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

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 12/08/04 

Date: Wed, 08 Dec 2004 00:24:09 -0800

Bilge wrote:
> Eugene Stefanovich:
> >Bilge wrote:
> >> Eugene Stefanovich:
> >> >Bilge wrote:
> >>
> >> >> Fine, then accept the fact that your concept of what ``c'' means
> >> >> differs from poincare invariance says it means. So, whether or not
> >> >> your theory is poincare invariant, depends upon whether I consider
> >> >> your theory independent of your interpretation of it or consider your
> >> >> theory to be defined by your interpretation of it. In the latter case,
> >> >> your theory is galilean invariant.
> >> >
> >> >In my approach, "c" means the speed of light or photons. This speed
> >> >is directly measured in experiment.
> >>
> >> In that case, your parameter `c' has nothing to do with the parameter
> >> `c' in the poincare group from which the relation E^2 = (pc)^2 + (mc^2)^2
> >> is obtained. So, your theory is still galilean invariant except that for
> >> some weird reason, photons propagate at a fixed velocity that has nothing
> >> to do with anything else in your theory.
> >>
> >> >Parameter "c" also enters in
> >> >the commutation relations of the Poincare group (which is the group
> >> >of transformations between inertial observers). The theory is invariant
> >> >with respect to these transformations. However, this does not prevent
> >> >me from saying that measurements of particle observables on the
> >> >"instant time hyperplane" are possible.
> >>
> >> Yes, it does. Primarily, the fact that you can't do it physically. You
> >> have two alternatives, neither of which is very appealing. (1) Your theory
> >> is gailiean invariant, (2) your theory is poincare invariant but isn't
> >> constructed from the connected component of the irreducible representation,
> >> since using your ``lorentz'' transforms, you can perform a lorentz
> >> transform and reverse the helicity of the photon.
> >
> >I am lost totally, because I do not understand what is the basis of your
> >conclusions. I always have a feeling that you are discussing some other
> >theory, not mine. In my theory, inertial observers are connected by
> >transformations forming the Poincare group. The commutators of the
> >Poincare algebra contain parameter c = 300000 km/s.
>
> So, how do you arrive at the conclusion that interactions can
> propagate infinitely fast?

First I take the Hamiltonian of QED (with infinite counterterms to
ensure correct S-matrix). Then I apply "dressing transformation" to this
Hamiltonian. This transformation preserves the accurate S-matrix and
makes the Hamiltonian finite. Then I analyze the interaction terms in
the "dressed" Hamiltonian. Let's first take the biggest term in
interaction between two electrons. I write it through creation and
annihilation operators of electrons omitting irrelevant factors

V = e^2 \int dk k^{-2} a^+(p_1-k) a^+(p_2+k) a(p_1) a(p_2)

This operator can be also written in terms of particle observables
(positions) of two electrons

V = 1/|r_1 - r_2|

which is instantaneous Coulomb interaction. Of course, there are
more terms of higher perturbation orders (power of e^2) and
relativistic corrections (proportional to powers of c^{-2}).
The important point is that all these terms can be written in the
general form

W = C \int dk f(p_1, p_2, k) a^+(p_1-k) a^+(p_2+k) a(p_1) a(p_2)

or through particle observables as

W = W (p_1, p_2, r_1, r_2)

So, all these terms correspond to instantaneous interactions between
two particles. There is no retardation present anywhere in this picture.

There also terms responsible for particle creation and annihilation,
like bremsstrahlung a^+a^+c^+aa. They are also relevant to the
discussion of interaction between electrons. Electron 1 may emit
a real photon due to bremsstrahlung term a^+a^+c^+aa, and this photon
may interact with electron 2 via Compton interaction term a^+c^+ac.
This correspond to the retarded part of interaction 1-2 transmitted
by real photons. Thus, total interaction between charged particles
has two contributions: one is instantaneous, another propagates with
the speed "c".

