Look there! Is that the position of the Masked Quantum Damsel?
neur...@yahoo.com.au
First, I thank you Sir Jay for your intervention.
By your:
> ... allowing me to miscomprehend even more aspects of nature
> than I had before and then spread them like wildfire on the
> internet. ;-)
you receive the honor of being the first knight to make _me_ laugh.
But the Queen is still furious (more so because the King
humiliated her by releasing me). Thus, I must creep about
the palace in mask and cloak to avoid attention. Also, the
King has recently begun to look at me in a most un-fatherly
way, so the mask and cloak are doubly necessary. I think I
must soon flee into exile.
- - - - -
This damsel was disappointed that (being imprisoned) she
could not accept the invitation to the Grand Ball of
Velocity Operators, proclaimed by Don Juan the Passionate,
where much dancing was to occur. (Ooh, this damsel swoons at
his passion, his tenacity, and truly hopes he will one day
visit her bower.) However, I do not understand why there was
also a call for the dancing to stop. Alas, this damsel is
dismayed to see how the dancing has degenerated into most
un-gentlemanly fisticuffs, which should be moved outside to
the jousting arena.
It seems to this damsel that much disagreement about velocity
operators is caused by controversies about which is the
correct position operator "Q". Apparently, all agree that
V = [H,Q], but what then is the correct "Q"?
Don Juan the Passionate and our avian hero Super-Hans both
seem to advocate the coordinate x, i.e., Q = x. (At least,
such is this damsel's impression. Please do not be angry
with her if she has misrepresented either of you.) But alas,
they seem to disagree on other matters.
While imprisoned, this damsel wisely used her time to read
many ancient scrolls, from which it became clear that the
problems with using coordinate x as a position operator are
long-debated, sometimes heatedly. Others, (many now long
dead, but including the still-living fearsome Sage of Vienna),
favour the Newton-Wigner position operator, which I denote
here as "Q_nw". (For the Klein-Gordon field, part of the
rationale is that ordinary "x" is not a self-adjoint
operator - see (eg) Schweber's explanation in "An
Introduction to Relativistic Quantum Field Theory",
pp61-62.) Strange's textbook seems not to discuss such
questions, and Feynman's was not in the dungeon library.
For general-spin fields, the reasons behind Q_nw are more
complicated and it takes a messier form, but there seems no
shortage of literature advocating "Q_nw" as physically more
reasonable than "x". Clearly "the literature" has not yet
settled to a consensus about such things. (Thus it seems that
"understanding" is also a local feature of the world, and
cannot be analytically continued across boundaries between
different domains.)
This damsel is becoming more persuaded that "Q_nw" is
physically more desirable than "x", in which case
V = [H, Q_nw] must be chosen as velocity operator. Of
course, I must also hasten to add that all of this is for
free theories only. When interactions are introduced, we
cannot even find a suitable NW position operator, so it
seems, nor an exact Foldy-Wouthuysen-like transformation.
But even for interacting field theories, x is still
undesirable, as the Reeh-Schlieder theorem suggests. This
damsel was most hurt that Super-Hans ignored her earlier
question about this (sniff, sob).
This damsel also observed that Don Juan and Prince Charles
entered battle over the question of commutativity of
velocity components. Prince Charles objected that \alpha_i
does not commute with \alpha_j and therefore does not
correspond to our conventional notions of velocity, where
all components are simultaneously measurable. Don Juan made
a counter-thrust with a mini-seminar in relativistic
physics, saying:
> ...two electron relativistic interaction ...
> Here one usually starts from classical action with the
> well-known magnetic term proportional to (v v').
>
> And next one substitutes the velocities by operators
>
> v_classical --> v_operator
but did not mention how Poisson brackets go over to Quantum
commutators (though I'm sure knows this full well).
Pursuing this line further...
If we take the classical v^i = dH/dp_i (those are partial
derivatives), then the classical Poisson bracket is
{v^i, v^j}_PB = dv^i/dq^k dv^j/dp_k - (... i<->j ...)
(with implicit summation over k). Using v^i = dH/dp_i, this
becomes
{v^i, v^j}_PB = d^2H/dq_k dp_i d^2H/dp_j dq_k - (... i<->j ...)
For a free classical theory (no acceleration, dH/dq_i = 0),
we find this Poisson bracket is zero (so presumably its
quantum analog should be zero too). However, for more general
Hamiltonians with interactions, the bracket is no longer zero,
so in this case one presumes that components of the velocity
observables should not commute in the quantum theory either.
The original "dance" was in the context of the free Dirac
theory, so presumably the former case must be the one which
is applicable: in a free theory the velocity components
should commute. This turns out to be so if Q_nw is used (for
details, see my previous Schweber reference), which
strengthens the case for Q_nw.
This damsel also notes that theories involving interaction
via a gauge field typically involve singular Lagrangians.
Hence there is no straightforward way to obtain velocities
from momenta, and if we start from a representation based on
the latter, obtaining velocity operators is problematic,
and the whole discussion is moot.
This damsel invites all brave knights of great agility to
a dance on these subjects (though it must be held secretly
far from my palace). I'll try to find more girls to attend
the dance, so that the boys don't starting fighting so
quickly due to lack of damsels. I'm sure my sisters will
come. Maybe I can find some others.
----
LOL from The Masked Quantum Damsel.
Oh No
>This damsel was disappointed that (being imprisoned) she
>could not accept the invitation to the Grand Ball of
>Velocity Operators, proclaimed by Don Juan the Passionate,
>where much dancing was to occur. (Ooh, this damsel swoons at
>his passion, his tenacity, and truly hopes he will one day
>visit her bower.) However, I do not understand why there was
>also a call for the dancing to stop. Alas, this damsel is
>dismayed to see how the dancing has degenerated into most
>un-gentlemanly fisticuffs, which should be moved outside to
>the jousting arena.
The ball is not over. The dancing may have been interrupted with minor
fisticuffs, and this should stop, but I would be honoured if the dashing
damsel would mark me on her card for the projection postulate polka.
I am all envy. Who is this mysterious masquerader "Q_nw"?
> in which case
>V = [H, Q_nw] must be chosen as velocity operator. Of
>course, I must also hasten to add that all of this is for
>free theories only. When interactions are introduced, we
>cannot even find a suitable NW position operator, so it
>seems, nor an exact Foldy-Wouthuysen-like transformation.
>But even for interacting field theories, x is still
>undesirable, as the Reeh-Schlieder theorem suggests. This
>damsel was most hurt that Super-Hans ignored her earlier
>question about this (sniff, sob).
For a non-interacting theory, I do not think there is much difference
between Q_nw and x, but when interactions are introduced and a
semiclassical approximation used, replacing the photon field operator,
A, with it's expectation <A>, it becomes necessary to use NW if one
seeks compatibility with the Lorentz force law.
>This damsel also observed that Don Juan and Prince Charles
>entered battle over the question of commutativity of
>velocity components. Prince Charles objected that \alpha_i
>does not commute with \alpha_j and therefore does not
>correspond to our conventional notions of velocity, where
>all components are simultaneously measurable.
Indeed, the use of alpha plays havoc with the projection postulate. It
should be given up without a fight, albeit on the jousting field rather
than at the ball (not least because it means I shall have to revise my
website :-/). A velocity operator which leads to only speeds of +-c for
electrons remains physically meaningless, but I have come to think that
the fault lies with the projection postulate, not with alpha. A turn-
around, I agree, but this is a dance after all. It does not appear
necessary to require that an observable operator creates an eigenstate
of the observable, but merely that the classical value of the observable
is given by the expectation of the operator.
>----
>LOL from The Masked Quantum Damsel.
I hope she will continue to dance, however uncertain her position.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Juan R.
> This damsel was disappointed that (being imprisoned) she could not
> accept the invitation to the Grand Ball of Velocity Operators,
> proclaimed by Don Juan the Passionate, where much dancing was to occur.
> (Ooh, this damsel swoons at his passion, his tenacity, and truly hopes
> he will one day visit her bower.) However, I do not understand why there
> was also a call for the dancing to stop.
There was not. I was really asking, how many more dancing before accepting
that theory says us?
> It seems to this damsel that much disagreement about velocity operators
> is caused by controversies about which is the correct position operator
> "Q". Apparently, all agree that V = [H,Q],
Well Oh No still maintained this wrong recently:
(\blockquote
There has been some confusion recently over the velocity operator, but as
Igor has pointed out recently in s.p.r. the correct velocity operator is
p/m (in any representation).
)
Igor has already recognized his mistake in s.p.r. but now he is doing
others.
> but what then is the correct
> "Q"?
>
> Don Juan the Passionate and our avian hero Super-Hans both seem to
> advocate the coordinate x, i.e., Q = x. (At least, such is this damsel's
> impression.
