Virtual Particles: Creation & Destruction in 10-43 Seconds?

[A. Neumaier]

I'm still waiting on your thoughts why Frank Wilczek would say in http://arxiv.org/PS_cache/hep-th/pdf/9803/9803075v2.pdf" on page 3 that the association of forces with the excange of 'virtual' particles is a general feature of quantum field theory.

I comment on his lecture in my theoretical Physics FAQ; see
http://arnold-neumaier.at/physfaq/physics-faq.html#virtual2
In the FAQ, you can also find a lot more information.

 

[tom.stoer]

This interpretation is not needed for the working of QFT and had done more damage than good.

Please make this large and bold!

 

[A. Neumaier]

What Does it mean when virtual particles travel "Backwards in time" does it mean literally?

No. It just means that one can draw Feynman diagrams where internal lines go backward in time. It has no meaning beyond that. In particular, interpreting a Feynman diagram as a process that happens in time is illegitimate.

I also see that virtual particles travel faster than c?

Since they don't travel but are drawn, virtual particles neither travel backwards nor faster than light. Such statements are only figures of speech, in an attempt to give life to lifeless drawings.

 

[A. Neumaier]

Please make this large and bold!

How?

 

[tom.stoer]

The diagrams were actually the device that made his approach intelligible and superior to the approaches by Tomonaga and by Schwinger.

The mistake was to sell the internal lines to the public as ''virtual particles''. This interpretation is not needed for the working of QFT and had done more damage than good.

 

[tiny-tim]

How!

 

[A. Neumaier]

A.Neumaier, this is still physics, not mathematics. As they like to say in the theoretical physics community, too much rigor soon leads to rigor mortis.

Even physics must be backed up with formulas; there is no substance in modern physics without lots of mathematics behind.

One cannot understand quantum mechanics without having understood the underlying mathematical machinery, at least on a level where one can translate claims into the formal language. (Though this is still far from mathematical rigor, which would be needed for an even deeper understanding.)

 

[tiny-tim]

Hi Dynamic Sauce!

What Does it mean when virtual particles travel "Backwards in time" does it mean literally?

I also see that virtual particles travel faster than c?

I've just replied in your https://www.physicsforums.com/showthread.php?t=452491".

 

[danlew]

They don't.

Virtual particles are not real.

Simple, no?

Are you saying they are not real to a humans eye?

 

[kexue]

I comment on his lecture in my theoretical Physics FAQ; see
http://arnold-neumaier.at/physfaq/physics-faq.html#virtual2
In the FAQ, you can also find a lot more information.

Well, I can not understand how anybody would say after reading this lecture that 'virtual' particles are a damage to understanding and describing nature, but whatever.

But I was referring to the survey article which is directed at physicists and what he says there on page 3. A.Neumaier, Tom, Tiny Tim, any thoughts?

It goes like this
With the correspondence of fields and particles, as it arises in quantum field theory, Maxwell’s discovery corresponds to the existence of photons, and the generation of forces by intermediary fields corresponds to the exchange of virtual photons. The association of forces (or, more generally, interactions) with exchange of particles is a general feature of quantum field theory.

 

[tom.stoer]

I think we discussed this example already, but I would like to come back to it. Look at the function

f(x) = \frac{1}{1-x} = \sum_{n=0}^\infty f_n x^n

Of course we all know that for |x| < 1 the coefficients are all equal to 1 and that for |x| > 1 this Taylor expansion at 0 is undefined.

Now let's talk about the question how "real" the function f(x) itself is and how real the coefficients are - especially if we do NOT specify for which value of x we want to calculate f(x).

I would say that the coefficients are an artifact of the Taylor expansion only, that for a different Taylor expansions we get different coefficients but we still have the same function f(x), so the function f(x) is the "real object" I am interested in. Of course we can find some funny names for the coefficients, we can use them in our calculations, we can speculate how "real" they are - but the object that really counts is f(x).

In the same sense the "physical object" object we are really interested in is (e.g.) an S-matrix-element. Of course we may calculate the S-matrix using perturbation theory,but this is an artifact of the calculation. If we would be clever enough to calculate it directly nobody would care about the expansion and nobody would try to find interesting names for the coefficients.

Of course the coefficients have a name (Taylor coefficients) and there are applications of Taylor expansions where Taylor expansion is really all we need. But there are other functions (with cuts, Riemann sheets, torus compactification, modular forms, ...) where Taylor expansion is closed to nonsense; it does not make sense, it does not solve your problem, it's the wrong tool, it hides reality and all relevant mathematical properties, ...

Another intersting function you may want to study is

g(x) = e^{-1/x^2}

How does its Taylor expansion at x = 0 look like?

 

[A. Neumaier]

the generation of forces by intermediary fields corresponds to the exchange of virtual photons. The association of forces (or, more generally, interactions) with exchange of particles is a general feature of quantum field theory.

