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Download a new book on quantum mechanics and relativity.

Download a new book on quantum mechanics and relativity. - Physics Forum

Download a new book on quantum mechanics and relativity. - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.

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Old 11-08-2004, 06:09 PM
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Default Download a new book on quantum mechanics and relativity.

"Eugene Stefanovich" <[Only registered users see links. ]> a écrit dans le message de
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When performing this measurement, you don't change
the state of your wallet. Your example implicitly interpret
quantum collapse to be a simple change in the state of
knowledge of the observer, not a change of the quantum
state of the observed system.

This interpretation would comply with a classical measurement.
It cannot comply with a quantum measurement because such a
measurement changes the quantum state of the observed system.

Indeed, when you perform the z-spin measurement of a spin 1/2
particule which is in an +1/2 y-spin pure state for instance, you don't
change only your state of knowledge. You also drastically change
the spin state of the observed particle which suddenly swaps
from a +1/2 y-spin pure state to a +1/2 or -1/2 z-spin state.


Yes there is. I could provide more details preferably about the
Greenberg Horn and Zeilinger thought experiment which is
even more puzzling (and less difficult to grasp) than the
violation of Bells inequalities.

Now you have noticed that I don't assume Poincaré group covariance
to be satisfied by any phenomena without any exception (because
I rather believe quantum collapse to be an objective phenomena).

Poincaré group covariance would amount to assume the principle
of relativity provided that two events that are really coincident are
really coincident whatever the observation artifact that can plague
the conclusion of some observer (be it inertial or not).

Bernard Chaverondier

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Old 11-08-2004, 06:46 PM
Eugene Stefanovich
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

Ken S. Tucker wrote:

I don't understand what you wrote here. Integral of zero?
In my approach points and events are associated with real physical
systems, i.e., particles. For example, I say that particle is prepared
at a point in space if its state is an eigenstate of the position

The exact definition of the mass operator is

M = + \frac{1}{c^2} \sqrt {H^2 - P^2 c^2}

where H is energy, P is momentum. This operator commutes with
all generators of the Poincare group. Therefore, the measured values
of mass (you can call it rest mass) are the same for all observers.

[Only registered users see links. ]

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Old 11-08-2004, 08:04 PM
Eugene Stefanovich
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

bernard.chaverondier wrote:

You are correct if you assume the Copenhagen interpretation of quantum
mechanics in which wavefunction describes the STATE of the physical
system and measurement results in the change of state. In this
interpretation, the particle's state before the measurement was
"really" the mixture of two possibilities (z-spin +1/2 and -1/2).
The measurement forced a unique result, i.e., changed the state.

My interpretation
of quantum mechanics is different (it is detailed in subsection
3.3.3). I think that wavefunction describes our KNOWLEDGE about
the state of the system. Then measurement does not change the state,
it simply changes the information we have about the system.
Then in your example, I can assume that before the measurement of
the z-projection of spin, the particle already had a definite value
of this projection (+1/2 or -1/2). The measurement just revealed this
preexisting value without changing the state of the particle.

Experimentally, these two interpretations are indistinguishable,
because from the experimental point of view, it is pointless to
talk about particle properties before measurement.


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Old 11-08-2004, 10:52 PM
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

"Eugene Stefanovich" <[Only registered users see links. ]> a écrit dans le message de news:
[Only registered users see links. ]...



No. The Copenhagen interpretation assumes collapse to be
a change in the state of knowledge of the observer as well.


No. Before the measurement takes place it's not a mixture.
On the contrary, as far as no measurement has not been
performed it's a coherent superposition ie a pure quantum state.
The reduced density operator of the spin state of the particle
is that of a pure spin state

Rhô = |+,y><+,y|

When the entanglement of the particle with the polarizer takes place,
then the spin state of the particle changes. The off-diagonal terms
of the reduced density matrix of the particle vanish (that's the
deterministic decoherence process that is now well known and has
been checked experimentally for more than 10 years) and instead
of being in a pure state (a rank 1 projector) the reduced density
operator of the spin state of the particle becomes a mixed state
(a weighted sum of rank 1 projectors in the preferred
Hilbert basis of the z-spin observable)

Rhô' = (|+,z><+,z|+|-,z><-,z|)/2

This is because the measurement process begins by a deterministic
propagation of EPR intrincation between the observed particle
and the polarizer and then an intrication takes place between the
particle, the polarizer and their environment. After this entanglement
has taken place, the measurement process is not finished because
actually, the density operator of the whole system comprising
the particle, the polarizer and the environement is still in a pure state.

