Non-Perturbative QFT without Virtual Particles

[rogerl]

Hi,

Is there a non-perturbative version of Quantum Field Theory that doesn't require virtual particles? If there is. Why is it not taught or emphasized so we have to completely do away with virtual particles which many pop-sci books claimed to be real when many experts here state these are just a mathematical artifacts of Perturbation Theory in QFT.

Is our current QFT just approximation or temporary effective theories? Meaning the Perturbation Theory inherent in it and the resulting virtual particles approach just a temporary model pending the right QFT theory? Is there justification that the right QFT theory is Non-Perturbative and hence doesn't require any virtual particles or perturbative expansion??

 

[meopemuk]

Hi,

Is there a non-perturbative version of Quantum Field Theory that doesn't require virtual particles? If there is. Why is it not taught or emphasized so we have to completely do away with virtual particles which many pop-sci books claimed to be real when many experts here state these are just a mathematical artifacts of Perturbation Theory in QFT.

Is our current QFT just approximation or temporary effective theories? Meaning the Perturbation Theory inherent in it and the resulting virtual particles approach just a temporary model pending the right QFT theory? Is there justification that the right QFT theory is Non-Perturbative and hence doesn't require any virtual particles or perturbative expansion??

rogerl,

there is a version of QFT that does not use virtual particles. See discussion in https://www.physicsforums.com/showthread.php?t=474666 [/URL]. However, this version still uses perturbation theory in calculations. Perturbation theory is not an inescapable feature of QFT. We need to resort to perturbations simply because otherwise calculations become too complicated to be doable. Our mathematics is not powerful enough to do QFT non-perturbatively in most cases.

Eugene.

 

[Demystifier]

There is at least one nonperturbative method in QFT without virtual particles: lattice QFT (used especially for QCD).

 

[rogerl]

rogerl,

there is a version of QFT that does not use virtual particles. See discussion in https://www.physicsforums.com/showthread.php?t=474666 [/URL]. However, this version still uses perturbation theory in calculations. Perturbation theory is not an inescapable feature of QFT. We need to resort to perturbations simply because otherwise calculations become too complicated to be doable. Our mathematics is not powerful enough to do QFT non-perturbatively in most cases.

Eugene.[/QUOTE]

Is it because of lack of computer processor speed that is why our mathematics can't do a QFT nonperturbatively? Or simply we don't math for it yet? And in case it can be done. Then virtual particles are no longer there? I just want to know if the concept of virtual particles are just dependent on the calculation method used. If virtual particles are just multivariate integrals and this mathematical procedure can be replaced by other that has the same or better prediction. Then it is when we can categorically state virtual particles are not real. However, if there will be no way to create non-perturbative QFT ever, then it's possible multivariate integrals are part of QFT, hence virtual particles are really there just like how the Dirac Equation predicts antimatter because it's in the math. We sure can't replace Dirac Equation by something that can't produce the antimatter, isn't it? So can we replace all the math of QFT by something that don't use perturbation and yet produce the same or better prediction??

 

[A. Neumaier]

Is it because of lack of computer processor speed that is why our mathematics can't do a QFT nonperturbatively? Or simply we don't math for it yet? And in case it can be done. Then virtual particles are no longer there? I just want to know if the concept of virtual particles are just dependent on the calculation method used.

Real quantum field theories are not exactly solvable; so one needs approximate methods. Apart from lattice QFT (which has no perturbation expansion and hence neither Feynman diagrams nor virtual particles), all approximation techniques use perturbation theory in one form or another. Each such approximation technique has (in each gauge) its own variety of virtual particles, with _different_ properties and _different_ associated integrals, which cannot be converted into each other.

If virtual particles are just multivariate integrals and this mathematical procedure can be replaced by other that has the same or better prediction. Then it is when we can categorically state virtual particles are not real. However, if there will be no way to create non-perturbative QFT ever, then it's possible multivariate integrals are part of QFT, hence virtual particles are really there just like how the Dirac Equation predicts antimatter because it's in the math. We sure can't replace Dirac Equation by something that can't produce the antimatter, isn't it? So can we replace all the math of QFT by something that don't use perturbation and yet produce the same or better prediction??