What Wilczek says here is correct, but what you want it to imply doesn't follow.
This doesn't make virtual particles real. To say that a virtual particle is exchanged is just saying that there is a diagram in which this particle carries an internal line. But as the name says, the exchange is not real but virtual (on paper, in the mind of those telling or reading the story). It is figurative speech only. The correspondence referred to by Wilczek is one in the formulas, not one of processes that happen in space and time.
The latter cannot even be translated into a meaningful formal statement that could be checked for mathematical consistence.

There are infinitely many diagrams with all the possible exchanges and exchanges between exchanged particles, etc. They are all part of a perturbation calculation, not of something really happening. You cannot have at the same time one particle exchanged and 2 particles exchanged and 3 particles exchanged etc for any number of particles, although this is what is needed to compose the Coulomb force perturbatively from virtual particles.

Unlike superpositions of 1,2,3, etc. real particles, which one has in observable coherent states, there is no way to interpret the presence of the infinitely many exchanges as a superposition of 1,2,3 virtual photons. For in order to say this meaningfully, one needs virtual particle states that could be superposed, and these don't exist, not even virtually.

 

[kexue]

A.Neumaier, yes 'virtual' particles are no 'real' particles, we all knew that 300 posts before.

My last words are again the words of Frank Wilczek. (Why not put these up in the PF FAQ?)

It comes down to what you mean by "really there". When we use a concept with great success and precision to describe empirical observations, I'm inclined to include that concept in my inventory of reality. By that standard, virtual particles qualify. On the other hand, the very meaning of "virtual" is that they (i.e., virtual particles) don't appear *directly* in experimental apparatus. Of course, they do appear when you allow yourself a very little boldness in interpreting observations. It comes down to a matter of taste how you express the objective situation in ordinary language, since ordinary language was not designed to deal with the surprising discoveries of modern physics.

It is a matter of taste. Allow you a very little of boldness or don't. Or don't bother at all. That's all there is, really.

I leave the discussion on that. Good night

 

[PhanthomJay]

(..snip..)It comes down to what you mean by "really there" (..snip..).

One thing is certain, I'll be re-reading Hawking's full chapter on "What is Reality" (in his book, "The Grand Design") tonight...

 

[A. Neumaier]

It comes down to what you mean by "really there".

It is a matter of taste. Allow you a very little of boldness or don't.

Except that this boldness (or writing things in bold) doesn't help the slightest in understanding.

If one clearly distinguishes between reality and virtual reality, one finds that the physics of the former is much more rational than that of the latter, where everything goes, and
where (as the Wikipedia article on virtual particles shows) inconsistent statements stand undisputed side by side.

On the other hand, those accustomed to the view that virtual particles are ''really there'' have later a difficult time unlearning it when they want to get real understanding and
want to work with the concepts. Below the surface talk, nothing but internal lines in diagrams is associated with the concept. No states, no positions, no motion (forward or backward in time), no times, no creation or annihilation - nothing.

But of course, everyone has a choice about what to regard as real. Let the esoterically minded choose the fantastic view, unconstrained by the requirements of formal consistency.

 

[lightarrow]

...
Another intersting function you may want to study is

g(x) = e^{-1/x^2}

How does its Taylor expansion at x = 0 look like?

I could be wrong, but the case you aree discussing here doesn't seem to me a good metaphor of virtual particles being real/not real: here the Taylor expansion in 0 doesn't exist; in the case of virtual particles, the superposition instead does exist (mathematically, I mean).

 

[A. Neumaier]

I could be wrong, but the case you aree discussing here doesn't seem to me a good metaphor of virtual particles being real/not real: here the Taylor expansion in 0 doesn't exist; in the case of virtual particles, the superposition instead does exist (mathematically, I mean).

Superpositions of virtual particles do not exist in any sense, since virtual particles have no associated states.

On the other hand, the function f(x):={ 0 if x=0; e^{-1/x^2} otherwise} is infinitely often differentiable and has a Taylor series, but the Taylor series does not converge to the function since f(x) is not analytic. -
The point here is that one must distinguish between objects that have an intrinsic meaning (such as f(x) or the S-matrix) and auxiliary quantities used to compute them
(here perturbation series, that need not even be related to the function value).

Thus one should avoid calling something existing that has a mathematical meaning in some calculus for computing something that exists in the usual (non-mathmetical) sense.

 

[tom.stoer]

@A.Neumeier: so it seems you like my example :-)

What I wanted to show is that something that is rather familiar - the Taylor expansion - may fail completely in certain cases. But whereas the Taylor expansion fails the object "f(x)" still does exist!

"f(x)" and its Taylor expansion are simplified examples for "S-matrix" and its perturbation series. I hope it becomes clear that - even if the analogy is not perfect - virtual particles may fail to be a reasonable concept in the same way as the Taylor expansion may fail. But why should a rather limited mathematical tool be as real as physically measurable entities w/o such limitations?

