Re: Download a new book on quantum mechanics and relativity.

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 11/16/04 

Date: Tue, 16 Nov 2004 12:17:39 -0800

Bilge wrote:
> Eugene Stefanovich:
> >Bilge wrote:
>
> >> Look eugene, you can't assert something and simultaneously assert the
> >> opposite. Are you claiming that no such transformation between the inertial
> >> and accelerated coordinates exists? It should, since that is what you are
> >> transforming.
> >
> >We are probably using different terminologies, and that confuses you.
> >Let me give you a short glossary of terms I am using.
>
> I believe we've already discussed the fact that you tend to use
> terminology in ways that aren't standard.
>
> >Inertial reference frame (or observer) - the reference frame moving
> >without acceleration and rotation. All inertial frames are equivalent
> >according to the principle of relativity.
>
> >Inertial transformation - a transformation between inertial frames.
> >There are 10 kinds of inertial transformations: time translation (1),
> >space translation (3), rotation (3), boost (3)
> >Poincare group - the group whose elements are inertial transformations.
> >The Lorentz group is a subgroup (rotations + boosts) of the Poincare
> >group.
>
> Why are you telling me this?
>
> >Physical system - an object which can be observed and measured by
> >inertial observers. I consider only isolated physical systems
> >(no external fields).
>
> >Observable - a quantity describing physical system. In QM, observable
> >is described by a Hermitian operator. Example: Newton-Wigner position
> >operator corresponds to the observable of position. Time is not a
> >property of physical system, so time is not observable. Time is a
> >parameter (time translation parameter) labeling inertial frames
> >of reference
>
> For any set of measurements, you must choose a set of observables
> which commute. For example, you cannot simultaneously choose x and p
> to be observables of your system. You don't seem to distinguish between
> expectation values, <A> = <\phi|A|\phi>, and eigenvalues, A|\phi> = a|\phi>.
>
> >Event - a physical process which is well localized in both space and
> > time. A good example of an event is a collision of two particles.
>
> An event is a spacetime point. I've already told you that. The
> usage is standard in the literature and in textbooks, so there's
> no excuse for ambiguity here.
>
> >Events have well-defined meaning only in the classical limit of theory.
> >In the quantum case the localization in both space and time is not
> >well-defined, in particular, due to the spreading of wave packets.
>
> Then don't use the term ``event'' when you aren't referring to
> a classical limit.
>
> >Boost transformations - transformations of observables of a physical
> >system measured in two inertial frames connected by boost. Boost
> >transformations of observables for parts of an interacting system
> >(i.e., r_1, r_2, p_1, p_2, in the system consisting of two interacting
> >particles) are non-linear and interaction-dependent.
>
> If you're going to go to the effort of defining something for clarity,
> you should not just write down a trivial definition that doesn't address
> the issue. The issue here is how you interpret that physically, which
> is related to your idiosyncracies about fields.
>
> >Lorentz transformation - a particular case of boost transformations
> >in which observables transform by linear Lorentz formulas. There are two
> >examples of observables transforming by Lorentz:
> >1) the total energy-momentum 4-vector (E, P_x, P_y, P_z) of isolated
> >physical system
> >2) Space and time coordinates of events with non-interacting particles
> >(e.g., intersections of particle trajectories) also transform by
> >Lorentz. Here position is taken as expectation value of the
> >Newton-Wigner operator, and time is the label of inertial frame.
>
> That represents particular choices of observables.
>
> [...]
> >> Since you don't use the standard definition of am event, either, what
> >> does that mean?
> >
> >See my definition of 'event' above
>
> Your definition above is rather meaningless. The ``collision'' of
> two particles is not an event unless you take it to occur at spacetime
> point. If you do thatm your dream of describing whatever evolution
> you think you'll be describing goes out the window.

Have you noticed, that there is no spacetime or spacetime points among
my definitions. I simply do not have those concepts. I still can
describe time evolution and other physical effects.

>
>
> [...]
>
> >I call 4-vector a quadruple of quantities which transform according
> >to above linear laws wrt boost. The spatial part of 4-vector must
> >be also a 3-vector, i.e. have usual linear transformations wrt
> >rotations. The time-like part of a 4-vector is 3-scalar.
>
> That makes a four-vector a geometric quantity.

