Go Back   Science Forums Biology Forum Molecular Biology Forum Physics Chemistry Forum > General Science Forums > Physics Forum
Register Search Today's Posts Mark Forums Read

Physics Forum Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


A version of QED without ultraviolet divergences

A version of QED without ultraviolet divergences - Physics Forum

A version of QED without ultraviolet divergences - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


Reply
 
LinkBack Thread Tools Display Modes
  #11  
Old 08-03-2004, 05:56 AM
Bilge
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences

Eugene:

I don't know what your deal is with qed, weinberg-salam, and qcd,
but the theories have been proven renormalizable (t'hooft-veltman).
I gave you an online reference. Did you read it?


Since you haven't posted anything, that remains a dubious claim.


``Effective theory'' means the parameters are scale dependent,
e.g., e and m are functions of the momentum transfer, so in effect,
that means a renormalized theories are effective theories.


SU(5) -> SU(3) x SU(2)_L x U(1)_Y -> strong, weak and E-M interactions.

Where I've used SU(5) as a unifying group only because it's the smallest
gauge group which can ``hold everything'', so-to-speak. In the unification
limit, the fields to become massless and and are expected to be asymptot-
ically free. so the no-interaction theorms no-longer apply.


That sounds like a foldy-wouthuysen transform. One makes a unitary
transform to the dirac hamiltonian that uncouples the large and small
components of the wave-function, which can be iterated in powers of
1/m to whatever precision one wishes. [e.g., ``Relativistic Quantum
Mechanics'']


That is not what renormalization means.


Define ``improved time resolution'' and how you think a particular
observable will be manifest in an experiment and consistent with
quantum mechanics.


So far, that is still vague. For example, do you mean by measuring
``spin projections?'' Of what? The spins in a collision between two
electrons, for example, is either a spin 1 triplet or a spin 0 singlet.


Reply With Quote
  #12  
Old 08-03-2004, 07:38 AM
Eugene
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences



Bilge wrote:


The renormalizability means that ultraviolet divergences can be
cancelled in all orders of the perturbation theory for the S-matrix.
This comes at a high price, though. The counterterms introduced
in the Hamiltonian by the renormalization are infinite. This is
a common problem of all renormalized theories.


See

E. V. Stefanovich, `` Quantum Field Theory without Infinities''
Ann. Phys. (NY) 292 (2001), 139-156.


The only similarity to the Foldy-Wouthuysen transform is that both are
unitary.



You can do renormalization in two ways. One way is to keep the original
finite Hamiltonian of QED and define some rules for subtracting
infinities from the matrix elements of the S-operator. This does not
look like a sound theory, just a collection of prescriptions.
Another (equivalent) approach is to add certaint counterterms to the
Hamiltonian, so that the S-matrix is still calculated by normal
quantum-mechanical formulas. This approach I like better.
But the problem is that in order to eliminate divergences from the
S-matrix the counterterms in the Hamiltonian must be infinite.
In renormalized QFT the masses and charges are formally infinite.
This is reflected in the presence of infinite counterterms in the
Hamiltonian.



Suppose that you want to describe collision of two electrons at very low
energies, e.g., some fraction of eV. You can apply usual
non-relativistic quantum mechanics, describe the electrons by their
wave functions, write the Hamiltonian with 1/r interaction term there,
solve the non-stationary Schroedinger equation, and obtain how the wave
function changes with time. From this wave function you can calculate
the time-dependent probability distributons for observables of
each electron (position, momentum, and spin projections).
From this solution, you can take asymptotic
limits at infinite past and infinite future where the electrons do
not interact anymore and move as free particles, and derive the S-matrix.