 

[lightarrow]

Superpositions of virtual particles do not exist in any sense, since virtual particles have no associated states.

Thank you.

On the other hand, the function f(x):={ 0 if x=0; e^{-1/x^2} otherwise} is infinitely often differentiable and has a Taylor series, but the Taylor series does not converge to the function since f(x) is not analytic. -
The point here is that one must distinguish between objects that have an intrinsic meaning (such as f(x) or the S-matrix) and auxiliary quantities used to compute them
(here perturbation series, that need not even be related to the function value).

Thus one should avoid calling something existing that has a mathematical meaning in some calculus for computing something that exists in the usual (non-mathmetical) sense.

Probably it's already been asked, but is it possible to compare the non reality of virtual particles with the non reality of the wave represented by a wavefunction?

 

[tom.stoer]

Thank you.Probably it's already been asked, but is it possible to compare the non reality of virtual particles with the non reality of the wave represented by a wavefunction?

No.

Of course you could start a (philosophical) discussion regarding the reality of the wave function, but the mathematical status is different as (I am referring to A. Neumeier now)

Superpositions of virtual particles (in QFT) do not exist in any sense, since virtual particles have no associated states.
whereas
Superpositions of wave functions (in QM) do exist; wave functions are in one-to-one correspondence with states.

 

[A. Neumaier]

@A.Neumeier: so it seems you like my example :-)

Not really - I tried to rescue it. But it introduces an additional complication in what is
already a problematic thing...

"f(x)" and its Taylor expansion are simplified examples for "S-matrix" and its perturbation series. I hope it becomes clear that - even if the analogy is not perfect - virtual particles may fail to be a reasonable concept in the same way as the Taylor expansion may fail. But why should a rather limited mathematical tool be as real as physically measurable entities w/o such limitations?

But experiment shows that in case of QED, the additional problems of exp^{-1/x^2} are absent since the perturbation series indeed approximates the function.

 

[tom.stoer]

QED yes; QCD below Lambda, no!

 

[A. Neumaier]

QED yes; QCD below Lambda, no!

In my understanding, the situation in QCD is different - here the divergent perturbation series is asymptotic to a physically meaningless object since the S-matrix must feature asymptotic bound stats rather than quarks, but the computed S-matrix has quarks as asymptotic states.

But this kind of nonexistence seems to me quite different from the misconvergence in your example, where a meaningful function is approximated by a meaningful (and correct) asymptotic expansion, which just happens to be convergent but to a different function.

But you probably know much more about QCD specific things than I, so please explain
in which sense the QCD situation is analogous to your example.

 

[tom.stoer]

Correct, is not possible to construct few-particle states from vacuum, quarks and gluons and to study their scattering below Lambda b/c quarks and gluons are "the wrong d.o.f." in this regime. But of course we should try something like Bogoljubov trf., dressing, integrating out d.of.s, ... in order to derive an effective theory.

The situation in QCD is even worse as in my exp(-1/x²) example as the physically reasonable operators are something like <Delta|J|proton> = <0|Delta* J Proton|0> where now |0> is the Fock vacuum and Delta, J and Proton are operators consisting of infinitly many Fock space operators.

I think we better stop discussing exp(-1/x²).

 

[Bararontok]

I read the replies and many of them say that the virtual particles are in fact no more than mathematical tools used to make calculations in quantum field theory but that they are not necessary to perform the calculations since there are other methods for making the calculations. Perhaps I have confused the virtual particles of mathematical calculations with the concept of field particles. When a gauge boson is within a field, is it appropriate to call it a field particle?

 

[tom.stoer]

In quantum field theory all "particles" ara quanta of the field.

 

[Bararontok]

Okay then, thank you for the clarification. It seems that there are only two distinctions for the boson; when the boson is within the force field, it is referred to as a field particle and when it travels freely in space outside the range of a force field, it is referred to as radiation.

 

[tom.stoer]

Okay then, thank you for the clarification. It seems that there are only two distinctions for the boson; when the boson is within the force field, it is referred to as a field particle and when it travels freely in space outside the range of a force field, it is referred to as radiation.

Unfortunately it seems that there is no such clarification. Your ideas are inspired by classical field theory, not by quantum field theory. The boson "is" somehow the force field. It is never outside the range of a force field; it is the quantum of the force field is is therefore never isolated from that field.

 

[Bararontok]

But what about when Fermions emit radiation? And I know that fields such as the electromagnetic field of a magnet have only a finite range so any particle that escapes the field no longer interacts with it. The same can be said when an object escapes the gravitational field of a planet, the object is no longer subject to the planet's gravity. And what about individual photons that are not part of a stream of radiation?

Of course I am not saying that the bosons are not the quanta of the fields, I am just saying that it is possible for a single boson to travel individually without being surrounded by other bosons.

 

[jtbell]

But what about when Fermions emit radiation?

Then we have real photons, not virtual ones.