There are only 2 examples of 4-vectors in my approach. They were
mentioned above: 1) space and time coordinates of events with
non-interacting particles and 2) total energy-momentum 4-vector of
isolated system. I agree, that geometrical representation is possible
in these two particular cases. I disagree that geometrical
representation can be applied to everything.

>
> [...]
> >> Earth to eugene. I would say that, except that I'm responding to you and
> >> you don't use standard terminology consistently. Personally, I think that
> >> it's a real pain in the ass to have to say everything so awkwardly so
> >> why don't you just try to get the point rather than finding another way
> >> to avoid it, ok?
> >
> >I explained my terminology above. I use it consistently.
>
> No, you don't.
>
> [...]
> >> from the fact that they are covariant as written. So, for all intents
> >> and purposes, you can reduce special relativity to the first postulate
> >> alone and relegate anything having to do with light to a theory of E&M.
> >
> >I asked you so many time times... Please prove the Lorentz
> >transformations from the 1st postulate alone, and nobody gets hurt!
> >If you do that, I'll agree with you that relativity is all about
> >geometry, I'll start to believe in Minkowski spacetime.
>
> I already told you, just admit you can't obtain a simple finite
> transformation from an infinitesimal one and I will. However, I'm
> not particularly interested in having to write everything in
> notation you'll accept and I'm not about to try figuring out what
> that is. If you aren't going to be satisfied with (t,x,y,z)
> written compactly as x^u, then you might as well not admit to
> anything, since I'm not going to accomodate your notational
> quirks just to keep t and x separate.

I do not insist on
using my notation. You can use any notation as long as you explained
what means what.

>
> >The point is that you cannot prove that. I am not going to believe you
> >until you actually prove this statement, or show me where such a proof
> >is written.
>
> The point is, that it's straight forward to prove. Just transform
> some arbitrary function, \tau(x,t) -> \tau'(x,t) = \tau(x',t')

That does not look like a proof to me. Try again.

[...]
>
> [...]
> >>
> >> Let's get one thing straight here. A frame of reference is a coordinate
> >> system, not a physical system.
> >
> >I don't have word "coordinate system" in my glossary.
>
> That is your problem, not mine. You also say you don't have fields,
> but you use field operators. Coordinates are an n-tuple of numbers.
> If you have time and space, then I can write the time and space as
> an n-tuple of numbers.

I have fields as abstract mathematical objects, and the arguments of
fields (t,x,y,z) are just 4 real parameters not related to real time and
space in my approach.

For real events (physical processes localized in space and time) I can
define the physical observable of position R (expectation value of the
  Newton-Wigner operator) and the value of time t (the time label of the
frame of reference in which the event happens).

[...]
>
> [...]
> >> >I agree that charge, momentum, etc. are conserved in QED. I do not agree
> >> >that interactions propagate instantaneously in (usual renormalized) QED.
> >> >I asked (a few times already) to prove this statement.
> >>
> >> I cam't prove that interactions propagate instantaneously in qed, since
> >> interactions don't propagate instantaneously in qed. You should know that,
> >> since you posted a lot of articles arguing about it when I attempted to
> >> convince you that qed required the interaction to propagate at `c', not
> >> instantaneously. You know, when someone implies that I said something
> >> completely opposite to what I've actually said, I get the distinct
> >> impression that person is an *** whose only reason for posting is to
> >> argue by misstating what I've said if it's necessary to switch their
> >> argument and avoid agreeing with me.
> >
> >Don't be mad at me, I've just misspoken. I asked you to derive the
> >speed of interaction in QED. I believe, you think that the speed is c.
> >Then prove it!
>
> It's obvious. The interaction is carried by A_u. A_u is a four-vector.
> It satisfies the klein-gordon equation.

This does not look like a proof to me. Try again.

>
> >> >I bet you cannot
> >> >do that, because in QED there is no well-defined Hamiltonian, without
> >> >Hamiltonian you cannot describe time evolution, without time evolution
> >> >you cannot say whether interaction is retarded or instantaneous.
> >>
> >> Dont be an imbecile. First of all, in _any_ lorentz invariant nothing
> >> propagates instantaneously. If it did, the theory wouldn't be lorenzt
> >> invariant.
> >
> >Wrong. You cannot prove that.
>
> Didn't you tell me that gauge invariance was trivial when I obtained
> the the transformation for p_u, which gave me A_u and showed how A_u
> had to transform? Maybe you should go back and look at it again.

