# Short distance divergences in thermal states

1. Jul 6, 2009

### DrFaustus

A question for the more theoretically oriented forum users... but please feel free to suggest an answer to the puzzle anyways.

In the book of selected works by Glimm & Jaffe, on page 4 (they discuss the massive scalar interacting filed in 1+1 dimension and give an overview of the divergences that occur in QFT), one finds the following:

"[...] In a similar but less obvious fashion, the ultraviolet divergences are forced by the invariance of the theory under Lorentz rotations."

(The "less obvious" refers to the "infinite volume" divergences that arise because of translation invariance of the theory, i.e. momentum conservation)

And I'd be fine with accepting that without actually seeing the "less obvious" reasons. What is puzzling me are really the claims one finds essentially in all the treatments on field theory at finite temperature (like LeBellac's book) that ultraviolet divergences at finite temperature are of the same nature as in the vacuum. But finite temperature (KMS) states are NOT Lorentz invariant. There is a preferred frame, the one in which you cannot "feel any temperature flow".

How is that possible? The two statements seem to be in contradiction...

2. Jul 6, 2009

### ytuab

(This is my personal opinion. OK?)

I think the basic problem exists in the "quantum mechanics(QM)".

In QM, we can not image the motion of the particle concretely (spin of the electron, uncertainty principle, the zero orbital angular momentum of S orbital.....).

So QM had to get away into " the world of numerical formulas and symbols" from this real world.

In this real world, we can easily use the relativistic theory for each particle's motion.
But in "the world of numerical formulas" like QM, it is much more difficult (we must restrict the formulas, and divergent problems (infinite bare charge and mass , infinite momentum and energy...) will occur).

If we use the nonrelativistic theory, we must give up the relativistic mass change and the spin-orbital interaction (of Dirac equation). (Of course we can use the nonrelativistic theory for approximation.)

To avoid such problems, we must go back to the real (classical mechanical) world (like Bohr model .....).

Last edited: Jul 6, 2009
3. Jul 7, 2009

### DrFaustus

ytuab -> I think I see your point, but unfortunately does not solve my doubts. The thing is that the problem I'm facing is, basically, of a purely mathematical/constructive nature. In other words, if insisting in constructing Lorentz invariant theories yields ultraviolet divergences, how comes that when the symmetry is somehow broken, the ultraviolet problem is not solved? Mind you, it need not have any physical interpretation. Think of it as a purely mathematical problem if you wish. That is, I define certain objects and some relations between them and verify a property ("ultraviolet divergence") and then want to see where is this coming from. And encounter the above "contradiction"...

4. Jul 7, 2009

### ytuab

I'm sorry to speak my personal opinion, (but I believe the classical mechanical model like Bohr model, which you probably think strange.)

Please explain more about "how comes that when the symmetry is somehow broken, the ultraviolet problem is not solved?"

I think when the symmetry (Lorentz invariance) is broken, the ultraviolet problem will be solved.
In Lorentz invariant theory,(like QED), in the vacuum, there are infinite "virtual" photons and particles (which have every momentum and energy (including infinite momentum and energy) at any condition (even at finite temperature)).
(photons and particles are interacted)

The ultraviolet divergent problems are caused by the infinite momentum and energy of them (the virtual process).
So the bare charge and mass of the electron are infinite to cancel it (by using the renormalization theory).

In QFT (nonrelativistic theory), we treat the actually-existing particles(their total number may be infinite), and the vacuum polarization is not occurred.
so there is a natural cutoff (there is a upper and lower limit of momentum and energy of the particles).

(I'm sorry if I misunderstand your question.)

Last edited: Jul 7, 2009
5. Jul 8, 2009

### Bob_for_short

Maybe you can find some answers in "Independent research" forum

6. Jul 9, 2009

### DrFaustus

ytuab ->
What I mean is simply that if there is a logical implication "Lorentz symmetry $$\Longleftrightarrow$$ ultraviolet divergences" then if I remove the assumption, the consequences should also fall. Unless the implication is not equivalent to the assumption and Lorentz symmetry is only sufficient for ultraviolet divergences to arise.

Which is precisely what I'm puzzled about. That is, I'm trying to make sense of the statment from my first post.

bob -> I appreciate your effort to solve the many problems of QFT, but as I'm still a PhD student trying to work my way to my degree, I'm not sure an unconventional view on the subject would in any way help me.

7. Jul 9, 2009

### ytuab

May I ask you one more question?

You said "finite temperature (KMS) states are NOT Lorentz invariant". What do you mean by that?

I think there is no relation between temperature and Lorentz invariant theory(QED).
When I care about the temperature, this is nonrelativistic theory.

As I said, In Lorentz invariant theory, "the vacuum polarization" is occurred.
The vacuum polarization means that infinite "virtual" photons and particles which have infinite momentum and energy are occurred in the vacuum even at finite temperature (the temperature is not relevant in Lorentz invariant theory.)

The temperature has the relation with the actually-existing (not virtual) particles.

Last edited: Jul 9, 2009
8. Jul 9, 2009

### Bob_for_short

I showed that divergences appear due to too bad initial approximation - strongly and permanently interacting (bound!) particles (electrons and photons) are initially considered as decoupled.

