantum field theory.

And as I said, in order to do non-perturbative quantum field theory you have to use the path integral approach, where you compute not with quantized fields, but with classical fields, were virtual particles per definition do not appear. But to do non-perturbative calculations you have to integrate over all paths, even over virtual ones.

So asking for non-perturbative calculation with virtual particles, does not make sense. What I can show though, is that every non-pertubative calculation which has to be carried out by path integral approach, implies, of course, integrating over virtual paths. Per definition.

 

[nismaratwork]

No Tom, this is exactly the point! This is the basic idea of QFT. It demands virtual particles to work, they are essential to quantum field theory.

And as I said, in order to do non-perturbative quantum field theory you have to use the path integral approach, where you compute not with quantized fields, but with classical fields, were virtual particles per definition do not appear. But to do non-perturbative calculations you have to integrate over all paths, even over virtual ones.

So asking for non-perturbative calculation with virtual particles, does not make sense. What I can show though, is that every non-pertubative calculation which has to be carried out by path integral approach, implies, of course, integrating over virtual paths. Per definition.

This sounds like a logical contradiction for the theory, and therefore the math... I could be wrong, but it seems as though you're shooting your own view down. Really, if you're saying that there's no mathematical representation that you can offer to support your view, just say it flat out, no frills.

 

[tom.stoer]

No Tom, this is exactly the point! This is the basic idea of QFT. It demands virtual particles to work, they are essential to quantum field theory.

Wrong; historically and from textbooks one could get the impression that QFT is about perturbative methods and virtual particles. But this is history! QFT is about quantizing fields (calculus is not about Taylor expansion, either).

And as I said, in order to do non-perturbative quantum field theory you have to use the path integral approach, ...

Wrong; please check the link in my last post.

But to do non-perturbative calculations you have to integrate over all paths, even over virtual ones.

Wrong; please check our last posts. You ALWAYS have to integrate over ALL paths. And there are no "virtual paths".

So asking for non-perturbative calculation with virtual particles, does not make sense.

OK. But as non-perturbative methods are more fundamental tham perturbative ones, virtual particles are not fundamental, either.

What I can show though, is that every non-pertubative calculation which has to be carried out by path integral approach, implies, of course, integrating over virtual particles. Per definition.

You mix up different concepts. For perturbative calculations you have to integrate over virtual particles. Simple example: Electron-electron scattering at tree level: it is a perturbative process involving one virtual photon.

 

[kexue]

Ok, I think we reached a somewhat dead point here. I'm sure you can't hear the word virtual no more as much as I do! I think I have stated my case as best as I can. And yes I could be wrong. Maybe at least we could agree that this is not an easy question. May others contribute to this sticky thread, and shall it reach 10000 posts! I leave (sorry!) with one last quote from Leonard Susskind (could not contain myself and had to write him an email). I kinda like it, I'm more than sure others do not, but here it goes.

I will give you an answer, I am virtually sure it would have been Feynman's answer. All particles are virtual. At least all particles that begin at a source and end in a detector. All photons that we detect are radiated at some finite place and are absorbed at some finite place. In other words they are exchanged between two systems. It could be the filament of a light bulb and the retina of your eye.

We usually pretend that some particles come in from infinity, and others go out to infinity. Those are the ones we call real. But as I said, the particles produced by
a source and are then detected (or are otherwise absorbed) are virtual. I hope that helps.

 

[tom.stoer]

Of course if you accept the fact that all detected particles must be virtual particles by definition, then you end up with having only virtual particles and no real particles in your theory. That's correct.

But - what do we learn from that? If we have "virtual particles" only, why not call them "particles" w/o the "virtual"?

I think this does not help

 

[kexue]

Of course if you accept the fact that all detected particles must be virtual particles by definition, then you end up with having only virtual particles and no real particles in your theory. That's correct.

But - what do we learn from that? If we have "virtual particles" only, why not call them "particles" w/o the "virtual"?

I think this does not help

So you finally agree that we should not differentiate between "virtual" and "real" particles?

Tom, this exactly is the point!

 

[tom.stoer]

Please check my post #25:

I think the existence of virtual particles is more a question about "existence" than about "virtual particles". Let me explain:

First of all virtual particles "exist" as they are (as a mathematical tool) able to describe a real process accessable experimentally.

They do not exist in the same sense as the "real" out-states exist b/c an out-state is always "one single particle" whereas a virtual particle is an integral over a "collection of particles" described by propagators. So there is a huge difference.

