I Are virtual particles real or just math filler

[friend]

Well, that's easy. What's called "virtual particle" in popular science books is symbolized by internal lines of Feynman diagrams, and they stand for free-particle Green's functions (in usual perturbation theory; sometimes they can have a different meaning, e.g., in the context of resummation schemes like the $\Phi$-derivable approximation or the functional RG methods), but that doesn't matter too much on the level of this discussion.

What you've shown here is for perturbation theory. What about the virtual particles that are supposed to be everywhere, the ones that get ripped apart by black hole horizons or Unruh acceleration, or that are supposed to be involved in the Casimir effect, etc? What is the math for these?

 

[bhobba]

What about the virtual particles that are supposed to be everywhere, the ones that get ripped apart by black hole horizons or Unruh acceleration, or that are supposed to be involved in the Casimir effect, etc? What is the math for these?

As has been explained in many many threads they don't exist. They are simply representations of integrals. Everything you cite above can be explained without them.

This has been explained to you many times - the following simply being the latest:
https://www.physicsforums.com/threads/can-particles-be-absorbed-into-a-field.855789/#post-5368838

I gave you a link to John Baez's paper before:
https://www.physicsforums.com/insights/struggles-continuum-part-5/
'Each of these diagrams is actually a notation for an integral! There are systematic rules for writing down the integral starting from the Feynman diagram.'

Please please read it and post with any queries you have so it can be put to rest once and for all.

Thanks
Bill

 

[friend]

I'm sorry, but I hear every Professor that gives a lecture invoking them to explain things. Perhaps that is just a tool, but I have seen them show equations where they sum up all the zero point frequency modes and give this as the reason that the calculated vacuum energy is so many orders of magnitude greater than what is measured. I've heard professionals teach about virtual particles, virtual paths, and even virtual geometries. From what I can gather, virtual objects are the differential parts of the path integral that are being summed up in superposition. They aren't observable in and of themselves, but they are the basis of the integral. We sum up differential parts in other integrals of physics, and those differential parts can't be observed either.

What I think is going on is that this aversion to virtual particles is being fueled by the faith that quantum mechanics cannot be explained. And any attempt to do so is misguided. For if there were to be any explanation of QM, it would have to necessarily involve virtual particles just as classical physics is explained in terms of differential parts. So I think that your attempt to definitively rule out virtual particles is misguided.

 

[Orodruin]

They aren't observable in and of themselves, but they are the basis of the integral. We sum up differential parts in other integrals of physics, and those differential parts can't be observed either.

Already the "are not observable" should ring the warning bells. They are also not the basis of the path integral, but only a very convenient way of computing it (approximately). Much like you might use partial integration or the Leibniz rule to perform computations in calculus.

That being said, they are a very useful tool and I know many people like to think in terms of them.

 

[bhobba]

That being said, they are a very useful tool and I know many people like to think in terms of them.

Exactly.

To Friend - did you read what John Baez wrote? He stated it clearly - they are representations of integrals. Why exactly won't you accept it? Why do you chose instead to worry about what others say? Here you get the real deal - but for it to be of value you must take it on board. You will get nowhere constantly saying others say different. They are not being careful. We are. It's that simple.

For if there were to be any explanation of QM, it would have to necessarily involve virtual particles just as classical physics is explained in terms of differential parts. So I think that your attempt to definitively rule out virtual particles is misguided.

You pretty well admit you haven't studied an actual QFT textbook. Until you do you will not have the background necessary to reach conclusions like the above. It's wrong - but since you seem to doubt what we say here I don't know what to say. I tell you its wrong - but because you won't accept it it won't make any difference. You come here seeking to learn - but won't accept what those you have chosen to learn from say. In science you have two choices. Either you believe what those that have studied it tell you or you read the textbooks yourself. There is no middle path of reading what others say then using that as ammunition to challenge those that tell you different. That leads nowhere.

The challenge I have for you is to study an actual text:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

When you have done that then we can discuss if you still think they are real.