At the end of the measurement process, the EPR correlation of
the particle with the polarizer (as a Stern and Gerlach for instance)
and with the environment is broken and the particle goes back to
a pure state, but it's not any more a pure y-spin sate. It's a pure
z-spin state.

Quantum measurement has nothing to do with a classical measurement
where only the state of knowledge of the observer changes.


They are and have been distinguished thanks to the _experimental_
study of decoherence. The quantum measurement modifies the state
of the observed sytem (when it is not in an eigenstate of the observable).
This point is now well known and has been proven experimentally.


Yes there is. I could provide more details preferably about the
Greenberg Horn and Zeilinger thought experiment which is
even more puzzling (and less difficult to grasp) than the
violation of Bells inequalities.

Now you have noticed that I don't assume Poincaré group covariance
to be satisfied by any phenomena without any exception (because
I rather believe quantum collapse to be an objective phenomena).

Poincaré group covariance would amount to assume the principle
of relativity provided that two events that are really coincident are
really coincident whatever the observation artifact that can plague
the conclusion of some observer (be it inertial or not).

Bernard Chaverondier
[Only registered users see links. ]
Compatibility of Alain Aspect experiment interpretation
as an action at a distance with a formulation of relativist
invariance of phenomena that satisfy this invariance in
the framework of Aristotle space-time SE(1)xSE(3)/SO(3)
and the compatibility of possible instantaneous transfer
of information thanks to EPR effect with an explicitly non
local and deterministic interpretation of quantum measurement.

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Old 11-08-2004, 11:10 PM
Eugene Stefanovich
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

bernard.chaverondier wrote:

Thank you for correcting my language. When I wrote "mixture" I
actually meant "coherent superposition", i.e., pure quantum state,
not mixed state.

So, you admit that the physical state of the particle changes when
measurement takes place. This contradicts your earlier statement
that only information has changed.

I am not interested what is the state of the system after measurement
has been performed. This would be needed if we want to predict
results of repetitive measurements on the SAME system. In my book,
I try to avoid this question. I assume that after measurement has been
performed, this copy of the system is discarded.

Could you give me a reference to these experiments?



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Old 11-09-2004, 01:40 AM
Ken S. Tucker
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

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


I'm trying to help you define the nature of
your manifold, with a very simple question.

"0" is a point, "x" can be anything you want.

Every variable you introduce needs to be
inter-defined by some standard. I'm ok with

$ 0 dx = a unit length which is relative,

but I suspect you're exploring

$ 0 dx = invariant constant

which is fine, so pick one.

How is that measured (not defined),
(third time I've asked)

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Old 11-09-2004, 02:39 AM
Eugene Stefanovich
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

Ken S. Tucker wrote:

The integral of zero is zero no matter how you slice it.
You have some explaining to do when you wrote ""0" is a point"
0 is a number, point is a location in space. They cannot be equal.
In summary, all this looks like complete nonsense to me.

I already wrote you about one type of mass measurement
(with the spring, you remember?). Another type of measurement
would be an elastic collision with an object having a known mass
and velocity (momentum). By measuring velocities of two objects
before ad after the collision we can figure out the unknown mass
by applying the energy and momentum conservation laws.

Any other experimental procedure for determining the mass, known from
school physics will also do. I don't think your purpose was
to examine my knowledge of mechanics. Then what was the point of your

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Old 11-09-2004, 10:24 AM
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

Eugene Stefanovich:

You are still missing the point here. Inertial frames, straight lines,
force, etc., are ill defined concepts in and of themselves. They are
_defined_ by geometry (obviously a straight line has to be a geometric
concept). When you say that the lorentz transforms are ``modified'',
in effect, you are redefining a a non-inertial trajectory to be
an inertial one. That is why your explanation sounds like you're
trying to do general relativity, except in this case, you can't
transform the force away.