Last but not least there is a ontological paradox. First of all virtual particles do not exist as no experimentalists cares about them; what is prepared is an in-state, what is detected is an out-state, so virtual particles are not accessable experimentally. But as soon as a particle is detected it interacts with the measuring device; this interaction is described via a Feynman diagram and in this Feynman diagram the out-state becomes an internal line, a virtual particle so to speak. If existence is related to observation everything that exists can only be described by virtual particles b/c out-states = real particles do not interact and are therefore never observed by construction!​

So it seems that I already came to this conclusion some time ago :-)

The key issue is: if you restrict yourself to "virtual particles" in the textbook sense you are far away from modern physics. If you instead insist on virtual particles being real you run into a paradox or a tautology. If you try to base your understanding of QFT on virtual particles you may run into the problem to mix up "perturbation expansion" with "definition of quantum field theory".

Virtual particles don't help us to understand QFT; they confuse us! Once we have understood QFT and once we did some perturbative calculations we may safely talk about them (like the Nobel prize winners you have cited). But from an educational perspective it's b..llsh..

Btw.: you seem to see the PI formalism as more fundamental than the canonical approach. Let me say that it's the other way round: the PI approach was originally derived by Feynman using the canonical approach and there are many physicists today rating the canonical approach as the more fundamental one.

 

[haushofer]

It has nothing to do with renormalization, but with the definition of the perturbation series itself...

Thanks, that's clear :)

 

[xepma]

I have not followed the whole threat, so forgive me for barging in here, but...

No Tom, this is exactly the point! This is the basic idea of QFT. It demands virtual particles to work, they are essential to quantum field theory.

And as I said, in order to do non-perturbative quantum field theory you have to use the path integral approach, where you compute not with quantized fields, but with classical fields, were virtual particles per definition do not appear. But to do non-perturbative calculations you have to integrate over all paths, even overvirtual ones.

So asking for non-perturbative calculation with virtual particles, does not make sense. What I can show though, is that every non-pertubative calculation which has to be carried out by path integral approach, implies, of course, integrating over virtual paths. Per definition.

Are you claiming that non-perturbative approaches can only be carried out using a path integral approach? Because that statement is simply wrong.

Take for instance Conformal Field Theory in two dimensions. They are an example of integrable quantum field theories: quantum field theories which contain an infinite number of symmetries. The symmetries give rise to an infinite number of constraints -- the Ward identities (a special class of these are known as Knizhnik-Zamolodchikov equations). It's the quantum version of a Noether current.

The constraints basically come down to differential equations which are satisfied by the correlation functions. Let me state that differently:

Every correlator satisfies a linear differential equation (with respect to the coordinates of the fields in the correlator). And by solving the differential equation you obtain an expression for the correlator.

Everything is exact, everything is non-peturbative. It's an interacting theory, and it's strong coupling. But most of all: there is absolutely no reference to intermediate states which are summed over. There are _no_ virtual particles.

 

[TrickyDicky]

Kexue, I still don't quite understand your point of view, first I thought you were confusingly mixing real and virtual particles but then you start bringing up a "field centered" view that seems to make distinctions between real and virtual particles useless, and this could be interesting, I always thought QFT was deriving towards a too "particle centered-high energies" perspective that seems to be a dead alley (well, let's wait til we have any surprise from the LHC), but then ,do you suggest that virtual particles are the quanta of some field?

 

[unusualname]

I think kexue just wants to know whether these "mathematical artifacts" are really popping in and out of existence, regardless of the arguments for and against mathematical rigour

eg: It's confirmed: Matter is merely vacuum fluctuations

I mean, we don't have to be able to observe them, I guess the point is that there should be a correct and simplest way to describe nature mathematically, and we should then be able to run a simulation of that mathematical model on a computer and have a "look" at the scale of protons and see what it looks like.

 

[tiny-tim]

I thought about posting more quotes, but then, what for?

so that we can read them

Only if a photon lives forever, moves forever, it would be on-shell. Every photon that gets created and absorbed is not.

do you have a reference for that (and not an email)?

 

[kexue]

Are you claiming that non-perturbative approaches can only be carried out using a path integral approach?

Kaku and especially Maggiore claim in their textbooks that non-pertubative calculations do not work in canonical quantization, since an exponential of an operator is defined by its Taylor expansion.

The rest I wrote was admittely wild speculation. All I know is, that when we got a path integral, either in qm or in qft, we have to integrate over all possible paths. In qm that would be paths that a classical particle never could take, paths that do not obey special relativity, i.e. faster than light, backwards in time, whatever. Similiar wild paths are taken when we integrate over field configurations. I called them freely virtual paths.

My reasoning was (proabably naive and wrong) that these "crazy" paths correspond in some sense to the virtual particles in the canonical quantization calculations.

