Question about Virtual Particles

[PeterDonis]

If all the paths contribute, but none except one are not observed

This doesn't happen. If all the paths contribute, it's because no path is observed; only the start and end points are observed. If you are actually observing every point along the path, then only that path contributes--just as if, in the double slit experiment, you measure which slit the particle goes through, there is no interference.

 

[TrickyDicky]

But the fields are either never "virtual" or always "virtual", depending on how you look at it. They are never directly observable (so always "virtual" in that sense).

I agree with this. I'm not arguing that "virtual particles" exist, rather that the distinction real vs virtual is basically arbitrary, so that if we decide the latter are mathematical artifacts this applies to the former too, and similarly if it is decided they have an ontological meaning.

The expansion in perturbation theory is an expansion of the amplitude in powers of the coupling constant, not powers of the field--the field doesn't even appear (so fields are never "virtual" in that sense).

I was actually referring to the appearance of the "virtual particles" effects when applying the name "virtual fields", in the sense that is not localized in the way dots in a screen are.
Note in any case that the coupling constant is related to the strength of the interaction, the EM force field in this case although this restricted sense of field is admittedly not the usual sense of field in QFT.

 

[TrickyDicky]

This is not correct. The correct statement is that quarks and gluons are confined. But confinement does not prohibit you from looking at diagrams that have external quark or gluon lines. For example, the standard diagram for a weak interaction decay has external quark lines (the only internal line at lowest order is the W particle).

You mean in the form of hadrons, I was obviously referring to single particles.
But I believe we are arguing the same, the distinction internal-external line is to a great extent arbitrary and with very little bearing on ontology, contrarily to the way is usually or often claimed.

 

[PeterDonis]

You mean in the form of hadrons

No; weak interaction decay diagrams don't show hadrons, they show quarks and leptons. The fact that the quarks involved are bound into hadrons is irrelevant to the diagram, because as far as the weak decay is concerned, the quarks do not behave like bound particles.

the distinction internal-external line is to a great extent arbitrary

Not really. An external line corresponds to a particle that is required to be on the mass shell. An internal line corresponds to a particle that can be off the mass shell--more precisely, it corresponds to an integral over a whole range of possible particles with momenta that are off shell.

 

[ddd123]

How is this interpreted wrt, say, a hydrogen atom? The bound electron doesn't really exchange photons with the proton since they are off shell and are thus only mathematical artifacts?

 

[TrickyDicky]

No; weak interaction decay diagrams don't show hadrons, they show quarks and leptons. The fact that the quarks involved are bound into hadrons is irrelevant to the diagram, because as far as the weak decay is concerned, the quarks do not behave like bound particles.

This only shows again how arbitrary and removed from what is observed can diagrams be drawn.

Not really. An external line corresponds to a particle that is required to be on the mass shell. An internal line corresponds to a particle that can be off the mass shell--more precisely, it corresponds to an integral over a whole range of possible particles with momenta that are off shell.

You probably know that the concept of on vs off the shell is not clear cut in many situations.

 

[PeterDonis]

You probably know that the concept of on vs off the shell is not clear cut in many situations.

Can you give an example?

 

[PeterDonis]

How is this interpreted wrt, say, a hydrogen atom? The bound electron doesn't really exchange photons with the proton since they are off shell and are thus only mathematical artifacts?

That's a possible interpretation, yes. But note carefully what it does not mean. It does not mean the electron is not really bound to the proton. It just means that our model of the binding interaction as being mediated by the exchange of virtual photons is not necessarily "true"; it's just the best we can do at this time. That's because we can't actually solve the equation that describes the binding interaction in closed form; we can only approximate it by a power series expansion in the coupling constant. Since the terms in this series get smaller for higher powers of the coupling constant (at least, they do for this particular interaction), we can compute answers accurately enough to compare with experiment. This is called "perturbation theory", and "virtual particles" are just particular mathematical factors that appear in the various terms of the expansion.

On the "virtual particles aren't real" interpretation, the virtual particles are therefore just artifacts of the particular model we are using; a better model would give us a way to solve the equations in closed form without having to do a power series expansion as an approximation. We know this is possible in principle because there are cases in which we can actually do it--i.e., where we can obtain non-perturbative closed-form solutions.

 

[TrickyDicky]

Can you give an example?

Sure, any "real" particle is always slightly off shell as it is in its way between interactions, in this sense free particles are the ones that "don't exist".
And viceversa you gave the example of how a virtual particle like a quark can be drawn as an external line(on mass as long as we consider them as pairs or triplets) even if it is always confined.