Now suppose that electrons collide with the energy of 1 GeV.
Clearly, the non-relativistic theory does not apply anymore.
My point is that the only difference between 1 eV and 1 GeV situations
is that the 1/r Coulomb interaction is not sufficient to describe
the latter. For the 1 GeV scattering we must take into account all
corrections to the Coulomb potential. These corrections include
1) relativistic corrections (like spin-orbit interactions)
2) radiative corrections (like those responsible for the Lamb shifts)
3) corrections which change the number of particles:
bremstrahlung - emission of photons
creation of various particle-antiparticle pairs
etc.
The rest of the theory should look the same as for the 1eV case.
We still should be able
to describe the system by a wave function (this time not the wave
function in the 2-electron sector, but a wave function which can
have components in many different sectors of the Fock space.
We still should be able to form the time evolution operator from
the interacting Hamiltonian. We still should be able to obtain
a time-dependent wave function and extract from it time-dependent
probability densities of observables of different particles.
We still should be able to take asymptotic limits and derive the
S-matrix from the more general dynamical theory.

Actually, in some sense the high energy problem is even simpler
than the low-energy problem. Because the collision of the electrons
and all these emission processes happen so fast that we see it
as a point interaction, and no experimental technique is capable
of resolving how the wave functions change during this short time
of collision. So, a long time ago (I think it started by Heisenberg)
we decided to simplify our theory and focus only on asymptotic limits,
i.e., on the S-matrix.

S-matrix calculations are much simpler than calculations of the full
time evolution, and we achieved a great success in calculating the
S-matrix. This is what QFT is all about.
However, it seems to me, we forgot that S-matrix is not
a complete description of nature. There is underlying dynamics
and time evolution of the wave function. A complete description
is given only by the Hamiltonian. My theory allows us to write
a finite Hamiltonian for the system of two electrons which includes
the Coulomb 1/r term plus all corrections listed above. This
Hamiltonian does not give much benefit in calculations of the
S-matrix or energies of bound states (old theory is doing just
fine there).

Probably, we will never observe interacting time-resolved dynamics
in 1 GeV collisions. But I am sure that with improved accuracy
we should be able to see the tiny effect of, e.g., radiative corrections
on the time dependence of observables of colliding 1eV electrons.

The most important lesson of all this is that usual rules of quantum
mechanics (wave function, time evolution, Hamiltonian, etc.) apply
equally well at 1 eV and 1 GeV energies. The same Hamiltonian applies
at all energies, and in order to describe creation and annihilations of
particles one does not need to add to quantum mechanics such
dubious concepts as fields or infinite masses and charges.



You are talking about the total spin of the system. This observable is
not so interesting, because it is conserved. I was talking about
spin projections of individual electrons. To predict the time evolution
of their probabilities one needs a dynamical theory.

Eugene

Reply With Quote
  #13  
Old 08-03-2004, 10:30 PM
Bilge
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences

Eugene:

You've ignored the ``price'', which I've already stated and referenced.
The energy at which the divergences become an issue is on the order of
10^250 MeV.


The hamiltonian formalism is not manifestly covariant. Time and
space are not treated on equal footing.


On line reference?



That's because you are looking at it as a collections of prescriptions,
rather than what the collection represents.


Those are parameters. What is measured is the renormalized mass
and charge.



The observables are not the independent spins of the electrons. The
observables are the eigenvalues for the two electrons as a single object.


Or, you can turn the interaction on and off at finite times.

[...]

[...]


Huh? We see things as point interactions when resolving them would
violate the uncertainty relations.


Why not? It contains all of the observables.


Basically, it sounds like you want a classical equation of motion.



Uh, I'm not sure I understand this. We already see the radiative
corrections, or else the radiative corrections wouldn't be there.


Sure, but you are asking different questions. In particular, you aren't
asking relativistic questions. You are only asking for relativistic
corrections in a particular frame.


Only if you limit yourself to distances larger than the classical
electron radius.



That is the only observable there is.


You seem to have forgotton that the individual spins are not good
quantum numbers.


Reply With Quote
  #14  
Old 08-03-2004, 11:26 PM
Eugene
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences



Bilge wrote:


This is not correct. You will encounter divergences in calculations of
any low-energy property, like electron's magnetic moment or Lamb's
shift. In order to get these numbers finite and accurate, the QED
Hamiltonian must contain infinite counterterms.




True, the Hamiltonian formalism is not MANIFESTLY covariant,
i.e., time and space are not equal. However, this formalism is still
RELATIVISTICALLY invariant. My definitions are:

Manifest covariance: All observables transform according to universal
linear Lorentz transformations, i.e., as 4-vectors, 4-tensors, etc.