People use what seems best for their particular purposes. Lattice gauge theory predicts some things much better than perturbative techniques. The latter are much better at other things.

 

[rogerl]

Real quantum field theories are not exactly solvable; so one needs approximate methods. Apart from lattice QFT (which has no perturbation expansion and hence neither Feynman diagrams nor virtual particles), all approximation techniques use perturbation theory in one form or another. Each such approximation technique has (in each gauge) its own variety of virtual particles, with _different_ properties and _different_ associated integrals, which cannot be converted into each other.

People use what seems best for their particular purposes. Lattice gauge theory predicts some things much better than perturbative techniques. The latter are much better at other things.

What can perturbative QFT do that Lattice QFT can't? If Lattice QFT can't do it all and Perturbative QFT is still the complete formulation. Then multivariate integrals can't be dispensed with.

Or let's take the example of Dirac Equation, is there other calculation method that can put negative energy state away or dispense the existence of the antimatter? If none, then antimatter is part of nature and real. Similarly, if there is no calculation method that can put multivariate integrals (hence virtual particles) away, then virtual particles may be as real as antimatter. So the important question is. Is Lattice QFT complete or can it never do everything that perturbative QFT does?

 

[meopemuk]

If virtual particles are just multivariate integrals and this mathematical procedure can be replaced by other that has the same or better prediction. Then it is when we can categorically state virtual particles are not real. However, if there will be no way to create non-perturbative QFT ever, then it's possible multivariate integrals are part of QFT, hence virtual particles are really there just like how the Dirac Equation predicts antimatter because it's in the math.

Feynman had a very bright idea to represent certain QFT integrals as graphs with external and internal lines and vertices. Then there was a not very bright idea to call internal lines "virtual particles". Well, one certainly has the right to call a line drawn on paper (or a factor in an inegral) "virtual particle", if one chooses. But then one should understand that this is merely a figure of speech, and one should not expect to find these "virtual particles" in experiments and measure their properties.

The fact that Feynman integrals allow us to calculate certain observable properties very accurately does not prove the "existence of virtual particles". Feynman integrals also contain the integration sign $$\int$$. But nobody is trying to detect integration signs in experiments. My proposal is to drop the term "virtual particle" altogether and use a less suggestive term "propagator" instead. Then, hopefully, you will not be tempted to identify propagators with anything measurable.

Eugene.

 

[rogerl]

Feynman had a very bright idea to represent certain QFT integrals as graphs with external and internal lines and vertices. Then there was a not very bright idea to call internal lines "virtual particles". Well, one certainly has the right to call a line drawn on paper (or a factor in an inegral) "virtual particle", if one chooses. But then one should understand that this is merely a figure of speech, and one should not expect to find these "virtual particles" in experiments and measure their properties.

The fact that Feynman integrals allow us to calculate certain observable properties very accurately does not prove the "existence of virtual particles". Feynman integrals also contain the integration sign $$\int$$. But nobody is trying to detect integration signs in experiments. My proposal is to drop the term "virtual particle" altogether and use a less suggestive term "propagator" instead. Then, hopefully, you will not be tempted to identify propagators with anything measurable.

Eugene.

If non-perturbative QFT can be developed that don't use multivariage integrals, then we can say with finally virtual particles are just mathematical artifact of perturbation theory. But then about Dirac Equation, isn't it that antimatter is just a result of a sign, that is, a negative sign. If a negative sign is enough to produce Antimatter, then integral sign may as well produce virtual particles.

 

[meopemuk]

If non-perturbative QFT can be developed that don't use multivariage integrals, then we can say with finally virtual particles are just mathematical artifact of perturbation theory. But then about Dirac Equation, isn't it that antimatter is just a result of a sign, that is, a negative sign. If a negative sign is enough to produce Antimatter, then integral sign may as well produce virtual particles.

Sometimes mathematics provides useful clues to physics (like in the case of Dirac equation -> negative sign -> antiparticles) and sometimes it doesn't (like in the case of Feynman diagrams -> internal lines -> "virtual particles"). Each case is different, and there is no general rule.

Eugene.

