Question about Virtual Particles

[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.

 

[haushofer]

Perhaps people, including the popular literature, should stress more that when talking about QM or QFT, metaphores/analogies are used to translate the math into tangeble concepts.

 

[PeterDonis]

"QFT for the gifted amateur" by Lancaster, sidenote 2 in page 348

I don't have this textbook handy but I'll see if I can get hold of a copy.

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).

Except that a bare particle does not have to be on shell, since bare particles can correspond to internal lines, and a dressed particle that corresponds to an external line is on shell once you realize that the physically measured mass is the renormalized mass, not the bare mass.

 

[TrickyDicky]

Except that a bare particle does not have to be on shell, since bare particles can correspond to internal lines, and a dressed particle that corresponds to an external line is on shell once you realize that the physically measured mass is the renormalized mass, not the bare mass.

I wasn't implying that a bare particle has to be on shell, as a matter of fact that is why I referred to the renormalized dressed particle, because after renormalization one can no longer refer to on and of shell in the same way we do with the pre-renormalized Feynman diagram. See the definition of dressed particle in wikipedia:"dressed particle refers to a bare particle together with some excitations of other quantum fields that are physically inseparable from the bare particle. For example, a dressed electron include the chaotic dynamics of electron-positron pairs and photons surrounding the original electron." I figured it might aid as a graphical heuristic to get an idea of what "slightly off shell" means, if it doesn't work for you or find it misleading just disregard it.

 

[JK423]

TrickyDicky and PeterDonis are you still disagreeing on the reality of virtual particles or on the definions of off/on shell?

 

[yoron]

No - because virtual particles don't actually exist. They are simply an artefact of the mathematical methods used called perturbation theory and something called a Dyson Series:
http://en.wikipedia.org/wiki/Dyson_series

Thanks
Bill

I'm with you bhobba, virtual particles are a artifact. you're not the only one reaching that conclusion. And it makes a lot of sense to me, it simplifies.

 

[yoron]

Yes but there are no apparent forces in GR, at least that's the only way I can interpret the "appear". A thermal bath can't just appear, either it's there or it isn't. Otherwise you may be appearing to me this moment and fall into a Jabberwock if I accelerated in a certain way wrt you.

Relativity is based on observer dependencies, it's not based on gold standards.

 

[Gerinski]

So quantum fluctuations are real and they have observable properties similar to as if they were particles, it's just that they should not be called particles in the popular literature

 

[TrickyDicky]

TrickyDicky and PeterDonis are you still disagreeing on the reality of virtual particles or on the definions of off/on shell?

Never disagreed on the reality of virtual particles that I'm aware of; wrt on/off shell I think it is just an issue about words. QFT is admittedly hard to translate to english.

 

[ddd123]

Can't we reverse the point and consider "particles" are just another way of quantum modes to manifest? Otherwise we should start saying: photons are one thing, electron-proton binding in a hydrogen atom completely another thing - not so good. Entia non sunt multiplicanda praeter necessitatem.

 

[PeterDonis]

TrickyDicky and PeterDonis are you still disagreeing on the reality of virtual particles or on the definions of off/on shell?

I think it's more a question of how to translate the math into ordinary language, as TrickyDicky said. The ultimate answer to that is that scientific theories are not formulated in ordinary language; they are formulated in math. The math involved is unambiguous (though there are differences of opinion on how rigorous it is). But not all questions we can ask in ordinary language necessarily have answers in the math. For example, the question "are virtual particles real?" doesn't have an answer in the math; there's no mathematical property that corresponds to "real" or "not real". My personal preference is to stop asking those kinds of questions, but not everybody wants to do that.

 

[Vanadium 50]

For example, the question "are virtual particles real?" doesn't have an answer in the math

No, but questions like "can they be counted?" have answers, and I think most people would argue that counting is a property one would like real things to possess.

 

[bhobba]

So quantum fluctuations are real and they have observable properties similar to as if they were particles, it's just that they should not be called particles in the popular literature

How you reach that conclusion from this thread beats me.

If they were real all ways of doing the QFT calculations would have them - but they don't.

Thanks
Bill

 

[bhobba]

My personal preference is to stop asking those kinds of questions, but not everybody wants to do that.

I think only some want to ask that. As you understand the mathematical language of the more advanced areas of physics better you recognise the math is the theory and in math you can't express questions like that. Also very deep ideas like Noethers Theorem and the role of symmetry simply can't be expressed properly otherwise.

Thanks
Bill

 

[yoron]

Actually I do think you have a excellent point in lifting up 'thermal baths' ddd, " A thermal bath can't just appear, either it's there or it isn't. Otherwise you may be appearing to me this moment and fall into a Jabberwock if I accelerated in a certain way wrt you."

and one I'm wondering about too. But to me it is about causality? Observer dependencies suddenly seems only half of the answer.

"
Causality means that cause precedes effect : an ordering in time
which every observer agrees upon.

Q: Whether it is possible to change the order of cause and effect just
by viewing two events from a different frame.

A: two events can only be cause and effect if they can be connected to
one another by something moving at speed less than or equal to the
speed of light.

Two such events are said to be causally connected."

In a wider context causality is, to me at least, what joins the universe we see, no matter (local) observer dependencies? Seems to belong in the same category as tachyons to me, when thinking of it from causality? Maybe it's not belonging to this thread though.

 

[JVM]

Since QFT is simply a perturbation theory it means that we don't know what happens in these intermediate stages of a process, and although we use these intermediate processes, it is definitely what actually happens. The problem is that we don't have an exact description, QFT at this moment is just an approximation.

Maybe you can compare it with a Taylor series, if you're not really home in the subject, where a function may be described by some finite sum up to a certain approximation, it does not mean that the function is equal to this finite sum.
In QFT a process is approximated by some individual feynman diagrams, but this is only an approximation and is definitely not the real process.

Hopefully someone will find out what really happens, until then we are sadly limited to these approximations (which are incredibly correct in describing our world, and are not to be underestimated).