Relativistic invariance: Transformations of observables furnish a
representation of the group of transformations of inertial observers,
which is the Poincare group.

My point is that manifest covariance is an artificial condition which
cannot be satisfied in interacting theories (It can be satisfied either
approximately, or for some truncated theories, like the S-matrix
approach). Relativistic invariance is the true law of nature
and must be satisfied exactly.



[Only registered users see links. ]



That's right. But these infinite parameters are present in the
Hamiltonian, which makes the Hamiltonian infinite and unsuitable for
studying time evolution.


The observables referring to the two electrons as a single object are
rather uninteresting: total energy, momentum, spin are conserved
quantities. The center-of-mass position moves uniformly along
straight line independent on reactions which may happen inside the
system. Conserved quantities are conserved, that's it.


S-matrix does not contain the time-dependence of observables,
only the relationship between asymptotic limits.


I talk about the wave function and quantum-mechanical operator -
Hamiltonian. If needed, we can take the classical limit and obtain
classical equations of motion and particle trajectories as an
approximation.



We see radiative corrections to static things like Lamb's shift or
electron's magnetic moment. Nobody has observed yet the radiative
corrections to time-dependent wave functions, or time-dependent
observables.
..

No, the Hamiltonian I am proposing is applicable at all distances and
energies. Of course, this Hamiltonian is approximate since it does
not take into account the possibility of creation of muons, pions,
W-mesons, neutrinos etc. But it correctly captures all electromagnetic
interactions.

You may be right if you define as GOOD only conserved quantum numbers.
But if you limit yourself only to such conserved observables, the
physics becomes trivial: everything is conserved, nothing happens.


Reply With Quote
  #15  
Old 08-04-2004, 12:58 AM
Bill Hobba
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences


"Eugene" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
referenced.

First when doing renormalization every textbook, including Weinberg, works
with the Lagrangian not the Hamiltonian eg in Weinberg see chapter 11 page
472. I had not really thought about it but Bilge is correct working with
the Hamiltonian would not be covariant - the Lagrangian is.

Second I have had a chance to review renormalization and my take is that
your concentration on counter terms is simply not warranted. Indeed page
525 of Weinberg discusses the floating cutoff method. The key is that all
physical quantities depend entirely on the cut off. IMHO this is a better
approach for thinking about basic issues because we use the conceptually
simpler momentum - or equivalently energy - cutoff rather than dimensional
regularization. The disadvange is that such a cutoff destroys gauge and
Lorentz covariance but that would seem a matter of mathematical convenience
rather than understanding fundamentally what is happening. This is what I
mean when I say that now we have the Electoweak theory that operates at
higher energies the need to address the infinites of QED would seem rather
moot because as we let the cutoff approach infinity we will certainly be
moving into physics where it is no longer really applicable.

The other point is that I am again struck by the mathematical complexity of
the technique just as much as when I first studied it. It seems one of
those things, at least with me, that familiarity does not breed contempt.
To be blunt I hate it - I keep saying is nature really that hard? The
answer of course is unfortunatly - yes. But as Einsten said - she is not
malicious - the problems seem tractable.


But if you are willing to give up covariance then the floating cutoff I
mention above would seem the more natural approach.


In the floating cutoff method there are no infinities - everything is
finite. So fundamentally what is your problem?

Bill

low
wave
object.
corrections
of
evolution


Reply With Quote
  #16  
Old 08-04-2004, 01:43 AM
Eugene
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences



Bill Hobba wrote:


We agreed already that Lagrangian and Hamiltonian approaches are
equivalent. Lagrangian approach tries to keep the manifest covariance.
This is an artificial condition, as I said earler. I prefer the
Hamiltonian approach, because the Hamiltonian has a direct physical
meaning as operator of energy and the generator of time translations.
What is the physical meaning of Lagrangian?

In the Hamiltonian approach
the relativistic invariance is preserved if we take care to add
interaction terms to the boost operator and keep the Poincare
commutation relations.