 

[Chalnoth]

Feynman had a very bright idea to represent certain QFT integrals as graphs with external and internal lines and vertices. Then there was a not very bright idea to call internal lines "virtual particles". Well, one certainly has the right to call a line drawn on paper (or a factor in an inegral) "virtual particle", if one chooses. But then one should understand that this is merely a figure of speech, and one should not expect to find these "virtual particles" in experiments and measure their properties.

The fact that Feynman integrals allow us to calculate certain observable properties very accurately does not prove the "existence of virtual particles". Feynman integrals also contain the integration sign $$\int$$. But nobody is trying to detect integration signs in experiments. My proposal is to drop the term "virtual particle" altogether and use a less suggestive term "propagator" instead. Then, hopefully, you will not be tempted to identify propagators with anything measurable.

Eugene.

Except in QFT, there is no clear distinction between virtual particles and real particles. A real particle is simply a virtual particle taken to asymptotic infinity. In fact, this is exactly what happens in the Unruh effect and in Hawking Radiation.

That the number of virtual particles is different depending upon the mathematical representation isn't troubling at all to me, considering that the number of particles isn't conserved, and a variety of various things we usually consider to be "real" can change dramatically depending upon your representation (such as the number of dimensions).

 

[rogerl]

Except in QFT, there is no clear distinction between virtual particles and real particles. A real particle is simply a virtual particle taken to asymptotic infinity. In fact, this is exactly what happens in the Unruh effect and in Hawking Radiation.

That the number of virtual particles is different depending upon the mathematical representation isn't troubling at all to me, considering that the number of particles isn't conserved, and a variety of various things we usually consider to be "real" can change dramatically depending upon your representation (such as the number of dimensions).

After reading the following long thread for 4 hours:

https://www.physicsforums.com/showthread.php?t=75307
"Are virtual particles really there?"

I tend to believe these are just mathematical artifact of perturbation theory. That is why, I'm looking for non-perturbative approach to QFT that won't need any virtual particles. Of course if the latter is impossible, and perturbative is the only thing possible, then I wonder if Dirac Equation antimatter being negative sign can be likened to virtual particles being related to integral sign hence real like antimatter..

About a real particle is simply a virtual particle taken to asymptotic infinity. We are talking about off-mass shell and on-mass shell and whether it can be modeled in space and time. Real particles have wavefunction but virtual particles don't. The internal or external lines in the Feynman diagram is not important.

About Hawking radiation, what is emitted is real particle-antiparticle pair as in:

What happens is that the high energy of the gravitational field of the black hole creates real particle-antiparticle pairs -- that before that event it could be viewed in terms of Feynman diagrams as virtual is completely irrelevant to his argument. One of the particles is swallowed by the black hole, the other is radiated away. As a result, the black hole loses radiation, hence total energy, and its effective mass decreases because of mass-energy equivalence. (by Neumaier)

 

[Chalnoth]

From my point of view, everything in quantum mechanics is just a mathematical model anyway, from electrons to photons to quarks. Nobody has ever seen, touched, or felt any of these things. They only appear as components in the theory. So saying that "they only appear in the math" is, to me, just absurd: that's the case with everything in quantum mechanics.

 

[rogerl]

From my point of view, everything in quantum mechanics is just a mathematical model anyway, from electrons to photons to quarks. Nobody has ever seen, touched, or felt any of these things. They only appear as components in the theory. So saying that "they only appear in the math" is, to me, just absurd: that's the case with everything in quantum mechanics.

But there is no other way to model stuff like electron, photon, quark without them existing. That is. You can't find another calculation method where they don't exist. In virtual partricles. They can be made to disappear when non-perturbative approach is used in QFT. But then, is it really true that non-perturbative approach can replace our current perturbative QFT. That is the question of this thread. If no non-perturbative approach is successful, then it is possible virtual particles are as real as antimatter.

 

[friend]

I guess the question is: how do you actually prove that a particle can change its properties without resulting from an interaction with another particle? We know that they do as we have seen in particle accelerators. Otherwise, we may not be measuring closely enough. I think the non-perturbative approach is just taking all the virtual interactions into account in the aggregate. But if a particle requires interaction with another particle to change properties, such as momentum and spin, then it would seem all the interactions of virtual particles predicted by the perturbation expansion are real.