Thanks
Bill

 

[A. Neumaier]

I hear every Professor that gives a lecture invoking them to explain things.

This is because in explaining things to people without a thorough grounding in math you cannot explain much without using gross simplifications and imagery in place of the real thing. But if you come to this forum to learn you are expected to realize that the views created for the general public are different from the views physicists have when doing real work.

One talks informally as if virtual particles were real since it is a quick way of conveying superficial information. But the word ''virtual'' (which is opposite to ''real'') is added everywhere to signal that this is only a figure of speech. Once one tries to substantiate in which way the virtual particles could be thought of as real, the whole concepts dissolves into nothing but a metaphor for multivariate integrals. Please read https://www.physicsforums.com/posts/5334791/bookmark and the link posted there, where this is carefully explained.

 

[cremor]

Any commentary on direct sampling of electric-field vacuum fluctuations?

http://science.sciencemag.org/content/350/6259/420

 

[vanhees71]

Without having read the paper, I can only say that for sure they don't measure the vacuum. The very fact that they measure something means that there is a measurement apparatus present, and that's not vacuum. There are quantum fluctuations of the electromagnetic field, but they manifest themselves always only at the presence of charges, because we cannot detect anything without having the interaction of the electromagnetic field with matter consituting a measurement device, and this matter consists of electrically charged particles.

 

[friend]

I feel like I'm arguing the existence of God, something that is the cause of everything else but not in and of itself observable. What is it, then, that we are integrating in the path integral? The integrand in the path integral is the exponential of the complex Action integral. And this exponential of the Action can be broken up into exponentials of differential Actions. What do these exponentials of differential Actions mean if not virtual particles?

I think the problem comes in because the path integral involves an infinite number of integrations. In the development of classical physics the integrals are along a path or throughout a space which seem more intuitive. So we don't question what the integrand means in those classical integrals, and we feel that the integrand in those classical integrals do have intuitive physical meaning. They are differential objects described in terms of force, velocity, and acceleration on infinitesimal bits of matter and charge that we then have no trouble integrating to get overall energies and distances, etc. But these differential bits are not any more observable than anything in the path integral. Nobody observes these bit of mass or charge or these differential displacements. But nobody argues that they are not real because it seems more intuitive to integrate them to get observables.

We still have differential bits in the path integral; these are called virtual to stress that they are not observable, but that's not something new. The difference here is that we are using an infinite number of integrations to take into account every possible combination of the differential, virtual effects from one point to another in a continuum. That together with the fact that we are summing up complex numbers to get a superposition of all these virtual effects makes the path integral less intuitive. But if we are going to understand what's going on in the math, we're going to have to get a better idea of what these differential, virtual processes are just as we do in the classical picture. Then we can take every combination of them in superposition to get the observables that we can measure. It seems we are doing some of that when we describe the Lagrangian in terms of interacting terms of quantum fields and coupling constants and the like, that exist in the Action integral inside the path integral. And then these quantum fields are described by particle number at each point. Some of these particles are real and others cancel out in superposition and are described as virtual.

 

[Orodruin]

I feel like I'm arguing the existence of God, something that is the cause of everything else but not in and of itself observable. What is it, then, that we are integrating in the path integral? The integrand in the path integral is the exponential of the complex Action integral. And this exponential of the Action can be broken up into exponentials of differential Actions. What do these exponentials of differential Actions mean if not virtual particles?

I do not understand your desire to interpret more into the virtual particles than there is to it. The path integral is perfectly well defined without the introduction of virtual particles as an integral over all possible field configurations (with a given appropriate measure). Expanding the exponential in an asymptotic series is essentially only a trick we use to compute this integral because it is generally very difficult to compute it analytically in other ways. Once you have made the asymptotic expansion, the Feynman rules, including the virtual particles, are only a means of keeping track of the terms in this series. This does not change the fact that the path integral itself is well defined without virtual particles.

The path integral itself is the same type of integral which appears in normal quantum mechanics (where you integrate over actual paths and not field configurations). The situation is similar for ghost fields which do appear in Feynman diagrams due to what is essentially a mathematical trick for rewriting the path integral in a way which handles gauge invariance in a pleasant manner.