Introduction of the cutoff (regularization) is just a first step in
renormalization. This step makes integrals convergent, so you can
apply regular mathematics. The second step is to add counterterms to
the Hamiltonian to make sure that masses (poles in the propagators) and
charges (coefficients of scattering amplitudes) calculated from
the Hamiltonian come out in agreement with experiment. Without this step
you'll end up with infinite masses and charges of particles in the
S-matrix. But with this step you put infinite masses and charges in the
Hamiltonian. So, the main trick of renormalization is to move around
these infinities. In QFT we don't care about time evolution, so it is
considered OK to hide these infinite quantities in the Hamiltonian.
The third step is to increase the momentum cutoff to infinity. So, we
end up with the infinite Hamiltonian and finite and accurate S-matrix.



Actually, renormalization is not as difficult as usually presented.
Take a look at my unpublished paper, where I explain it on a simple model

[Only registered users see links. ]

Regards.
Eugene.

Reply With Quote
  #17  
Old 08-04-2004, 04:27 AM
Bill Hobba
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences


"Eugene" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
works
page
with
page
all
better
dimensional
convenience
I
rather

Not AFAICS with the technique outlined on page 525 of Weinberg (Wilsons
method). The process AFAICS is as follows.

Suppose we have a lagrangian L(m, c) - m is the mass c the coupling
constant. We introduce a cutoff in the lagrangian so that we have L(m, c,
y) - m is the mass, c the coupling and y the cutoff and all physical
quanties can be calculated without any divergences (eg after Wick Rotation
we impose an arbitrary cutoff in momentum so the integral is no longer
divergent - this breaks covariance and introduces mathematical difficulties
which is why dimensional regularization is used - but we are looking at
principles here). We note for renormalised theories the miracle that the
lagrangian with a cutoff is of the form L(m(y), c(y)) - ie the cutoff can be
absorbed into a redefinition of the mass and coupling constant. This is
where the counter terms come in - for the new mass we note it is of the form
(mr + dela m) times something; mr is the renormalized mass and delta m times
something is the counter term. The point being that all physical quantities
calculated from the lagrangian are finite - the counter term (s) are simply
a byproduct of the process. And in Wilsons method he shows that for
renomalisable theories: (to quote from Wienberg):

'At cutoff y << y0 all the couplings therefore approach an N dimensional
surface J with coordinates G(y) which is independent of both the initial
surface and y0'
'Changes in y with y much less than y0 will change the couplings but the
couplings will remain close to J'
'But we can invert the relation G(y) in terms of the conventional parameters
and y. In this way we can justify the usual renormalization program: all
physical quantities are expressed in a cutoff-independant way in terms of
the conventional renormalized couplings and masses'

For it to work theories do not even have to be renormalisable - it is great
when it occurs and we have a way to express our physical quintiles in a
cutoff independant way - but in principle we could impose some large but
finite cutoff we do not exceed and still obtain finite answers. In fact my
understanding is that is exactly what effective field theories of gravity
do.



From where I sit that is not the main trick - the main trick is that
everything still remains finite regardless of the value of the cutoff and
there exists some cutoff beyond which the physical quantities do not really
depend on - this means we do not have to worry about infinities at all. The
absorption into a redefinition of mass and the coupling constant that is the
renormalization 'shell' game is simply a mathematical byproduct.

Bill


In QFT we don't care about time evolution, so it is
of
contempt.
not


Reply With Quote
  #18  
Old 08-04-2004, 06:32 AM
Eugene
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences



Bill Hobba wrote:



Sorry, I have a hard time to understand what Weinberg is talking about
in Section 12.4. Can we discuss renormalization in terms of the
conventional approach? I trust Weinberg when he is saying that both
approaches are equivalent. I think I understand the conventional
approach pretty well. This is the one outlined on page 473.
Though, I would prefer to substitute the Lagrangian with the Hamiltonian
there. The Hamiltonian (before renormalization) has the form of eq.
(8.4.22) - (8.4.25).