>
> >There is a
> >representation of the Poincare group (and algebra) in the Hilbert space
> >of each physical system. The photon is one such system. The operator
> >of velocity |V| = |P|c^2/H for the photon (as well as for any massless
> >particle) has just one eigenvalue c.
>
> I'm not buying that as an operator. The eigenvectors of 1/H are not
> the as the eigenvectors of H.

Eigenvectors of 1/H are the same as eigenvectors of H

> You also can't possibly expect me to
> accept a velocity defined with absolute value signs and finally,
> the absolute value signs mean your function isn't even analytic.

That's OK, I don't have a rule that observables should be
expressed by analytical functions. The free Hamiltonian is not an
analytical function of momentum

H = + \sqrt{M^2c^4 + P^2c^2}

>
> >This means that photons move with the speed of light.
>
> It doesn't mean anything of the sort. What it means is that you
> knew the answer you wanted to get, and then asserted that any
> arbitrary expectation value could be written in terms of its
> operators.
>
> >The (Coulomb and magnetic) interactions between
> >charged particles are described by potentials depending on instantaneous
> >positions and velocities of involved particles. These are
> >instantaneous potentials, and interactions propagate faster than light.
> >There is no direct connection between the speed of propagation of
> >light "c" and the speed of interaction.
>
> Well, you're helping make my point here, since there is apparently
> nothing in your theory in which `c' plays any role apart from your
> wish to say that photons propagate at that speed.

'c' plays a role in commutation relations of the Poincare algebra,
and 'c' is the speed of propagation of free photons.

>
> >> If the signals propagate instaneously, why do I need to bother
> >> using wires that are the same length? Since you've asserted that
> >> the interaction propagates instantaneously, the interaction at
> >> the detector ought to show up instantaneously at the other end
> >> of the wire and so we can skip waiting for the TEM wave of the
> >> coax to propagate its way down the wire. Then by definition,
> >> every signal that arrives at the same time happened at the same time.
> >
> >The speed of propagation of signal along the wire (=c) is related to
> >the speed of propagation of EM wave (=flow of photons) along the
> >wire.
>
> You assume a transmission line, in which case the speed is due to
> the dielectric, not the wire. (By the way, if the interaction propagates
> instantaneously, why do you suppose the charges in the dielectric only
> react as if the signal propagated at a finite speed?) But, forget the
> signal due to the transmission line. When I apply an electric field
> to a conductor, that electric field causes charges to move. If the
> charges move instantly, why don't the charges at the far end of the
> wire move instantly?

Now you asking me to describe how electromagnetic interactions and light
propagate in macroscopic bodies. I don't know yet the full answer from
the point of view of my theory. I am sure that this involves some
complicated interplay between instantaneous Coulomb and magnetic forces
and photons emitted, absorbed, and again reemitted by the atoms of the
matter. If you move electrons on one end of the metal wire, the
electrons (and nuclei) on the other end will not feel their changed
Coulomb
interaction, because the screening in metal is very strong. But, if you
move electrons on one side of the dielectric (e.g., shine some light on
it), it might well be that
the other side will respond instantaneously. Though this response will
be much smaller than that due to the propagation of light through the
dielectric. I cannot do much better than simply speculate at this point.

However, the traditional theory is not doing a good job in this area
either. You probably know about experiments by Nimtz, Chiao, and others
who demonstrated superluminal tunneling of light and microwaves in
various structures. No matter how you slice it, superluminal propagation
cannot be consistent with the traditional theory. I have some ideas
(they are not well developed yet) how my approach can explain this
behavior. That's a fascinating topic, but I am afraid it will lead us
too far from the basic ideas of the theory, which we are still
struggling to understand.

>
> >As I said many times, this speed is equal to the speed of light.
> >This speed has no direct relationship to the speed of propagation
> >of the Coulomb and magnetic interaction between charged particles,
> >which is infinite.
>
> So the force on a charge in a B-field should be what,
> F = q(E + (v/oo) x B))?

I can write you a (approximate) formula for the potential energy of
interaction between two charges

V (r_1, v_1, r_2, v_2) = q_1q_2/r
                         +q_1q_2/(c^2 r)[v_1.v2 + (r.v_1)(r.v_2)/r^2]

where r = r_1 - r_2
The first term is usual Coulomb potential. The second term is Darwin
potential. You can obtain the magnetic force by taking a gradient
of this potential.