 

[friend]

I do not understand your desire to interpret more into the virtual particles than there is to it.

I'm not trying to interpret more than there is to it. I'm trying to understand what is to it. I don't understand how you could have missed my point. It seems obvious and unavoidable that we need to understand what's inside the integrals just as we do in classical mechanics. If some of that involves virtual effects then we need to understand that as well.

We may be using the term "virtual" in different ways. You seem to be using them to refer only to terms in a perturbation expansion. I think I may be using them more generally as mathematical artifacts that exist everywhere and must be taken account of in calculations. I don't think your use of virtual particles in a perturbation expansion take into account other uses of virtual in such things as the Casimir effect, that assume they are all over the place and not just hiding in a the calculation of some particular observable.

 

[A. Neumaier]

What do these exponentials of differential Actions mean if not virtual particles?

It is a fallacy that each part of a formula means something else than what the formula actually says.

the Feynman rules, including the virtual particles, are only a means of keeping track of the terms in this series. This does not change the fact that the path integral itself is well defined without virtual particles.

Yes. The path integral is (in certain cases) well-defined, but as you say, the virtual particles are only a means of keeping track of the terms in this series. They don't have more meaning in the path integral itself than the terms $x^n/n!$ in the expansion of the exponential function have for the exponential function itself! They even have less meaning since most individual Feynman diagrams evaluate to infinity if taken by themselves, and only well-chosen combinations in the formal expansion lead to a well-defined numerical result.

we don't question what the integrand means in those classical integrals, and we feel that the integrand in those classical integrals do have intuitive physical meaning.

We also don't question what the integrand means in a path integral, and the integrand in a path integral does have intuitive physical meaning. But once you are singling out particular contributions to the path integral appearing in a perturbative expansion and declare them to have physical meaning by themselves it is like saying that $x^n/n!$ has an intrinsic meaning for the exponential function. But the exponential function can be defined in many other ways, e.g., as the limit $\lim_{n\to\infty}(1+x/n)^n$, where these terms are completely absent - so they cannot have an intrinsic meaning. Similarly, the path integral can be defined in other ways, e.g., as a formal limit of lattice approximations, and if you do that, virtual particles are completely absent - so they cannot have an intrinsic meaning.

But I won't argue that again; read and think about the link in post #4 of this thread! If after having digested that you still want to argue, you are incurable and I won't answer anymore.

 

[friend]

It is a fallacy that each part of a formula means something else than what the formula actually says.

Yes. The path integral is (in certain cases) well-defined, but as you say, the virtual particles are only a means of keeping track of the terms in this series. They don't have more meaning in the path integral itself than the terms $x^n/n!$ in the expansion of the exponential function have for the exponential function itself! They even have less meaning since most individual Feynman diagrams evaluate to infinity if taken by themselves, and only well-chosen combinations in the formal expansion lead to a well-defined numerical result.

We also don't question what the integrand means in a path integral, and the integrand in a path integral does have intuitive physical meaning. But once you are singling out particular contributions to the path integral appearing in a perturbative expansion and declare them to have physical meaning by themselves it is like saying that $x^n/n!$ has an intrinsic meaning for the exponential function. But the exponential function can be defined in many other ways, e.g., as the limit $\lim_{n\to\infty}(1+x/n)^n$, where these terms are completely absent - so they cannot have an intrinsic meaning. Similarly, the path integral can be defined in other ways, e.g., as a formal limit of lattice approximations, and if you do that, virtual particles are completely absent - so they cannot have an intrinsic meaning.

But I won't argue that again; read and think about the link in post #4 of this thread! If after having digested that you still want to argue, you are incurable and I won't answer anymore.

