Zitterbewegung and the Time Operator
Jay R. Yablon
time *operator*.
In the paper linked below, I have analyzed this new operator in relation
to its Zitterbewegung.
http://jayryablon.files.wordpress.com/2008/07/zitterbewegung.pdf
Once this is done, we also learn a few new things about the
Zitterbewegung of the usual space operators.
Comments appreciated.
Thanks,
Jay.
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm
Pmb
"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message
news:6f55slF...@mid.individual.net...
> In a recent post dated 7/27, I showed how to derive a Lorentz-covariant
> time *operator*.
>
> In the paper linked below, I have analyzed this new operator in relation
> to its Zitterbewegung.
>
> http://jayryablon.files.wordpress.com/2008/07/zitterbewegung.pdf
This article makes no sense since in relativistic quantum mechanics both
position and momentum are demoted from being operators to being variables.
This allows both time and space to be treated in he same way, i.e. as
variables. You have also made the mistake of thinking that 4-velocity is
something that can make sense in the Dirac equation. Its easy to write
things down in a document, its much more difficult to give them and the
resulting interpretation, a valid meaning. You can's simply write down
velocity "v" and expect it to be meaningful. You have not even provided an
explanation of how dx/dt is to be interpreted. Nor did you mention what the
interpretation of what the field that the Dirac equation acts on means
Jay R. Yablon
news:tuKdnV_Pou5tPBDV...@comcast.com...
there.
I respectfully refer you to the original FW paper at:
http://jayryablon.files.wordpress.com/2008/07/foldy-wouthuysen-original-paper.pdf
as well as a WIKI article at:
http://en.wikipedia.org/wiki/Zitterbewegung
I do not believe I am doing anything differently than is shown in those
articles, other then perhaps using the classical symbols for velocity,
position, etc. with "hat's" on them to show correspondence to what they
are "representative" of, to use the FW terminology. That does not at
all change the substantive content.
For example, in FW Table 1, the symbol I denote as the velocity operator
v-hat is given by:
v-hat == \alpha = d x-hat /dt (1)
and the scalar velocity is given by the frequently-appearing term:
v == p / E_p (2)
Also, in the WIKI article, for example, in the fourth equation, there is
reference to:
hbar (dx_k/dt) = i[H,x_k] = \alpha_k (== v-hat_k) (3)
all of which are 4x4 Dirac-style operators. The (== v-hat_k) is my
notational addition, just to be clear about the "classical" name of each
operator, or, in FW language, what each operator is "representative" of.
As to physical interpretation, which I will agree I have not discussed
yet, none of the "operators" is itself a direct observable. Rather, one
uses any operator O in the usual manner to compute its
statistically-observable expectation value according to:
<O> = $ <psi^CT| O |psi> (4)
I would like to know how what I am doing, aside perhaps from the
additional notation using x-hat and v-hat and t-hat, is at all different
from what is done in these two links.
Jay.
Juan R.
> "Pmb" <som...@someplace.com> wrote in message
> news:tuKdnV_Pou5tPBDV...@comcast.com...
(...)
> http://jayryablon.files.wordpress.com/2008/07/foldy-wouthuysen-original-
paper.pdf
>
>
(...)
> For example, in FW Table 1, the symbol I denote as the velocity operator
> v-hat is given by:
>
> v-hat == \alpha = d x-hat /dt (1)
>
> and the scalar velocity is given by the frequently-appearing term:
>
> v == p / E_p (2)
This is not a scalar but a vector and it is an operator
v-hat == p / E_p (2')
This is usually named the mean velocity operator.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Pmb
Thank you Jay. Please give me time to review the material that you just sent
me. I will then get back to you at that time. Thank you.
Best wishes
Pete
Jay R. Yablon
I look forward to your further comments. I am still wondering, frankly,
about the combination of expressions in (2.17). If this is true, it
means that nobody previously has reduce the Zitterbewegung expressions
down as tightly as they can be reduce down, but my hesitance which has
been on my mind today arises because in (2.16) we are multiplying
(taking the 3-D dot product of) x-hat_l by dx_l/dt. I'd be curious of
you or anyone sees anything wrong in that calculation -- I can't quite
put a finger on it but am a bit uneasy.
The rest of the calculation I am comfortable with as mathematically
correct.
Best wishes to you as well,
Jay.
Jay R. Yablon
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.07...@canonicalscience.com...
> Jay R. Yablon wrote on Mon, 28 Jul 2008 09:00:24 -0600:
>
>> "Pmb" <som...@someplace.com> wrote in message
>> news:tuKdnV_Pou5tPBDV...@comcast.com...