 

[ParticleGrl]

But there is no other way to model stuff like electron, photon, quark without them existing.

The problem is that there is no real difference between, say, a "virtual" photon and a "real" photon. We measure photons when they interact with something in a detector, so the photon is always an internal line in a feynman diagram. It might be an extremely long lived state, but its still internal.

Also- how do you unambiguously define something like a quark?- think about the Altarelli-Parisi equations, there is a non-zero probability to "find a gluon inside a quark." You can find pions inside protons, quarks inside photons. Everything is all mixed up.

 

[rogerl]

I guess the question is: how do you actually prove that a particle can change its properties without resulting from an interaction with another particle? We know that they do as we have seen in particle accelerators. Otherwise, we may not be measuring closely enough. I think the non-perturbative approach is just taking all the virtual interactions into account in the aggregate. But if a particle requires interaction with another particle to change properties, such as momentum and spin, then it would seem all the interactions of virtual particles predicted by the perturbation expansion are real.

https://www.physicsforums.com/showthread.php?t=75307
Thread title: "Are virtual particles really there?"

Here's what I read at the bottom of thread page #6 to give you a main idea of their arguments which is of the line that virtual particles are just mathematical artifact of perturbation theory

"The calculational tool represented by Feynman diagrams suggests an often abused picture according to which “real particles interact by exchanging virtual particles”. Many physicists, especially nonexperts, take this picture literally, as something that really and objectively happens in nature. In fact, I have never seen a popular text on particle physics in which this picture was not presented as something that really happens. Therefore, this picture of quantum interactions as processes in which virtual particles exchange is one of the most abused myths, not only in quantum physics, but in physics in general. Indeed, there is a consensus among experts for foundations of QFT that such a picture should not be taken literally. The fundamental principles of quantum theory do not even contain a notion of a “virtual” state. The notion of a “virtual particle” originates only from a specific mathematical method of calculation, called perturbative expansion. In fact, perturbative expansion represented by Feynman diagrams can be introduced even in classical physics [52, 53], but nobody attempts to verbalize these classical Feynman diagrams in terms of classical “virtual” processes. So why such a verbalization is tolerated in quantum physics? The main reason is the fact that the standard interpretation of quantum theory does not offer a clear “canonical” ontological picture of the actual processes in nature, but only provides the probabilities for the final results of measurement outcomes. In the absence of such a “canonical” picture, physicists take the liberty to introduce various auxiliary intuitive pictures that sometimes help them think about otherwise abstract quantum formalism. Such auxiliary pictures, by themselves, are not a sin. However, a potential problem occurs when one forgets why such a picture has been introduced in the first place and starts to think on it too literally."

-----------------

The essence of it all is that when you write a non-perturbative approach to Quantum Field Theory, the virtual particles disappear.

Anyway. Here's the relevant passage " perturbative expansion represented by Feynman diagrams can be introduced even in classical physics [52, 53], but nobody attempts to verbalize these classical Feynman diagrams in terms of classical “virtual” processes", Pls. adress this part, how do you refute it?

 

[Lapidus]

https://www.physicsforums.com/showthread.php?t=75307
Thread title: "Are virtual particles really there?"

Here's what I read at the bottom of thread page #6 to give you a main idea of their ...

Thanks for the link, Rogerl!

What about https://www.physicsforums.com/showpost.php?p=3007871&postcount=8" which says

Whether virtual particles are real or not is a moot question.

Here's the idea. In quantum mechanics nothing is really real unless you can observe it or measure it. In order to be observable, a particle has to have some minimum amount of energy for some minimum amount of time; this comes out of the uncertainty principle that says the product of those two things has to be bigger than a certain number.

So it's possible to conceive of a particle whose energy is not big enough or whose lifetime is not long enough to permit a true quantum measurement, but still both of them could be greater than zero. The world could be full of such particles, and the measurements would never show it.

Well, quantum field theory takes those particles seriously. It says they interact with observable particles, for example they make the electron which emits and absorbs them a bit heavier, and a bit more sluggish in motion, than it would be if they didn't exist.

