Are virtual particles really there?

In summary: In this case, the particle is said to be "virtual." The idea of virtual particles is a way of accounting for the fact that we can't always observe things as they happen. In the case of electrons scattering off of each other, for example, we can't observe the individual photons that are created. But we can calculate how many photons would be emitted if we did observe the event, and that number is called a "virtual photon." Virtual particles are a way of saying that the photons that we can't see are still there, they just haven't been observed yet.
  • #211
tom.stoer said:
I fully agree.

The problem with the perturbation series is that you can't take all orders into account b/c you can't calculate them. That's why a perturbation series "becomes" an asymptotic series when adding more and more terms. But that's of no relevance here, it simply reflects that perturbation theory is mathematically (partially) ill-defined and therefore all our "physical reasoning" is "handwaving".
Isn't the conventional view that the perturbation series is a non-analytic expression which is the limit of an analytic formula we don't know yet (akin zeta function regularization)?
 
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  • #212
How can the perturbation series which is a Taylor series in the coupling constant be a non-analytic expression?
 
  • #213
tom.stoer said:
How can the perturbation series which is a Taylor series in the coupling constant be a non-analytic expression?
Euh, a mathematical requirement is that the perturbation series converges. None of the known perturbation expansions in physics (at least in 4-d) enjoy this property. So analytic = Taylor + nonzero convergence radius; I guess, this was merely a matter of language.
 
  • #214
tom.stoer said:
The problem with the perturbation series is that you can't take all orders into account b/c you can't calculate them. That's why a perturbation series "becomes" an asymptotic series when adding more and more terms. But that's of no relevance here, it simply reflects that perturbation theory is mathematically (partially) ill-defined and therefore all our "physical reasoning" is "handwaving".

Well, there are two different levels involved: the conceptual and the computational.

On the computational level, the work to compute the coefficient of g^n grows at least exponentially with n, in a way that only a few coefficients can actually be calculated.
This doesn't affect the convergence radius of the series, which is a conceptual quantity, independent of computational difficulties. Moreover, in principle, arbitrarily many coefficients could be computed with enough patience and memory.


On the conceptual level, the perturbation series is defined at all orders, and after renormalization it is a mathematically well-determined power series. Its radius of convergence is most likely zero (this is not known for QED, but can be proved in many simpler situations). Thus, conceptually, the perturbation series is infinite and asymptotic.

It is _not_ the case that perturbation theory is mathematically ill-defined since all mathematics happens on the conceptual level. There are even mathematical rigorous books about it; e.g., one by Salmhofer. (I discuss the mathematical status of perturbation theory in detail in Chapters B4 and B5 my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html )

The hand-waving is _not_ at the level of perturbation theory but in pretending that the perturbation theory specifies the solution at the physical value of g. It doesn't.
Every asymptotic power series is the Taylor expansion of infinitely many different functions. What is missing (and cannot be provided as long as no mathematically satisfying nonperturbative definition of QED, etc. exists) are quantitative statements about the error made when only a few terms are taken into account.
 
  • #215
Careful said:
Isn't the conventional view that the perturbation series is a non-analytic expression which is the limit of an analytic formula we don't know yet (akin zeta function regularization)?

The perturbation series is simply an infinite series
sum_k a_k g^k (*)
with numbers a_k that ultimately grow faster than exponential in k. The latter makes
the convergence radius zero, and the series asymptotic. An asymptotic series does _not_ define a function. But for arbitrary asymptotic series (*) - no matter what its coefficients - there are infinitely many _different_ functions that are infinitely often differentiable for small nonnegative g, whose Taylor expansion is (*). None of these is an analytic function in a neighborhood of g=0. But in many cases, there are functions that are analytic in the intersection of such a neighborhood with a cone whose apex is at g=0.

All this has nothing to do with zeta regularization or dimensional regularization, etc.;
the latter is used to define the a_k in terms of QFT. Constructing a function whose expansion is (*) is rater done by means of tools such as the Borel transform.
 