``Appearance'' has no relevance. If I rotate a stick about the y
direction, its projection on the x-axis changes. Rather than pretend the
z-direction doesn't exist and conclude the stick gets shorter, I use the
postulate of relativity, to conclude that the length of a stick does't
depend upon the direction it's oriented, postulate that my geometry
is three dimensional, not two dimensional and construct the coordinate
transforms, such that the length of the stick is invariant, l^2 = x_i x_i,
x -> x_i' = (I_i'i + a_i'i)x_i so that l^2 = x_i' x_i' = x_i x_i.
Relativity is no different apart from the fact that i runs from 0-3
instead of 1-3 so that a_ij has 6 components instead of 3. A rotation
in x-t changes the angle the stick makes on the x and t axes. The slope
\beta = tanh(A) is called the velocity.

Non-relativistic quantum mechanics is poincare invariant, so whatever
it is you think follows from poincare invariance, a finite value for
the constant `c' is not one of those things.

No, it isn't. For example, the ten generators,

d/dt, \grad, (x_i d/dx_j - x_j d/dx_i) and (t d/dx_i + m x_i)

represent time and space translations, rotations and galilean boosts.
(incorporate the i\hbar's if you want quantum versions). So, poincare
invariance holds, with c->\infinity. If you don't believe the last term
is a boost:

(-i\hbar t d/dx_i + m x_i)\phi = (tp - mx)\psi = m(t (p/m) - x)\psi

The quantity, p/m is the classical velocity, so I get m(vt - x).

It might be non-trivial, but you're going beyond non-trivial to physically
incorrect. If you define a particular representation, you have to stay
within the context of what you've defined. For example, given the generators
for boosts and translations, [L^ab, p^c], if the commutator is unchanged by
a unitary transform, it's obvious that either or both generators can
be transformed. That doesn't mean any physical results differ. It means
that you have to know what you are measuring.


I referenced bjorken and drell for you. They don't reference light
pulses. Light is an electromagnetic effect. It's speed of propagation
in a theory, depends upon the theory. If you don't assume lorentz
invariance, what is your theoretical rational for assuming the number
`c' is finite? In special relativity, time and space are treated on
equal footing and since `t' is just a coordinate, it's natural to
set c = 1 for the same reason no one defines a 45 degree angle in
the x-y plane as dy = b tan(\theta) dx. Whether or not light propagates
along the 45 degree line in the x-t plane is question that should
be addressed theoretically in a theory of E&M. It has nothing to do
with poincare invariance or lorentz invariance. I can create a poincare
invariant theory in which light does not propagate at `c'.


Don't be an imbecile. I demonstrated above that poincare invariance
isn't sufficient to even define your energy-momentum relation, so
if it didn't come from invariance under a spacetime translation,
where _did_ it come from? The energy and momentum form a locally
conserved current if the lagrangian is invariant under spacetime
displacements. I realize you have a personal issue with lagrangians,
but that doesn't alleviate you from having to prove the relationships
you lift from derivations using lagrangians if you are going to
declare the lagrangian formalism invalid.


I just told you below.

Don't be dense. You seem to have a mental block against the rather
general and widely held concept that an approximation to an exact solution
is still an approximate solution and seem to think that the only
approximations allowed are the ones that are the least physically
appropriate to the solution being sought. I can obtain classical E&M as
the lowest order approximation to qed. Therefore, it's a valid
approximation, even if I don't muster all of the machinery of feynman
diagrams to do the calculation. Since a particle trajectory is a classical
concept, the appropriate way to calculate one is via the classical
reduction of qed to classical E&M. Throwing math at a problem without
first figuring out what physics is involved is silly. Do you believe it's
necessary to solve a problem with spherical boundary conditions using
cylindrical coordinates and cartesian coordinates to prove you should
calculate the same numbers for a given coordinate?