Those are some interesting points. But if virtual particles are a way of understanding processes in some calculations, then the bottom line is that if they can lead to calculations, then use them. That's what they do in the Casimir effect, isn't it? I'm not arguing that they are necessarily real. I'm perfectly content to say that they are mathematical artifacts. For we haven't proven that our mathematical description of physics is unique, have we? There might be other math that results in the same answers. So perhaps we should start showing the math we are referring to and stop arguing about undefined words. (What is a virtual particle? Sheeeesh)

 

[A. Neumaier]

use of virtual particles in a perturbation expansion take into account other uses of virtual in such things as the Casimir effect,

It is precisely the same use, once you look at the calculations done.

Only how one talks about them may differ from application to application and from author to author since virtual particles as ''real'' objects (rather than wiggles on paper) are limited in their properties only by the fantasy of the respective authors.

 

[A. Neumaier]

if they can lead to calculations, then use them

They don't lead to calculations. They are a pictorial way to talk about calculations without having to display the details. And they are indeed heavily used in this way. Just don't mistake this use as being more than figurative speech!

 

[bhobba]

I feel like I'm arguing the existence of God, something that is the cause of everything else but not in and of itself observable. What is it, then, that we are integrating in the path integral? The integrand in the path integral is the exponential of the complex Action integral. And this exponential of the Action can be broken up into exponentials of differential Actions. What do these exponentials of differential Actions mean if not virtual particles?

When you do a Fourier analysis on a function and express the function as an integral what does that function mean? That too is integrating over complex exponentials. Does that mean it contains virtual particles that are real?

Thanks
Bill

 

[bhobba]

I'm not trying to interpret more than there is to it. I'm trying to understand what is to it.

If you want to that you must study a textbook.

Thanks
Bill

 

[vanhees71]

It is precisely the same use, once you look at the calculations done.

Only how one talks about them may differ from application to application and from author to author since virtual particles as ''real'' objects (rather than wiggles on paper) are limited in their properties only by the fantasy of the respective authors.

One should, however, stress that the Casimir effect is (of course) NOT proof of the existence of "vacuum fluctuations" but of quantum mechanical charge and em.-field fluctuations. Without any matter there's no Casimir effect. The usually treated way in introductory textbooks describing two conducting plates in terms of a boundary-value problem is in fact the em. coupling to $\infty$ limit (leading to an ideal conductor as an idealized model for the plates) of the true affairs. See

http://arxiv.org/abs/hep-th/0503158

 

[cremor]

Without having read the paper, I can only say that for sure they don't measure the vacuum. The very fact that they measure something means that there is a measurement apparatus present, and that's not vacuum. There are quantum fluctuations of the electromagnetic field, but they manifest themselves always only at the presence of charges, because we cannot detect anything without having the interaction of the electromagnetic field with matter consituting a measurement device, and this matter consists of electrically charged particles.

They used electro-optic sampling with following setup: Electro-optic sampling of an electric-field waveform by an ultrafast probe pulse, consisting of an EOX (electro-optical crystal), a quarter-wave plate (λ/4), a Wollaston polarizer (WP), and a differential photocurrent detector (DD).

Obviously you have a point. What perplexes me is how can this research be titled as it is? The only way to understand it is umm... 'We tuned into the void to see what the fabric of reality itself translates into when looked at with an optical kaleidoscope?'

 

[vanhees71]

Well, I must admit that I couldn't understand the paper just from reading, because I'm not an expert in quantum-optics. At least they could have written out their three-letter acronyms. I'd have to dig through a lot of literature before being perhaps able to do so.

Obviously the referee was not very strict in letter through this paper in the published form! I don't say that there's anything wrong with the core physics, but despite the misleading title at least the introduction, I'd have rejected right away, because neither of the given examples for radiation correction effects of QED are "vacuum fluctuations". Rather they are indeed quantum fluctuations of charges and the em. field: There's no Casimir effect without charges (google for Jaffe and Casimir effect to find a nice treatment). The Lambshift of the hydrogen lines are quantum effects on the Coulomb-bound state energies (vertex corrections, photon polarization, including also QCD corrections) etc. etc.