>
> >> Uh, why any of this matter? If the electromagnetic interactions
> >> propagate instantaneously, then at every point in the universe,
> >> every observer agrees with the definition of t = t_0.
> >
> >Again, in my theory, the flow of photons (what is usually called
> >"transverse electromagnetic wave") has no direct relation to the
> >electric and magnetic interactions between charged particles.
> >Photons move with the speed of light c. Interactions propagate
> >instantaneously.
>
> I'm not talking about photons. I'm talking about the interactions,
> which according to you move a lot faster than photons, so I can
> skip all of the non-sense about matching cable length.

I told you, I don't have a description of currents and EM waves
propagating in matter in my theory yet. However, I know well that
signals propagate along wires with about the speed of light. This is
enough for me to propose the detector array for instantaneous
measurements.

[...]
>
> >In addition to direct Coulomb and magnetic
> >interactions between charged particles, there is also an interaction
> >mediated by real (not virtual!) photons and propagating with
> >the speed of light.
> >When charged particle A accelerates, it emits bremsstrahlung photons,
> >they propagate with the speed "c", reach charged particle B and cause
> >its acceleration by the "Compton effect".
>
> Why does it do that in your theory? I can't see a single reason that
> an accelerated charge should radiate. (I actually can't see how a
> charge can even move in your theory, but I digress.)

This is simple. If you find time to read my book, you'll find in
Table 12.1 several examples of interaction terms in the Hamiltonian.
One important contribution comes from the bremsstrahlung term, e.g.

a^+a^+c^+aa (1)

(I omit here all factors, momentum dependence, etc.)
This term describes interaction in which two electrons meet and
produce a photon. When two electrons meet they normally accelerate,
so acceleration leads to the emission of photons.

Just a note: I didn't pull all these operators from my imagination.
They follow in a rigorous way from the usual Hamiltonian of QED
by using the dressing procedure.

>
> >This kind of indirect interaction works in radio communications.
>
> Yes, it does and in the standard theory of E&M, it's easy to understand
> why. In your theory, I see no reason it should work at all except by
> fiat.

The bremstrahlung photon emitted in the above interaction (1) (or other
interactions leading to acceleration of electrons in the emitting
antenna) will propagate with the speed of light until it reaches the
receiving antenna where it will interact with the electrons there.
The strongest interaction is due to the Compton term

a^+c^+ac

which will induce the movement of electrons in the receiving antenna
registered as current. That's how radio works.

>
> >This retarded interaction prevails at large distances. The propagation of
> >the electrical signal along the wire is of this second kind (retarded).
> >It is not related to the direct Coulomb interaction between charged
> >particles.
>
> What do you suppose causes radiation? Magic?

Acceleration of charges causes emission of photons. In my theory, this
effect is described by the bremsstrahlung terms in the Hamiltonian.

>
> [...]
> >> Detectors cannot do anything without changing the energy of the
> >> particles.
> >
> >Why does it matter?
>
> The system is in a different state after the measurement.

So what? I performed my measurement, I got the information I need.
I don't care what is the state of the system after that. There is no
place in my book where I use repeated measurements (i.e., measurement
after measurement).

>
> >> your past light cone. But, I don't understand the point of this elaborate
> >> scheme. If the interactions propagate instantaneously, then all you
> >> have to do is move one charge and you'll receive a signal from every
> >> detector instantanerously, regardless of how long the wires are or where
> >> in the universe the charges are - just like galilean relativity.
> >
> >See above.
>
> Your theory is galilean (except for your inclusion of an anomaly called
> radiation that doesn't seem to have anything to do with the rest of your
> theory). You can't escape the fact that instantaneous propagation of
> an interaction means global simultaneity.
>
> Perhaps you should create two theories. A theory of photons and a theory
> of electromagnetism, since you don't think photons have any connection to
> charges, or if you do, I can't possibly figure out what you think it is.

