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

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 11/22/04 

Date: Sun, 21 Nov 2004 21:23:53 -0800

Bilge wrote:
> Eugene Stefanovich:
> >Bilge wrote:
>
[...]
>
> >No fields needed. I can do it even better than QED, because I can
> >calculate time evolution, but QED can't.
>
> You also believe force and velocity are hermitian operators.
>

That's right.

> [...]
>
> >If you like, you can say that the Coulomb interaction between two
> >particles e^2/|r_1 - r_2| happens in this way: particle 1 being at point
> >r_1 creates field \phi(r) = e/|r_1 - r| in all space.
> >Particle 2 being at point r_2 feels this field, so that the energy of
> >the system is changed by the amount e \phi(r_2).
>
> No, you can't - at least not if you want a relativistic description.
> You cannot separate \phi from A covariantly.

I used the Coulomb potential just as a simple example. In order to have
a relativistic description, one needs to include all other potentials.
In particular, the Darwin's potential

V(r,v_1,v_2) = e^2/r[(v_1.v_2) + (r.v_1)(r.v_2)/r^2]

where r= |r_1 - r_2|, as well as spin-orbit, bremsstrahlung, and
infinite number of other terms.

>
> >I do not like this description, because the same situation may be
> >described without invoking fields at all: I can just say that the
> >potential energy of the two particle system depends on the interparticle
> >distance as e^2/|r_1 - r_2| + magnetic terms + spin-orbit + ...
>
> You can't obtain the spin from that term.

Particle spin is obtained naturally in Wigner's theory of irreducible
unitary representations of the Poincare group (1939). A good reference
is section 2.5 in Weinberg's book vol. 1.

>
> [...]
> >> Physics doesn't depend upon choosing a particular representation for
> >> a group. What does depend upon the choice of representation is what
> >> you can address easily. You are attempting to address physics that is
> >> not naturally contained in your choice of representation.
> >
> >We have discussed this point already. Physics does depends on the choice
> >of representation of the Poincare group. Take 2 particles which are
>
> No, it doesn't.
>
> [...]

I love how you dismiss my explanation without even thinking about it.

> >>
> >> > The assumption of the instant form of dynamics (interactions present
> >> > in the Hamiltonian and boost operator only) is a separate postulate
> >> > of my theory. This variant agrees with experiment better than other
> >> > possibilities (point form, front form, or the general form with
> >> > interaction operators in all generators)
> >>
> >> So, your interactions affect the energy but not the momentum?
> >> Doesn't that tell you something?
> >
> >I don't see anything wrong with it.
>
> Not, if you want to use virtual particles. Otherwise it violates
> E^2 = p^2 + m^2, since E is no longer related to p such that m is
> the mass of your particle.

Interaction terms are added to the TOTAL observables E and M
of physical
system. In the instant form, both E and M are modified, so that
E^2 = P^2 + M^2. The operators e_i and p_i of individual particles
are not modified by interaction. They satisfy
e_i^2 = p_i^2 + m_i^2

[...]
>
> [...]
> >> Then you'll have to abandon your theory entirely. A gauge transformation
> >> is phase transformation. Any U(1) transformation is a phase transformation.
> >> U(1) invariance is gauge invariance. You cannot have electrons (or any
> >> other particles) unless your transformation leaves the phase unobservable.
> >
> >I don't quite understand your logic. My theory does not have the
> >property (phase) you yourself call unobservable. I think that's a
> >plus for a physical theory rather than minus.
>
> Then your theory cannot explain interference.

We are talking about different phases here. In gauge theory, phase
transformations are applied to quantum fields. There are no quantum
fields in my approach (except as convenient mathematical objects),
their phase has no physical meaning, and is not observable.

The interference is related to the phase differences of wave functions.
Phases of wave functions of particles are present in my approach.
They are observable in some circumstances (e.g., interference).
I think you understand the difference between quantum field and wave
function?