 

[Gerinski]

What about this? In minute 38:00 of his lecture, Krauss says that over 90% of the mass of the proton is due to virtual particles. Does this not make them a real thing?

 

[Vanadium 50]

And that's exactly the problem with trying to learn physics by taking a line here and a line there from different popularizers.

 

[PeterDonis]

any "real" particle is always slightly off shell

No, it isn't. Off shell means it violates the energy-momentum relation. No real particle that's ever been observed does that.

 

[PeterDonis]

In minute 38:00 of his lecture, Krauss says that over 90% of the mass of the proton is due to virtual particles.

He is using "virtual particles" in yet a third sense, not yet used in this thread. What he really means is that over 90% of the mass of the proton is not due to the rest masses of the valence quarks that compose it. "Virtual particles" here is just a label for "all the other stuff".

 

[Nick666]

He actually says "over 90% of the proton mass comes from fluctuations in empty space"

 

[bhobba]

He actually says "over 90% of the proton mass comes from fluctuations in empty space"

That isn't quite what's going on - its more complicated than that:


The trouble with popular explanations, including what I posted, is it is at best a half truth not to be taken literally. Although having seen that video its rather good.

Thanks
Bill

 

[TrickyDicky]

No, it isn't. Off shell means it violates the energy-momentum relation. No real particle that's ever been observed does that.

Apparently you are missing my point. My point is that off-shell and on-shell is a convention of the formalism, and in that sense yes, external lines are defined as on-mass, no in-between case of ever slightly off-shell, and this is a practical formalism as it leads to calculations that result in good approximations to what is observed. But if one is going to make any ontological comment to this, like if that makes them real, or if they really exist or are math artifacts, one has to realize that the on-shell particle of the formalism is not "real" at all, and to be observed it must be considered as off-shell to some degree. Because the completely on-shell particle that appears in the formalism is an idealized entity that comes from infinity and goes to infinity, and you'll admit that what we detect has a finite source, detection and route between source and detection. So we come full circle and conclude that "real" particles are the actual mathematical artifacts which only function is approximate what happens in reality and the "virtual particle" perturbative corrections are what allow us to describe the finite interactions.
So it is always about useful conventions or formalisms, certainly not a debate about virtual vs real, in any case about the contradictory concept of particle itself. It is more than likely that a future theory will completely give up the concept as fundamental.

 

[PeterDonis]

off-shell and on-shell is a convention of the formalism

Off-shell might be considered this way, in the sense that we only integrate over off-shell momenta for internal lines in Feynman diagrams, so if we're not using that formalism, we're not really considering off-shell particles.

But on-shell is not a convention of the formalism, because on-shell particles are actually observed. See below.

to be observed it must be considered as off-shell to some degree

Then please show me the experimental results that have observed off-shell particles.

the completely on-shell particle that appears in the formalism is an idealized entity that comes from infinity and goes to infinity

That doesn't mean actually observed particles are off-shell; it just means that, in this particular formalism, we only know how to extract answers if we let the limits of integration go to infinity. That is a limitation of the formalism that does not in any way correspond to a physical condition on observed particles. The theory most certainly does not say that a particle can only be on-shell if it comes in from minus infinity and goes out to plus infinity.

 

[TrickyDicky]

On-shell particles are defined by the formalism to be the ones we observe as dots,clicks, etc, if there can be anything more conventional you tell me.
Now, have you ever observed a photon on infinite route between asymptotic source and detector? I don't think so. See? We have two contexts here.
"The theory most certainly does not say that a particle can only be on-shell if it comes in from minus infinity and goes out to plus infinity."
We are talking about QFT, right? What does S-matrix say in relation to interacting particles then in your opinion?

 

[PeterDonis]

On-shell particles are defined by the formalism to be the ones we observe as dots,clicks, etc, if there can be anything more conventional you tell me.

So you think direct observables are "conventional"? That's a very...interesting...use of terminology.

have you ever observed a photon on infinite route between asymptotic source and detector?

Irrelevant. See the last paragraph of my post #87.

What does S-matrix say in relation to interacting particles then in your opinion?

See the last paragraph of my post #87. You don't appear to understand what an "approximation" is. S-matrix theory, in which we take the limits of integration to infinity in order to compute matrix elements, is an approximation. Nobody claims that it means particles are only on-shell "at infinity" or that they can only be observed "at infinity".

 

[TrickyDicky]

So you think direct observables are "conventional"? That's a very...interesting...use of terminology.

This would indeed lead to an interesting discussion. Not going to happen.

Irrelevant.