  • #216
A. Neumaier said:
The perturbation series is simply an infinite series
sum_k a_k g^k (*)
with numbers a_k that ultimately grow faster than exponential in k. The latter makes
the convergence radius zero, and the series asymptotic. An asymptotic series does _not_ define a function. But for arbitrary asymptotic series (*) - no matter what its coefficients - there are infinitely many _different_ functions that are infinitely often differentiable for small nonnegative g, whose Taylor expansion is (*). None of these is an analytic function in a neighborhood of g=0. But in many cases, there are functions that are analytic in the intersection of such a neighborhood with a cone whose apex is at g=0.
Thank you for rehearsing what I already knew. However, this is hardly the point for instance, consider the series

\sum_n n (g - 1)^n where you are interested in the behavior of this function in a neighborhood of zero. Your argument would be, there is no such unique function, since many C^{infinity} functions could give rise to this formal power series. However, you might introduce another parameter s and consider

\sum_ n n^{-s} (g - 1)^n then for Re(s) > 1, this function is a legitimate power series expansion for an analytic function (this is similar to the trick of dimensional regularization invented by 't Hooft). Now, we want to have this nonperturbative expression A(s,g) and analytically continue to s = -1 if possible.

A. Neumaier said:
All this has nothing to do with zeta regularization or dimensional regularization, etc.;
the latter is used to define the a_k in terms of QFT.
Sure, but nobody forbids you to apply such trick twice no ? Do you have any legitimate objections against that ?

Careful
 
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  • #217
Careful said:
consider the series
\sum_n n (g - 1)^n
where you are interested in the behavior of this function in a neighborhood of zero. Your argument would be, there is no such unique function, since many C^{infinity} functions could give rise to this formal power series. However, you might introduce another parameter s and consider

\sum_ n n^{-s} (g - 1)^n then for Re(s) > 1, this function is a legitimate power series expansion for an analytic function (this is similar to the trick of dimensional regularization invented by 't Hooft). Now, we want to have this nonperturbative expression A(s,g) and analytically continue to s = -1 if possible.

Sure, but nobody forbids you to apply such trick twice no ? Do you have any legitimate objections against that ?

If you can do this, it just defines _one_ of the possible functions f(g).
How do you know that it picks the right one among the infinitely many possible ones?

Nobody forbids you to regularize sum_n a_n (g-1)^n by considering instead
sum_n a_n phi(n)^{-s} (g - 1)^n
with an arbitrary function phi(n) asymptotically equivalent to n up to bounded factors,
sum it, and analytically continue it to s=0 if possible.

Do you always get the same answer? If not, why should you trust the choice phi(n)=n more than any other choice?


The reason one can trust zeta regularization for computing the a_k in QFT lies deeper.
In perturbative QFT, under appropriate conditions, it is known that the coefficient a_k is
defined by a certain renormalization limit which must exist if the theory is to make sense at not too high energies (the results cannot depend on the large energy behavior);
hence all mathematically justified ways of obtaining that limit are equivalent. One
therefore knows that zeta regularization must produce this limit, too.

(The same happens in your example: The sum equals (g-1)/(2-g)^2 in some local
sector with apex g=0; so you only need the assumption of continuity of your function
to get the value -1/4. Zeta regularization therefore yields the same limit.
But consider instead the series sum_n n! (g - 1)^n, which looks more like QFT series.
Now you can't perform zeta regularization!)


Unfortunately, there is no such result that would ensure a limit representation of the
QFT perturbation series. There is no known nonperturbative, physically motivated definition of the function f(g) whose power series is the perturbation series
sum_k a_k g^k. Purely mathematical recipes are inherently ambiguous; one needs
additional properties to select the right one.

In lower dimensions, Borel summation often works, and a proof that the nonperturbative QFT produces the Borel summation of the perturbation series (and not one of the infinitely many alternatives) is available (only) in some cases of 2D QFT. But the Borel summation breaks down for QED because of renormalon effects related to the Landau pole.

In any case, even if some regularization method would produce a finite result for a QED
perturbation series, one needs, for every particular way to define the sum, additional reasons or experiments that prove that this is the definition Nature obeys. For there are infinitely many other qualifying functions that match theory as currently developed.
 
  • #218
A. Neumaier said:
If you can do this, it just defines _one_ of the possible functions f(g).
How do you know that it picks the right one among the infinitely many possible ones?