That's somewhat of a dilemna, eugene. First of all, you have stated on
numerous occasions that your theory does not contain the quantity \rho. I
don't know what your view of the (three-) current density J is, but that
expression definitely contians \rho:

__ __
\rho = \psi\gamma^0\psi, \psi = \gamma^0\psi^{\dagger}

See, the basic problem I have, is that you deny these quantities exist
in your theory, and then you not only say the quantities exist in your
theory, but the expressions you claim represent those quantities turn out
to be the same ones used in theories to which you object, based upon the
kinds of quantities you say don't have any physical meaning. If you wonder
why I am rude, this sort of sophistry is precisely why. Every time I say
something about a charge density, your theory doesn't have one. When I say
that's problem, then all of a sudden, your theory has one, and it just
happens to be the same one found in every textbook. When I the one of two
or more mutually contradictory answers I'm given depends upon the question
I ask, I tend to get the impression that the person is trying to bullshit
me, hoping I won't notice. I consider that rude - even more so than
outright abuse, since at least outright abuse isn't an attempt at

In order to prove charge conservation, you need to show explicitly
that (1) the current is conserved, and the integral over all space
is a constant so that the time derivative of the integral of the
charge density is zero. That's usually easy to do, except that you
obviously cannot assume that j^u is invariant under a lorentz transform,
because you have additional terms in your lorentz transforms that
correspond to the interactions.


That obviously requires you to find and interpret D(...) to not
just satisfy momentum conservation, but to satisfy the expression you
want to use in relating p and E. Also, you still keep forgetting that
those theorems you quote aren't likely to be valid if you insist that
real physics is representation dependent.


Then why did you refer to what I wrote a while back in terms of the
unitary transformtion of p_u as if it were obvious and inconsequential
and that I didn't know what physics it contained?

Well, that's unfortunate, because in my opinion, the problem with your
theory is precisely that your misinterpretation of the ``mathematical
trick'' of dressing your charges. Personally, I think your motivation
has little to do with E&M and everything to do with having an excuse
to say special relativity is wrong. Otherwise, you wouldn't be so adamantly
opposed to suggestions that your idea about boosts is simply a misinter-
pretation of your theory. You seen to prefer having the entire theory
tied to your interpretation.

Non-observation of something is not a legitimate reason to discard
terms that the theory says exist.

But you don't prove it. In fact, you don't even give a reason for
charge to exist. If I called your charge lepton and baryon number,
it would mean the same thing.

But theories are supposed to explain experimental facts and if you
say something is the charge, you are supposed to explain why it's the
charge you claim. To put it another way, angular momentum, momentum,
energy, baryon number, lepton number, etc. are all conserved charges.
Each is associated with a symmetry. There is a conserved charge associated
with invariance under a rotation. That charge is called angular momentum.
In a gauge theory, the electrical charge is the charge associated with
invariance under a U(1) phase transformation (or actually, it's the
integral of a conserved current.) The four-momentum is the conserved
current associated with invariance under spacetime displacements.
The conserved charge is the integral of that current, and is called
the mass.

It really doesn't matter if it makes sense to you. It makes sense to
enough people that the standard model was based upon gauge invariance.
It's really quite straight forward, too. Invariance under U(1), SU(2)
and SU(3) phase transforms gives the electromagnetic, weak and strong

What you call observables, so far, have been things which quantum
mechanics does not call observables, so I don't see that as being
a problem.

So what allows you to discard those terms on theoretical grounds?
Simply discarding terms you don't like is ok for phenomenological
models which are fit from data (optical models, for example), but
not for a fundanental theory.

I have no difficulty whatsoever deriving the spin interaction for
myself. I don't need to look it up. That isn't the issue. Even the
dressing transformation in your theory is not the issue as far as it
goes. Your interpretation of that is the issue. First of all, if you
don't accept the concept of a B field, then it's rather hard to explain
a term like S.B, which is, after all, the term relevant to a stern-gerlach
experiment. Of course, you don't _have_ to use B, since a B-field is
produced by charged particle dynamics, but the fact that you believe
your theory refutes lorentz invariance, doesn't allow you to simply
use results derived from assumptions that include lorentz invariance.
Reading articles in which the assumptions include an assumption you
explicitly reject, doesn't address the question here.