I don't advocate that. I have said that if you are working beyond the Zb
scale the x is not the instantaneous x but an averaged one (without Zb).
And the velocity then is that Strange calls the averaged velocity.
Also I have said that in the momentum representation it is -i@/@P but
In any case the definition of velocity is the same...
> Please do not be angry with her if she has misrepresented either of
> you.) But alas, they seem to disagree on other matters.
>
> While imprisoned, this damsel wisely used her time to read many ancient
> scrolls, from which it became clear that the problems with using
> coordinate x as a position operator are long-debated, sometimes
> heatedly. Others, (many now long dead, but including the still-living
> fearsome Sage of Vienna), favour the Newton-Wigner position operator,
> which I denote here as "Q_nw". (For the Klein-Gordon field, part of the
> rationale is that ordinary "x" is not a self-adjoint operator - see (eg)
> Schweber's explanation in "An Introduction to Relativistic Quantum Field
> Theory", pp61-62.) Strange's textbook seems not to discuss such
> questions, and Feynman's was not in the dungeon library.
>
> For general-spin fields, the reasons behind Q_nw are more complicated
> and it takes a messier form, but there seems no shortage of literature
> advocating "Q_nw" as physically more reasonable than "x". Clearly "the
> literature" has not yet settled to a consensus about such things. (Thus
> it seems that "understanding" is also a local feature of the world, and
> cannot be analytically continued across boundaries between different
> domains.)
>
> This damsel is becoming more persuaded that "Q_nw" is physically more
> desirable than "x", in which case V = [H, Q_nw] must be chosen as
> velocity operator.
Hum but if you work in the Newton Wigner representation for the position,
you may also work in the Newton Wigner representation for the Hamiltonian
V = [H_nw, Q_nw]
and then the velocity is not c \alpha.
I already remarked several times that the problem with Dirac Hamiltonian
is on that it is linear in momentum.
Any modern serious attempt I know to solve localization problems of
relativistic physics start from using a nonlinear Hamiltonian.
I also remarked consistency problems with Hans usage of a position
operator in quantum field theory. I had a discussion about this with Igor
in s.p.r.
I also cited here a work (in the several flavors of QED post), the RQD
book, where the Newton Wigner operator is used.
> This damsel also observed that Don Juan and Prince Charles entered
> battle over the question of commutativity of velocity components. Prince
> Charles objected that \alpha_i does not commute with \alpha_j and
> therefore does not correspond to our conventional notions of velocity,
> where all components are simultaneously measurable.
I also remarked this in the past in a discussion with Hans. Effectively
matrices alpha don't commute and this implies that the vector velocity
that Hans uses
v = c \alpha
is not observable. I explained that Hans 2D chiral equation is more or
less fine, but the direct generalization to 4D is not possible because the
components v_x v_y v_z are not measurable *at once* and the vector
v = v_x + v_y + v_z is not observable.
However, the *magnitude* of the vector is well defined and it is
|v|^2 = c^2
as is well-known.
The fact that the velocity vector is not observable is the last reason
which Feynman did not compute the eigenvalues using v but using v^2.
Because only the scalar part of the vector velocity eigenvalue is well
defined. Hans did not understood this and accused Feynman from using
tricks :-)
Of course, Oh no is right that c alpha does not correspond to "to our
conventional notions of velocity" but this was also explained. What Oh no
call conventional velocity is the velocity after eliminating the Zb
effect.
This is well explained in Paul Strange book. Where he explains the
physical and mathematical differences between the instantaneous velocity
operator and the averaged velocity operator.
I wrote next formal recipe
c = v + v_Zb
many many many many times. And also explained in what spatio-temporal
scales of measurement you may ignore the Zb effect and use a conventional
notion of velocity.
> Don Juan made a
> counter-thrust with a mini-seminar in relativistic physics, saying:
>
> > ...two electron relativistic interaction ... Here one usually starts
> > from classical action with the well-known magnetic term proportional
> > to (v v').
> >
> > And next one substitutes the velocities by operators
> >
> > v_classical --> v_operator
>
> but did not mention how Poisson brackets go over to Quantum commutators
> (though I'm sure knows this full well).
Well in another discussion (I think was with Peter) I explained that the
right definition of velocity operator is v = [H,x] and this also works for
classical physics
v = {H,x}
The usual classical definition v = @H/@p is easily derivable from above
more rigorous and complete form using Poisson brackets.
As can be observed most of above was said. But some people continue
dancing about the subject, trying to obtain some magical way to evade one
or several of those questions.
I ask again how many more dancing?
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Jay R. Yablon
> First, I thank you Sir Jay for your intervention.
> By your:
>
> > ... allowing me to miscomprehend even more aspects of nature
> > than I had before and then spread them like wildfire on the
> > internet. ;-)
>
> you receive the honor of being the first knight to make _me_ laugh.
>
> But the Queen is still furious (more so because the King
> humiliated her by releasing me). Thus, I must creep about
> the palace in mask and cloak to avoid attention. Also, the
> King has recently begun to look at me in a most un-fatherly
> way, so the mask and cloak are doubly necessary. I think I
> must soon flee into exile.
>
>
> ----
> LOL from The Masked Quantum Damsel.
It seems tha the King has now restricted my participation in this
newsgroup over the next several days, by visiting a hardware crash
upon the motherboard of my computer and forcing me to work on borrowed
laptops without the resources I usually have available. I hope that
this is not in retaliation for obtaining release of the dainty damsel,
though my palace sources sugest it could be.
So, as the damsel and I lay low for a period of time, we will look to
others to carry the ball along, pendng the day of our triumphant
return from exile and a restoration of peace through the land.
Jay. ;-)
neur...@yahoo.com.au
Prince Charles the Valiant responded:
> The ball is not over. .... I would be honoured if the
> dashing damsel would mark me on her card for the projection
> postulate polka.
This damsel fears she has not yet studied that polka
adequately. Of late, there has been much discussion of
decoherence and projection-related matters by the Master of
Zurek, and some time ago the fearsome Sage of Vienna issued
an open challenge concerning detailed models of quantum
measurement. Some dismissed the Sage's challenge too glibly
(in this damsel's humble opinion), and it remains largely
unanswered.
But perhaps that polka should be danced at a separate ball?
(Too many different dances simultaneously in the same
ballroom is rarely wise.)
>> This damsel is becoming more persuaded that "Q_nw" is
>> physically more desirable than "x",
> I am all envy. Who is this mysterious masquerader "Q_nw"?
Oh Prince Charles, are you teasing me? You surely know the
answer full well.
But Q_nw is indeed mysterious. He swears he is a complete
man, but whenever I interact with him it seems he is not
all there, as if he were only created perturbatively.
> For a non-interacting theory, I do not think there is much
> difference between Q_nw and x,
Do you have a copy of Newton & Wigner's original paper?
Newton, Wigner: "Localized States for Elementary Systems".
Rev Mod Phys Vol 21, No 3, pp 400-406
(or perhaps the Schweber book I mentioned in my previous post?)
The differences seem quite significant, at least in this
damsel's eyes.
> ... the use of alpha plays havoc with the projection
> postulate. ... A velocity operator which leads to only
> speeds of +-c for electrons remains physically meaningless,
> but I have come to think that the fault lies with the
> projection postulate, not with alpha. A turn- around, I
> agree, but this is a dance after all. It does not appear
> necessary to require that an observable operator creates an
> eigenstate of the observable, but merely that the classical
> value of the observable is given by the expectation of the
> operator.
Oh calamity! The dreaded slimy octupus of quantum
measurement draws near and threatens to engulf the city.
I feel a need to maintain a rigorous distinction between
"observable operator" and "act of measurement".
The former is a mapping between states in Hilbert space.
The latter is a mapping from a Hilbert space operator to
a number (ie a linear functional over an algebra of
observable operators).
> I hope she will continue to dance, however uncertain her
> position.
She dances here! She dances there!
And sometimes she collapses hard upon her pretty little tooshie.
----
LOL from the Masked Quantum Damsel!
Hans de Vries
> Do you have a copy of Newton & Wigner's original paper?
>
> Newton, Wigner: "Localized States for Elementary Systems".
> Rev Mod Phys Vol 21, No 3, pp 400-406
Hi, princess.
They made it available on your special request :^)
http://prola.aps.org/pdf/RMP/v21/i3/p400_1
Regards, Hans.
Oh No
>
>Prince Charles the Valiant responded:
>
>> The ball is not over. .... I would be honoured if the
>> dashing damsel would mark me on her card for the projection
>> postulate polka.
>
>This damsel fears she has not yet studied that polka
>adequately. Of late, there has been much discussion of
>decoherence and projection-related matters by the Master of
>Zurek, and some time ago the fearsome Sage of Vienna issued
>an open challenge concerning detailed models of quantum
>measurement. Some dismissed the Sage's challenge too glibly
>(in this damsel's humble opinion), and it remains largely
>unanswered.
