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Why don't virtual particles cause decoherence?

by Mukilab
Tags: decoherence, particles, virtual
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mattt
#127
Feb27-13, 09:56 AM
P: 125
Quote Quote by tom.stoer View Post

(I do not want to go through all the discussion regarding virtual particles, I simply refer to a perfect summary written by Arnold Neumaier) http://www.mat.univie.ac.at/~neum/ph.../unstable.html
Totally agree with everything A. Neumaier wrote there (and like Tom, in the future I will simply refer to that link).
JK423
#128
Feb27-13, 09:59 AM
P: 381
Quote Quote by mattt View Post
Totally agree with everything A. Neumaier wrote there (and like Tom, in the future I will simply refer to that link).
It's not that simple; as you can see in another thread "Source of virtual particles in space?" not everyone agrees with what Neumaier says. So...
tom.stoer
#129
Feb27-13, 05:06 PM
Sci Advisor
P: 5,366
Anyway, it's irrelevant for decoherence
Mukilab
#130
Feb28-13, 12:07 PM
P: 73
Quote Quote by tom.stoer View Post
I guess we should come back to the question



My first answer was



The discussion over the last couple of days did not change anything; the first answer is still correct.

Let me summarize some additional ideas



But all this is not directly relevant for the original question b/c virtual particles are completely irrelevant in the context of decoherence: they are not present in the full theory; they do neither introduce the above mentioned factorization of H nor the partial trace; and they are not states in any Hilbert space Hsystem, Hpointer or Henvironment .

Last but not least: nobody would assume that any approximation like a Taylor series (or green dwarfs) do introduce additional effects which are not already present in the full theory w/o the approximation (w/o green dwarfs). So if the theory w/o virtual partices green dwarfs) already contains decoherence (gravity) it would be silly to say that decoherence (gravity) is due to virtual particles (green dwarfs). This changes if the theory cannot be formulated w/o virtual particles (w/o green dwarfs), or if the formulation is conceptally simpler (in the sense of Ockham's razor) using virtual particles (green dwarfs).

I am not an expert regarding green dwarfs, but I know that perturbation theory is incomplete and misses relevant non-perturbative effects. So I can't see any reason to rely on the interpretation of partially unphysical artifacts due to an incomplete approximation instead of using the full theory.
Thanks, but if virtual particles don't 'exist' or they are not necessary in many quantum theories, why do they exist?

Please understand that I do not understand the large majority of scientific jargon on this thread, would it be possible for you to try to explain this as if to someone that is slightly above a layman (I have knowledge of most concepts, just not the mathematics behind them, which I do not understand). Thank you.
tom.stoer
#131
Feb28-13, 01:57 PM
Sci Advisor
P: 5,366
Virtual particles in the sense I define them - as internal lines in Feynman diagrams - are a mathematical tool used for an approximation - so-called perturbation theory. Unfortunately most quantum field theories like QED, QCD etc. are very complicated and we do not have the mathematical tools to solve them exactly. But fortunately we have several approximations. Perturbation theory is used for weak coupling, where it makes sense to start with free, non-interacting particles and to add small corrections for interactions. Interestingly this works very well in many cases, especially for scattering experiments (but there are other problems like QCD bound states, e.g. protons, neutrons, ... where this approximation is useless).

If you would have mathematical tools to solve quantum field theories exactly, there would be no reason to introduce perturbation theory, there would be no name for the mathematical artifacts, and we would not have these discussions. Before studying QCD I was working on two-dimensional models, fields living on a line = one space dimension + one time dimension. These models a rather simple, a good starting point for beginners. There are exactly solvable models with bound states, there are other approximations, and thefore no reason to use perturbation theory. In QCD there are tools to study non-perturbative aspects, tools like chiral effective theories, lattice gauge theories, ... All these tools do not require Feynman diagrams and therefore - using these tools - there is nothing which we call 'virtual particle'.

In addition there are mathematical reasons against perturbation theory. We know that strictly speaking it is ill-defined, it is something which does not exist mathematically, but nevertheless it seems to work in a very restricted sense. And there are applications where this ill-defined math does produce correct results which agree with experiments (strange, isn't it? we can prove that it does not work, but using it seems to work ...). Now what I am saying is that we can START with a formulation w/o any approximation and w/o virtual particles. Then we have to introduce approximations, but doing this we CREATE several problems, or we apply approximations outside their scope of applicability, so the approximation BREAKS DOWN. Doing this we have mathematical artifacts - virtual particles - but due to the problems we introduce there seems to be no good reason to believe in virtual particles to be more than just limited tools.

Does this help?
rkastner
#132
Feb28-13, 05:25 PM
P: 161
I would not be quick to dismiss something as only a 'mathematical artifact' just because it is part of an approximation. This is an interesting methodological question that deserves careful study. Remember that the number 'e' is a limiting value for many types of series expressions. It's possible that these 'approximating' series do have specific meaning in themselves; the fact that they have a well-defined limit at a specific irrational number does not negate that the series terms might have physical content if the argument of the exponential has physical content. Remember that the perturbative expansion of QFT has as an argument for the exponential the action of the field, which certainly has physical content. It's a leap to say categorically that terms in the expansion of a quantity with physical content do not themselves have physical content. Now of course we have to be careful about what the physical quantities are. In the earlier example of the radioactive atom, the exponential argument is not a field, it's just a number (decay rate). So that comparison was not a good one to the QFT case.

I appreciate the interesting discussion here but I should let you know that I am currently swamped with various obligations and may not be able to check in for a while. Thanks again everyone for your interest. I hope you will visit my website which presents preview material from my book and explores some of the ideas we've discussed here.

http://transactionalinterpretation.org

Best wishes
RK


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