Taking some essential part of their interaction into account in the initial approximation improves the perturbative series behaviour. Electron interacting with the quantized EMF has a smeared charge (a natural cut-off in coordinate and in the momentum spaces), so no divergences appear.

Finite temperature is not different from zero temperature case since it signifies the initial and final photon state have non-zero populations. In QED there is no limitations or particularities about that. In fact, the correct QED approach takes the initial and final non-zero photon state populations automatically - in the inclusive picture that is the only picture to make sense.

Last edited: Jul 9, 2009
9. Jul 9, 2009

### Naty1

Au contraire!!!,

As if Copernicus, Einstein, Planck, de Broglie, Witten, 'T Hooft, Susskind to name a few, solved problems "conventionally"...good grief, man, don't use that as any criteria!!!!!

That may be the most "fantastical" (yes, I know not a real word) comment I have seen in this forum so far!!!

10. Jul 9, 2009

### malawi_glenn

Are you a professor and know how one does science?

11. Jul 9, 2009

### Bob_for_short

My approach is not unconventional or fantastic. It is clear and practical.

You asked a fundamental question, not resolved so far. What did you expect as an answer?

I answer with certainty because I am very well aware of this problem.

Last edited: Jul 9, 2009
12. Jul 9, 2009

### DrFaustus

I'll try to be as clear as possible. The reason for my wanting a conventional answer is very simple. Actually very pragmatic. Since I want to get a PhD recognized in the "conventional physics" comunity, I better understand conventional physcs damn well before going on to consider alternative explanations and speculations. (And this is coming from someone who doesn't mind at all new and alternative ideas.) Trust me, the examination commitee will not appreciate unconventional explanations that openly contradict the established knowledge. And just by the way, it is "established knowledge" for a reason.

Naty1 -> By what you say I will assume you are not too familiar with the works of the people you mentioned and the state of physics at the time they were or are working. I can assure you that except for Einstein, DeBroglie and probably Copernicus, the others have been solving problems in a pretty conventional way. Brilliantly, but conventional science neverthelss. (And I'm still not sure what does Susskind have to do with the likes of Planck, Einstein, Witten... ) But I'm pretty pleased of having made your day with the most fantastical post on here.

Bob -> I will have a look at your work, but more out of curiosity than as the answer I'm looking for as I have two remarks on what you said. One, the fact that I or you don't know how the above puzzle is resolved in a conventional scientific way does not mean it has not been resolved or that it has no solution at all. I'm pretty sure both Glimm or Jaffe would know how to answer. Two, you mention photons and electrons and hence QED. In whatever approach you might want, but you're still talking about one specific model. My question, on the other hand is not tied to a specific model. It is supposed to be a feature of QFT in general. And incidentally comes from a discussion of scalar quantum fields with polynomial interaction in 2D. So it's not even tied to the dimensionality where the theory is formulated. If your work can be extended so to include every possible quantum field, regardless of the type or the dimension in which you formulate the theory, then it is a valid alternative and I'm sure it can be published. If it's model dependent, then you'll have to expand your results to really claim a novel perspective on old problems.

ytuab -> Finite temperature states (any physical state in thermal equilibrium) are not Lorentz invariant in the sense that there exists a preferred reference frame. Consider a huge box of gas at constant temperature. The preferred frame is the one in which you don't feel any "flow of warm gas" in your face. Also, if you want to describe an extremly hot gas for which the particles are relativistic, then you'll have to tackle field theory at finite temperature. So it's defeinitely not something intrinsically non relativistic. And finally, if you have not done so, please read my reply to bob_for_short. I'm NOT asking a question specifically in the case of QED. The claim is supposed to hold for any QFT model. Like $$\varphi^4$$, for instance.

13. Jul 9, 2009

### ytuab

EDIT: I'm very sorry. You are speaking 1+1 dimension? I'm misundestanding. Please tell me in 1+1 dimension, does the ultraviolet divergence occur? Is it as same as 4 dimension ? I'm speakin in 4D as follows,

------------------------------------------------------------------------------------------------
I think ultraviolet divergence (Lorentz invariant theory) is caused by the vacuum polarization.

Of course this world is at finite temperature, but Lorentz invariant.
I didn't say "extremely hot gas" means Lorentz invariant. ("extremely hot energy" is caused by the "virtual" vacuum).

But when the ultraviolet divergence is solved using the renormalization theory, in this actual vacuum(observable), the temperature is finite.

And also in $$\varphi^4$$ (Lorentz invariant theory), the ultraviolet divergence is caused by the infinite momentum and energy. And when we use the renormalization theory, the temperature becomes finite and it keeps Lorentz invariant.
(the virtual process)

Last edited: Jul 10, 2009
14. Jul 10, 2009

### Bob_for_short

The very first time the electron charge smearing was considered in T. Welton's publication in Phys. Rev. 1948, and I refer to it since he made a lot of good estimations.

Currently the UV divergences are removed with renormalizations, and Glimm or Jaffe know it. Just their explanation is too wrong.

The problem is not in QED but in how QED and other theories are constructed - the field interaction is a product of fields involved that contain the self-action. Quantization of any QFT is similar to QED so everything in them is alike, not model-specific.

I just wanted you to know that sometimes a theory can be reformulated so that no divergences appear at all.

Last edited: Jul 10, 2009