>
> (...)
>
>> http://jayryablon.files.wordpress.com/2008/07/foldy-wouthuysen-original-
> paper.pdf
>>
>>
> (...)
>
>> For example, in FW Table 1, the symbol I denote as the velocity
>> operator
>> v-hat is given by:
>>
>> v-hat == \alpha = d x-hat /dt (1)
>>
>> and the scalar velocity is given by the frequently-appearing term:
>>
>> v == p / E_p (2)
>
> This is not a scalar but a vector and it is an operator
A rose by any other name . . . ;-) Sorry, I used sloppy language. v
above is a velcoity vector with three components, one for each spatial
coordinate. I tend to refer to those as "scalar" components, though one
needs to be careful with "scalar" because that is also often taken to
mean an invariant.
>
> v-hat == p / E_p (2')
>
> This is usually named the mean velocity operator.
Yes. One of the discussion that we are having in related threads . . .
enter Princess Neuropulp . . . is whether "mean" is a good choice of
language by FW, because that can be confused with the observable
expectation value which this is not. I think they are really talking
about a velocity operator for which the zitterbewegung acceleration
operator vanishes, so that have in essence "smoothed out" its effects
and so call it a "mean." I would be more comfortable, for example,
calling this a "zitterbewegung-free" operator.
Jay.
>
>
>
> --
> http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
>
Hans de Vries
> In a recent post dated 7/27, I showed how to derive a Lorentz-covariant
> time *operator*.
>
> In the paper linked below, I have analyzed this new operator in relation
> to its Zitterbewegung.
>
> http://jayryablon.files.wordpress.com/2008/07/zitterbewegung.pdf
>
Hi, Jay
I'm close to posting an update which shows that under
the sole assumption that:
The spinor part of the field is transformed identically
as the phase part (exp(-i\phi) . In other words. They
have the same velocity, then:
For a plane wave [H,L], [H,S] and [H,v] are all zero.
That is: Spin and Orbital angular momentum are
conserved separately and the plane-wave does
not experience acceleration.
Regards, Hans
Pmb
"Hans de Vries" <Hans.de....@gmail.com> wrote in message
news:3b205117-47fe-468e...@l64g2000hse.googlegroups.com...
Hans - You have still neglected to tell me how I ca obtain a copy of your
book. Is it inomplete and not ready to be read or something like that?
Since this is about the third time I have asked about this I will assume
that you are choosing not to rspond for one reason or another. I'm not even
convinced that this is the case but I don't want to be a pest and keep
asking. You might find that irritating and I don't want to irritate you. I
enjoy your presence on the internet and appreciate your contributions. :)
Pete
======================================= MODERATOR'S COMMENT:
This moderator has got immediate access to all links Hans has provided :-)
Peter
I would like to guess that this should be expected for interactionless
particles, correct?
Thank you,
Peter
Jay R. Yablon
"Hans de Vries" <Hans.de....@gmail.com> wrote in message
news:3b205117-47fe-468e...@l64g2000hse.googlegroups.com...
I think I saw somewhere that this is precisely what happens in FW.
Best,
Jay.
Hans de Vries
> Hans - You have still neglected to tell me how I ca obtain a copy of your
> book. Is it inomplete and not ready to be read or something like that?
>
> Since this is about the third time I have asked about this I will assume
> that you are choosing not to rspond for one reason or another. I'm not even
> convinced that this is the case but I don't want to be a pest and keep
> asking. You might find that irritating and I don't want to irritate you. I
> enjoy your presence on the internet and appreciate your contributions. :)
>
> Pete
>
> ======================================= MODERATOR'S COMMENT:
> This moderator has got immediate access to all links Hans has provided :-)
Hi, Pete
Please be assured that I'm flattened rather than annoyed by
your interest! The book is just still "Work in progress" and
incomplete so there are no paper copies or complete versions.
Many chapters which are sufficiently stable are available here:
http://www.physics-quest.org/
Regards, Hans
Pmb
> Hi, Pete
>
> Please be assured that I'm flattened rather than annoyed by
> your interest!
We had discussed relativity in that other forum and I and a very smart and
well educated friend or mine, came to have a great deal of admiration for
you. :) I'm hoping I can have the privilege to pick your brain more in the
future?! :) In particular I have started studing Quantum Field Theory
(QFT). Many people have told me that QFT can't be learned through self
teaching.