Nobody forbids you to regularize sum_n a_n (g-1)^n by considering instead
sum_n a_n phi(n)^{-s} (g - 1)^n
with an arbitrary function phi(n) asymptotically equivalent to n up to bounded factors,
sum it, and analytically continue it to s=0 if possible.
Sure, but doesn't the same comment apply when you use similar tricks to regularize the coefficients ? :biggrin:

Point is that we don't know what interacting QFT really is and the fact that you have to resort to such ''magic'' at any level of the calculation shows that something is wrong with it.


A. Neumaier said:
The reason one can trust zeta regularization for computing the a_k in QFT lies deeper.
In perturbative QFT, under appropriate conditions, it is known that the coefficient a_k is
defined by a certain renormalization limit which must exist if the theory is to make sense at not too high energies (the results cannot depend on the large energy behavior);
hence all mathematically justified ways of obtaining that limit are equivalent. One
therefore knows that zeta regularization must produce this limit, too.
And who says that is correct? It is well known in non-commutative models that we have IR/UV mixing in ''QFT''. So, are you simply cutting out such possibilities by hand?

A. Neumaier said:
(The same happens in your example: The sum equals (g-1)/(2-g)^2 in some local
sector with apex g=0; so you only need the assumption of continuity of your function
to get the value -1/4. Zeta regularization therefore yields the same limit.
But consider instead the series sum_n n! (g - 1)^n, which looks more like QFT series.
Now you can't perform zeta regularization!)
That's true, but now I could pull another magic trick out of my hat. For example, by Stirlings formula n! = f(n) e^{n log(n) - n} where f(n) is bounded. Now replace this by
f(n) e^{s(n log(n) - n)} and this is super well convergent for Re(s) < 0. Or, I could do something which is ''close'' to zeta regulation, I could pick

f(n) e^{ - s n^{-s} log(n) - n^{-s}}.
 
  • #219
Careful said:
Sure, but doesn't the same comment apply when you use similar tricks to regularize the coefficients ? :biggrin:

Point is that we don't know what interacting QFT really is and the fact that you have to resort to such ''magic'' at any level of the calculation shows that something is wrong with it.



And who says that is correct? It is well known in non-commutative models that we have IR/UV mixing in ''QFT''. So, are you simply cutting out such possibilities by hand?

No; I had added the qualification ''under appropriate conditions''.

Careful said:
That's true, but now I could pull another magic trick out of my hat. For example, by Stirlings formula n! = f(n) e^{n log(n) - n} where f(n) is bounded. Now replace this by
f(n) e^{s(n log(n) - n)} and this is super well convergent for Re(s) < 0. Or, I could do something which is ''close'' to zeta regulation, I could pick

f(n) e^{ - s n^{-s} log(n) - n^{-s}}.

One could simpler take the regularizing factor n!^{-s}?
This sort of trick works for arbitrarily fast growing a_n.

But what assures you that both tricks give the same result?

The problem is that there are lots of magic tricks and infinitely many variations of them,
which gives the possibility of getting infinitely many different results. One now needs a theory that shows that all these results are the same, or that tells which of the many different results are appropriate for the physics. No such theory currently exists for 4D QFT.

The only way to settle the question properly is to _derive_ from the action formalism
and appropriate canonical quantization a convergent definition of f(g), and then show that this f(g) expands into the asymptotic expression. Then one can introduce into the
good expression for f(g) an additional analytic dependence on a parameter s and expand the extended function, to see what sort of regularizing recipe one needs.
 
  • #220
A. Neumaier said:
The problem is that there are lots of magic tricks and infinitely many variations of them,
which gives the possibility of getting infinitely many different results. One now needs a theory that shows that all these results are the same, or that tells which of the many different results are appropriate for the physics. No such theory currently exists for 4D QFT.
Right, that's what I said, we don't know what QFT really is. You don't need mathematics to answer that question, you require new physics !

A. Neumaier said:
The only way to settle the question properly is to _derive_ from the action formalism
and appropriate canonical quantization a convergent definition of f(g), and then show that this f(g) expands into the asymptotic expression.
No, that's the only way YOU can think of. I stopped thinking in terms of action principles long time ago.
 