>
>But perhaps that polka should be danced at a separate ball?
>(Too many different dances simultaneously in the same
>ballroom is rarely wise.)
>
>
>>> This damsel is becoming more persuaded that "Q_nw" is
>>> physically more desirable than "x",
>
>> I am all envy. Who is this mysterious masquerader "Q_nw"?
>
>Oh Prince Charles, are you teasing me? You surely know the
>answer full well.
>
>But Q_nw is indeed mysterious. He swears he is a complete
>man, but whenever I interact with him it seems he is not
>all there, as if he were only created perturbatively.
I am relieved to hear that he is without his vital bits, as I could
hardly describe, in the presence of a blushing damsel, how I might
otherwise remove them.
>> For a non-interacting theory, I do not think there is much
>> difference between Q_nw and x,
>
>Do you have a copy of Newton & Wigner's original paper?
>
>Newton, Wigner: "Localized States for Elementary Systems".
>Rev Mod Phys Vol 21, No 3, pp 400-406
I have it now, with thanks to Hans. I am struck by the challenges faced
by those early adventurers when so little of the kingdom was charted.
Let us not forget that to measure a particle, one must interact with it,
and that the only interaction available to us in qed creates
antiparticles as well as particles.
>> ... the use of alpha plays havoc with the projection
>> postulate. ... A velocity operator which leads to only
>> speeds of +-c for electrons remains physically meaningless,
>> but I have come to think that the fault lies with the
>> projection postulate, not with alpha. A turn- around, I
>> agree, but this is a dance after all. It does not appear
>> necessary to require that an observable operator creates an
>> eigenstate of the observable, but merely that the classical
>> value of the observable is given by the expectation of the
>> operator.
>
>Oh calamity! The dreaded slimy octupus of quantum
>measurement draws near and threatens to engulf the city.
Let me at him! Ball or not, what kind of prince would not rush to the
task of dissevering its legs in defence of the realm?
>I feel a need to maintain a rigorous distinction between
>"observable operator" and "act of measurement".
>The former is a mapping between states in Hilbert space.
>The latter is a mapping from a Hilbert space operator to
>a number (ie a linear functional over an algebra of
>observable operators).
Good distinction. May we require that, to say that an observable
operator has a well defined value in a given state, an act of
measurement should necessarily yield that value as the expectation of
the operator?
We need some such statement in order to say that <A> is the classical
e.m. vector field. This is weaker than the projection postulate, which
requires an eigenstate (in which the value is trivially given by the
expectation). It has been shown by Hans that alpha gives the expectation
p/E given a plane wave state, which justifies the description of alpha
as a velocity operator.
Such a statement does raise issues concerning the description of Hilbert
space in terms of a basis consisting of the eigenstates of an
observable. Such a description is surely only physically meaningful for
an observable obeying the projection postulate. At this point the
octopus might try to join the ball and lead us in a merrie waltz, but
perhaps we may strike at his heart, since it is only required, in the
first instance, to define Hilbert space in terms of a basis of position
states for a single particle.
>> I hope she will continue to dance, however uncertain her
>> position.
>
>She dances here! She dances there!
>
>And sometimes she collapses hard upon her pretty little tooshie.
Good heavens, what do antipodean princesses wear to a ball?
http://www.net-a-porter.com/Shop/Designers/Tooshie
Oh, you mean your mode of transport
http://encyclopedia.farlex.com/tooshie
(good, the internet, innit?)
neur...@yahoo.com.au
> ... I was really asking, how many more dancing before
> accepting that theory says us?
Alas, this damsel sometimes finds it difficult
to understand your sentences (though of course such
understanding is not necessary in the bedroom, where
a sexy Spanish accent is more than enough).
My attempted translation of the above is "...how long will
people continue to dance before accepting what current
theory is telling us." (?) If my translation is accurate, I
suspect the answer is "until theory becomes sufficiently
complete and satisfactory in all aspects that everyone is
satisfied with it". Sadly, questions about quantum
localization (which cause the disagreements about velocity
operators) are still an unresolved area of research.
>> Apparently, all agree that V = [H,Q],
> Well Oh No still maintained this wrong recently ...
>
> (\blockquote
> There has been some confusion recently over the velocity
> operator, but as Igor has pointed out recently in s.p.r. the
> correct velocity operator is p/m (in any representation).
> )
>
> Igor has already recognized his mistake in s.p.r. but now he
> is doing others.
Oh Don Juan! You sent my sisters into uncontrollable giggling
with your last remark above. (I think I should send you a
booklet about English slang and idiom.) I didn't know that
Iggy-babes was so promiscuous!
>> Don Juan the Passionate and our avian hero Super-Hans both
>> seem to advocate the coordinate x, i.e., Q = x. (At least,
>> such is this damsel's impression.
> I don't advocate that. I have said that if you are working
> beyond the Zb scale the x is not the instantaneous x but an
> averaged one (without Zb). And the velocity then is that
> Strange calls the averaged velocity.
So (for the benefit of other readers), let me summarize a
few things before I continue...
In Strange's section 7.3 "Zitterbewegung", what he calls
"standard velocity" (7.35) and "average velocity" (7.41) are
both operators. One can relate them to "standard position"
and "average position" operators respectively via
commutation with the Hamiltonian. The "average position
operator" is essentially equivalent to the Newton-Wigner
position operator. Strange explains how these two things are
related via a Foldy-Wouthuysen transformation. So far, this
is all standard, well-known, and mathematically
uncontroversial.
BTW, I think this also shows that Strange's "average
position operator" is a distinctly different type of
mathematical entity to Super-Hans's "average velocity" which
seems to be more of a scalar expectation value. (Super-Hans
said this was not so in earlier correspondence with this
damsel, but later reports seem to contradict this. I'll
leave that little puzzle for later, though.)
I have no problem with the mathematical aspects of Strange's
presentation. But I start to have serious objections to the
*interpretation* when he says:
>> [quote: Strange, p208, 2nd paragraph]
>> ....Firstly there is the average velocity (7.41) and
>> secondly there is very rapid oscillatory motion which
>> ensures that if an instantaneous measurement of the velocity
>> of the electron could be made it would give c.
(This interpretation also appears in many other textbooks.)
My objection to the above lies in the words "...if an
instantaneous measurement of the velocity of the electron
could be made...". I think it actually _departs_ from
quantum mechanics, since the only contact between QM theory
and experiment lies in the expectation values of operators,
their variances, their eigenvalue spectra, (and maybe
higher-order moments). That is, if I prepare a system in
state "psi" and I wish to measure a property "O", all that
quantum theory can predict for me is the expectation
<psi|O|psi>, also <psi|O^2|psi> and higher moments, and all
possible expectations for other psi's, from the operator's
eigenvalue spectrum. In this context, "instantaneous
measurement of the velocity of the electron" is revealed as
a mental fiction, and not merely a technological limitation,
as I explain below...
Later (p209, last paragraph), Strange says:
>> To see the Zitterbewegung, we would have to perform a
>> measurement of the velocity of the electron (involving two
>> exceedingly accurate measurements of the electron position)
>> over a timescale of order ... 3x10^-13 seconds!
He then dismisses it by saying that this is not a feasible
experiment. (Again, many other textbooks say similar things.)
My objection to all such glib dismissals is that they ignore
the question of whether such a hypothetical measurement even
makes sense as a Gedanken experiment in quantum theory. The
first measurement of position inevitably changes the
electron's momentum, so what is the point of the second
measurement of position? We cannot use such an experiment to
measure the electron's original velocity (just as we cannot
use sequential Stern-Gerlach experiments to measure all the
components of an electron's original spin). All we can do at
such energies are scattering experiments.
This damsel begins to think that such _interpretations_ of
Zitterbewegung are not really physics, but more like what
teenagers sometimes do when they're alone in their
bedrooms. (Hmm, I should think of a more polite phrase for
that, shouldn't I? Maybe I'll call it "frolicing".)
> Also I have said that in the momentum representation
> [the position operator] is -i@/@P ....
For Klein-Gordon, the operator "x" (in position
representation) or "-i@/@P" (in momentum representation)
suffers from the disease that it is not self-adjoint, and
therefore cannot correspond to a physically observable
property in a relativistic quantum theory. I'm reasonably
sure that similar objections apply for spin-1/2, (but I
haven't done the calculation personally).
The simplest "cure" for this disease is to take instead the
self-adjoint part of the operator (which happens to coincide
with Q_nw in the Klein-Gordon case, if I remember
correctly).
>> This damsel is becoming more persuaded that "Q_nw" is
>> physically more desirable than "x", in which case V = [H,
>> Q_nw] must be chosen as velocity operator.