I'd like to discuss some particular things in E-mail with you if you're
interested? If so then E-mail me at physics_world[at]yahoo[dot]com. I will
then E-mail you back with my working E-mail address.
>The book is just still "Work in progress" and incomplete so there are no
>paper copies or complete versions.
>
> Many chapters which are sufficiently stable are available here:
> http://www.physics-quest.org/
I am aware of that web site. I was not aware that the book was not competed.
I was unsure as to whether you had completed it and are merely constructing
a web site for it or if you are posting chapters as you complete them. I was
hoping that you had an earlier version of the chapters that I could read
while the new ones were being constructed. But I can read as we go. That'd
be great! Thanks Hans!
Best wishes
Pete
Hans de Vries
> "Hans de Vries" <Hans.de.Vries...@gmail.com> wrote in messagenews:3b205117-47fe-468e...@l64g2000hse.googlegroups.com...
> > That is: Spin and Orbital angular momentum are
> > conserved separately and the plane-wave does
> > not experience acceleration.
>
>
>
Maybe this link of you?
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=rep1&type=pdf
Regards, Hans
Jay R. Yablon
(2.15) and (2.20) need to be corrected. The "zitter" motion remains but
takes on a simpler form dependent only on the rest mass an not on the
entire Hamiltonian.
On the other hand, though I had some doubts earlier today, the result
(2.17) is correct and one can deduce this from (2.16) which I know is
correct and can be calculated straight from the usual zitterbewegung
without the time operator or anything else -- just take the dot product
of the usual zitterbewegung with dx_l/dt and then reduce.
The question whether one can go from (2.16) which I am quite certain is
calculated correctly to (2.17) is trickier, but answers in the
affirmative. For an ordinary vector, say (picked at random to
illustrate the point):
A^i = (2,3,4)
B^i = (1,0,6)
x_i=(1,3,5)
one has A^i x_i = B^i x_i = 31, even though A^i <> B^i, so that A^i x_i
= B^i x_i does *not* imply that A^i = B^i.
But, this is because A^i and B^i are NOT linearly independent.
When, as in (2.16), each of the gamma^i IS linearly independent (and
that is part of the definition of the Dirac group elements), then it
turns out that (2.16) implies that (2.17) must be true, that is, one can
no longer mess around with the selections of A^i (= z-hat^i) and B^i
(=gamma^i) as in the above example, precisely because each of the
gamma^i is linearly independent. That is, there is nothing that one can
add to gamma^1 to arrive at gamma^2, but there is something that one can
add to, say, A^1=2 to arrive at A^2=3.
So, (2.17) is correct, and it represents a simplification of the usual
zitterbewegung that does not depend on anything other than straight
mathematical calculation.
But, as I said, the integrals all need to be fixed.
Will try to post a corrected update soon.
Jay.
Jay R. Yablon
Pmb
> Thanks Pete,
>
> I look forward to your further comments.
Please keep in mind that I'm just learing relativistic quantum mechanics and
Field theory as we speak. So if I say that something doesn't make sense then
the problem may very well be with me. My main priority right now is to
review quantum mechanics in case I go back to graduate school in the fall so
I won't be able to comment on your paper because I can't spend the time
reading it and the ones you cited and learn re-qm at the same time. rel-qm,
while I'm trying to learn it through self-study, is something that is also a
low priority for me now. I will say this though; from what I've learned
about rel-qm so far is that time and position are treated on the same
footing on order to adhere to the principles of quantum mechanics. In
non-relativistic quantum mechanics time is treated on a different footing
than position. The second derivates with respect to position appear in
Schrodinger's equation whereas time appears as a first derivative. position
is an operator whereas time is a variable. To bootstrap up to relativistic
quantum mechanics we have two posible alternatives (1) demote position from
an operator to a variable or (2) promote time to an operator. It is my
uncerstanding that it is the later which was chosen. For this reason I am
confused as to why there is a position operator in that 1950 paper!? Also
what is the physical meaning of a time operator? In your paper you defined a
time operator in a very strange way. Normally an operator corresponding to
an observable is defined by describing its action of a ket. What is the
physical meaning of the eigenvalues of the time operator?
Also this quantity that the call a "velocity operator" doesn't seem right to
me. In quantum mechanics the momentum that is spoken of is not linear
3-momentum but is canonical mometum. In the 1950 article the authors have a
section in that paper called "Dirac particle in an external electromagnetic
field" in which the magnetic field is different than zero. This results in
the canonical momentum p being a functon of the magnetic vector potential.