Furthermore, QFT says that the virtual particles are the ones that carry the forces. For example with photons, the "real" photons make light, and other forms of electromagnetic radiation, but the virtual photons carry the electric force; a charged particle is charged because it emits virtual photons. And the other bosons, that carry the weak and strong forces, behave the same way. Real particles interact with each other by exchanging virtual bosons.

This is the story quantum field theory tells, and the justification, the reason you should at least consider beliving in it, is that it makes fantiastically correct predictions. That bit above where I said that interacting with virtual particles made the electron sluggish? It's called the anomalous moment of the electron, and the prediction, based on virtual particles, matches experiment to six decimal places.

Nicely put! And I think pretty much the standard view on virtual particles.

The way I see it, virtual particles are like other people's emotions -- we can't isolate them and put them in a box, but thinking about them is very helpful for making sense of the world. so: are other people's emotions real?

The problem is using the word "real". what does "real" mean? Especially in quantum physics! What a physicist cares about is whether the model matches observation. Quantum field theory, with it's virtual particles, matches observation very very well. end of story.

 

[rogerl]

Thanks for the link, Rogerl!

What about https://www.physicsforums.com/showpost.php?p=3007871&postcount=8" which says



Nicely put! And I think pretty much the standard view on virtual particles.

The way I see it, virtual particles are like other people's emotions -- we can't isolate them and put them in a box, but thinking about them is very helpful for making sense of the world. so: are other people's emotions real?

The problem is using the word "real". what does "real" mean? Especially in quantum physics! What a physicist cares about is whether the model matches observation. Quantum field theory, with it's virtual particles, matches observation very very well. end of story.

The reason it's not end of story is because they can replace our QFT with non-perturbative approach that can totally remove any existence of the "virtual particles" which are nothing more than multivariate integrals.

According to Tom in the same thread you mentioned at page 13 (message 207) (yeah, i read the whole thread more than 3 hours):

Kexue,

Instead of repeating the same view umpteen times which is obviously in vain, we should try to get a different perspective. There are essentially two, namely
1) the perspective of physicists during decades where virtual particles were used in calculations significantly advancing science
2) the perspective of physicists of physicists today where perturbation theory obviously meets its limits

First of all two comments: your statement that “… virtual particles are primarily not defined by perturbation theory” is simply wrong.

Then you didn’t understand my statement that “in calculus nobody would say that Taylor expansion is calculus; it's just one tool that applies to a certain problem space and fails dramatically for others”. Both perturbation theory and Taylor expansion are two rather limit tools in a much broader context; that’s what I wanted to say. Looking at the Taylor expansion of 1/(1-x) = 1+x+x²+x³+… and concluding that the function 1/(1-x) is (equivalent with) the entire set of Taylor coefficients {1, 1, 1, …} is wrong. In the same sense the perturbation expansion is not the theory itself!

Now let’s change perspective: one can look at the problem regarding virtual particles from an entirely different point of view, namely from rating progress in fundamental physics:

40 years ago: standard model (theoretically) established: (perturbative) renormalizability of QED, QCD and GSW model; Ok, fine.

Since then the progress (or the points were progress got stuck) is mostly related to non-perturbative methods (or to the lack of knowledge regarding methods beyond perturbation theory).

QCD scale Lambda indicates breakdown of perturbation expansion for low-energy phenomena;
Deep-inleastic scattering / nucleon structure functions F(x,Q²): Q² dependence captured perturbatively (scaling violations), x-dependence entirely non-perturbative;
Confinement, chiral symmetry breaking, QCD vacuum: all treated via non-perturbative methods;
Theta-vacuum, instantons, (merons, sphalerons, …): non-perturbative;
Complete understanding of anomalies (relation to Atiyah-Singer index theorem): non-perturbative;
Canonical quantization of QCD; in the meantime entirely non-perturbative w/o any reference to perturbation expansion or virtual particles at all;
Hadron masses, form factors: from lattice calculations, non-perturbative;
Sponataneous symmetry breaking, Higgs-like mechanisms: non-perturbative;
Perturbative renormalizibity (order by order) is fine, but the perturbation series as a whole does not converge; see my example regarding Taylor expansion; unfortunately the situation with perturbation expansion is much more serious as the radius of convergence is strictly speaking zero (asymptotic series / radius of convergence shrinks to zero in g when higher orders are taken into account)