> Hum but if you work in the Newton Wigner representation for
> the position, you may also work in the Newton Wigner
> representation for the Hamiltonian
>
> V = [H_nw, Q_nw]
>
> and then the velocity is not c \alpha.
Exactly.
In any representation there are two distinct operators "Q"
and "Q_nw". In Dirac representation, Q=x (or id/dp in
momentum basis). The question becomes "which of these makes
more physical sense to be called a "position operator"?
(Currently, both have problems - which is why the subject
of Localization remains an active research topic.)
> I already remarked several times that the problem with Dirac
> Hamiltonian is on that it is linear in momentum.
>
> Any modern serious attempt I know to solve localization
> problems of relativistic physics start from using a
> nonlinear Hamiltonian.
Using canonical transformations that mix position and
momenta, one can greatly change the form of Hamiltonians,
even to the point where the original and final versions of
the Hamiltonian become unrecognizably different. Yet, both
describe the same underlying dynamics (because the
transformation was canonical). Therefore, one needs to start
from an even deeper place (such as the dynamical group, so
one can be sure whether one is/isn't dealing with the same
dynamics in disguise.)
> I also cited here a work [...] post), the RQD book, where
> the Newton Wigner operator is used.
This damsel is getting confused among all the various posts.
Which RQD book do you mean?
> I wrote next formal recipe
>
> c = v + v_Zb
>
> many many many many times.
I hope you can see from my earlier reply above that I have
studied Strange's stuff on Zitterbewegung carefully (and
quite a few other sources as well in the past).
> And also explained in what spatio-temporal scales of
> measurement you may ignore the Zb effect and use a
> conventional notion of velocity.
For me, the question is really whether the v and v_Zb (or
rather, the Hilbert space operators corresponding to them)
make sense as self-adjoint operators *separately*, and
whether such separation is representation-independent (ie
whether they stay separate under canonical transformations).
If the Zitterbewegung part of the velocity operator is not
well-defined as a self-adjoint operator, distinct from other
velocity operator(s), in a representation-independent way,
then it cannot correspond to an observable feature of
physics in RQM. (This problem is _distinct_ from the issue
of not having the technology to perform experiments at the
required energy scale.)
> Well in another discussion (I think was with Peter) I
> explained that the right definition of velocity operator is
> v = [H,x] and this also works for classical physics
>
> v = {H,x}
Yes, but to be clear we should first write v = {H,q}, where
"q" denotes some physically relevant and sensible position
operator, and then the issue is to "define q". The simple
and standard definition "q=x" from non-relativistic QM
displays severe problems when we try to use it in
relativistic QM.
> As can be observed most of above was said. But some people
> continue dancing about the subject, trying to obtain some
> magical way to evade one or several of those questions.
> I ask again how many more dancing?
It's not "dancing to evade questions". It's really
"expressing concern" that *none* of the proposed definitions
of "position operator" in relativistic quantum theory
(and/or QFT) are entirely satisfactory.
----
LOL from the Masked Quantum Damsel!
"More reading, less frolicing."
Oh No
>It's not "dancing to evade questions". It's really "expressing concern"
>that *none* of the proposed definitions of "position operator" in
>relativistic quantum theory (and/or QFT) are entirely satisfactory.
>
and tries to quantise it. Starting with a (non-relativistic) Fock space
we can simply write down
Sum |x>x<x|
When one takes this into a relativistic qft, one has to recognise that
all physical operators (including those required for measurement of
position) must be a consequence of interactions. So, we can only use
field operators in the corresponding expression for a position operator
Sum psi_bar x psi
This works if one starts with a pure particle (or antiparticle) state,
but, of course, it means that, if sufficient energy is available, which
it must be for a sufficiently precise measurement, then pair production
will mess up the measurement. But this is surely just a fact, not
actually a problem.
Myrddin, son of Morfryn
neur...@yahoo.com.au
> So, [in relativistic qft with interactions] we can only use
> field operators in the corresponding expression for a
> position operator
>
> Sum psi_bar x psi
This looks like the standard lifting of a single-particle
operator to Fock space (modulo some normal-ordering,
presumably). Thus, it's only acceptable if the
single-particle operator is itself acceptable. But ordinary
'x' is not a satisfactory position operator even in simple
free relativistic QM. (It's not self-adjoint.)
In any case, you've only included Dirac fields there, right?
(Ie psi). So that operator is not defined on the entire
interacting multiparticle Hilbert space. To make it so,
you'd need to include the other fields, and all the possible
bound states, in the decomposition of unity. How 'bout we
include just the photon field as well? Then there's serious
trouble just defining a position operator for the single-photon
sector, (nevermind lifting it to a multiphoton Fock space).
> if sufficient energy is available, which it must be for a
> sufficiently precise measurement, then pair production will
> mess up the measurement. But this is surely just a fact, not
> actually a problem.
In that case, if we are to have a satisfactory position
operator applicable in an _interacting_ theory, such effects
should be incorporated into our foundational Lie algebra.
Eg, a physically correct position operator should not commute
with the number operator in general.
> Myrddin, son of Morfryn
Morfryn?? Isn't that what they gave me in the Houses
of Healing when I recently injured my tooshie in a riding
accident? Was it named eponymously after your sire?
----
LOL from the Masked Quantum Damsel (now confined to her bed).
Oh No
>Sorcerer Myrddin wrote:
>
>> So, [in relativistic qft with interactions] we can only use
>> field operators in the corresponding expression for a
>> position operator
>>
>> Sum psi_bar x psi
>
>This looks like the standard lifting of a single-particle
>operator to Fock space (modulo some normal-ordering,
>presumably).
Yes, of course. Remiss of me to omit the colons.
> Thus, it's only acceptable if the
>single-particle operator is itself acceptable. But ordinary
>'x' is not a satisfactory position operator even in simple
>free relativistic QM. (It's not self-adjoint.)
You confuse me. It looks self adjoint to me. Is there some subtle
distinction in formulation I have missed? Perhaps I am too brief. If I
write a bit more fully
x = Integral_d^3 : psi_bar(x) x psi (x) :
x on lhs is the proposed position operator, while on rhs it is just the
parameter for the integration (such horrible overloads of notation seem
unavoidable in qft, but they should not upset us too much).
>In any case, you've only included Dirac fields there, right?
>(Ie psi).
yes.
> So that operator is not defined on the entire
>interacting multiparticle Hilbert space. To make it so,
>you'd need to include the other fields, and all the possible
>bound states, in the decomposition of unity. How 'bout we
>include just the photon field as well?
The operator acts as an identity on states of other particles. I am
happy if we include just electrons (and positrons) and photons. I am not
concerned about bound states, as the operator acts at given time (time
being a parameter, as in non-relativistic qm). I have a model in which
particles act as free particles between interactions. Interactions are
given by the usual minimal interaction operator in qed. This means I do
not need interacting fields. I guess this is the rub. If one wants to
set the thing up in a Fock space formulation, one must show that the
usual stack of issues are dealt with. I believe I have done that, but,
this would extend the discussion hugely. Can we just pretend for now
that it is done, or can be done?
>Then there's serious
>trouble just defining a position operator for the single-photon
>sector, (nevermind lifting it to a multiphoton Fock space).
Since I cannot measure the position of a photon (but only the position
of an electron where it was annihilated) I do not see a need to define a
position operator for a photon. I am happy if the only operator leading
to classical quantities is A.
> > if sufficient energy is available, which it must be for a
> > sufficiently precise measurement, then pair production will
> > mess up the measurement. But this is surely just a fact, not
> > actually a problem.
>
>In that case, if we are to have a satisfactory position
>operator applicable in an _interacting_ theory, such effects
>should be incorporated into our foundational Lie algebra.
>Eg, a physically correct position operator should not commute
>with the number operator in general.
I do not know if there is a way to construct a number operator from the
fields which counts just electrons (as distinct from counting lepton
number, which is conserved). Nor do I think there should be a way,
because there must always be an indeterminate number of virtual
electron-positron pairs.
>> Myrddin, son of Morfryn
>
>Morfryn?? Isn't that what they gave me in the Houses
>of Healing when I recently injured my tooshie in a riding
>accident? Was it named eponymously after your sire?
I am not familiar with the preparation.
>LOL from the Masked Quantum Damsel (now confined to her bed).
Sad, and I have no invitation to visit her bower.
Casting his cloak about himself, Myrddin vanishes into the mists of time
Cl.Massé
>>> ....Firstly there is the average velocity (7.41) and
>>> secondly there is very rapid oscillatory motion which
>>> ensures that if an instantaneous measurement of the velocity
>>> of the electron could be made it would give c.
<neur...@yahoo.com.au> a écrit dans le message de
news:77d60963-3cbb-45cb...@j22g2000hsf.googlegroups.com...
> (This interpretation also appears in many other textbooks.)