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Old 11-09-2004, 11:52 PM
Ken S. Tucker
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

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

What's your constant of integration?

0 = point
x = length
x^2 = area
x^3 =volume

What's you geometry?

So does everything else, that's why we help.

Ok your equation above is LORENTZ INVARIANT,
but that does not apply in your thoery.

That is all conservation of energy in spacetime,
based directly on SR.

If you intend to use {H^2 - P^2 c^2}
then you are using SR, and therefore
accepting it as is, otherwise you must
find a different means to derive it.

Too prove yet another inconsistency. So far
you've used more Duct Tape than Red Green has
to jury rig a foundation that couldn't support
a fly.

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Old 11-10-2004, 12:02 AM
Eugene Stefanovich
Posts: n/a
Default Download a new book on quantum mechanics and relativity.

Bilge wrote:

In general relativity the curved spacetime affects all objects in
the same way (equivalence principle), so one can talk about
geometry. In my approach, the boost transformations are different for
different systems: for a non-interacting group of particles they are
familiar Lorentz transformations; for interacting particles, the
transformations are nonlinear and interaction-dependent.
So, I refuse to talk about 4D geometry in my approach. Usual 3D geometry
is fine, though, because space translations and rotations are

My point is that boost transformations are dynamical
(interaction-dependent). When interactions in the system are weak
(in most cases they are; the strength of interaction can be roughly
determined by the ratio of the mass defect to the total mass of the
system) then boost transformations reduce to linear Lorentz
transforms which act AS IF they were "rotations" in the 4D spacetime.
So, simple geometrical picture is OK for weak interactions,
but it is just an approximation. That's what confused people for last
99 years. That's why modern theoretical physics is stuck in the
dead end of "geometrical" concepts like strings, loop quantum
gravity, etc. The idea (formulated by Minkowski) that physics is
(in large part) a geometry, was so beautiful that it shaped most of
theoretical thinking for the last 99 years. GR, gauge invariance,
strings, etc. are all derivatives of this idea.

I disagree. In non-relativistic quantum mechanics, the interaction is
usually introduced in the Hamiltonian only, while generators of
space translations, rotations, and boosts remain kinematical.
This is OK if generators satisfy the Lie algebra of the Galilei group.
This is not OK if the generators satisfy the Lie algebra of the Poincare
group. Non-relativistic quantum mechanics is only Galilei-invariant.

Let us check the Poincare properties of your generators. One of the
commutators should be [K_x,P_x] = -H. It is easy to check that this
commutator does not hold

P_xK_x\psi=d/dx(t d/dx + mx)\psi = t d^2/dx^2 \psi + m\psi + mxd/dx \psi
K_xP_x \psi = (t d/dx + mx)d/dx\psi = t d^2/dx^2 \psi + mx d/dx \psi

(K_xP_x - P_xK_x) \psi = -m \psi

I don't quite understand your point here. I hope you are not trying to
say that all representations of the Poincare group are unitarily

But they do not derive Lorentz transformations either. They just take
them as given.

Let me make it clear. I do assume Poincare group properties of
transformations between different inertial frames of reference.
The number 'c' is a parameter participating in the Poincare group
multiplication law and in commutation relations of the Poincare algebra.
If this parameter is set to infinity, the Poincare group degenerates
to the Galilei group. This is the non-relativistic limit.

However, the properties of inertial frames (the Poincare group)
are not sufficient to predict
how PHYSICAL SYSTEM will be seen from different reference frames
(this includes boost transformations of the properties of the system).
In order to do that, you need to have some idea about the
physical system. The 2nd postulate
(invariance of the speed of light) gave Einstein this additional piece
of information and allowed him to derive Lorentz transformations for
events associated with light pulses. However, there is no similar
piece of information related to other physical systems (e.g., massive
particles with interactions). Einstein's solution was to extend Lorentz
transformations to all physical system and to declare boosts
kinematical, and eventually to introduce the Minkowski spacetime as a
universal background.