You are absolutely right that I broke the identity of light and
electromagnetic wave introduced by Maxwell. In my approach, photons are
massless particles moving according to quantum laws. Electric and
  magnetic interactions between charges are transmitted by instantaneous
"potentials" whose values have nothing to do with wave functions of
photons. The photons can be emitted or absorbed by charges when two or
more charges are close together and accelerate. All this is governed by
a single Hamiltonian acting in the Fock space of the system. I wrote
a few simple terms of this Hamiltonian in my book. More work is
needed to calculate higher order terms.

Contrary to what you are saying, my theory is not Galilean, because
the above Hamiltonian is the time translation generator belonging to a
representation of the Poincare group in the Fock space. So, the theory
is perfectly Poincare invariant.

[...]
>
> [...]
> >> >This collision affects particle B due to the Coulomb (+magnetic)
> >> >interaction between A and B. Looking at the trajectory of particle B
> >> >one can notice a small cusp there which marks the time when B
> >> >"became aware" of the fact that trajectory of A has changed.
> >>
> >> According to you, thise happen at the same time, so neither
> >> can cause the other.
> >
> >No, I do not give time order to these events, they occur simultaneously.
>
> Then neither event can be the cause of the other. Period.
>
> >Though I can say that the impact of particle C is what caused
> >particle A to change its trajectory and particle B to change its
> >trajectory as well.
>
> No, you can't say that. They happened at the same time, therefore
> at most they are correlated, but correlation does not imply cause
> and effect.

These subtleties can be left to philosophers.

>
> [...]
>
> >> The ``standard theory'' gives that result for c -> \infty, except that
> >> it can be done without having to invoke some artificial assumptions
> >> about cause and effect.
> >
> >Let us agree to call "standard theory" the one with c=300000 km/s.
> >The limit c -> \infty belongs to approximate versions of the standard
> >theory.
>
> >In my (exact) approach, the speed of light is c=300000 km/s, and the
> >speed of propagation of the Coulomb and magnetic interactions is
> >\infty.
>
> Your approach isn't exact, or even correct. You have start with an
> interaction, j_u A^u, and you arbitrarily decide that the same interaction
> propagates at different speeds.

I started with interaction j_u A^u and I transformed it to the form
of superposition of "potentials"
(I write "potentials" in quotes, so you will be not tempted
to interpret them as scalar and vector potentials of Maxwell theory)
depending on instantaneous positions and velocities of particles
(see Coulomb + Darwin potential above). By doing that,
I haven't modified
any verified prediction of the theory (S-matrix). The procedure is
exact, though expressed in a perturbative setting.

>
> >> Then obviously your theory is not poincare invariant. As soon as
> >> I perform an experiment in which I don't measure any photons, the
> >> `c' in your theory serves no purpose and has no reason to be in any
> >> equation. You have `c' in your theory for one purpose - to have photons
> >> that propagate at `c', and in your theory, photons serve no puropse
> >> either. They have no theoretical reason to exist in your theory.
> >
> >I don't buy your arguments.
>
> It makes no difference if you buy them or not. Explain what `c' means
> in your theory, other than the speed of a photon.

Originally, 'c' was a parameter of the Poincare group (section 2.2).
Later, I find that
'c' is the speed of propagation of massless particles and the limiting
speed for massive particles (sections 7.1 and 7.4).
'c' has no relationship to the speed of
propagation of Coulomb and magnetic interactions (chapter 12).

>
> >The Poincare group properties and their dependence on "c" do not depend
> >on the existence or non-existence of photons. The Poincare group would
> >be valid even if photons had non-zero mass and moved slower than "c".
>
> That's true and that exactly my point. If you had photons that didn't
> propagate at `c', `c' would be meaningless in your theory, since it
> appears no other physics depends on it.

'c' will still be a parameter in the multiplication law of the Poincare
group.

