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

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 12/22/04 

Date: Wed, 22 Dec 2004 09:29:46 -0800

Bilge wrote:
> Eugene Stefanovich:
> >Bilge wrote:
> >> Eugene Stefanovich:
> >
> >>
> >> [...]
> >> >
> >> >H = H_0 + V(r_1, p_1, r_2, p_2)
> >> >
> >> >This means that the interaction depends on instantaneous positions and
> >> >momenta of both particles. From this I conclude that interparticle
> >> >interactions are instantaneous.
> >>
> >> That is an erroneous conclusion. What you've written implies nothing
> >> about propagation times.
> >
> >This means that the force acting on particle 1 (derivative of V wrt r_1)
> >depends on the position and momentum of particle 2 at the same instant.
>
> In other words, you have a potential that can't commute with any
> operator that's a function of either a position or momentum operator,
> since it depends explicitly upon both the momentum and position.

The potential energy V generally does not commute with observables
of individual
particles r_1, p_1, r_2, and p_2, this is why the time evolution of
these observables is non-trivial, as it should be when interaction is
present. For example

dp_1(t)/dt = i [H, p_1(t)] /= 0

which means that there is a non-zero force acting on particle 1.

> The potential is non-local and can't conserve momentum

This potential commutes with the total momentum operator, so
momentum is conserved. For example, the Coulomb potential commutes with
P

[P, 1/|r_1 - r_2|] = 0

> and therefore
> your theory formally doesn't exist, since there are no eigenstates
> of your full hamiltonian, even in principle.

There are eigenstates. Again, in the simplest case of Coulomb potential
V, the eigenstates and eigenvectors are given by well-known solutions
for the hydrogen atom.

>
> >Isn't it the same as saying that interaction is instantaneous. There is
> >no delay.
>
>
> Yes, it says that your theory has an absolute time, isn't poincare
> invariant and if your potential wan't explicitly momentum dependent,
> you _could_ make the theory galilean invariant.
>
[...]

>
> >we can translate these measurements to
> >statements about the probability distribution for one electron
> >|\Psi(r,t)|^2.
>
> So, are you trying to tell me that the wavefunction depends upon
> how precise the measurement is or are you telling me that there's no
> such thing as a measurement that isn't an ensemble of measurements?
> Make a choice. Quantum mechanics postulates that measurements are
> hermitian operators. |\Psi|^2 is not an hermitian operator. It's a
> wavefunction upon which you operate with hermitian operators. How
> many such sub-measurements are needed before you call the total
> a measurement? I was under the impression that a measurement was
> a one shot affair in which I measured something.

This is not correct. Suppose that a particle is described by a wave
function delocalized in space \psi(r). Suppose you make your one-shot
measurement of position. You'll get some definite answer r_0.
Suppose you prepare the particle in the same state again, and measure
its position again. You'll get a definite but differerent result r_1.
If you continue this process ad infinum, you'll get different values of
position each time. However after many such one-shot measurements you'll
realize that the probability of finding position in the vicinity of
value r is given by real number |\psi(r)|^2. Full information obout the
state of quantum system can be obtained only by repeated measurements in
ensembles.

[...]
>
> [..]
> >> What do you think can't be calculated (and don't just say ``evolution).
> >> Give a specific example of something you've calculated that agrees with
> >> experimental data and why you think qed can't obtain the result. So
> >> far, I haven't been able to decide whether your idea of evolution is
> >> completely unphysical or just wrong, so give an example as I've described.
> >
> >Recently, there were many experiments performed for measuring time
> >evolution of electronic wave packets formed from excited atomic levels.
> >One recent example is quant-ph/0407105. These experiments probe
> >properties beyond those contained in the S-matrix (scattering
> >cross-sections and energies of bound states).
>
> You have to be kidding.
>
> (1) That isn't even a relativistic situation. Calculate d<x>/dt.
>
> (2) The only place relativistic effects would be included is
> possibly in calculating the the excitation energies of
> the atomic level structure,

Radiative corrections to the energies of atomic states can be
calculated by QED. However, radiative corrections to atomic
wavefunctions are beyond reach of QED.