Looking at quantum gravity: failure of perturbative quantum gravity (instead asymptotic safety which is a non-perturbative renormalization group approach; LQG: entirely non-perturbative from the very beginning)

Looking at string theory: the progress regarding perturbative string theory is tremendous, but there is essentially one big road block, namely that the proof of perturbative finiteness seems to be out of reach; no commonly accepted definition of a measure beyond two loops! Same problem as above, name divergence of perturbation series suspected

…

Conclusion:

“Reality” of virtual particles seems to be directly related with their usefulness in calculations. As soon as more advanced methods are developed, other concepts become “real”, whereas older (limited) methods fade away.

Questions to you:
Can something be “real” if it is limited to a rather narrow domain of problems?
Would you agree that in that case we simply “made it real” as we get used to it?
Would you please select a non-perturbative definition of a quantum field theory (e.g. lattice gauge theory w/o any gauge fixing), check some of its equations and show us the definitions of “particles”, “real particles” and “virtual particles” (quarks, gluons, hadrons)?

If from the very beginning of QFT non-perturbative methods would have been available, neither Feynman diagrams nor the term “virtual particles” would have been invented.

Comment? Anyway. What I ask in this thread is where is the non-perturbative QFT?? Can they make complete prediction and can it replace current QFT? If it can, then virtual particles are refuted. But if it can't be done, then virtual particles may be more than math.

 

[A. Neumaier]

What can perturbative QFT do that Lattice QFT can't? If Lattice QFT can't do it all and Perturbative QFT is still the complete formulation. Then multivariate integrals can't be dispensed with.

If we had computers that are nearly infinitely fast, so that lattice QFT could be done with 100 x 1000 x 1000 x 1000 lattices instead of the tiny lattices used today, we could completely dispense with perturbation theory. But perturbation theory is very useful especially in QED, where the coupling constant is tiny, since it gives extremely high accuracy results with a limited amount of computations. E.g., lattice QED cannot produce highly accurate magnetic moments.

Or let's take the example of Dirac Equation, is there other calculation method that can put negative energy state away or dispense the existence of the antimatter?

In QED, there are no negative energy states. But there are antiparticles in QED. This has nothing to do with perturbative or not.

So the important question is. Is Lattice QFT complete or can it never do everything that perturbative QFT does?

In terms of principle, lattice QFT seems to be fully expressive of all QFT predictions.
But in terms of practicality of the calculations, it is very inefficient in almost all areas where perturbation theory can be made to work.

 

[A. Neumaier]

Anyway. Here's the relevant passage " perturbative expansion represented by Feynman diagrams can be introduced even in classical physics [52, 53], but nobody attempts to verbalize these classical Feynman diagrams in terms of classical “virtual” processes",

In particular, classical perturbation theory for a canonical field theory produces _precisely_ the tree diagrams form those of QFT, including their internal lines. The loop diagrams provide the quantum corrections.

 

[rogerl]

In particular, classical perturbation theory for a canonical field theory produces _precisely_ the tree diagrams form those of QFT, including their internal lines. The loop diagrams provide the quantum corrections.

Quantum corrections? Why, Can't the canonical field produce virtual particles too? Hasn't anyone describe it as such? What's the refutation arguments that canonical field can't produce virtual particles?

 

[rogerl]

If we had computers that are nearly infinitely fast, so that lattice QFT could be done with 100 x 1000 x 1000 x 1000 lattices instead of the tiny lattices used today, we could completely dispense with perturbation theory. But perturbation theory is very useful especially in QED, where the coupling constant is tiny, since it gives extremely high accuracy results with a limited amount of computations. E.g., lattice QED cannot produce highly accurate magnetic moments.

In QED, there are no negative energy states. But there are antiparticles in QED. This has nothing to do with perturbative or not.

In terms of principle, lattice QFT seems to be fully expressive of all QFT predictions.
But in terms of practicality of the calculations, it is very inefficient in almost all areas where perturbation theory can be made to work.