My approach is different. I do not make an assumption that Lorentz
transformations (which worked fine for light pulses and non-interacting
systems) will also work for interacting systems. I DERIVE boost
transformations for interacting systems and show that they are different
from linear Lorentz transformations.

I agree with your general point that the principle of relativity and
Poincare group properties are not directly
related to the properties of light.
The fact that c is the speed of propagation of light and, at the same
time, a parameter
in the Poincare algebra commutators, can be probably considered as a

I don't understand your point here. One thing I can say, though, is that
operators of TOTAL energy and momentum of any isolated system are
defined in my approach as equal (up to a factor) to the generators
of (representatives of) the time and space translations in the Hilbert
space of the system. Therefore, the conservation of energy and momentum
are direct consequences of translational invariance. The definition of
the energy-momentum relation

H^2 = M^2 c^4 + P^2 c^2

(it requires also the definition of the (Casimir) operator of mass M.)
is also a direct consequence of the Poincare group properties.
The Poincare group properties are sufficient to derive all kinds of
relations between total observables of isolated system (including
linear Lorentz transformations for momentum-energy or position-time).

The Poincare group properties are not sufficient to derive
time translation and boost transformations for separate PARTS of
an interacting system. This requires explicit construction of the
representation of the Poincare group in the Hilbert space of the
system. Different representations define different interactions in the
system, and therefore, different time translation and boost

I bet you cannot do that. QED Hamiltonian is infinite (it must contain
infinite counterterms if you want to obtain finite and accurate
S-matrix). Even if you disregard these infinities, the QED Hamiltonian
contains trilinear terms, like

e -> e + \gamma
0 -> e + p + \gamma

These terms will remain even in the classical (\hbar-> 0) limit.
These processes are completely unphysical. The Hamiltonian of QED is
sick, and taking its classical limit will not heal it.

You can probably write something formally resembling Maxwell's
equations as a classical limit of QED, but I don't believe you
can get a viable dynamical theory this way without handwaving.

I am sorry that you got an impression that I am trying to deceive you.
I am not. This is true that my approach both has and has not the
quantities like \psi and \rho. Let me explain that.

I do have field quantities, like \psi and \rho in my approach only
as convenient mathematical entities. For me they are just certain
combinations of creation and annihilation operators of particles.
Using these combinations it is easy to write the interaction Hamiltonian
in a compact form and it is easy to prove the relativistic invariance
of QED a'la Weinberg. That's it. I don't have any other use of \psi
and \rho. I basically work with creation and annihilation operators

So, if you ask me: "Do \psi and \rho have any role in physical
interpretation of your theory?" My answer is "No".
If you ask me: "Do you use \psi and \rho in your mathematical
formalism?" My answer is "Yes". If you simply ask
"Do you have \psi and \rho in your theory?" my answer could be
"Yes" or "No" depending on how I (mis)understood the context
of your question.

For electron-positron system, my definition of charge can be formally
written as

Q = -e \int (a^{\dag}(p) a(p) - b^{\dag}(p) b(p))

(a and b are electron and positron operators, respectively).
This charge operator has a clear interpretation as a sum of charges
of electrons (-e) plus sum of charges of positrons (+e). This
operator explicitly commutes with the full interacting Hamiltonian.
(though this operator is not a 4-th component of any 4-vector)
This is enough for me to say that charge is conserved in my theory.

You are talking about different "charge" which is calculated as a
spatial integral of \rho(t,x). This "charge" is expressed differently
through creation/annihilation operators

Q' = -e \int (a^{\dag}(p) a(p) + b(p)a(-p) + a^{\dag}(p)b^{dag}(-p) -
b^{\dag}(p) b(p))

I am not quite sure what is the physical interpretation of this

I don't think I understood your point. Let me just give you an example.
The Coulomb interaction in my approach is represented by the operator V
having the form

V = \int dq'_1 dq'_2 dq_1 dq_2 |q'_1 - q_1|^{-2}
\delta ( q'_1 + q'_2 - q_1 - q_2)
a^{\dag}_{q'_1} a^{\dag}_{q'_2} a_{q_1} a_{q_2}.