Hmmmm, you are basing your point on direct observability and when I point out to you that you are not actually observing what you are claiming you observe you find it irrelevant, that's amusing.

Nobody claims that it means particles are only on-shell "at infinity" or that they can only be observed "at infinity".

You do. You just don't realize it.
I

 

[PeterDonis]

you are basing your point on direct observability and when I point out to you that you are not actually observing what you are claiming you observe

Um, what?

You do. You just don't realize it.

Please elaborate; I don't see where I have made these claims, or anyone else for that matter.

 

[TrickyDicky]

Ok, if you insist.
I was trying to explain to you that there are two planes in this discussion, let's call them the formal and the actual, it is well known that they contradict each other, it's an accepted feature of QED(it's technically called Haag's theorem), the good news is that we get really great results so we basically pretend the contradiction is not there and move on, but it's still there. Now you are ignoring my point about these two planes in relation with the concept of particle, and yet you are using it to attack my assertions from one of the planes contradicting it from the other plane, if I say something from the formal plane you go to the actual one and viceversa, that's just too easy, because as I said there is a basic contradiction between them. I doubt you are aware of it (don't consider you so mean) but it leads nowhere.
Now you could have manifested from the beginning that you don't agree with my distinction if that was the case and that would have just ended the discussion, but you chose to not even acknowledge it and instead nitpick anything I said in the way I described above.

 

[PeterDonis]

there are two planes in this discussion, let's call them the formal and the actual, it is well known that they contradict each other, it's an accepted feature of QED(it's technically called Haag's theorem)

I am well aware of Haag's Theorem and the school of thought that says it undermines the theoretical foundation of QFT. In fact I acknowledged that our QFT-based model of scattering is incomplete when I said that taking the limits of integration to infinity in order to compute matrix elements is an approximation.

However, none of this has anything to do with your claim that actually measured particles are off shell. If all you had said was that, theoretically, we don't have a rigorous way to compute matrix elements and detection probabilities at finite times using QFT--all we have is the approximation where we take the limits of integration to infinity--then I would not object (except to point out that the school of thought about Haag's Theorem that I referred to above is only one school of thought--it is by no means a mainstream position in QFT). But you said more than that: you said that actually measured particles are off shell. That's the statement I objected to. They're not; if they were, they would violate conservation laws.

 

[TrickyDicky]

you said that actually measured particles are off shell. That's the statement I objected to..

Only what I said exactly was(and I'm by no means the first to use this expression) "slightly ever off shell",not simply off shell(remember you asked me for an example of the meaning of on/off shell not being completely unambiguous), and that in the actual plane, in the formal plane they are of course on shell. So if that is your only objection we're done.

 

[PeterDonis]

and that in the actual plane, in the formal plane they are of course on shell

I think you have this backwards; they are on shell in the "actual plane", correct?

(I also wouldn't say they are "slightly off-shell" in the "formal plane"; I would say that in the model they are on-shell as an approximation, and beyond that we don't have a model from which we can extract answers.)

 

[TrickyDicky]

I think you have this backwards; they are on shell in the "actual plane", correct?

No, the formalism is the one that equates external lines with on shell particles, and we are talking about external lines in this particular case(real particles). What I call the actual plane is the realization that physically the asymptotic limit at infinity is an idealization required by the formalism and particles are really absorbed after finite time(therefore internal in a small amount) and in that sense one could refer to them as "slightly off shell". The fact is none of the characterizations is fully satisfactory due to the problems of the concept of particle itself.

 

[PeterDonis]

particles are really absorbed after finite time(therefore internal in a small amount)

The parenthesized statement is not correct. The fact that the particles are detected at finite time does not make them "internal". You have a misunderstanding of how perturbation theory works. I suggest actually looking at the math.

in that sense one could refer to them as "slightly off shell"

No, you can't, because, once more, "off shell" means "violates conservation laws". The theory does not predict that any observed particle violates any conservation laws.

 

[JK423]

The question of the reality of virtual particles arise a looot of times in a number of threads, and of course the reason is the usual treatment by textbooks.
People mean different things with the word *virtual*. Some mean the artifacts of perturbation theory, others mean the on/off shell particles.
Whatever the meaning one gives to the word 'virtual', in order to assess whether it is real or not he/she should ask the following question:

Is this object that i have in my mind (and i call it 'virtual particle') described by a quantum state that evolves in time?
A) NO: Then sorry it cannot be real. It is something that cannot even be described by quantum mechanics. The artifacts of perturbation theory are included in this case.
B) YES: Then perhaps it is real perhaps it is not, we can talk about it. If it has an evolving quantum state then you can probably interact with it using some probe during the time of its existence. If you can do that, then yes, it is 'real'. Whatever 'real' means in quantum mechanics.