>
> My objection to the above lies in the words "...if an
> instantaneous measurement of the velocity of the electron
> could be made...". I think it actually _departs_ from
> quantum mechanics, since the only contact between QM theory
> and experiment lies in the expectation values of operators,
> their variances, their eigenvalue spectra, (and maybe
> higher-order moments). That is, if I prepare a system in
> state "psi" and I wish to measure a property "O", all that
> quantum theory can predict for me is the expectation
> <psi|O|psi>, also <psi|O^2|psi> and higher moments, and all
> possible expectations for other psi's, from the operator's
> eigenvalue spectrum. In this context, "instantaneous
> measurement of the velocity of the electron" is revealed as
> a mental fiction, and not merely a technological limitation,
> as I explain below...
In QM, there is a difference between prediction and actual measurement. The
theory can't predict the outcome of each single measurement, only the
statistical parameters of many measurements, as long as the possible values
of single measurements. The Dirac equation predicts that if a single
instantaneous measurement of the velocity is performed, it's magnitude is c.
That is, the set of possible values, or the spectrum, is the sphere of
radius c.
> My objection to all such glib dismissals is that they ignore
> the question of whether such a hypothetical measurement even
> makes sense as a Gedanken experiment in quantum theory.
It's the same as for the measurement of position, which also require
infinite energy. Yet, position continue to be honorable for everybody.
The problem can be solved by using operators describing approximate
location,
that is, the associated projection occurs onto a finite space region. If
that is done with the velocity operator, the Zitterbewegung isn't
observable.
> The
> first measurement of position inevitably changes the
> electron's momentum, so what is the point of the second
> measurement of position?
It gives the velocity *between* both position measurements, not before the
first one, and that's the common understanding of the velocity concept.
> We cannot use such an experiment to
> measure the electron's original velocity.
And we don't care too, since an electron remains an electron.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
neur...@yahoo.com.au
on a little longer...
>> ... ordinary 'x' is not a satisfactory position
>> operator even in simple free relativistic QM.
>> (It's not self-adjoint.)
Sir Charles the Valiant wrote:
> You confuse me. It looks self adjoint to me. Is there some
> subtle distinction in formulation I have missed?
My apologies. I neglected to include the words "positive
energy" in the incantation. (The NW position operator is
derived under this assumption.) With a +ve energy projector
in place, (and viewed in momentum basis) id/dp is not
self-adjoint. (Is that enough to acknowledge the difference
in formulation? Since your discrete formulation is not the
same as axiomatic QFT, lots of things are different, yes.
Eg, Reeh-Schlieder relies on analytic continuation, among
other things.)
But now, with the mention of "positive energy", an old dance
might begin again about the role of negative energy, etc,
etc. And then we'd have to talk about whether an "electron
remains an electron" or has an ongoing secret affair with
a positron alter-ego. (Yawn.) I think I must sit that one
out.
> Sad, and I have no invitation to visit her bower.
She is falling asleep anyway, and with an injured tooshie,
she is in no condition for anything else.
----
(....dainty snoring....)
[The MQD has a dream about a dance request from the
mysterious Cl.Masse from the mythical land of Liberty and
Profitability. Princess Spank-A-Lot is very much in favor of
Profitability from bondage - which is perhaps incompatible
with Liberty. But every time she tries to rise and accept
the dance, she finds she does not know the correct steps and
the whole world seems strange. Perhaps the world will make
more sense when she wakes. All the boring old images of
swords-and-cups, rockets-and-spacestations,
cars-and-garages, prancing stallions, etc, etc, are still
swirling around. The subconscious mind REALLY needs to find
some new material...]
Oh No
>The Masked Quantum Damsel grows weary, but struggles
>on a little longer...
>
>>> ... ordinary 'x' is not a satisfactory position
>>> operator even in simple free relativistic QM.
>>> (It's not self-adjoint.)
>
>Sir Charles the Valiant wrote:
>
>> You confuse me. It looks self adjoint to me. Is there some
>> subtle distinction in formulation I have missed?
>
>My apologies. I neglected to include the words "positive
>energy" in the incantation. (The NW position operator is
>derived under this assumption.) With a +ve energy projector
>in place, (and viewed in momentum basis) id/dp is not
>self-adjoint.
No wonder you say he is not a full man.
>(Is that enough to acknowledge the difference
>in formulation?
Indeed, severe differences.
>Since your discrete formulation is not the
>same as axiomatic QFT, lots of things are different, yes.
>Eg, Reeh-Schlieder relies on analytic continuation, among
>other things.)
>
>But now, with the mention of "positive energy", an old dance
>might begin again about the role of negative energy, etc,
>etc. And then we'd have to talk about whether an "electron
>remains an electron" or has an ongoing secret affair with
>a positron alter-ego. (Yawn.) I think I must sit that one
>out.
I am very convinced by the interpretation of Feynman and Stueckelberg
that it is an electron going backwards in time.
Cl.Massé
> My apologies. I neglected to include the words "positive
> energy" in the incantation. (The NW position operator is
> derived under this assumption.) With a +ve energy projector
> in place, (and viewed in momentum basis) id/dp is not
> self-adjoint. (Is that enough to acknowledge the difference
> in formulation? Since your discrete formulation is not the
> same as axiomatic QFT, lots of things are different, yes.
> Eg, Reeh-Schlieder relies on analytic continuation, among
> other things.)
I won't leave you waiting longer for the answer. It is in textbooks (e.g
Davydov _Quantum Mechanics_ §61), but not necessary in the most impressive
parts.
x is self-adjoint, of course, but it isn't even. Some physicists don't find
it palatable that torchons et serviettes be in the same basket, so they
require that the operator change the positive energy solutions to positive
energy solutions, and the same for the negative one ones. Taking the even
velocity operator [alpha] instead, the craved for velocity operator +/- p/E
sees light. The even position operator [x] has, in addition to x, a fast
oscillating term with frequency 2m. Yet the eigenfunctions of [x] are no
longer delta functions, but have a spatial extension of the order of the
Compton wavelength 1/m.
All that dancing, to realize that the Zitterbewegung, if attempted to get
rid of, gives a spatial extension to the electron, inside which the
Zitterbewegung is still raging, and besides sticks to the ruler, as if the
road were moved under the electron in order to always have the same reading.
I see a big drawback to this approach: the white line is blurred, and the
cops have enough difficulties already measuring the velocity density.
Now, let's be serious. The standard model says that the leptons can't have
a mass. So everything quiets down again, since the frequency becomes zero.
The speed of the leptons is c, and nobody screams about that.
Nevertheless, it seems natural that every particles, photons included, be so
urbane as to have a same behavior in common. As there is a single
fundamental constant of speed dimension, it's good that everyone drives at
c, that would avoid many accidents. We don't know what is mass, but we see
that its effect, be it self-interaction or interaction with the Higgs or
other, is to continuously modify the velocity of the electron. I don't see
any reason for being crossed with that, to the contrary: a bound state of
two massless particles has a mass, and its local speed is always c, while
its average one isn't.
> But now, with the mention of "positive energy", an old dance might begin
> again about the role of negative energy, etc, etc. And then we'd have to
> talk about whether an "electron remains an electron" or has an ongoing
> secret affair with a positron alter-ego. (Yawn.) I think I must sit that
> one out.
The electron remains an electron, it is its state that changes. But we are
only interested by its state between the two position measurements, which
must be infinitely close in time.
Juan R.
> In Strange's section 7.3 "Zitterbewegung", what he calls "standard
> velocity" (7.35) and "average velocity" (7.41) are both operators. One
> can relate them to "standard position" and "average position" operators
> respectively via commutation with the Hamiltonian. The "average position
> operator" is essentially equivalent to the Newton-Wigner position
> operator. Strange explains how these two things are related via a
> Foldy-Wouthuysen transformation. So far, this is all standard,
> well-known, and mathematically uncontroversial.
In my opinion it was but...
> BTW, I think this also shows that Strange's "average position operator"
> is a distinctly different type of mathematical entity to Super-Hans's
> "average velocity" which seems to be more of a scalar expectation value.
What?
> I have no problem with the mathematical aspects of Strange's
> presentation. But I start to have serious objections to the
> *interpretation* when he says:
Notice that interpretation varies with the theoretical framework. It is
not the same in RQM than in QFT for instance.
>>> [quote: Strange, p208, 2nd paragraph] ....Firstly there is the average
>>> velocity (7.41) and secondly there is very rapid oscillatory motion
>>> which ensures that if an instantaneous measurement of the velocity of
>>> the electron could be made it would give c.
>
> (This interpretation also appears in many other textbooks.)
Yes.