  • #221
Careful said:
Sure, but doesn't the same comment apply when you use similar tricks to regularize the coefficients ? :biggrin:

Point is that we don't know what interacting QFT really is and the fact that you have to resort to such ''magic'' at any level of the calculation shows that something is wrong with it.

For the coefficients, the situation is different since we know that the theory must be a limit of _any_ theory that is cut off at high energies. Therefore any regularization that in the limit only modifies the high energy portion and leads to a finite answer must arrive at _the_ answer. And this is born out by the many different schemes that were tried and agreed.

We do not have a similar situation for the whole series. Indeed, this series is often divergent already in the regularized case.

There is nothing wrong with renormalized perturbative QFT applied to situations where
perturbation theory is expected to be applicable - except that we have not the slightest control over the error terms at the physical value of the coupling constant.
 
  • #222
Careful said:
Right, that's what I said, we don't know what QFT really is. You don't need mathematics to answer that question, you require new physics !


No, that's the only way YOU can think of. I stopped thinking in terms of action principles long time ago.

How do you define QED without an action formalism? Or at least (as in the causal approach) an inter-action formalism? I didn't mean a particular ''action principle'' like Schwinger's but the derivation from first principles in terms of actions and interactions.

The quest to understand QED can only mean to find a nonperturbative definition of the S-matrix that is manifestly convergent. This must involve better mathematics; the physics of QED cannot be changed. Thus new physics cannot help.

Of course there are those who think that QED cannot be rescued as a consistent theory,
and that one needs to add in the standard model ... and gravitation ... and superstrings
and ...; or consider noncommutative deformations of QED. All this would require more
(and perhaps new) physics.

But I never found the arguments for any of this convincing. And since 40 years of research on new physics didn't help advance the understanding of QED I think it is a
dead end. New physics is needed for new theories but not for a better understanding of the best theory we have.
 
  • #223
A. Neumaier said:
For the coefficients, the situation is different since we know that the theory must be a limit of _any_ theory that is cut off at high energies. Therefore any regularization that in the limit only modifies the high energy portion and leads to a finite answer must arrive at _the_ answer. And this is born out by the many different schemes that were tried and agreed.
Yes, but such strategy can never ever lead to a fundamental theory in the UV, no? Don't you find that remotely unsatisfying?


A. Neumaier said:
There is nothing wrong with renormalized perturbative QFT applied to situations where
perturbation theory is expected to be applicable - except that we have not the slightest control over the error terms at the physical value of the coupling constant.
That's a nice way of saying you don't know the physics. What is the real deal then ?? The Feynman path integral expanded around a Gaussian theory where you throw away all terms after third order in the expansion?
 
  • #224
A. Neumaier said:
How do you define QED without an action formalism? Or at least (as in the causal approach) an inter-action formalism? I didn't mean a particular ''action principle'' like Schwinger's but the derivation from first principles in terms of actions and interactions.
You can, but I am not going to tell you how on a public forum.

A. Neumaier said:
The quest to understand QED can only mean to find a nonperturbative definition of the S-matrix that is manifestly convergent.
Euh, why do you think the S-matrix is something fundamental ?

A. Neumaier said:
This must involve better mathematics; the physics of QED cannot be changed. Thus new physics cannot help.
Of course, new physics can help. You might look for an alternative formulation which makes QED asymptotically free.

A. Neumaier said:
Of course there are those who think that QED cannot be rescued as a consistent theory,
and that one needs to add in the standard model ... and gravitation ... and superstrings
and ...; or consider noncommutative deformations of QED. All this would require more
(and perhaps new) physics.
Well, there exist options apart from those possibilities.

A. Neumaier said:
But I never found the arguments for any of this convincing. And since 40 years of research on new physics didn't help advance the understanding of QED I think it is a
dead end. New physics is needed for new theories but not for a better understanding of the best theory we have.
I refer you to my first comment: how will you be able to give a unique high energy physics ?!

Careful
 
  • #225
Careful said:
Yes, but such strategy can never ever lead to a fundamental theory in the UV, no? Don't you find that remotely unsatisfying?

The question so far wasn't about a fundamental theory in the UV.
Independent of the lack of such a fundamental theory, there is a big difference between regularizing the coefficients of the perturbation series and regularizing the whole series, and this was the discussion point.