In this case, the function D(...) is |q'_1 - q_1|^{-2},
which leads to the 1/r
interaction potential in the position representation.

I don't understand this point either. The physics is representation
dependent. If you represent the Poincare group in the Hilbert space
of the system by non-interacting operators H_0, P_0, J_0, K_0
(just sums of 1-particle terms) you get dynamics of a non-interacting
system. If the Poincare group is represented by interacting operators
H= H_0 + V, P_0, J_0, K = K_0 + W, you get dynamics of an interacting
system, i.e., something very different physically.

By the way, the operator V can be taken (in a good approximation) to
be the above Coulomb operator.

The dressing transformation and the dynamical character of boosts are
closely related. When I work with QED, I recognize that both Hamiltonian
and the boost operator K must contain interaction terms. When I do
"dressing", I do it for both H and K (otherwise, the Poincare
commutation relations will be violated). When I obtain the final dressed
Hamiltonian with interaction terms V in the form shown above
(the Coulomb interaction is an example), I realize
that these interactions are
instantaneous. Without knowing that boosts are dynamical this would be
a disaster, due to the contradiction between instantaneous signaling and
causality. However, the dynamical character of boosts fixes this
problem: instantaneous interaction remains instantaneous in all frames
of reference, the cause and effect never change their time order in the
moving frame.

If a theory predicts something that has not been (and will never be)
observed, there is something wrong with the theory. An alternative
theory which does not have such predictions would be desirable.

I do not have explanation of charge in my theory. My approach
is not intended to
be a "theory of everything". I just accept as given that charge of the
electron is -e, charge of the positron is +e, etc., and that the total
charge of a system of particles is the sum of particle charges.
Am I doing something wrong?

Yes, in my approach I can write the operator of charge as

Q = e (B - L)

where B is operator of the baryon number, L is operator of
the lepton number.

You are preaching Langrangean gauge field theory to me. I am doing
things differently:

1. Physical system is described by the Hilbert space h
2. Inertial transformations of observers form the Poioncare group.
3. Quantum relativistic description of the dynamics of the
physical system is achieved by construction of a unitary
representation of the Poincare group in the Hilbert space h.
4. Operators of total momentum P, angular momentum J and energy
H are just multiples of the generators of the Poincare group
representation in h. The conservation of P and J follows directly
from their Poincare commutators with H.
5. I DEFINE the operator of total charge Q as the sum of charges
of particles. I confirm that Q commutes with the interaction
Hamiltonian H of QED.

This logic is different from the logic of Lagrangean gauge field theory,
but it is still a viable logic. It does not involve fields and gauges.
(fields are only used as mathematical tools for compact writing
of the Hamiltonian H, as I wrote above).

I acknowledge the great success of the standard model in formulating
interaction Hamiltonians of E&M, weak and strong interactions.
I don't know how to do it myself, so I rely on the Hamiltonians
provided by the standard model. My theory starts only after getting
this Hamiltonian from SM. I prefer not to discuss how this
Hamiltonian was
derived. I prefer to discuss physical results stemming from
this Hamiltonian.

What about trajectories of interacting particles r(t)? They make perfect
sense in the classical limit of quantum theory. Can QED calculate
trajectories of two charged particles in the classical limit?

If I obtain operator V by "dressing" the QED interaction Hamiltonian
(as I do), such forbidden terms are discarded automatically.

So far, I was able to develop my approach at the level of direct
interparticle interactions. This would be silly to apply these
formulas directly to the description of a charged particle in
accelerator or in the Stern-Gerlach apparatus. This would require
direct summation of interactions of the test charge with all moving
charges in the electromagnet's wires. I think that this description
can be simplified by (approximately) describing the collective
action of all these moving charges by some kind of "field". I haven't
done that yet, but I don't think it's difficult (as long as
trajectories of charges in the wires are predetermined).
The resulting "field"
will be similar to your field B, and the movements of test
charges and spins
in this "field" will be very close to those predicted by standard


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