 

[TrickyDicky]

The parenthesized statement is not correct. The fact that the particles are detected at finite time does not make them "internal". You have a misunderstanding of how perturbation theory works. I suggest actually looking at the math.
No, you can't, because, once more, "off shell" means "violates conservation laws". The theory does not predict that any observed particle violates any conservation laws.

I'll quit here, you can take a look at this if you wish : http://physics.stackexchange.com/questions/17087/slightly-off-shell?rq=1

 

[PeterDonis]

you can take a look at this if you wish : http://physics.stackexchange.com/questions/17087/slightly-off-shell?rq=1

This discussion thread is not an acceptable source by itself, and I don't see any references to actual papers, textbooks, or other scientific writings, except for one link to a 1975 paper that, as far as I can tell (it's behind a paywall so I can only read the abstract), does not support your position.

 

[Mark Harder]

Wiki also says:

"The longer a virtual particle exists, the more closely it adheres to the mass-shell relation. A "virtual" particle that exists for an arbitrarily long time is simply an ordinary particle.
However, all particles have a finite lifetime, as they are created and eventually destroyed by some processes. As such, there is no absolute distinction between "real" and "virtual" particles. In practice, the lifetime of "ordinary" particles is far longer than the lifetime of the virtual particles that contribute to processes in particle physics, and as such the distinction is useful to make."

Re. The lifetimes of real and virtual particles. There is an uncertainty relation between energy and time. The more certain you are of the time of some observation or event or lifetime of a quantum state, the less certain you can be about its energy. If a particle exists for a long time, and you know it, then its energy may be very precisely specified. Under those conditions, it makes sense to say the energy of the system is conserved. But suppose you consider a quantum state that exists for a much much shorter time. Then you would have only a vague idea of what its energy is. So a particle could appear out of nothing, seemingly violating conservation of energy, but only for this extremely short time. When the particle disappears again, energy disappears, again violating energy conservation. But the energy change over the lifetime of the appearance/disappearance event is zero! This, as I understand it, is the basis of that heuristic picture of Hawking radiation. For the briefest of instants, a particle-antiparticle pair winks into existence at the boundary of a black hole. Sometimes, before the pair recombines and disappears, one of the pair is sucked away by the hole. Leaving its partner to wander away, sometimes. I suppose one weakness of that explanation is that as long as the 2 particles exist, energy has been created out of nothing. Perhaps, in some way or another, the energies of the 2 particles cancel each other out?

 

[TrickyDicky]

This discussion thread is not an acceptable source by itself, and I don't see any references to actual papers, textbooks, or other scientific writings, except for one link to a 1975 paper that, as far as I can tell (it's behind a paywall so I can only read the abstract), does not support your position.

I didn't link it as a formal referece, just as an example where it is explained( the stack exchange has been linked many times in this site with no one complaining that I know of) since you were talking as if I was the one who came up with this concept, just for you to get some understanding of it. Off the top of my head you can also see the concept explained in a textbook that's been recommended here many times: "QFT for the gifted amateur" by Lancaster, sidenote 2 in page 348. I have seen it in other QFT textbooks that I don't have at hand if not maybe called "slightly off shell" which is not an standard term, but neither are the term on/off shell themselves if one is going to be strict about it.
You can think heuristically of an 'on shell particle slightly off shell' as a renormalized particle i.e.: a "dressed particle" in the sense that it is obtained from a "bare particle" (on shell) with the radiative corrections by virtual particles(off shell).

 

[Gerinski]

So, the popular assertion that a pair of particle + its antiparticle can spontaneously appear out of the vacuum is false?
And does not quantum tunneling also involve borrowing energy from the vacuum which can be thought of as virtual?

 

[JK423]

@Gerinski:

Yes, that assertion is wrong. And you will not find it in any 'good' textbook on QFT.

The reason why quantum tunneling happens is because the particle that tunnels does not have a well defined energy beforehand, i.e. it is in a superposition of states with different energies, including energies as high as to overcome the barrier. Nothing spooky (except the superposition).
Therefore nobody borrows from anywhere. There is no violation of energy conservation, not even for 'small times' as some popular assertions go..

 

[mattt]

Advise:

first: study it all in its mathematical rigorous (if possible) formulation (it will take many many long years of hard study).

second: only after that, read popularizations (pop-science books, famous quotes, whatever...) if you like, but now knowing perfectly well what actually lies beneath.If you ever want to (really) understand some of it, never start with pop-books.