But I _do_ find the presentation of current QFT unsatisfying; that's one of the reasons I continue studying it.


Careful said:
That's a nice way of saying you don't know the physics. What is the real deal then ?? The Feynman path integral expanded around a Gaussian theory where you throw away all terms after third order in the expansion?

Nobody currently knows the real deal (the real physics), and all are groping in the dark.

But I think the answer is not in a different perturbation theory but in a better understanding of the mathematical concepts underlying QFT. The spaces are wrong,
too large and too small at the same time - this causes the UV and IR problems.
One can see this in much simpler examples than QED.

The difficulty is to find out what needs to replace Fock spaces; the knowledge obtainable from the study of more accessible 1D and 2D cases doesn't easily extrapolate to 4D.

I am patiently studying all paths tradition has followed, both dead ends, partial successes, and minority routes, and slowly putting things together to a coherent picture. One day I'll write a book on coherent quantum mechanics that explains QED in a satisfying, nonperturbative way.

But this day hasn't yet come; so today you can easily accuse me of not knowing the physics. Nevertheless I am surprised how you reach such sweeping conclusions when you hardly know me. As if a few posts on this forum reveal the extent or lack of my knowledge. And I'd be surprised if you knew more about nonperturbative QED than I do.
 
  • #226
Careful said:
You can, but I am not going to tell you how on a public forum.

Well, you can find my email address on my website http://arnold-neumaier.at/

Careful said:
Euh, why do you think the S-matrix is something fundamental ?

It must be definable in any fundamental theory, and in this sense it is fundamental.
It is not fundamental in that it tells nothing about what happens at finite times.
Thus a fundamental theory must be able to do more than define an S-matrix.

Careful said:
I refer you to my first comment: how will you be able to give a unique high energy physics ?!

Nobody knows how to give a unique high energy physics; so it is unfair to require this from me.

My aim is to give a unique, well-defined QED. That it differs from reality at high enough energies is well-known physics, and not my concern.

I believe that once the techniques are developed that render QED consistent, consistency of the standard model and some version of quantum gravity will follow by a modest extension of these techniques together with what is already known about the handling nonabelian gauge theory.
 
  • #227
Careful said:
Of course, new physics can help. You might look for an alternative formulation which makes QED asymptotically free.

And how do you know that this new formulation is still QED? I.e., makes the same predictions?

If it does, the new formulation is just a new mathematical technique for interpreting the old QED.
If it doesn't, the new formulation is no longer QED, since it makes different predictions.

Thus progress in understanding QED as it is known since 1948 can come only from improving the mathematics of QED.
 
  • #228
A. Neumaier said:
The question so far wasn't about a fundamental theory in the UV.
I know, but in physics, you have to focus on the whole picture; not just some part of it.


A. Neumaier said:
But I think the answer is not in a different perturbation theory but in a better understanding of the mathematical concepts underlying QFT. The spaces are wrong,
too large and too small at the same time - this causes the UV and IR problems.
One can see this in much simpler examples than QED.
What you write is correct, but in order to find this new formulation, you will need new physics too. You are probably aware of the axiomatic foundations of QFT by Weinberg. If you assume particle statistics, Poincare invariance of the S matrix, the cluster decomposition theorem and the fact that QED is about massless spin one particles; then you automatically end up at the perturbative formulation of QED in Fock space. No way around that, so some hidden assumption must be wrong, which will change quantum theory drastically.


A. Neumaier said:
The difficulty is to find out what needs to replace Fock spaces; the knowledge obtainable from the study of more accessible 1D and 2D cases doesn't easily extrapolate to 4D.
There isn't much room for moving away from Fock space, is there ? :smile:

A. Neumaier said:
But this day hasn't yet come; so today you can easily accuse me of not knowing the physics. Nevertheless I am surprised how you reach such sweeping conclusions when you hardly know me.
I didn't accuse you of anything: I just wanted to make clear what we know and what we don't.

A. Neumaier said:
As if a few posts on this forum reveal the extent or lack of my knowledge. And I'd be surprised if you knew more about nonperturbative QED than I do.
Perhaps you are trying to climb the wrong mountain, that's what I said. I certainly did not specialize in QED, but it often happens that ''non-specialists'' coming from a different field have a better view on a particular question than the ''specialist'' has. In mathematics, this is a well known phenomenon.