>
> (3) To the extent that any theory could be used to calculate the level
> structure precisely from first principle, I can't possibly see how
> your theory would simpler, or for that matter be much different,
> except that you handicap yourself with illogical philosophical
> constraints. You've chosen an example for which nothing precludes
> using qed, in principle.

QED does not have a well-defined finite Hamiltonian, so it cannot
deal with any situation where time evolution is involved.

> The issue is a practical one of
> tractability. Your theory offers no advantage over qed on a
> many-body problem. If anything, I'd say less, since at least qed has
> no philosophical restrictions against writing down a
> phenomenological potential in which the word ``field'' appears.
>
> (4) I have no idea how _you_ would model a laser pulse, since the photons
> in such a pulse have no well-defined energy (I believe they exploit
> that fact to obtain the coherence that gives the beats). I believe
> you would need write that as a continuous spectrum of operators,
> N_k, with a bandwidth fixed by the width of the pulse envelope.
>
>
> >In order to describe these properties quantum-mechanically one needs
> >a Hamiltonian.
>
> They use the schroedinger equation. Have you solved this example?
> If not, then you evaded the question I asked.

Their experiment can be described by using non-relativistic Schroedinger
equation. However, QED
cannot describe it, because there is no Hamiltonian in QED.
I agree that at the current level of such experiments, radiative
corrections to time evolution are redundant, they are way too small
to be observed. So, non-relativistic theory works fine. However,
with improvements of experimental precision,
we'll reach a situation when
radiative corrections to the time evolution and wavefunctions will
become important, then we'll need a theory which
1) allows for creation and annihilation of particles and
2) has a well-defined time evolution operator.
  QED is not such a theory, because it does not satisfy condition 2).
QED is OK for the S-matrix, scattering cross-sections, and energies of
bound states.

>
> >QED does not have a well-defined finite Hamiltonian, so QED in its
> >rigorous form is incapable of describing these effects.
>
> http://scholar.google.com,
>
> qed evolution operator atomic systems: 1100 references
>
> qed evolution operator plasma: 447 references
>
> I suppose your article is referenced a few times in the count, so
> round down by 10 or so articles.

I can probably find this combination of keywords in a phone book.
This does not prove anything.

>
> >Of course, you can use ordinary quantum mechanics with Coulomb
> >interaction between charges, and claim that this is a low-energy
> >limit of QED. But this would be a cheating, because nobody have shown
> >how finite Hamiltonian of quantum mechanics can be obtained from
> >infinite Hamiltonian of QED.
>
> (1) You haven't shown your perturbation series is finite, so using
> your logic, you can't either.

I have proven that
each term in the perturbation series for the dressed particle
Hamiltonian is finite. The entire series may diverge, I agree.
QED also does not have a proof that the S-matrix series converges.
My approach is not worse than QED in this sense.

>
> (2) QED (and electromagnetism) itself might really not be finite
> without including the weak and strong interactions. That doesn't
> preclude me from using the the theory at energies for which the
> contributions from the weak and strong interaction are irrelevant.

The important point of my approach is that electromagnetism is finite
for all energy scales without the need to invoke any outside
considerations, like weak interactions, strings, etc.

[...]
>
> [...]
> >> It's rather remarkable that such a simple requirement ends up
> >> requiring the existence of electric charge and the electromagnetic
> >> field. Are you suggesting phases are observables?
> >>
> >
> >You changed subject. We were talking about physical consequences of
> >the point form of dynamics. I told you that in the point form, an
> >interacting system will look differently for two observers shifted in
> >space wrt to each other. I told you that this effect has never been
> >observed before, which proves that interactions do not have point form.
>
> You failed to tell me what this alleged effect happens to be,
> so I wrote down the modified translation operators. What's your
> problem?
>

If a generator contains interaction, then the corresponding
transformation may have effect similar to that of time translation.
For example, we know that time translation of an unstable particle leads
to its decay. Then, if space translation generator contains interaction,
one should be able to observe decay "induced by translations". I.e.,
if observer close to the particle sees that the particle is not decayed,
the observer far away may conclude that the particle has decayed.
This effect has never been observed.

Eugene.

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