It seems to me that the goal of a velocity operator, v, was to be able to
write an expression of the form p = mv. If this is not true then of what use
is the velocity operator?
How does the definition of your time operator compare with the time operator
defined elsewhere?
Why is all this called a Zitterbewegug? From its definition in Wiki that
term doesn't seem to mean anthing different than "rapid motion" which is par
for the course in relativity.
Also, the velocity operator as the time derivative of the position operator
is the Schrodinger picture makes no sense since the position operator is not
time dependant. That kind of thing only seems to makes sense in the
Heisnberg picture.
I hope Hans can shed some light onto these concerns.
Hans?? Any thoughts?
Pete
======================================= MODERATOR'S COMMENT:
German verb "zittern" means a not regular oscillatory motion, eg, if you feel cold or fear
Juan R.
>> v-hat == p / E_p (2')
>>
>> This is usually named the mean velocity operator.
>
> Yes. One of the discussion that we are having in related threads . . .
> enter Princess Neuropulp . . . is whether "mean" is a good choice of
> language by FW, because that can be confused with the observable
> expectation value which this is not.
And as pointed in a previous message from mine (in the thread opened by
Peter) there is no confusion. One thing is the *mean value* associated to
an operator (either mean or instantaneous), and other different thing a
*mean operator*.
The physical meaning of the NW mean operator was also discussed.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Hans de Vries
> In particular I have started studing Quantum Field Theory
> (QFT). Many people have told me that QFT can't be learned through self
> teaching.
>
Hi, Pete.
Learning QFT through self teaching is indeed for the very brave.
But then, you are not someone who gives up easily.... :^)
Best Regards, Hans
Jay R. Yablon
> In particular I have started studing Quantum Field Theory (QFT). Many
> people have told me that QFT can't be learned through self teaching.
Some people might say that I am a good example for proving this point!
;-) Jay.
Juan R.
> In particular I have started studing
> Quantum Field Theory (QFT).
Then have in mind that QED becomes in three main flavors:
i) Wave flavor
The spacetime formulation of QED. This is the formulation given by
Feynman. It is based in a generalization of Dirac *wave* function adding
quantization of fields A_b.
Uses the Path integral approach for computing the amplitudes and Feynman
theory of positrons (i.e. backward in time electrons).
Space is a operator but time is a parameter. This gives inconsistencies
for questions like Lorentz invariance and the localization problems:
luminical speeds, non_analitic propagation outside light cones, and
others.
This is the formulation preferred in molecular physics and quantum
chemistry probably because looks close to non-relativistic quantum
mechanics and original relativistic quantum mechanics (Dirac theory).
Textbook:
http://www.amazon.com/Quantum-Electrodynamics-Advanced-Book-Classics/
dp/0201360756
ii) Field flavor
Uses Schrödinger equation for evolution. The fundamental elements are
fields and particles are a subproduct from second quantization.
The Dirac equation lost status as valid wave function and is here
interpreted like a functional for *field* operators.
Positrons (in general antiparticles) travel forward in time. This
formalism uses creation-destruction operators.
Position is downgraded to parameter. This introduces consistency with
special relativity but breaks with ordinary quantum mechanics where
position is an operator. This flavor avoids localization issues rather
than solving them.
This is the formulation preferred in particle physics probably because
avoid all traditional difficulties associated to Dirac wave theory and
lets easy generalization beyond QED: weak, strong...
Textbook:
http://www.amazon.com/Quantum-Field-Theory-Rev-Ed-Franz/dp/0471941867
iii) Modern flavor
Also uses Schrödinger equation for evolution. This is a more particle
'version' of the above field approach. Particles are taken like
fundamental blocks of nature and fields are used as formal blocks for
building cluster separable invariant interactions.
This is the formulation preferred by Weinberg and the emphasis on
particles rather than fields is probably based in Weinberg interest in
string theory.
Textbook:
http://www.amazon.com/Quantum-Theory-Fields-Vol-Set/dp/0521585554
All of above are limited and inconsistent at some extension.
iv) Alternative flavor
Any other was under development. Some few.
iv.a)
Based in Stueckelberg-Piron-Horwitz covariant wave theory. For instance
here position is an observable again and 'time' is also observable with
associated 'time' operator. Note: that 'time' is Einstein clock, which is
not time. I.e. the theory has two concepts of 'time': time and clocks
rate.
http://en.wikipedia.org/wiki/Relativistic_dynamics
The classical version is 'revised' here
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-
limitations_20.html
This approach solves difficulties with field theories.
iv.b)
Based in extension of Weinberg modern flavor with more emphasis in
particles and abandonment of exact Lorentz invariance (Special Relativity
is recovered as special case). This RQD approach solves well-known
troubles of QED with time evolution and does via a dressing approach that
eliminates virtual and bare particles.