I am not sure what is the connection between the gauge invariance and
the speed of propagation of interactions. I have a counterexample to
your statement. In my theory, interaction propagates instantaneously,
and the relativistic invariance is satisfied.

>
>
> >> Second in qed, the photon is what carries the electromagnetic
> >> interaction.
> >
> >Do you call "photon" the wavy (internal) line in Feynman diagram?
> >You probably know that this "virtual particle" is off-shell. It has
> >some unusual properties, like imaginary mass.
>
> (1) Observable results are always the covariant sum of a set of diagrams,
> at each order,
>
> (2) what you are calling a ``mass'' is the longitudinal momentum
> (or polarization, if you like) transferred in the coulomb
> interaction considered in isolation from the other components,
> for which cannot be separated covariantly. It vanishes when
> coupled to a conserved current.

This does not explain why you believe that virtual photon propagates
with the speed of light. Calling it a "photon" does not make it speed
equal to 'c'. Some proof is needed here.

>
> [...]
> >> For the simple reason that the energy-momentum relation is conserved
> >> current, which is lorentz invariant and instantaneous interactions
> >> represent energy propagating over spacelike intervals. You can't have
> >> conservation of energy and momentum if energy and momentum can appear
> >> acausally.
> >
> >I have a theory in which momentum and energy are conserved, and
>
> Separately. How do you form a scalar product E^2 - (pc)^2, if
> you don't have a metric? If you have a metric, why are you making
> a big deal out the word ``spacetime?''

I write this combination E^2 - (pc)^2, because this is a Casimir
operator of the Poincare group. It commutes with all generators,
so it's an invariant called "mass squared". Note, that the
invariance is true
only if E and p are total energy and total momentum, respectively.
For individual particles in the interacting system E_i and p_i
do not form a 4-vector.

>
> [...]
> >> No one has ever observed a creation or destruction operator, either,
> >> but that didn't stop you creating a number operator did it? Velocity
> >> isn't an observable, but you have no problem with that. If you don't
> >> like off-mass shell, use the word ``accelerated.'' While acceleration
> >> is also not an observable, that doesn't seem to present a problem for
> >> you.
> >
> >Creation and destruction operators are useful mathematical tools.
>
> Then perhaps you shouldm't talk about those either.

Why? They provide a useful notation for writing operators in the Fock
  space.

>
> >Number operator corresponds to measurable observable - the number of
> >particles. Velocity is observable. It can be measured. There is a
> >Hermitian operator of velocity. Virtual particles have not been
> >measured.
>
> There is not an hermitian operator for the velocity (other than
> \alpha in the dirac equation which has eigenvalues +/- c).

I wrote you a well-defined operator of velocity V = Pc^2/H
which has eigenvalues less that 'c' for massive particles.
What's wrong with that?

>
> [...]
>
> >> You simply choose to ignore whatever you don't want
> >> to acknowledge based on personal idiosyncracies. So far, you have
> >> not been able to tell me how to even observe the spin of anything
> >> without referencing something you deny exists.
> >
> >Stern-Gerlach apparatus is fine for that.
>
> Two problems with that. (1) The stern-gerlach apparatus uses the
> word ``field'' as in B-field. You don't consider fields to exist,
> therefore there is no such thing in your theory, at least not
> until you derive it without a field.

In my theory I have direct interparticle forces. Strictly speaking
the (spin-dependent) force acting on electron in the Stern-Gerlach
apparatus is the sum of forces acting on this electron from
all charged particles (moving electrons and nuclei at rerst) in the
apparatus. I can probably (approximately) represent this force as a
"field" varying in space. I don't see problem with that.

> (2) Your consider your entire
> interaction to consist of the coulomb potetial, j_0 j_0/|x-x'|.

No. The QED interaction operator I am using is

V = -\int dx j(x).a(x) + \int dx dx' j_0(x) j_0(x')/(8 \pi |x-x'|)

I would prefer to write this interaction operator using
creation and annihilation operators only, and thus avoiding
the field quantities j(x) and a(x). This would make the formula
very long.

