Re: why can't fields be quantized too?



tfleming1@xxxxxxxxxxxxx wrote:
Bruce Scott TOK wrote:

Physics said the energy levels of fields are quantized
as in photons but the field itself is not quantized.
What would happen if you quantize the fields too?

You get QED.


-- ciao, Bruce

drift wave turbulence: http://www.rzg.mpg.de/~bds/


finish the sentence: you get QED with all its inaccuracies

What is an "inaccuracy" of QED?


and singularities, including renormalizations!!

Renormalization works quite fine, where is your problem?

Care to calculate e.g. the anomalous magnetic moment of the electron
without QED?



those people who say dirac "did it already" (quantized the fields) are
wrong!!

Well, partly. Dirac started the work only, it was completed by Feynman, Tomonaga and Schwinger.



what dirac did was to quantize the EQUATIONS, not the fields
> which remained continuous as in the older classical days;
> the fields of QFT are STILL continuous today!!

Word salad and utter nonsense. Saying that equations are quantized makes no sense at all! Thanks for showing that you have not the *faintest* clue what you are talking about.


a look at heitler amongst others will confirm
W. Heitler, The quantum theory of radiation, (Dover, New York, NY,
1984)

Heitler shows in no way that "the fields of QFT are continuous".


what self-field theory does in effect is to use a discontinuous, and
discrete form of fiel , photons which can themselves be particles
too!

Err, and what's the difference to QED? If you did not notice: QED also uses photons, and in QED, photons also are particles. I don't understand why you inserted the word "themselves" and "too" here, both make no sense in the sentence.



it was this that lead me to TRY to use the photon as a massive
entity

With what mass, specifically? You *are* talking about the *rest* mass here, aren't you?



and following this analytic path yielded photonic chemistry

And that's what, precisely? And: can it explain the observed electron densities of molecules?


[snip]


> it should be realised that NO-ONE has previously solved Maxwell's
equations DIRECTLY;

Balderdash. A lot of solutions can be found in any textbook on electrodynamics.



quantum mechanics uses DERIVED wave equations with potentials,

And what exactly is the problem with that?


(that are integrals of the fields),

And fields themselves, may I remind you. Scalar fields.


constants of
integration (ambiguities and hence  'renormalizations' near
singularities),

Renormalization has little to do with ambiguities and constants of integration, and saying "near singularities" makes absolutely no sense here.



and lagrangians.

Lagrangians were already used in classical mechanics about 100 years previously, so what is your problem with them?


Like QED and QCD that use modified lagrangians, SFT uses
modified Maxwell-Lorentz field equations, by adding in the capability
of particles to move with additional degrees of freedom to those
exhibited by EM,

Word salad.


thus we obtain strong nuclear and weak nuclear SFT
(SNSFT and WNSFT).

Feel free to explain even *one* actual experimental result with these theories. Quantitatively.

You could start with the breaking of Bjorken scale invariance.


so there is a parallelism between QFT and SFT that is
analogous to FEM and FDM!!

That looks like a false analogy, since both FEM and FDM are legitimate methods, whereas only QFT is a sensible theory.


SFT also fixes up the coulomb assumption

And that is?


which is incorrect at the
atomic domain- you need to calculate the fields between centres of
motion not between charge points.

Why?

And: centres of motions of *what*?


this causes the inaccuracy known as
the uncertainty principle,

Plain nonsense. Read up on "hidden variables", Bell's inequality etc.


when you use point-to-point you always have
a non-linear field form,

Huh?


but with centres of motion you effectively
'get a chance to linearize'
the system of equations hence you can solve the equations directly
rather than having to use lagrangians etc a la QFT

Word salad.


Bye, Bjoern


.



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