 

[friend]

Thinking again about the math of virtual particles. They are supposed to come in pairs, but together they don't result in anything permanent. What kind of math would do that? What kind of math leads to complete annihilation or cancellation for two particles that both start at the same point at the same time and both end at the same different point at the same time? Presumably they are described by some sort of amplitude with a magnitude and phase. So I don't think you'd multiply the two amplitudes because even if you got the phases to cancel, you can't get the magnitude to equal zero. So I think we're talking about adding the amplitudes in superposition to try to get cancellation. But even here you can't have one being the complex conjugate of the other because that does not guarantee that the two vectors/amplitudes are 180° out of phase in order to cancel. Is there some way to make sure the two amplitudes are 180° out of phase even though they start at the same place at the same time and end at a different place at the same time?

 

[bhobba]

Thinking again about the math of virtual particles. They are supposed to come in pairs, but together they don't result in anything permanent. What kind of math would do that? What kind of math leads to complete annihilation or cancellation for two particles that both start at the same point at the same time and both end at the same different point at the same time?

Obviously math you do not understand.

Its simply terms in what's called a Dyson series:
https://en.wikipedia.org/wiki/Dyson_series

Your continual harping on about it will not change anything and simply leads to posts that repeat the same thing over and over.

Thanks
Bill

 

[DelcrossA]

Is there a paper explaining Hawking Radiation without succumbing to virtual particles similar to Jaffe's paper on the Casimir Effect? Unfortunately, Gravitation by MTW (the only GR textbook I have) doesn't cover hawking radiation and a cursory glance at google all mention pair-creation.

Or, if there is a specifcally good textbook treatment of it, I'd appreciate it.

 

[Vanadium 50]

Is there a paper explaining Hawking Radiation without succumbing to virtual particles similar to Jaffe's paper on the Casimir Effect?

Yes. Hawking's original paper.

 

[sheaf]

Yes. Hawking's original paper.

Available from project Euclid:

 

[friend]

Thinking again about the math of virtual particles. They are supposed to come in pairs, but together they don't result in anything permanent. What kind of math would do that? What kind of math leads to complete annihilation or cancellation for two particles that both start at the same point at the same time and both end at the same different point at the same time? ...

The transition amplitude for a particle to go from |x> to |x'> is
&lt; x&#039;|U(t)|x &gt; \,\,\, = \,\,\,{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{im{{(x&#039; - x)}^2}/2\hbar t}}.
I take this as true even for a virtual particle. The antiparticle is said to travel backwards in time between the same two points. So its transition amplitude would be
&lt; x|U(t)|x&#039; &gt; \,\,\, = \,\,\,{\left( {\frac{{ - m}}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{ - im{{(x - x&#039;)}^2}/2\hbar t}}
by simply replacing t with -t. (Or take the complex conjugate). Yes, you can argue that these transitions are not measurable at these specific points since the |x> basis is a continuous spectrum. Granted! But bear with me because I'm trying to prove just that. I'm just considering the transition from some generic point to another generic point for a virtual particle pair that is said to go from some point to another.

The minus sign comes out of the square-root as the complex number i. So, we get
&lt; x|U(t)|x&#039; &gt; \,\,\, = \,\,\,i{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{ - im{{(x - x&#039;)}^2}/2\hbar t}}
This is not a transition from one point by a particle and then back again by the antiparticle. The two transitions happen at the same time. And a real particle might interact with one or the other, and we can't say which. So we can consider these two virtual particles to be in superposition with each other. And then the expectation value for measuring these particles would be
| &lt; x&#039;|U(t)|x &gt; {|^2}\,\, + \,\,\,| &lt; x|U(t)|x&#039; &gt; {|^2}
As seen from the above, the only difference between these terms is the complex number i in the antiparticle. After squaring it the only difference would be a minus sign, and the sum would be zero. This is what we are told, that they exist as wave functions but have zero expectation value of ever being measured. So you could fill space with as many virtual particle pairs as you like, even infinitely many, and it would not be noticeable by any observer.

Did I get my math right?

 

[bhobba]

I take this as true even for a virtual particle.

Since virtual particles are not real why you assume that beats me.