> You can't derive the interaction related to the spin from that.
> If you want the interaction via the magnetic moment, you need
> the entire current operator, \gamma^u.

[...]
>
> [...]
> >> If your theory was poincare invariant, you wouldn't have interactions
> >> that propagate instantaneously.
> >
> >Your statement is flat wrong. I have an action-at-a-distance theory
> > which is relativistically (Poincare) invariant. Such theories are
> >known for many years (see, for example, Bakamjian and Thomas,
> >Relativistic Particle Dynamics, II, Phys. Rev. 92 (1953), 1300).
> >The main difference of my approach wrt most BT-type theories is
> >that I do not assume the conservation of the number of particles.
>
> You must be confused. Your theory is quantized on a hypersurface
> of constant t (or else, it isn't in instant form). That means you
> specify the positions of all of your particles and on that surface
> and let them evolve with an interaction. That says nothing about
> the propagation of the interaction.

The force acting on any given particle depends on positions and momenta
of other particles at the same time instant. This is
action-at-a-distance.

>
> [...]
> >> That is not how experiments work. I propose that you explain precisely
> >> how to count a nanoamp of beam current on a target and all of the outgoing
> >> particles into 4\pi and tell me how to determine if a single charge was
> >> destroyed or created. If you can't do that, do something even better.
> >> Prove that your operator represents the electric charge and tell me
> >> what observables I can measure fairly easily that would show whether
> >> or not your theory is correct regarding that being the electric charge as
> >> well as whether or not it's conserved. That's what every other theory
> >> does.
> >
> >I define total charge as the sum of charges of individual particles
> >in the system.
>
> What is the charge on a particle called the neutron or the neutrino
> according to your theory? Your theory apparently doesn't define charge
> in a way that differs from the way it defines particle.

The charges of particles are not derived in my theory. They are put in
theory according to experimental data. So, the charges of neutron and
neutrino are zero.

>
> >Current is how many charges cross the surface in unit time.
>
> Is this another of those misstatements? J is a current density.
> The (halliday & resnick) meaning of current, I = \integral J.dS,
> isn't what we're talking about here.
>
> >If you have a different definition of charge, then what is it?
>
> It's the integral over all space of the conserved current arising
> from the U(1) symmetry of the lagrangian. For the U(1) transform,
> \Psi -> \Psi\exp(ia) =~ (1+ia)\Psi, the variation in the lagrangian
> is,
>
> \delta L = (dL/(d\Psi)) \delta\Psi + (dL/d(d_u\Psi))\delta d_u\Psi + h.c.
>
> \delta\Psi = ia\Psi, \delta d_u\Psi = ia d_u\Psi
>
> Pluggig all of that in and using the fact that
>
> d_u [ dL/d(d_u\Psi)) \Psi] = d_u [ dL/d(d_u \Psi)] \Psi
>
> + (dL/d(d_u \Psi)) d_u \Psi
>
> gives me:
>
>
> \delta L = [(dL/(d\Psi)) - d_u (dL/d(d_u \Psi))] \delta \Psi
>
> + d_u [ dL/d(d_u\Psi)) \Psi] + h.c.
>
>
> The terms on the first line in [ ]'s vanish by virtue of the euler
> lagrange equations. The second term is the conserved current via
> noether's theorem (once you include the hermitian conjugate).
> The integral over all space of the spatial part is the charge.

Thanks for this dervation. I do things differently. I just write

j^u = -e \psi^{\dag} \gamma^0 \gamma^u \psi

and plug this expression in the interaction Hamiltonian above.
The field quantity j^u has no direct connection to the observable
of charge Q, which is just the sum of charges of individual particles.

>
>
> [...]
> >> Well, how come neutrinos don't have electric charge? Your definition
> >> of electric charge is perfectly consistent with the definition of a
> >> neutrino. It's a lepton, just like the electron. Let's not forget
> >> about quarks. Let's especially not forget about neutrons. The strong
> >> interaction doesn't distinguish between neutrons and protons, so the
> >> only way to distinguish between them is by their electric charge and/or
> >> magnetic moment (the neutron does have a magnetic moment), But, your
> >> definition of charge cannot distinguish between neutrons and protons.
> >
> >I just know that neutrinos and neutrons are neutral.
>
> Your operators do not make such a distinction. Your ``definition''
> of charge is synonymous with your definition of a number operator for
> a spin 1/2 particle. How would you propose to deal with the neutron
> magnetic moment?