Infinitely fast? Since nothing is infinitely fast, It means lattice QFT can't be done. If it can't done, and perturbation theory is here to stay forever and part of QFT. Then who knows, the multivariate integrals may be describing a physical process that is actually there although unmeasurable.. virtual particles..

To prove there is no virtual particls. Lattice QFT must be proven to be true. Since we don't have infinitely fast computer, it can't be done... so lattice QFT is not practical at all (?)

 

[Lapidus]

"virtual particles" which are nothing more than multivariate integrals.

But these multivariate integrals describe very, very well the anomalous moment of the electron and a host of other things with a staggering precision. So some people, among them Professors and Nobel Prize winners, use the convenient word 'virtual particle' for it. Much the same as they use the convenient word electric field for components of an antisymmetric rank-two tensor that appear in some differential equations.

In quantum mechanics you can use the Schrödinger equation or the path integral method to calculate transition amplitudes. Does that refute one of the two approaches? Why should then non-perturbative calculations refute perturbative calculations? What matters are correct formulas that can predict the probabilty amplitude for a process.

 

[A. Neumaier]

Quantum corrections? Why, Can't the canonical field produce virtual particles too? Hasn't anyone describe it as such? What's the refutation arguments that canonical field can't produce virtual particles?

I don't understand your question.

A Lagrangian with parameters defines both a family of classical field theories and a family of quantum field theories. The latter depend on an additional parameter hbar; the limit hbar to zero gives the classical theory. In perturbation theory, the diagrams with k loops have a factor hbar^k. Thus in the classical perturbation theory, precisely the tree diagrams are present.

Thus virtual particles are either present in both or in neither classical field theory and quantum field theory.

 

[A. Neumaier]

Infinitely fast? Since nothing is infinitely fast, It means lattice QFT can't be done. If it can't be done, and perturbation theory is here to stay forever and part of QFT.

I said ''nearly'' infinitely fast, which means finite speed far, far larger than achievable in the forseeable future. The effort for lattice calculations scale with the number N of lattice points as O(N^k), where (if I remember correctly) k>=4. The accuracy obtainable scales with 1/sqrt(N). Current accuracy is at about 5%. So one can easily tell how long it would take to get the 12 digit accuracy of perturbative QED calculations -- longer than the age of the universe.

So perturbation theory is going to stay. But this doesn't prove the existence of virtual particles in any meaningful sense. Computational tools don't decide about existence.

 

[A. Neumaier]

In quantum mechanics you can use the Schrödinger equation or the path integral method to calculate transition amplitudes. Does that refute one of the two approaches?

This means that transition amplitudes exist in Nature in a deeper sense than the Schrödinger equation and the path integral, which are just computational tools.

 

[friend]

-----------------

The essence of it all is that when you write a non-perturbative approach to Quantum Field Theory, the virtual particles disappear.

Anyway. Here's the relevant passage " perturbative expansion represented by Feynman diagrams can be introduced even in classical physics [52, 53], but nobody attempts to verbalize these classical Feynman diagrams in terms of classical “virtual” processes", Pls. adress this part, how do you refute it?

As I understand it, perturbation theory is just the Taylor expansion for the non-perturbative formula, expanded around some coupling constant. But since when is a Taylor expansion of some function actually not equal to the function itself?

 

[A. Neumaier]

But since when is a Taylor expansion of some function actually not equal to the function itself?

Since ancient times: The function f defined by f(x)=0 if x=0, f(x)=e^{-1/x^2} otherwise has a Taylor expansion around zero that is identically zero. Other functions may have Taylor expansions that converge only at x=0.

But in a Taylor expansion of a function with physical meaning, the individual terms generally have no physical meaning. In the same way, the Feynman diagrams, which encode the expansion coefficients, have no physical meaning although the whole sum has.

 

[A. Neumaier]

The problem is that there is no real difference between, say, a "virtual" photon and a "real" photon. We measure photons when they interact with something in a detector, so the photon is always an internal line in a feynman diagram.

This is based on the unverifiable assumption that the lines in a Feynman diagram actually signal the beginning and end of existence of a particle at some point in space and time.

But in QFT, production and detection of particles is _always_ represented by external lines, not by internal ones.

