Re: Connes & Marcolli paper on renormalization
From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 10/27/04
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Date: Wed, 27 Oct 2004 15:56:32 +0000 (UTC)
Chris Oakley wrote:
>>The less serious problem is revealed by Haag's theorem. In a nutshell,
>>it says that interacting fields cannot have Lorentz transformation
>>properties. In my approach I do not have interacting fields at all.
>>The interaction in my Hamiltonian is written in terms of particle
>>creation and annihilation operators. From this Hamiltonian I can
>>(in principle) calculate S-matrix, bound states, time evolution,
>>whatever I like, without ever mentioning "interacting quantum fields".
>>So, the Haag's theorem does not apply.
>
>
> A corollary of this is that if you force the fields to have Lorentz
> transformation properties then the Hamiltonian must be the free one. I don't
> have a problem with this. If an interacting scalar field \phi(x) is composed
> from a free field \phi_0(x) through (e.g.)
>
> \phi(x) = \phi_0(x) + \int d^4x'd^4x'' f(x-x',x-x'') \phi_0(x')\phi_0(x'')
>
> for some invariant function f - the kind of approach I advocate - then you
> find that the free Hamiltonian works just as well as the time displacement
> operator as the interacting one (just apply it directly & you will see). You
> don't therefore need to build an interacting Hamiltonian at all.
That's not the way I would like to do things. In my view, fields (either
interacting or non-interacting) do not have any physical meaning. I
haven't seen any experiment measuring fields. I formulate my approach
in terms of particles. I express the Hamiltonian (generator of time
translations) through particle creation and annihilation operators. Once
I have interacting Hamiltonian, I can do all kinds of physical
calculations: bound states, time evolution, S-matrix, etc. Fields
are not needed.
>
>
>>The more serious problem is related to renormalization. You are right
>>that if we take usual QED Hamiltonian (e.g., (8.4.22) - (8.4.25) in
>>Weinberg's book) and calculate elements of the S-matrix, we'll get
>>infinite results. My idea is that this Hamiltonian H is simply wrong.
>>It is wrong, but not completely wrong. It contain's a grain of truth.
>>In order to find the correct Hamiltonian H'', one should perform
>>two steps:
>>
>>1. The first step is renormalization. We add infinite (mass and
>>charge renormalization) counterterms to H, so that in the process of
>>calculation of the S-matrix all infinities nicely cancel each other
>>and we get agreement with experiment to 10^{-11}%. I agree that
>>resulting Hamiltonian H' is infinite and ugly, but this is the only
>>way to calculate the S-matrix we have so far. We swept infinities
>>"under the rug", this rug is Hamiltonian. That's were QED stands today.
>>
>>2. The second step is "dressing transformation". That's what I suggest
>>in my book. We can apply a unitary transformation U to the Hamiltonian
>> H'
>>
>>H'' = U H' U^{-1}
>>
>>This transformation can be chosen so that
>>1) the good S-matrix obtained in p. 1. is preserved
>>2) H'' is finite.
>>
>>All infinities from H' "transfer" to U. We swept infinities under
>>another "rug". This time the rug is operator U. Since (in contrast
>>to Hamiltonian) U has no physical meaning whatsoever, we can now
>>throw this rug away and forget the problems with ultraviolet infinities:
>>We have finite Hamiltonian H'' with interactions written in terms
>>of particle creation and annihilation operators. We can choose H'',
>>so that there are no more self-interaction effects in vacuum and
>>one-particle subspaces. So, one can work with H'' just as we work
>>with non-relativistic Hamiltonians (though, everything is relativistic,
>>of course). The only difference is that H'' does not conserve the
>>number of particles.
>>
>>The Hamiltonian H'' is rather complex. It has interactions in
>>all perturbation orders starting from 2nd. It has Coulomb,
>>magnetic, spin-orbit, etc. interactions between charged particles.
>>It has interaction responsible for the Lamb's shifts in 4th and
>>higher orders. It has interactions responsible for the Compton
>>scattering, bremsstrahlung, pair creation, etc.
>>
>>It would be great if we could write operator H'' somehow from
>>"general principles". Then we could avoid the two bad-smelling
>>steps I wrote above. However, I simply don't know how it is possible.
>>
>>I think the original problem of QED dates way back to the end of 1920's
>>when Dirac, Born, Heisenberg, and others guessed the form of QED
>>Hamiltonian (from canonical quantization of Maxwell's theory).
>>Their guess was fortunate and unfortunate at the same
>>time. It was unfortunate that their interaction had self-interaction
>>effects, and we pay now with all these tricks (regularization,
>>renormalization, dressing, etc.) to get rid of these effects.
>>It was fortunate that their guess contained a grain of truth,
>>which gives us accurate S-matrix and a hope to finally find
>>the correct Hamiltonian H''.
>
>
> It may be that no-one has found a "clean" way to construct this interacting
> Hamiltonian simply because it cannot be done ...
I have done that. You are probably right that the road to the
Hamiltonian H'' is somewhat untidy (moving infinities around
is not a rigorous mathematical exercise). However, after H''
is obtained (I have clear recipes how to do that) we can forever
forget such terms as "regularization", "renormalization",
"ultraviolet divergences". We can just postulate H'' as the
correct Hamiltonian of the theory. Then doing calculations in
QED becomes not more difficult than in non-relativistic QM.
Eugene
www.geocities.com/meopemuk
> indeed, it was the belief
> that I personally could not do any better that led me to try something
> completely different.
>
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