Careful
 
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  • #229
A. Neumaier said:
And how do you know that this new formulation is still QED? I.e., makes the same predictions?
I hope not, because all predictions of QED are wrong if you don't cut off the perturbation series sufficiently fast. And how can you say that a theory which is asymptotically free is the same as QED while we don't know what QED is like in the high UV (as far as I know)?
 
  • #230
A. Neumaier said:
My aim is to give a unique, well-defined QED. That it differs from reality at high enough energies is well-known physics, and not my concern.
Like I said, you think too much like a mathematician. A physicist knows nature is not forgiving towards partial cures. But you misunderstand me: I am not asking something unfair from you, I am just telling you that the secret to a well defined QED may reside in directly adressing the UV question.
 
  • #231
Careful said:
I hope not, because all predictions of QED are wrong if you don't cut off the perturbation series sufficiently fast. And how can you say that a theory which is asymptotically free is the same as QED while we don't know what QED is like in the high UV (as far as I know)?

QED as defined in all textbooks is a theory that is well-known to make wrong predictions at high energies; for example it doesn't contain virtual contributions from quark-antiquark loops. So a formulation that is closer to reality than QED is not what currently counts as QED in the narrow sense, although it is of course QED in a wider sense.

So you aim at different things than I. You want to include new physics to change QED.
But to get a QED that is in accordance with reality at arbitrarily high energies this requires a theory that accounts for the standard model and for gravity, a theory of everything. Well, good luck!

Note that asymptotic freedom (a) does not solve the UV problems since the perturbation series still diverges, and (b) introduces severe additional infrared problems. So you get a theory that might work asymptotically for large energies but fails badly at the energies of ordinary life. You escaped Scylla but fall prey to Charybdis...

My aims are lower. I just want to find a formulation of QED that in perturbation theory fully agrees with the standard QED perturbation series at all orders, and for all energies in the range where weak forces, quarks, or gravitation play no significant role.
This is physically adequate and seems to me mathematically reachable.
 
  • #232
Careful said:
Like I said, you think too much like a mathematician.

I _am_ a mathematician, and I choose my topics differently.

Careful said:
A physicist knows nature is not forgiving towards partial cures.

This knowledge is spurious. If this were true, we wouldn't have a consistent classical relativity, classical mechanics of point particles, a consistent fluid dynamics, a consistent kinetic description of gases, and many other theories for approximate descriptions
(from a fundamental point of view) that are mathematical consistent.

The practice of physics tells the contrary: At almost every level of approximation, the proper tools produce a consistent mathematical model. It would be very exceptional if QED were different.


Careful said:
But you misunderstand me: I am not asking something unfair from you, I am just telling you that the secret to a well defined QED may reside in directly adressing the UV question.

I know that many physicists think that. But they tried to address the UV question directly in your sense (by changing the physics) for many years now, without much success. Always progress promised around the corner, but although a formidable machinery has accumulated, no promise redeemed.

I intend to address the UV question directly without adding new physics, by improving the
tools for studying QED as defined in the textbooks.

I read enough to know that this is a minority quest, but also enough to know that the chances are not worse than with the main stream outlook.
 
  • #233
A. Neumaier said:
You want to include new physics to change QED.
But to get a QED that is in accordance with reality at arbitrarily high energies this requires a theory that accounts for the standard model and for gravity, a theory of everything. Well, good luck!
No luck needed :tongue2:

A. Neumaier said:
This is physically adequate and seems to me mathematically reachable.
We will see about that.
 
  • #234
Careful said:
I know, but in physics, you have to focus on the whole picture; not just some part of it.

I adapt my focus to what I consider profitable. Adaptivity is almost always better than an injunction to follow rigid principles. The focus must move between the whole picture and the parts, and gain insights from both views.


Careful said:
What you write is correct, but in order to find this new formulation, you will need new physics too. You are probably aware of the axiomatic foundations of QFT by Weinberg. If you assume particle statistics, Poincare invariance of the S matrix, the cluster decomposition theorem and the fact that QED is about massless spin one particles; then you automatically end up at the perturbative formulation of QED in Fock space. No way around that, so some hidden assumption must be wrong, which will change quantum theory drastically.