Kexue, I still don't quite understand your point of view, first I thought you were confusingly mixing real and virtual particles but then you start bringing up a "field centered" view that seems to make distinctions between real and virtual particles useless, and this could be interesting,

I subscribe to what I arrogantly call the Feynman way of thinking, as described by the Susskind quote or what I was trying to convey more clumsy in post 111 and what I think is an important message to understand, that there is no qualitative difference between virtual and real particles, particles can be more or less "off-shell", but are never actually exactly on-shell.

Basically, this what I like to emphasize. Tom and others do not find that helpful, though I understand they admit it is a legal view.

And no, I do not think of virtual particles as little billard balls or interpret single Feynman graphs naively as a single physical processes.

And of course Tom's objections (first and foremost: where are the virtual particles in non-pertubartive calculations?!) are well taken, and to be honest I'm not in the position to argue with him. For that I know way to little quantum field theory.

 

[tom.stoer]

Kaku and especially Maggiore claim in their textbooks that non-pertubative calculations do not work in canonical quantization, since an exponential of an operator is defined by its Taylor expansion.

kexue, either you are misinterpreting something or this is simply wrong (sorry, Prof. Kaku). In canonical quantization there is at the very beginning no need to define an exponential of an operator. You just need the Hamiltonian H (which has a well defined exponential provided that H itself is well-defined). Later you may want to define the time evolution operator U, S- or T-matrix - and for that you need an exponential of iHt.

But it is this exactly this way you derive a path integral and for that we still do not know how to proof its existence. So the construction of H is fundamental to QFT, even for the PI; you can't bypass H or T in order to derive the PI.

[You may check textbooks where they use some thing they call H, sandwich it between some states and derive the PI; in a very last step they calculate the Gaussian integral to go from the Hamiltonian to the Lagrangian PI; in order to do that you need some H - and you need H to have a rather specific form: it has to be quadratic in the momenta; none of these steps can be justified mathematically w/o having first constructed H; what they do instead is to write down Z[J] and call it the PI; so I am sorry - I don't trust textbooks here as they omit well-known obstacles; what about Weinberg? he seems to be rather serious about those topics, but I haven't studied his books, unfortunately]

Note: regardless what we believe about H or Z, up to now no perturbation expansion has been used.

when we got a path integral, either in qm or in qft, we have to integrate over all possible paths. In qm that would be paths that a classical particle never could take, paths that do not obey special relativity, i.e. faster than light, backwards in time, whatever. Similiar wild paths are taken when we integrate over field configurations. I called them freely virtual paths.

I think you got the main idea; the confusion is due to the fact that you called these non-classical paths "virtual".

Assume for a moment that you have a way to define this procedutre rigorously; I think lattice gauge theory is a rather good example. Then your PI will collect contributions from non-classical paths. It that sense you are right. But then you have to make your last step drop the distinction between real and virtual. In lattice gauge theory all quarks and gluons "are" in a sense "virtual"; not even the bound state (e.g. the proton) is "real". But note: calling these "particles" "virtual" is misleading as the term is reserved for perturbation theory - but we didn't use perturbation theory here.

... what I was trying to convey more clumsy in post 111 and what I think is an important message to understand, that there is no qualitative difference between virtual and real particles, particles can be more or less "off-shell", but are never actually exactly on-shell.

As said I tend to agree here.

I think it was a long a thorny detour, but it was hopefulyl not in vain :-)

 

[unusualname]

I subscribe to what I arrogantly call the Feynman way of thinking, as described by the Susskind quote or what I was trying to convey more clumsy in post 111 and what I think is an important message to understand, that there is no qualitative difference between virtual and real particles, particles can be more or less "off-shell", but are never actually exactly on-shell.

I like this view, but I think the universe itself is "on-shell", but no subsystem within it is.

 

[tom.stoer]

now we have a sticky thread with 131 posts; that was of course not the intention for a sticky thread :-)

 

[nismaratwork]

So you finally agree that we should not differentiate between "virtual" and "real" particles?

Tom, this exactly is the point!

No, the point is that you're reading the right books and naming the right authors, but you don't understand the material. You know, it's easier to just take the time to learn and share ideas rather than try to convince a 'room' full of people that black is white.

As stickies for virtual particles go, there are MUCH better threads to sticky in which there isn't some endless battle for one fellow's ego and worldview.

 

[weejee]

I don't quite get this 'slightly off-shell' part.

To me, it sounds more natural to say that a real particle state contains small off-shell components. I guess it is a totally different situation compared to the case where a virtual particle becomes almost on-shell (which sounds more or less similar to slightly off-shell), which means that it is more of an actual transition rather than a perturbative correction.

 

[tom.stoer]

I don't quite get this 'slightly off-shell' part.