>
> [...]
> >> >p'_1 = f_1 (p_1, r_1, p_2, r_2, \theta) (1)
> >> >r'_1 = g_1 (p_1, r_1, p_2, r_2, \theta) (2)
> >> >p'_2 = f_2 (p_1, r_1, p_2, r_2, \theta) (3)
> >> >r'_2 = g_2 (p_1, r_1, p_2, r_2, \theta) (4)
> >> >
> >> >where f_i and g_i are certain non-linear functions.
> >>
> >> So what? Changing what you call your variables doesn't cause any
> >> physics to change. It just changes the names of your variables.
> >
> >No, it does not. I hope you agree that changes of observables to
> >different frames of reference are real observable changes, not just
> >name changes.
>
> Changing vriables does nothing but change the interpretation of
> the same quantities. Rotating p such that p'_1 = p_1 cos(A) + p_2 sin(A),
> just changes what I call p_1 and p_2.

In writing eqs (1) - (4) I didn't mean naming conventions.
p_1, r_1, p_2, r_2 are particle observables really measured by one
observer. p'_1, r'_1, p'_2, r'_2 are observables of the same particles
measured by a different observers. These observables have well-defined
meanings. They are not subject to any ambiguity or conventions.

>
>
> >When you apply time translation to the frame of reference,
> >you obtain changes of observables due to time evolution. They are
> >certainly observable.
>
> Physical objects don't become different objects just because a
> different observer looks at them. All that changes is perspecteive.
> [...]

Agreed. But the change of perspective is real. From the perspective
of different observers, the same particle has different position,
velocity, etc. The purpose of relativistic transformations is to
connect perspectives of different observers.

>
> >transformations. I believe, you consider Lorentz transformation observable?
>
> The object is to paramaterize your observers such that the observers
> are _not_reachable_ by a lorentz transform.

Sorry, I missed your point here.

>
>
> >> I'm saying that it doesn't matter what you call it. Orthogonal rotations
> >> are orthogonal rotations and behave like orthogonal rotations. So, the
> >> only difference between the ``spacetime'' of relativity and ``space and
> >> time'' in your theory is that you insert `` and '' between the `e' and `t'
> >> in spacetime. The algebra is identical.
> >
> >Thats right, the algebra is identical.
>
> Then you can't get different results.

Yes I can. There are more things in physics besides algebra of
transformations. You cannot derive all physics from the algebra alone.

>
>
> >> Apparently not, since what you consider the high point of your theory
> >> is violating the requirements of the algebra.
> >
> >Prove it!
>
> Prove that a circle is 2\pi radians to someone who is determined to
> not accept it. Same thing exactly.

Not exactly. You claim that dynamical laws of my theory violate
the Poincare commutation relations. I say that you are wrong,
and that Poincare commutators are valid everywhere. I don't know how
to solve our dispute unless you show me where the error is.

[...]
>
> [...]
> >> I consider that the pinnacle of hypocrisy coming from someone who
> >> consider creation/anihilation operators in fock the primary quantities
> >> for a field theory in which the fields are denied any physical meaning
> >> leaving the operators nothing to act on.
> >
> >Creation/annihilation operators act on the state vector of the system.
>
> Creation/anihilation operators act on creation/anihilation space,
> i.e., a field. They are field operators.

I respectfully disagree. C-A operators (as all other operators in
quantum theory) act on state vectors (these are vectors in the
Fock space). Quantum fields are combinations (linear in the case of free
quantum fields) of C-A operators.

>
>
> >Fields are linear combinations of creation and annihilation operators.
> >Fields also act on the state vector. Both creation and annihilation
> >operators and fields have no physical meaning. They are just convenient
> > mathematical objects. Using these objects it is easy to write
> >down expressions for operators of observables (momentum, energy, etc)
> >which do have direct physical meaning.
>
> Oh, really? Have you ever seen an angular momentum? How about an
> energy. Pleas describe what an angukar momentum looks like.

The operator of total angular momentum J is generator of rotations.
So, it is a basic operator of my approach together with H, P, and K.
I prefer to express all other operators through H, P, J, and K.
Although, if you like, I can write

J = R x P + S

where R is Newton-Wigner position operator and S is spin operator.

>

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