>
> [...]
> >> Do you use any
> >> concepts like position, energy, frequency, momentum, time, wavelength,
> >> wave vector, electron, proton, particle, photon, etc?
> >
> >Yes, all these are quite respectable properties.
> >
> >> Obviously, a
> >> particle for which the phase appears explicitly in <\Psi|\Psi> will
> >> not retain its identity.
> >
> >We just agreed that <\Psi|\Psi> =1 for all observers.
>
> Yes, but I feel obligated to make sure it's clear, since so far,
> what I thought was rather clear, apparently left out too much.
>
>
> >
> >> Since an electron in one frame is still an
> >> electron in another frame, an observer who chooses coordinates x',t'
> >> must also measure E and t such that the electron is still an electron.
> >
> >For different observer, the wave-function will be different, but the
> >electron will remain electron no matter what. That's because electron is
> >DEFINED as a system described by a certain irreducible unitary
>
> Good enough... Keep that in mind.
>
>
> >> A transformation that preserves the phase (like a U(1) transform) is
> >> a gauge transformation.
> >
> >Stop right here! I don't know what is U(1) transform.
>
> I've only said what it is a few dozen times. It's a unitary transform.
> U as in unitary. 1 as in one parameter (as opposed to SU(2) which is
> a matrix of the form a.\sigma).

I know the mathematical idea of the U(1) transform pretty well.
I do not know how
it is related to physics. That's something which has never been
explained neither by you nor by dozens of gauge theory books I read.
 From this I conclude that all gauge invariance stuff is just a clever
  mathematical trick, which leads to correct expressions for the
Hamiltonian and boost operators, but has no real basis in physics.

>
> >I know space translations, rotations, boosts, and time translations.
> >What's the physical meaning of the U(1) transform? Are you just changing
> >the phase of the electron's wave function by hand?
>
> No, I'm requiring that regardless of which observer measures the
> electron, that all observers agree that it's an electron. That requirement
> comes with a price: the existence of electric charge.
>
> >If you just multiply it by a unimodular factor, that's OK, because
> >it doesn't change the quantum state. If you multiply the wave function
> >by a position-dependent unimodular factor, then you change the quantum
> >state, and I would not allow you to do that.
>
> No, I don't change the state. In fact, the reason for a position dependent
> phase factor is that the phase factor has to vanish in a measurement,
> regardless of whether or not different observers agree on a phase
> convention. The entire point is to insure that _no_ observer sees anything
> but an electron and observers can be located at different points.
>
> >
> >> The quantity that reconciles the phases in the
> >> sane way that distance reconciles choosing coordinates, is the photon,
> >> (for a U(1) transform, it's the W+/- and Z for an SU(2) transform
> >> and gluons for an SU(3) transform).
> >
> >Bla-bla-bla. I read this stuff 100 times in all kinds of books.
>
> Then why are you having so much difficulty (by your own admission)
> understanding it? Don't tell me something is handwaving if you
> don't understand it.
>

I have a good suggestion. I suggest not to discuss gauge field
derivation of the Hamiltonian and boost operator. We may have different
views on this subject, but I didn't touch this subject in my book,
and I do not want to discuss it because it leads away from the main
ideas of my approach. Let us start from the point on which (I want
to believe) we both agree: from the expressions of the Hamiltonian and
9 other generators of the Poincare group in QED:

H = H_0 + \int dx j(x).a(x) + \int dx dy j_0(x)j_0(y)/(8 \pi |x-y|)
P = P_0
J = J_0
K = K_0 - \int dx x j(x).a(x) - \int dx dy x j_0(x)j_0(y)/(8 \pi |x-y|)
         - \int dx j_0(x) C(x)

where C(x) is a combination of creation and annihilation operators of
photons which can be found in Weinberg's lecture
"The quantum theory of massless particles", in Lectures on Particles
and Field Theory, vol. 2 (Prentice-Hall, Englewood Cliffs, 1964).

These expressions were derived by using gauge theory approach, but
I don't care about that too much. To me it is important that these
interacting generators satisfy the Poincare commutation relations
(relativistic invariance) and (after renormalization) produce
very accurate S-matrix. Let us take these expressions as a starting
point and work out their physical consequences. My claim is that
instantaneous propagation of electro-magnetic interactions follows
from these expressions, and I am going to defend this claim.

Eugene.

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