The faulty assumptions are exhibited by Haag's theorem that says that there is no interaction picture under the usual (i.e., Weinberg's) assumptions.

But there is no theorem that tells that quantum theory must change drastically when
some assumptions are slightly relaxed.

There are a number of places where Weinberg's treatment is only hand-waving, and hence mathematically either not conclusive or not precisely specified, which leaves a lot of room for improvement.

On the other hand, his argumentation shows that QED is essentially unique, and hence has a right to exist without tampering with it, like the complex numbers (another unique object).

The quest is to find the right mathematical underpinning that makes his statements rigorously correct without running into contradictions like Haag's theorem.




Careful said:
There isn't much room for moving away from Fock space, is there ?

There is lots of room. Functional analysis provides a much larger variety of spaces than physicists commonly dream of. And some of these spaces are used in 1D and 2D QFT to
get rid of the renormalization problems without introducing anywhere a mathematically dubious process.

Fock space captures only the perturbatively accessible part of a quantum field theory, and that's not very much. For example, how do you describe quantum versions of solitons in a Fock space? There is no room for these in a Fock space, but they are needed...


Careful said:
I didn't accuse you of anything: I just wanted to make clear what we know and what we don't.

Then you'd be more careful with the use of the word ''you'', that you used quite liberally in ways inviting misunderstanding.

Careful said:
Perhaps you are trying to climb the wrong mountain, that's what I said. I certainly did not specialize in QED, but it often happens that ''non-specialists'' coming from a different field have a better view on a particular question than the ''specialist'' has. In mathematics, this is a well known phenomenon.

But only when the nonspecialist sees interesting ''wrong'' mountains to be climbed,
whereas the specialist has the narrow conception of ''right'' mountains, and focuses
exclusively on these.

The typical specialist in QFT works along the lines you propose, while I am a ''non-specialist'' coming from a (very) different field. Thus you'd grant me that I might have a better view on the particular question of the existence of QED that your specialist view gives.
 
  • #235
I just mailed you, so we can move the discussion to a more specific plane. I agree with many things you just said, but I disagree it will be just all mathematics.

Careful
 
  • #236
Perhaps there is a misunderstanding of the meaning of virtual in the phrase virtual particle. Before Feynman diagrams, about 3 years after the discovery of the positron (~1932), physicists realized that real pairs of positrons and electrons could be created in strong Coulomb fields. One of these physicists, E. A. Eehling, wrote the following paper

Polarization Effects in the Positron Theory
E. A. Uehling
Phys. Rev. 48, 55 (1935) – Published July 1, 1935
Abstract
Some of the consequences of the positron theory for the special case of impressed electrostatic fields are investigated. By imposing a restriction only on the maximum value of the field intensity, which must always be assumed much smaller than a certain critical value, but with no restrictions on the variation of this intensity, a formula for the charge induced by a charge distribution is obtained. The existence of an induced charge corresponds to a polarization of the vacuum, and as a consequence, to deviations from Coulomb's law for the mutual potential energy of point charges. Consequences of these deviations which are investigated are the departures from the Coulombian scattering law for heavy particles and the displacement in the energy levels for atomic electrons moving in the field of the nucleus.


This paper refers to the existence of an induced charge that leads to a polarization of the vacuum and consequent deviations from Coulombs law at very small distances. The Uehling integral representing this effect is still correct, and is still used to represent this deviation from Coulomb's law at short distances. This effect is now called vacuum polarization. It is caused by real particles (positrons and electrons), NOT hypothetical particles whose existence is only inferred, which recombine very shortly after they are created. The attached thumbnail is the first page of a more recent article by Foldy and Erikson that discusses the consequences of vacuum polarization. The complete document can be found at (use local pdf copy button at bottom of page).

http://www.google.com/url?sa=t&sour...2MMlx093kKqgs7-GA&sig2=GEaIzSpX9tUh_cQBkR25-Q

The integral in Equation (1) is the Uehling integral. It includes no new particles (m represents the electron mass) and no new coupling constants. The accuracy of the Uehling integral has been verified in the measurement of muonic atom energy levels. So what is virtual?