Yes some your math is correct but you are atrociously mixing concepts.

This is not a transition from one point by a particle and then back again by the antiparticle. The two transitions happen at the same time. And a real particle might interact with one or the other, and we can't say which. So we can consider these two virtual particles to be in superposition with each other. And then the expectation value for measuring these particles would be

That's utter nonsense.

You need to study an actual textbook.

Thanks
Bill

 

[friend]

Since virtual particles are not real why you assume that beats me.

If virtual particles exist at all, then they have a wave function that guides the way they transition from place to place.

I've heard professors say that space itself is made of virtual particles. And I've come to understand this in my own way. So if a particle propagates it must be through this sea of virtual particles (=space). If virtual particles have the same kind of transition amplitudes as those found in the propagator, then this provides a method of propagation through that sea. The propagation proceeds as follows: There already exists virtual particle pairs everywhere, including near a real particle. If a virtual particle-antiparticle pair appears near a real particle, then the real particle can annihilate with the antiparticle of the virtual pair. This leaves real the virtual particle that did not annihilate with its original partner. Thus the real particle use the transition amplitude of the virtual particle to jump from one position to the next. And now that the "real-ness" has been handed off to the virtual particle that did not annihilate, it is now subject to another jump by annihilating with yet another virtual pair, and so on through the path normally described with the path integral. If there is no expectation whatsoever of ever observing a virtual particle pair, then this is as good an interpretation of what's going on in the path integral as any other. If we accepted virtual particles to begin with, then maybe that would have led to the path integral formulation much sooner. Who knows what other properties can be described with them.

That's utter nonsense.

This is a pretty safe comment. No speculation there. And you're not even saying I'm wrong.

You need to study an actual textbook.

That's always a good idea. All in all, you've not said anything, though a bit snarky on your part.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles. It's disturbing that just about every professor in the classroom as well as the popular stage uses virtual particles to describe what's going on. But non of them go into depth into the math. Maybe that's what you're frustrated with.

 

[bhobba]

If virtual particles exist at all, then they have a wave function that guides the way they transition from place to place.

Why are you starting with a falsehood? And that is not what a wave-function is.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles. .

This does:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

IIt's disturbing that just about every professor in the classroom

A number of professors post here and they don't.

This is my last post to you in this thread. Go away and study a textbook.

Thanks
Bill

 

[friend]

Go away and study a textbook.

Honestly, I doubt I'm going to find what I'm looking for in a textbook, though it may help. I'm looking into foundational issues, why QM is the way it is? What logic justifies QM to begin with. However, most textbooks give a record of the history of its development. And the math they use seems to be used only because it works. But in my opinion, that does not explain why it is the way it is. It only describes that it is that way and what the implication of that are to application.

 

[bhobba]

[QUOTE="friend, post: 5400050, member: 93840"But in my opinion, that does not explain why it is the way it is.[/QUOTE]

Why dindt you say that from the start.

Be enlightened
http://www.scottaaronson.com/democritus/lec9.html

If you want to pursue it start a new thread.

Thanks
Bill

 

[Orodruin]

And I've come to understand this in my own way.

This is generally a bad thing to do without any guidance. You will come out on the other side with several misunderstandings, such as the ones you have displayed in this thread and in the rest of the quoted paragraph.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles.

Then you have not looked very well. Essentially any introductory text on quantum field theory will cover this and the mathematics is rather straight forward given the required previous knowledge.

It's disturbing that just about every professor in the classroom as well as the popular stage uses virtual particles to describe what's going on. But non of them go into depth into the math.

This is also wrong. You will see professionals use this kind of language in popular science and perhaps in courses which do not go very deep into the underlying quantum field theory aspects. Through popular science and survey courses you will learn about science, you will not learn science.

And the math they use seems to be used only because it works.

This is the only reason to use anything in an empirical science. People who know QFT know what they are talking about when they mention virtual particles and they know how they enter into the mathematics. Just because you cannot figure it out on your own does not mean it is not already known.

With that said, I believe it is time to close this thread. The original question has been answered several times over.