Textbook online:
http://arxiv.org/abs/physics/0504062
iv.c)
My own. *Still under development*. I am trying to solve all difficulties
of above theories at once. Which is a very very difficult task!
The basic idea is that one starts with a wave function theory with time
being a *parameter* and position an *operator*. This is similar to
Stuckelberg-Horwitz-Piron theory and Dirac-Feynman original approach.
One can then introduce the observable for clocks. This gives an operator
for 'time' t_op somewhat as in Stuckelberg-Horwitz-Piron. However this
operator is not introduced as dynamical dof in wave function at the same
status that x_op. This eliminates difficulties with redundancies in multi-
clock theories and does the whole theory more economical. Also also does
unneeded the reparametrization of mass and can work with m instead M. In
the classical limit, this gives a 6N phase space whereas Stuckelberg-
Horwitz-Piron is 8N. N being the number of particles of course.
Interaction is introduced via DPI. This is similar to Stuckelberg-Horwitz-
Piron and Stefanovich. However, the covariant potentials of the former are
not used because x^0 is not variable and the latter potentials are not
used because the field based potentials on RQD are of class
V(t) = \Int d^3x (\bar \Psi) h(x,t) \Psi
where h(x,t) is that from field QED. The dressing transformation changes
this for eliminating divergences and other problems. But the fundamental
role of fields as building blocks for interactions remains.
A problem with field based interaction is that are local h(x,t) and cannot
give the correct limit h(R(t)), which is nonlocal and time implicit. I.e.
field based approaches cannot explain recent advances in understanding of
interactions (dualism)
http://prola.aps.org/abstract/PRE/v53/i5/p5373_1
and cannot give a complete many-body dynamics, because is built over
nonlocal time implicit potentials. Indeed the potentials in Stuckelberg-
Horwitz-Piron theory are of kind h(\rho(\tau)). And this explain why the
theory gives equations of motion cannot be obtained using field theoretic
methods. See Schieve monograph cited in
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-
limitations_20.html
for details at classical level of theory.
Due to different definition of evolution time and x^0 I cannot reuse their
nonlocal time implicit potentials. The quantum potentials I am using are
h(R(t)).
This is all work under development. I have just developed a technique that
shows how this new theory gives Chubikalo and Smirnov-Rueda dualism more
certain corrections and extend their work to quantum and gravitational
domains.
I have proved that new theory contains classical electrodynamics and
special relativity as special cases. E.g. the Lienard-Wiechert potentials
are recovered as approximation h(R(t)) --> h(x,t).
I have also proved in my work "Chubykalo and Smirnov-Rueda dualism:
Foundation and generalizations" (still under review) that in the quantum
case the one-photon interaction from Feynman QED (in Feynman gauge) arises
as special case from the new theory. I.e. QED interactions arise as
special case from a more general theory of quantum interactions.
Several difficulties and puzzles that Feynman notices in his book
(remember based in a generalization of Dirac wave theory) are
automatically absent in the new theory, which gives me some consistency.
But still more work is needed to prove this a valid way.
My advice when studying QFT is to follow the path QM --> RQM --> Feynman
QED --> field QFT --> Weinberg QFT --> current research state of the art.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
======================================= MODERATOR'S COMMENT:
Note that several notions are not common (at least not to this moderator ;-)
Pmb
Okay. I think I understand the problem now. I was thinking about the
demotion of the position operator to a variable. I now see that is what
happens in field theory, not in relativistic quantum mechanics.
I take it that the operator expresion you mentioned, i.e. dX/dt, is in the
Heisenberg representation and you simply assumed the reader knows this fact
and thus you didn't bother including "H" to denote the fact that the
operator is in the Heisenberg picture. Is that correct?
Thanks
Pete
Jay R. Yablon
. . .
>
> Okay. I think I understand the problem now. I was thinking about the
> demotion of the position operator to a variable. I now see that is
> what happens in field theory, not in relativistic quantum mechanics.
>
> I take it that the operator expresion you mentioned, i.e. dX/dt, is in
> the Heisenberg representation and you simply assumed the reader knows
> this fact and thus you didn't bother including "H" to denote the fact
> that the operator is in the Heisenberg picture. Is that correct?
>
> Thanks
>
> Pete
Pete, that is correct. Best, Jay.