I guess, the neutron magnetic momentum comes from the internal (quark)
structure of the neutron. I don't do strong forces (yet). So, in my
approach, the neutron will be just a neutral particle not
participating in electromagnetic interactions.

[...]
>
> [...]
> >> To paraphrase you, no one has ever observed that potential. All they've
> >> ever observed are some spectral lines inferred from a reading on a
> >> meter.
> >
> >That's not correct. One can measure trajectories of interacting charged
> >objects in the macroscopic world.
>
> That isn't any better. OK, no one has ever observed tht potential.
> All they've ever observed are some trajectories and spectral lines.
> However, using your own criteria, the trajectories are dubious, since
> no one has ever seen an electron. All that has ever been observed
> are counts in some histogram, from which one must infer existence.

One can see electron tracks in bubble chambers. That's good enough for
me. When I'm talking about trajectories, I don't mean electrons.
They are too small to have trajectories.
I am more inclined to talk about macroscopic particles, like charged
specks of dust or pieces of paper attracted to a charged comb.

> This is not a problem from my point of view, but since your argument
> against fields and every other conventional quantity your theory doesn't
> admit to using, is that no one has ever seen a field, I'm just applying
> your argument to other things no one has seen.

The issue is not only that no one has ever seen a field. It is even more
important that all physical effects attributed to fields (e.g.,their
action
on particle trajectories) can be equally well (or even better) described
without using the field concept. It is enough to have particles and
direct interactions between them, to get all there is in physics.
Fields are not just unobservable, they are redundant.

>
> >These trajectories are determined mainly by the Coulomb term, but the
> >magnetic interaction also has a sizeable contribution. This can be
> >measured, at least in principle.
>
> (1) To use your own criteria of observable (i.e., no one has ever
> seen a field''), without being hypocritical, no one has ever seen
> a ``coulomb term''. And since you've consistently obhected to any
> use if electric or magnetic fields, no has seen a magnetic interaction,
> either. That would seem to require a magnetic field.

Interactions (Coulomb or magnetic) are needed to explain the observable
effects: deviations of particle trajectories from straight lines,
scattering, bound states, etc. Interactions are theoretical concepts,
you cannot touch them. But they are very useful. On the other hand, my
theory is doing quite well without fields, so I say that quantum
fields are
reduced to merely convenient mathematical combinations of creation
and annihilation operators.

>
> >> So then your theory is galilean invariant and c = \infty. If you
> >> don't have a problem with that, neither do I.
>
> >Incorrect. My theory is Poincare invariant and c (the speed of light)
> >has its normal value of 300000 km/s, and interactions propagate
> >instantaneously.
>
> Since nothing can propagate faster than `c' in a theory which is
> poincare invariant, it would seem your claim is inconsistent. Obviously,
> if the interaction propagates instantaneously, then energy and momentum
> associated with that interaction propagates instantaneously. If energy and
> momentum propagate instantaneously, then E^2 - (pc)^2 = m^2 can't hold,
> since that requires energy and momentum to propagate at a finite speed.

I don't see any connection between E^2 - (pc)^2 = m^2 and the speed of
propagation of interaction.

>
> Furthermore, that is inconsistent with your earlier claims that
> your operators like p and x commute on a spacelike interval.

The commutators of position and momentum are

[R_i, P_j] = i \hbar \delta_{ij}

no "intervals" of any kind involved here.

> The
> only reason incommensurate operators can commute on spacelike
> intervals is that it's not possible for any sort of signal to
> propagate between the measurements. An instantaneous interaction
> violates the requirements for them to commute.

You are probably talking about the commutation of fields \psi(x)
at spacelike intervals. As I said before, the arguments of fields
(x) have no relationship to experimentally measured position and time.
The commutators of fields have no relationship to the speed of
interaction.

Eugene.

>
>
>

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