> My objection to the above lies in the words "...if an instantaneous
> measurement of the velocity of the electron could be made...". I think
> it actually _departs_ from quantum mechanics, since the only contact
> between QM theory and experiment lies in the expectation values of
> operators, their variances, their eigenvalue spectra, (and maybe
> higher-order moments). That is, if I prepare a system in state "psi" and
> I wish to measure a property "O", all that quantum theory can predict
> for me is the expectation <psi|O|psi>, also <psi|O^2|psi> and higher
> moments, and all possible expectations for other psi's, from the
> operator's eigenvalue spectrum. In this context, "instantaneous
> measurement of the velocity of the electron" is revealed as a mental
> fiction, and not merely a technological limitation, as I explain
> below...
Before continuing reading I would remark that both operators Strange use
are related to measurement via standard QM expressions
v_inst = <psi| v_inst |psi>
v_aver = <psi| v_aver |psi>
with
v_inst = v_aver + v_Zb
> Later (p209, last paragraph), Strange says:
>
>>> To see the Zitterbewegung, we would have to perform a measurement of
>>> the velocity of the electron (involving two exceedingly accurate
>>> measurements of the electron position) over a timescale of order ...
>>> 3x10^-13 seconds!
>
> He then dismisses it by saying that this is not a feasible experiment.
> (Again, many other textbooks say similar things.)
Yeah, because current technology is not that developed. Strange gives
bounds.
> My objection to all such glib dismissals is that they ignore the
> question of whether such a hypothetical measurement even makes sense as
> a Gedanken experiment in quantum theory. The first measurement of
> position inevitably changes the electron's momentum, so what is the
> point of the second measurement of position? We cannot use such an
> experiment to measure the electron's original velocity (just as we
> cannot use sequential Stern-Gerlach experiments to measure all the
> components of an electron's original spin). All we can do at such
> energies are scattering experiments.
If your argument was valid for dt = 10^-13 seconds then it would be also
above that scale.
Are you then claiming that we cannot measure velocities?
> This damsel begins to think that such _interpretations_ of
> Zitterbewegung are not really physics, but more like what teenagers
> sometimes do when they're alone in their bedrooms. (Hmm, I should think
> of a more polite phrase for that, shouldn't I? Maybe I'll call it
> "frolicing".)
Call it mental masturbation if you prefer. The point is that
i)
|v| = +-c is a prediction of Dirac theory.
ii)
No known experiment with velocities, concretely no experiment regarding
special relativity (Bertozzi, etc.), contradicts above *quantum* result.
iii)
That result is needed in order to interpret experimental results about
scattering, bound energies, etc. That is, the prediction in i) has been
*indirectly* verified.
>> Also I have said that in the momentum representation [the position
>> operator] is -i@/@P ....
>
> For Klein-Gordon, the operator "x" (in position representation) or
> "-i@/@P" (in momentum representation) suffers from the disease that it
> is not self-adjoint, and therefore cannot correspond to a physically
> observable property in a relativistic quantum theory. I'm reasonably
> sure that similar objections apply for spin-1/2, (but I haven't done the
> calculation personally).
This is the old issue of relativistic localization.
The position *vector* is not observable and, as consequence, the velocity
*vector* cannot be. We find here a similar decomposition
x_inst = x_aver + x_ZB
This is why position was downgraded to parameter in QFT. QFT avoided
localization issues, but the price was too high, one loses dynamical
information about particles. Luckily this is not a real problem in
particle physics where only scattering is studied, but is a real problem
in many body theory for finite times.
Of course, we already know examples of non-observable vectors since non-
relativistic quantum mechanics epoch: e.g. angular momentum. Therefore I
don't understand some of the claims that some people has been doing here.
> The simplest "cure" for this disease is to take instead the self-adjoint
> part of the operator (which happens to coincide with Q_nw in the
> Klein-Gordon case, if I remember correctly).
See above. The average position operator don't consider Zb effects. It is
a position operator that is only defined for scales of space and time
above the Zb.
>>> This damsel is becoming more persuaded that "Q_nw" is physically more
>>> desirable than "x", in which case V = [H, Q_nw] must be chosen as
>>> velocity operator.
>
>> Hum but if you work in the Newton Wigner representation for the
>> position, you may also work in the Newton Wigner representation for the
>> Hamiltonian
>>
>> V = [H_nw, Q_nw]
>>
>> and then the velocity is not c \alpha.
>
> Exactly.
And this is a well-known result that has next consequences.
i)
The nw quantities are not exact in the framework of Dirac theory. They are
only valid as smooth quantities for certain scales of space and time. E.g.
dt > 3x10-13 seconds as pointed.
ii)
The Hamiltonian here is nonlinear in momentum. This is the reason which
the resulting velocity is a function of momentum with magnitude less than
c.
iii)
This is not the velocity one may introduce in the potentials, QFT currents
etc. E.g. if the classical potential is the well-known
U = ee / r {1 - v v' / c^2}
then the Coulomb-Gaunt quantum potential is
U = ee / r {1 - alpha alpha'}
And this is the potential used by atomic and molecular physicists and
quantum chemists in Dirac theory.
Would I also cite here Breit, Bette Salpeter...?
> In any representation there are two distinct operators "Q" and "Q_nw".
> In Dirac representation, Q=x (or id/dp in momentum basis). The question
> becomes "which of these makes more physical sense to be called a
> "position operator"? (Currently, both have problems - which is why the
> subject of Localization remains an active research topic.)
Since that the Newton-Wigner operator follows from elimination of Zb
effects. It cannot be fundamental. Indeed it is acknowledged that Newton
Wigner is valid in a smooth sense over certain scales of space and time
but not accurately describe electron position.
>> I already remarked several times that the problem with Dirac
>> Hamiltonian is on that it is linear in momentum.
>>
>> Any modern serious attempt I know to solve localization problems of
>> relativistic physics start from using a nonlinear Hamiltonian.
>
> Using canonical transformations that mix position and momenta, one can
> greatly change the form of Hamiltonians, even to the point where the
> original and final versions of the Hamiltonian become unrecognizably
> different. Yet, both describe the same underlying dynamics (because the
> transformation was canonical).
Yes then one would be just changing the problems of place. I was really
alluding to *new* theories.
> Therefore, one needs to start from an even deeper place (such as the
> dynamical group, so one can be sure whether one is/isn't dealing with
> the same dynamics in disguise.)
The new theory may reproduce the old results and produce some new testable
result.
>> I also cited here a work [...] post), the RQD book, where the Newton
>> Wigner operator is used.
>
> This damsel is getting confused among all the various posts. Which RQD
> book do you mean?
See references in my post "Re: flavors of QED..."
>> Well in another discussion (I think was with Peter) I explained that
>> the right definition of velocity operator is v = [H,x] and this also
>> works for classical physics
>>
>> v = {H,x}
>
> Yes, but to be clear we should first write v = {H,q}, where "q" denotes
> some physically relevant and sensible position operator, and then the
> issue is to "define q". The simple and standard definition "q=x" from
> non-relativistic QM displays severe problems when we try to use it in
> relativistic QM.
Correct, the x in RQM is taking into account antiparticles, which are not
present in the classical theory (special relativity). In the classical
bracket the observable position arises from <Psi|v_aver|Psi> and the Zb
term 'vanishes'.
But this is not at all different from other quantum quantities spin, or
negative energies in Dirac theory that 'vanish' in the classical theory.
Moreover, the Hamiltonian in the classical v = {H,x} is not some classical
version of the Dirac Hamiltonian, you know. The Hamiltonian of special
relativity is the characteristic root one finds in textbooks.
The point was that v = [H,x] and v = {H,x} are the proper mechanical
definitions. I was perplexed seeing some people who think that the
definition of velocity is something freely adjustable to own personal
desires and each one chooses the definition more like.
>> As can be observed most of above was said. But some people continue
>> dancing about the subject, trying to obtain some magical way to evade
>> one or several of those questions. I ask again how many more dancing?
>
> It's not "dancing to evade questions". It's really "expressing concern"
> that *none* of the proposed definitions of "position operator" in
> relativistic quantum theory (and/or QFT) are entirely satisfactory.
Some examples of recent dancing (next is all wrong):
v = [H,x] is wrong definition of quantum velocity. Velocity is v = p/m
H = (c alpha p) is not linear in p therefore the velocity is not c
v = c alpha is observable because [alpha_x, alpha_y] /= 0 is irrelevant.
Int Psi* c alpha Psi d^3x = v < c
|v| < c and v^2 = c^2
Etc.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Jay R. Yablon
"Cl.Massé" <daniell...@gmail.com> wrote in message
news:48a57ff7$0$17376$426a...@news.free.fr...
> <neur...@yahoo.com.au> a écrit dans le message de
> news:c1e02e35-a4c9-4027...@z72g2000hsb.googlegroups.com...
>
> Now, let's be serious. The standard model says that the leptons can't
> have
> a mass. So everything quiets down again, since the frequency becomes
> zero.