As an example of its application, the second thumbnail shows the use of the Uehling integral to calculate the 4f and 3d atomic energy level shifts in two pionic atoms, titanium and calcium. (from Shafer, Pion Mass Measurement ...., Phys Rev Vol 163, pages 1451-1461 (1967).

Bob S
 

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  • #237
Bob S said:
Before Feynman diagrams, about 3 years after the discovery of the positron (~1932), physicists realized that real pairs of positrons and electrons could be created in strong Coulomb fields. [...]
So what is virtual?

Surely this was known to Feynman and others around 1948; nevertheless, they coined the name virtual particles to distinguish them from real particles. Why should someone want to call real particles virtual? They had very good reasons to call them virtual and not real.

But those who can't hear will never be healed from their illusions.
 
  • #238
A. Neumaier said:
Surely this was known to Feynman and others around 1948; nevertheless, they coined the name virtual particles to distinguish them from real particles. Why should someone want to call real particles virtual? They had very good reasons to call them virtual and not real.

But those who can't hear will never be healed from their illusions.

re: bolded.
It was, and yes I agree.
 
  • #239
A. Neumaier said:
Surely this was known to Feynman and others around 1948; nevertheless, they coined the name virtual particles to distinguish them from real particles. Why should someone want to call real particles virtual? They had very good reasons to call them virtual and not real.
This leads to the following questions:

1) Is Uehling's representation of the vacuum polarization effect (Feynman's bubble diagram) correct?

2) Is it accidental that the photon coupling constant α/3π in the Uehling integral is the same as the coupling constant in Dirac's model of the hydrogen atom?

3) And finally, is it accidental that the mass of the "virtual" particles in the Uehling integral is exactly the same as the mass of the "real" electron and positron?

Bob S
 
  • #240
Bob S said:
This leads to the following questions:

1) Is Uehling's representation of the vacuum polarization effect (Feynman's bubble diagram) correct?

2) Is it accidental that the photon coupling constant α/3π in the Uehling integral is the same as the coupling constant in Dirac's model of the hydrogen atom?

3) And finally, is it accidental that the mass of the "virtual" particles in the Uehling integral is exactly the same as the mass of the "real" electron and positron?

The parameters occurring in the integral are parameters of QED. The rest mass of a particle is _defined_ to be the parameter m of QED, the pole position of the S-matrix. And a similar recipe defines the coupling constant.

Thus this cannot depend on what one calls a real or a virtual particle. The latter is defined instead by the difference between external and internal lines in Feynman diagrams, i.e., between observable (real) and unobservable (virtual).
 
  • #241
A. Neumaier said:
Surely this was known to Feynman and others around 1948; nevertheless, they coined the name virtual particles to distinguish them from real particles. Why should someone want to call real particles virtual? They had very good reasons to call them virtual and not real.

But those who can't hear will never be healed from their illusions.

Have you ever read Feynman's QED Theory of light? Especially chapter 3?
 
  • #242
kexue said:
Have you ever read Feynman's QED Theory of light? Especially chapter 3?

You resurrected this thread to say THAT?! Shouldn't you be re-reading a LOT of material, and trying to get it right this time before you worry about the curriculum of others?
 
  • #243
virtual particles

A month ago, I sent emails to Frank Wilczek, Ed Witten, Gerad t'Hooft, David Pollitzer, Steven Weinberg, Leonard Susskind, Michael Peskin and Curtis Callan. I asked them if they think that 'virtual' particles are 'really out there' or just a mathematical artefact of perturbation theory.

Over the last weeks I asked the same question Philip Anderson, Roy Glauber, Shelly Glasow, Joe Polchinski, Howard Georgi, John Preskill, Mark Srednicki, Warren Siegel, Steve Carlip, Helen Quinn, Roman Jackiw, Juan Maldacena, Jogish Pati, Cumrun Vafra, Steven Gubser, Daniel Bjorken, Nathan Seiberg and a lot of others. Again, very kindly they all replied.

I received a wide range of beautiful answers! Are people in this forum interested that I post them all here?
 
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  • #244


Yes, I for one am interested. Of course you should ask them for permission before posting any personal correspondence.

p.s. Since I am very cynical I cannot stop myself asking: how do we know you are not making this up?
 
  • #245


I'd be interested. Please post them.
 

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