To me, it sounds more natural to say that a real particle state contains small off-shell components. I guess it is a totally different situation compared to the case where a virtual particle becomes almost on-shell (which sounds more or less similar to slightly off-shell), which means that it is more of an actual transition rather than a perturbative correction.

Please do interpretet too much here; these are only words, interpretations, ...

On-shell, off-shell real and virtual have a precise meaning in perturbation theory (and I think they are nearly meaningless in the non-perturbative regime). A particle with rest mass m is on-shell if it's 4-momentum p satisifies p² = m², otherwise it's off-shell. A real particle in a Feynman diagram (an in- or an out-state) is always on-shell, a virtual particle that is exchanged can be off-shell; in loops, where momentum-integrals survive the 4-momentum conservation constraint at the vertices, particles can be arbitrarily off-shell as one integrates over 4-momentum.

The propagator is usually something like 1/(p²-m²) which means that virtual particles must not be on-shell as that would mean they would always sit at the pole; therefore being off-shell and being virtual is the same.

You could e.g. use the naive electron-positron scattering at tree level to calculate the mass of the exchanged photon.

 

[weejee]

Please do interpretet too much here; these are only words, interpretations, ...

On-shell, off-shell real and virtual have a precise meaning in perturbation theory (and I think they are nearly meaningless in the non-perturbative regime). A particle with rest mass m is on-shell if it's 4-momentum p satisifies p² = m², otherwise it's off-shell. A real particle in a Feynman diagram (an in- or an out-state) is always on-shell, a virtual particle that is exchanged can be off-shell; in loops, where momentum-integrals survive the 4-momentum conservation constraint at the vertices, particles can be arbitrarily off-shell as one integrates over 4-momentum.

The propagator is usually something like 1/(p²-m²) which means that virtual particles must not be on-shell as that would mean they would always sit at the pole; therefore being off-shell and being virtual is the same.

You could e.g. use the naive electron-positron scattering at tree level to calculate the mass of the exchanged photon.

Right. Still, as is pointed out, all the particles we detect are not exactly in asymptotic states . They should have some amount of perturbative correction(off-shell components), in principle, although we don't really care about it when calculating the scattering matrix.

My point was just that calling this simply 'slightly off-shell' sounds misleading.

 

[tom.stoer]

If I understand you correctly you are referring to the fact that strictly speaking one cannot detect asymptotic plane wave states as they do (by definition) not ineract with a detector. I am not sure if one should try to fix this via "being slightly off-shell" rather than referring to the "measurement problem".

This would lead to a discussion "what are particles?" instead of our the discussion "what are virtual particles?"

 

[weejee]

If I understand you correctly you are referring to the fact that strictly speaking one cannot detect asymptotic plane wave states as they do (by definition) not ineract with a detector. I am not sure if one should try to fix this via "being slightly off-shell" rather than referring to the "measurement problem".

This would lead to a discussion "what are particles?" instead of our the discussion "what are virtual particles?"

Maybe I wasn't specific enough on expressing what I mean. Let me elaborate.

From what I know, in an asymptotic state, particles are assumed to be very far away from one another so that they don't interact. However, as this is only an idealization, the state that is actually detected should be (asymptotic state) + (correction terms due to inter-particle interaction). These correction terms should be off-shell.

 

[tom.stoer]

Something like that can be done, but it's not that particles are off-shell. I know the method of so-called distorted waves, which can e.g. be used to calculated scattering states, phase shifts, nuclean mass corrections in soliton models (Skyrme model and related models). The idea is not to use plane wave states but exact solutions of the full problem with the soliton included. These solutions are then used to construct the appropriate operators for the canonical quantization, especially H and T and to do the renormalization.

The difference to your ideas is that interaction taken into account is not due to the detector, so strictly speaking the asymptotic states are on-shell again.

 

[weejee]

Something like that can be done, but it's not that particles are off-shell. I know the method of so-called distorted waves, which can e.g. be used to calculated scattering states, phase shifts, nuclean mass corrections in soliton models (Skyrme model and related models). The idea is not to use plane wave states but exact solutions of the full problem with the soliton included. These solutions are then used to construct the appropriate operators for the canonical quantization, especially H and T and to do the renormalization.

The difference to your ideas is that interaction taken into account is not due to the detector, so strictly speaking the asymptotic states are on-shell again.

My point is that the detector is never placed at infinity, so that the "out-state" we measure is not exactly the asymptotic out-state, in which particles don't interact at all since they are infinitely far away. The difference between the actual and ideal out-states are basically perturbative corrections due to the interaction, and I guess it is OK to call just that part (however small that is) virtual or off-shell? However, as I've said, I disagree with calling the whole thing (the out-state we measure) "slightly off-shell".