> The speed of the leptons is c, and nobody screams about that.
>
> Nevertheless, it seems natural that every particles, photons included,
> be so
> urbane as to have a same behavior in common. As there is a single
> fundamental constant of speed dimension, it's good that everyone
> drives at
> c, that would avoid many accidents. We don't know what is mass, but
> we see
> that its effect, be it self-interaction or interaction with the Higgs
> or
> other, is to continuously modify the velocity of the electron. I
> don't see
> any reason for being crossed with that, to the contrary: a bound state
> of
> two massless particles has a mass, and its local speed is always c,
> while
> its average one isn't.
>
> --
> ~~~~ clmasse on free F-country
> Liberty, Equality, Profitability.
>
whether all the Zb stuff contains hidden within, some symmetry breaking
mechanism for fermion masses.
Let's take this from the top. In electroweak theory, we start with
massless vector bosons, we break symmetry and most of the bosons then
obtain mass. For fermions, we really know squat (nothing) about the
mass. The only thing we do know how to consider are massless fermions
which do travel at c and nobody would raise an eyebrow about that. The
question is how does mass really get generated, and in that process,
what happens to the Zb? Because for massless fermions, which is all we
really do know how to talk about, a "c" velocity is to be fully expected
and all else would be a surprise.
Jay.
Juan R.
> "Cl.Massé" <daniell...@gmail.com> wrote in message
> news:48a57ff7$0$17376$426a...@news.free.fr...
>> <neur...@yahoo.com.au> a écrit dans le message de
>> news:c1e02e35-
a4c9-4027-909...@z72g2000hsb.googlegroups.com...
But, and this *is* the problem, the mass in Dirac theory and QED is *not*
zero.
The attempt to imagine that fermions traveling to c may be massless
follows from a *misapplication* of special relativity formulae
H = (\sqrt m^2c^4 + p^2c^2)
implies v = c when m = 0 because then H = pc. Then you repeat your
misguided claims.
But the Dirac Hamiltonian *is*
H = c \alpha p + \beta mc^2
Would I emphasize that m above is the mass of the particle and it is non-
zero?
Regards.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Jay R. Yablon
"Juan R. González-Álvarez" <juanR...@canonicalscience.com> wrote in
message news:pan.2008.08...@canonicalscience.com...
the
same thing in different words. The problem revolves around the mass,
and
since our understanding of the true nature of rest mass is limited, this
in
all likelihood limits our understanding of the zb as well.
Jay.
Cl.Massé
message de news:pan.2008.08...@canonicalscience.com...
> But the Dirac Hamiltonian *is*
>
> H = c \alpha p + \beta mc^2
>
> Would I emphasize that m above is the mass of the particle and it is non-
> zero?
In the SM, the mass term arises from the interaction with the Higgs field,
after the symmetry breaking. The initial mass is zero.
Chalky
> Thus spake neurop...@yahoo.com.au
> >But now, with the mention of "positive energy", an old dance
> >might begin again about the role of negative energy, etc,
> >etc. And then we'd have to talk about whether an "electron
> >remains an electron" or has an ongoing secret affair with
> >a positron alter-ego. (Yawn.) I think I must sit that one
> >out.
>
> I am very convinced by the interpretation of Feynman and Stueckelberg
> that it is an electron going backwards in time.
At the risk of boring the masked damsel, could you clarify what you
mean by "it". Are you saying that "it" is a negative energy electron,
or a positron, or both?
I am a little confused here because this has echoes of a valiant
ancient battle between Sir Charles, another knight, and the subversive
blackguard John (Liberty) Bell. IIRC Bell argued (amongst other
things) that negative energy travelling backwards in time would look
the same, in some respects, as if it was positive energy travelling
forwards in time .
Aside.....
On Aug 13, 9:59 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> Thus spake neurop...@yahoo.com.au
>
> >Morfryn?? Isn't that what they gave me in the Houses
> >of Healing when I recently injured my tooshie in a riding
> >accident? Was it named eponymously after your sire?
>
> I am not familiar with the preparation.
A clue is given in the example provided for the definition of
eponymously at http://www.thefreedictionary.com/eponymously:
"the eponymous heroine"
For those not well rounded in the dark arts of physicks, Heroyne is
the sister of Morfryn.
Not that this has much to do with anything, but I have even heard
whispered in hallowed circles that one Nobel Prize winning Magician of
Physic attributed his success, in part, to having imbibed a physik
which sounded indistinguishable from the distilled essence of olde
English currencies (to my naive ears).
SCNR
Oh No
>On Aug 14, 8:55 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake neurop...@yahoo.com.au
>
>> >But now, with the mention of "positive energy", an old dance
>> >might begin again about the role of negative energy, etc,
>> >etc. And then we'd have to talk about whether an "electron
>> >remains an electron" or has an ongoing secret affair with
>> >a positron alter-ego. (Yawn.) I think I must sit that one
>> >out.
>>
>> I am very convinced by the interpretation of Feynman and Stueckelberg
>> that it is an electron going backwards in time.
>
>At the risk of boring the masked damsel, could you clarify what you
>mean by "it". Are you saying that "it" is a negative energy electron,
>or a positron, or both?
In the Feynman and Stueckelberg a positron is a negative energy electron
going backwards in time
>I am a little confused here because this has echoes of a valiant
>ancient battle between Sir Charles, another knight, and the subversive
>blackguard John (Liberty) Bell. IIRC Bell argued (amongst other
>things) that negative energy travelling backwards in time would look
>the same, in some respects, as if it was positive energy travelling
>forwards in time .
That is the Feynman-Stuckelberg interpretation
Chalky
> In the Feynman and Stueckelberg a positron is a negative energy electron
> going backwards in time
>
> >I am a little confused here because this has echoes of a valiant
> >ancient battle between Sir Charles, another knight, and the subversive
> >blackguard John (Liberty) Bell. IIRC Bell argued (amongst other
> >things) that negative energy travelling backwards in time would look
> >the same, in some respects, as if it was positive energy travelling
> >forwards in time .
>
> That is the Feynman-Stuckelberg interpretation
Thanks. We do seem to have consistency here then. You don't have a
formal reference for the Feynman-Stuckelberg interpretation, do you?
Of course, this does seem to make a nonsense of the principle of
causality (as usually expressed in its temporally biased form).
Oh No
>On Sep 1, 7:58 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> In the Feynman and Stueckelberg a positron is a negative energy electron
>> going backwards in time
>>
>> >I am a little confused here because this has echoes of a valiant
>> >ancient battle between Sir Charles, another knight, and the subversive
>> >blackguard John (Liberty) Bell. IIRC Bell argued (amongst other
>> >things) that negative energy travelling backwards in time would look
>> >the same, in some respects, as if it was positive energy travelling
>> >forwards in time .
>>
>> That is the Feynman-Stuckelberg interpretation
>
>Thanks. We do seem to have consistency here then. You don't have a
>formal reference for the Feynman-Stuckelberg interpretation, do you?
Unfortunately not. I have only ever found rather casual references in
text books - interpretation being frowned upon according to modern
thinking (I disagree that it should be). What I have done is incorporate
the interpretation in my treatment
http://www.teleconnection.info/rqg/TheDiracEquation#Antiparticles
>
>Of course, this does seem to make a nonsense of the principle of
>causality (as usually expressed in its temporally biased form).
>
Yes, I think it does. My view is that the usual temporal bias is
something we impose from our macroscopic viewpoint, and that it is not
reflected in the underlying structures of matter.
Chalky
> >Thanks. We do seem to have consistency here then. You don't have a
> >formal reference for the Feynman-Stuckelberg interpretation, do you?
>
> Unfortunately not. I have only ever found rather casual references in
> text books
I also recall something related about rotation of axes in Feynmann
diagrams, I think, but have no references to that, either. Any luck
there?
> - interpretation being frowned upon according to modern
> thinking.
1) Why? (beyond the obvious fact that the only way to enforce a dogma
is to suppress its antithesis)
[I am aware, of course, of a variety of traditional arguments to
support that point of view but have invariably found them weak.]
2) AFAICT there has always been an undercurrent of disapproval of the
idea of time being bidirectional. I even recall my SR tutor "proving"
that nothing could exceed c by showing that if it did it would travel
backwards in time, "which is impossible, QED". I had got the
impression that matters were gradually getting more relaxed, but you
seem to be implying the opposite?
Oh No
>On Sep 1, 11:25 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> >Thanks. We do seem to have consistency here then. You don't have a
>> >formal reference for the Feynman-Stuckelberg interpretation, do you?
>>
>> Unfortunately not. I have only ever found rather casual references in
>> text books
>
>I also recall something related about rotation of axes in Feynmann
>diagrams, I think, but have no references to that, either. Any luck
>there?