 

[friend]

But in a Taylor expansion of a function with physical meaning, the individual terms generally have no physical meaning.

Can it be mathematically equivalent but not physically equivalent? That doesn't sound right.

In the same way, the Feynman diagrams, which encode the expansion coefficients, have no physical meaning although the whole sum has.

Another way of saying this might be that you need to include every term in order for it to be physically equivalent. So you need to include every possible interaction (including with virtual particles) to get equivalence with a non-perturbative formulations.

It seems to me that if we can say that individually every kind of particle exists in some context (electrons, photon, quarks, etc), then you can't deny them just because they appear in a perturbative expansion that equates to a non-perturbative formula which you do accept.

 

[dm4b]

But in a Taylor expansion of a function with physical meaning, the individual terms generally have no physical meaning. In the same way, the Feynman diagrams, which encode the expansion coefficients, have no physical meaning although the whole sum has.

In some cases, don't Fourier expansions have physical meaning? I am thinking of musical notes, consisting of separate harmonics.

So, sometimes terms in a series can have actual physical meaning.

Maybe, in QFT, we stumbled across something with physical meaning, because of our need for an approximate solution to interacting fields. Then again, maybe we didn't.

Seems like the jury is still out on the issue, although not actually leaning in favor of virtual particles?

 

[Haelfix]

In theoretical physics, you are only able to ask as many questions as your theory allows you to ask.

In perturbative field theory, you simply are not permitted to ask what the empirical difference is between virtual and real particles. Any experiment (even in principle) that you could perform would fail to measure this distinction of 'virtualness' of a particle. You could try, for instance to scatter a photon (with the right wavelength) off such a beast, but you would quickly find that it merely changes the process (by putting it on shell) at least within the language of perturbation theory. This is most likely simply a statement that quantum mechanics tends to create complementarity between many of its objects.

Further, the operational difference in practise is very small. Consider a photon emmitted from some process on Alpha Centauri and "promptly" absorbed by your retina. Is that photon real or virtual? Well it was real for nearly its entire trip, until a last second interaction made it appear as if it was virtual.

And as others have pointed out, there is no concrete notion of these things in nonperturbative physics, so I would say its simply a matter of convenience and context whether you choose to admit them as physical entities (eg is it useful). The rest is philosophy.

 

[A. Neumaier]

In some cases, don't Fourier expansions have physical meaning? I am thinking of musical notes, consisting of separate harmonics.

So, sometimes terms in a series can have actual physical meaning

.
Sometimes.

But I was specifically talking about the terms in a Taylor series. The fact that the sum
$$1+\sum_{k>0}(-1)^{k-1}\frac{(2k)!}{(2^{2k}k!)^2}$$
equals the length of the diagonal of a square with side 1 (proof by Taylor expansion of sqrt(x) around x=1) doesn't imply that the terms in the sum have a geometric meaning. Not even when some such meaning could be found would it imply that there are virtual factorials that cause the diagonal to have the length sqrt(2).

 

[A. Neumaier]

In perturbative field theory, you simply are not permitted to ask what the empirical difference is between virtual and real particles. Any experiment (even in principle) that you could perform would fail to measure this distinction of 'virtualness' of a particle.

No. By definition, a scattering experiment contributing to the statistics for a cross section refers to external lines in Feynman's expansion. The standard Feynman diagrams have no interpretation at all outside a scattering framework.

Thus if you deny the prepared and detected particles the status of real particles, nothing is left to check the theory for correctness.

 

[Chalnoth]

Since ancient times: The function f defined by f(x)=0 if x=0, f(x)=e^{-1/x^2} otherwise has a Taylor expansion around zero that is identically zero. Other functions may have Taylor expansions that converge only at x=0.

But in a Taylor expansion of a function with physical meaning, the individual terms generally have no physical meaning. In the same way, the Feynman diagrams, which encode the expansion coefficients, have no physical meaning although the whole sum has.

This, to me, is like claiming that the electric field between the plates of a capacitor isn't actually made up of a collection of photons.

And one could equally say that the virtual particle-free formulations are merely describing the collective behavior of the virtual particles, instead of summing up their individual contributions.