Wouldn't you agree that the picture of physical world presented
by the renormalized QFT is rather odd? Vacuum is a "boiling soup"
of particle-antiparticle pairs which cannot be observed and which
have infinite energy. Particles, that are supposed to be the simplest
elements of nature, are "bowls of spagetti" made of pairs, photons, etc.
which cannot be observed as well. The interaction is described by
emission and absorption of virtual particles with imaginary masses,
which also never observed. I hate when a theory has so many ingredients
that are not observable.

My point is simple: the Hamiltonian (8.4.22) - (8.4.25) (plus
renormalization counterterms) is wrong.
It is not completely wrong. It captures correctly two important
things: It has correct spectrum of energies of bound states, and it
predicts correct scattering. The rest of it is wrong. All we need to
do is to perform a unitary transformation of this Hamiltonian
(such that the bound states and the S-matrix are not altered).
With the new transformed Hamiltonian everything becomes simple and
normal: vacuum is a state without particles and has zero energy.
Single particles are just single particles with experimentally
measured masses and charges. The interaction is described by
interparticle potentials (just as in usual non-relativistic quantum
mechanics with the addition of potentials which do not conserve the
number of particles: bremstrahlung, pair creation, etc.).

Anticipating your objections, I claim that all this can be done
without violating the relativistic invariance of the theory.

Eugene.

Eugene.

Reply With Quote
  #19  
Old 08-04-2004, 09:12 AM
Bill Hobba
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences


"Eugene" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
c,
Rotation
difficulties
the
can be
form
times
quantities
simply
parameters
all
of
great
my
gravity

The point I am making is that Wilson's approach (the effective field toper
approach), IMHO, resolves the infinites so it is no longer an issue.


Sure - just like I mentioned I found the math a bit yucky. That however is
just a personnel reaction and has nothing to with it as a theory.


As you know I disagree. I am waiting to get some comment on your paper
regarding radioactive decay and want to reserve farther discussion of you
ideas until then.

Bill


Reply With Quote
  #20  
Old 08-04-2004, 04:09 PM
Bilge
Guest
 
Posts: n/a
Default A version of QED without ultraviolet divergences

Eugene:


The renormalization in the one-loop approximation is given by

e_bare = e^2/(1 - k ln(q/q_0)), k = e^2/6\pi^2

I hunted down and provided you with a reference for this previously.


See above mentioned references, although if you bjorken & drell,
that should provide enough justification for either of those two
calculations.


QED gets the electron magnetic moment to 13 decimal places, which
is sufficient justification until such time as one can do better
with a different theory.

You seem to be claiming several completely different things in the
context of your hamiltonian. One is that your hamiltonian offers some
advantage over the lagrangian formalism. That may or may not be the case,
so I'm not going to comment on that. Another is that qed is a problem due
to divergent integrals. That isn't the case for the reasons I've mentioned
previously, which I'll restate briefly. (a) Renormalization becomes a
problem at the landau pole. That energy is extremely large; so large as to
be irrelevant, since (b) qed is not believed to be a fundamental theory,
and (c) since qed is only one part of a theory which is believed to
asymptotically free, there is no issue. The third claim you make is that
your hamiltonian is free from any infinities. I've only had time to look
briefly at your articles, but just to note an item or two, you start by
constructing a fock space. There exists a theorem which proves that an
interaction picture doesn't exist in a fock space. I also note that you
simply define some operators as ``unphysical''.



Reply With Quote
Reply

Tags
divergences , qed , ultraviolet , version


Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On

Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Version 8 of AiO Christiaan Karreman Protocols and Methods Forum 0 05-04-2006 07:27 AM
photologie forestière - version 2005 Hubert Roussel Forum Biologia 1 09-29-2005 02:50 PM
GNU units and units.dat; Units of Measurement and Unit Conversion James Redford Physics Forum 0 07-31-2005 12:08 PM
FW: A less garbled version = Ryder, Elizabeth F C Elegans Forum 0 12-04-2003 02:21 AM
How come some low-end thermometers are still made in mercury version? AC/DCdude17 Chemistry Forum 9 11-09-2003 01:09 AM


All times are GMT. The time now is 08:00 PM.


Powered by vBulletin® Version 3.8.4
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
Copyright 2005 - 2012 Molecular Station | All Rights Reserved
Page generated in 0.25176 seconds with 16 queries