>
fact that they can be read any way round you like, and there is no time
ordering of internal nodes. That is, they have a number of external
legs, any of which can be made to be the out states (usually drawn at
the top) and any of which can be made to be the in states.
>> - interpretation being frowned upon according to modern
>> thinking.
>
>1) Why? (beyond the obvious fact that the only way to enforce a dogma
>is to suppress its antithesis)
I'm afraid it does seem like that.
>[I am aware, of course, of a variety of traditional arguments to
>support that point of view but have invariably found them weak.]
>
>2) AFAICT there has always been an undercurrent of disapproval of the
>idea of time being bidirectional. I even recall my SR tutor "proving"
>that nothing could exceed c by showing that if it did it would travel
>backwards in time, "which is impossible, QED". I had got the
>impression that matters were gradually getting more relaxed, but you
>seem to be implying the opposite?
>
This is a little different. Exceeding c would mean going outside the
light cone. Backwards in time would mean directed into the past light
cone. The one is space-like, but the other is still time like. There
really are consistency implications if there are faster than light
effects, because a change in the ordering of interactions due to choice
of reference frame would lead to a change in the probability for the
outcome of a particular process. That is not legit.
Chalky
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
> >I also recall something related about rotation of axes in Feynmann
> >diagrams, I think, but have no references to that, either. Any luck
> >there?
>
> There are no axes in Feynman diagrams.
?? Both http://en.wikipedia.org/wiki/Feynman_diagram and
http://scienceworld.wolfram.com/physics/FeynmanDiagram.html
give time axes, in their examples, and the time axis of one is 90
degrees rotated relative to the time axis of the other.
> I think you are referring to the
> fact that they can be read any way round you like, and there is no time
> ordering of internal nodes. That is, they have a number of external
> legs, any of which can be made to be the out states (usually drawn at
> the top) and any of which can be made to be the in states.
Perhaps. This needs deeper thought on my part.
> >> - interpretation being frowned upon according to modern
> >> thinking.
>
> >1) Why? (beyond the obvious fact that the only way to enforce a dogma
> >is to suppress its antithesis)
>
> I'm afraid it does seem like that.
>
> >[I am aware, of course, of a variety of traditional arguments to
> >support that point of view but have invariably found them weak.]
>
> >2) AFAICT there has always been an undercurrent of disapproval of the
> >idea of time being bidirectional. I even recall my SR tutor "proving"
> >that nothing could exceed c by showing that if it did it would travel
> >backwards in time, "which is impossible, QED". I had got the
> >impression that matters were gradually getting more relaxed, but you
> >seem to be implying the opposite?
>
> This is a little different. Exceeding c would mean going outside the
> light cone. Backwards in time would mean directed into the past light
> cone. The one is space-like, but the other is still time like. There
> really are consistency implications if there are faster than light
> effects, because a change in the ordering of interactions due to choice
> of reference frame would lead to a change in the probability for the
> outcome of a particular process. That is not legit.
I think you have missed my point here too. The point was not in what
was being allegedly proven, but in the final step which was to pull a
separate dogma out of a hat as a statement of absolute fact, as if
that had proven something, or anything, logically. Your additional
comments are doubtless valid but merely go to confirm my immediate
impression that his was a sufficiently poor excuse for a rigorous
proof, that it was an insult to my intelligence.
Oh No
>On Sep 2, 7:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>> >I also recall something related about rotation of axes in Feynmann
>> >diagrams, I think, but have no references to that, either. Any luck
>> >there?
>>
>> There are no axes in Feynman diagrams.
>
>?? Both http://en.wikipedia.org/wiki/Feynman_diagram and
>http://scienceworld.wolfram.com/physics/FeynmanDiagram.html
>give time axes, in their examples, and the time axis of one is 90
>degrees rotated relative to the time axis of the other.
>
>> I think you are referring to the
>> fact that they can be read any way round you like, and there is no time
>> ordering of internal nodes. That is, they have a number of external
>> legs, any of which can be made to be the out states (usually drawn at
>> the top) and any of which can be made to be the in states.
>
>Perhaps. This needs deeper thought on my part.
It's a little naughty of Wiki and Wolfram to represent time as an axis,
and very naughty of Wolfram to draw in an (apparent) space axis. All we
really have is an in state and an out state, designated left and right,
or bottom and top respectively. An axis has more structure than the
diagram contains.
>>
>> This is a little different. Exceeding c would mean going outside the
>> light cone. Backwards in time would mean directed into the past light
>> cone. The one is space-like, but the other is still time like. There
>> really are consistency implications if there are faster than light
>> effects, because a change in the ordering of interactions due to choice
>> of reference frame would lead to a change in the probability for the
>> outcome of a particular process. That is not legit.
>
>I think you have missed my point here too. The point was not in what
>was being allegedly proven, but in the final step which was to pull a
>separate dogma out of a hat as a statement of absolute fact, as if
>that had proven something, or anything, logically. Your additional
>comments are doubtless valid but merely go to confirm my immediate
>impression that his was a sufficiently poor excuse for a rigorous
>proof, that it was an insult to my intelligence.
>
I just wanted to be sure that you didn't think the result might be wrong
simply because you had been given a weak argument. There are theories
involving tachyons, but they invoke subtleties, and still do not allow
transmission of information faster than light. I am not convinced such
theories are meaningful, but I don't want to start getting involved in
proving that!
In fact qft does give an amplitude for a particle to be created and then
annihilated at some point outside the light cone. However, the process
is balanced by the creation of the antiparticle at the second point and
its annihilation at the first, with the net result of a zero amplitude.
Such weirdness is unavoidable, and usually dealt with by trying not to
think about it. Imv it has to do with spacetime as an emergent, rather
than a fundamental, quantity.
Chalky
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>
>
>
> >On Sep 2, 7:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
> >> Thus spake Chalky <chalkys...@bleachboys.co.uk>
> >> >I also recall something related about rotation of axes in Feynmann
> >> >diagrams, I think, but have no references to that, either. Any luck
> >> >there?
>
> >> There are no axes in Feynman diagrams.
>
> >?? Both http://en.wikipedia.org/wiki/Feynman_diagramand
> >http://scienceworld.wolfram.com/physics/FeynmanDiagram.html
> >give time axes, in their examples, and the time axis of one is 90
> >degrees rotated relative to the time axis of the other.
>
> >> I think you are referring to the
> >> fact that they can be read any way round you like, and there is no time
> >> ordering of internal nodes. That is, they have a number of external
> >> legs, any of which can be made to be the out states (usually drawn at
> >> the top) and any of which can be made to be the in states.
>
> >Perhaps. This needs deeper thought on my part.
>
> It's a little naughty of Wiki and Wolfram to represent time as an axis,
> and very naughty of Wolfram to draw in an (apparent) space axis. All we
> really have is an in state and an out state, designated left and right,
> or bottom and top respectively. An axis has more structure than the
> diagram contains.
That is certainly true if you assume the axes are Cartesian or
otherwise rigid as in SR. I am not so sure it is true if you assume
the axes are Gaussian (ie clocks and measuring rods are allowed to
"flop about in any way whatsoever" [Authorised translation of
Einstein's words]).
I certainly find it helpful to see a visual indication of what is
supposed to be going on, even if I know that it could potentially be
looked at another way.
> >> This is a little different. Exceeding c would mean going outside the
> >> light cone. Backwards in time would mean directed into the past light
> >> cone. The one is space-like, but the other is still time like. There
> >> really are consistency implications if there are faster than light
> >> effects, because a change in the ordering of interactions due to choice
> >> of reference frame would lead to a change in the probability for the
> >> outcome of a particular process. That is not legit.
>
> >I think you have missed my point here too. The point was not in what
> >was being allegedly proven, but in the final step which was to pull a
> >separate dogma out of a hat as a statement of absolute fact, as if
> >that had proven something, or anything, logically. Your additional
> >comments are doubtless valid but merely go to confirm my immediate
> >impression that his was a sufficiently poor excuse for a rigorous
> >proof, that it was an insult to my intelligence.
>
> I just wanted to be sure that you didn't think the result might be wrong
> simply because you had been given a weak argument.
No chance of that. I might be a bit weak in pure maths, but I
certainly am not weak in pure logic.
You may recall a query some time ago about what Einstein actually said
about the constraint c, I think at spr. Turns out it was completely
ambiguous in the sense that there is no distinction between speed (the
scalar) and velocity (the vector) in Einstein's tongue (German).
Consequently Einstein's stated constraint admits the possibility of
cause/effect light cone inversion, since negative del time and
negative del distance still gives the same positive velocity as does
positive del time and positive del distance.
> There are theories
> involving tachyons, but they invoke subtleties, and still do not allow
> transmission of information faster than light. I am not convinced such
> theories are meaningful, but I don't want to start getting involved in
> proving that!
Neither do I