On the velocity operator and zitterbewegung.
Oh No
as Igor has pointed out recently in s.p.r. the correct velocity operator
is p/m (in any representation). The use of alpha introduces spin indices
which are not properly part of a measurement of velocity. The state on
measurement of velocity is not an eigenstate of alpha, but a
superposition of spin eigenstates. These eigenstates have velocity +-c,
but the superposition does not.
This superposition is not directly related to zitterbewegung (I am not
sure if anyone has drawn the inference that it is). Zitterbewegung can
be understood in quantum field theory, where it arises from the
commutator of the field operators, leading to the appearance of
"virtual" particle/antiparticle pairs in the propagator. Such terms drop
out given a state of an observable particle or antiparticle.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Hans de Vries
> There has been some confusion recently over the velocity operator,
If only people would just do the calculations....
The velocity operator [H,x] applied on a Dirac plane wave
simply gives the speed v.
The acceleration operator [H,v] applied on a free Dirac
plane wave simply gives zero.
http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
see sections 14.8 and 14.9
Regards, Hans
http://www.physics-quest.org
Oh No
enlightening result, and one which made sense out of a much
misunderstood topic.
Juan R.
> On Aug 6, 12:24รย pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> There has been some confusion recently over the velocity operator,
>
> If only people would just do the calculations....
>
> The velocity operator [H,x] applied on a Dirac plane wave simply gives
> the speed v.
For instance, the eigenvalues of the velocity operator (14.25) are well-
known to be c.
One can also compute <v^2> using an analogous to your (14.26) and the
result is
<v^2> = |v|^2 = c^2
You pretend since several threads ago
|v| < c
and
|v|^2 = c^2
But the well-known results are
|v| = +-c
and
|v|^2 = c^2
> The acceleration operator [H,v] applied on a free Dirac plane wave
> simply gives zero.
See Feynman monograph I pointed to know in the past.
He obtains the force for Dirac electron in an external field and then
explain why is not analogous to the classical electrodynamics expression.
> http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
>
> see sections 14.8 and 14.9
>
> Regards, Hans
> http://www.physics-quest.org
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
======================================= MODERATOR'S COMMENT:
I have allowed this post, although I think Hans has already shown what is wrong with it. CF
Juan R.
The Dirac theory of electrons predicts particles to travel at c. Is this a
right prediction? Is this a wrong prediction? I will not discuss this
point here, because by obvious motives...
However, I find really interesting all kind of fantastic dancing that
people is trying to avoid that prediction.
A short summary (from memory):
1)
One pretends that v = [H,x] is a wrong operator and that right operator is
v = p/m
2)
Other claim that v = [H,x] is the right operator and that this gives v for
free particles.
3)
I also found one who pretended linear Hamiltonians to be really non-linear
and thus giving a speed different from c.
4)
Another think that v = [H,x] follows from assuming certain QM picture and
that for other QM picture the result would be (v = 0).
5)
The Newton Wigner operator was claimed to be wrong and then a long mistake
corrected, but the author who was doing those strong claims not even
understood what the operator was measuring.
6)
Now I am reading all kind of fantastic stuff about how we get gravitation
thanks to alpha with the bonus both gravitation and quantum theory both
get united. More nonsense...
7)
...
I also explained that Dirac result is easily understood. For linear
Hamiltonians the velocity cannot be a function of momenta. It may be thus
a constant and the natural constant in a relativistic theory is c.
This is also true for photon Hamiltonians:
H = pc implies |v| = c
I am curious, how many more funny attempts and revolutionary claims I will
be reading before people starts to accept that Dirac theory predicts |v|
to be +-c?
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> 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).
Sure there has been some confusion... e.g. your velocity operator of above
v = p/m
continues being plain wrong.
The velocity operator is (in obvious units)
v = [H,x]
(...)
> Zitterbewegung can
> be understood in quantum field theory, where it arises from the
> commutator of the field operators, leading to the appearance of
> "virtual" particle/antiparticle pairs in the propagator.
In the abstract of this recent paper it is said the contrary
http://prola.aps.org/abstract/PRL/v93/i4/e043004
(\blockquote
We also find that quantum field theory prohibits the occurrence of
Zitterbewegung for an electron.
)
But I have not got the paper and have not studied it still. My current
understanding is that QFT *also* gives Zbw effect.
My current understanding is that the Zbw effect in QFT has a similar
rationale than in RQM: existence of antiparticles.
> Such terms drop
> out given a state of an observable particle or antiparticle.
Sure but the velocity operator for those observable particles
antiparticles continues being
v = [H,x]
and not your wrong (p/m).
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Oh No
>
>> Zitterbewegung can
>> be understood in quantum field theory, where it arises from the
>> commutator of the field operators, leading to the appearance of
>> "virtual" particle/antiparticle pairs in the propagator.
>
>In the abstract of this recent paper it is said the contrary
>
>http://prola.aps.org/abstract/PRL/v93/i4/e043004
>
>(\blockquote
> We also find that quantum field theory prohibits the occurrence of
> Zitterbewegung for an electron.
>)
>
>But I have not got the paper and have not studied it still. My current
>understanding is that QFT *also* gives Zbw effect.
>
>My current understanding is that the Zbw effect in QFT has a similar
>rationale than in RQM: existence of antiparticles.
Indeed.
>
>> Such terms drop
>> out given a state of an observable particle or antiparticle.
>
>Sure but the velocity operator for those observable particles
>antiparticles continues being
>
>v = [H,x]
>
>and not your wrong (p/m).
This is already answered. It is not directly related to Zitterbewegung.
Peter
> Oh No wrote on Wed, 06 Aug 2008 04:24:21 -0600:
>
> > 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).
>
> Sure there has been some confusion... e.g. your velocity operator of above
>
> v = p/m
>
> continues being plain wrong.
>
> The velocity operator is (in obvious units)
>
> v = [H,x]
>
> (...)
>
> > Zitterbewegung can
> > be understood in quantum field theory, where it arises from the
> > commutator of the field operators, leading to the appearance of
> > "virtual" particle/antiparticle pairs in the propagator.
>
> In the abstract of this recent paper it is said the contrary
>
> http://prola.aps.org/abstract/PRL/v93/i4/e043004
>
> (\blockquote
> We also find that quantum field theory prohibits the occurrence of
> Zitterbewegung for an electron.
> )
>
> But I have not got the paper and have not studied it still.
I guess I can email it to you in case of interest
P.
PS: I agree that v=p/m is a special case only (body subject to velocity-
*in*dependent potential)
Oh No
>
>The Dirac theory of electrons predicts particles to travel at c. Is this a
>right prediction? Is this a wrong prediction? I will not discuss this
>point here, because by obvious motives...
It seems you do, but it is clearly a wrong prediction. The issue, if
there still is one, is what is wrong with it.
>
>However, I find really interesting all kind of fantastic dancing that
>people is trying to avoid that prediction.
>
>A short summary (from memory):
>
>1)
>One pretends that v = [H,x] is a wrong operator and that right operator is
>v = p/m
There is no pretence. Clearly v = p/m is correct as it gives the correct
results for measurement of velocity. As I pointed out, [H,x] has a spin
index, which is not a part of measurement of velocity.
>2)
>Other claim that v = [H,x] is the right operator and that this gives v for
>free particles.
Study the calculation carefully. You are free to find an error in it, if
there is one. You have not done this. You cannot refute it simply by
claiming that the result is wrong.
>3)
>I also found one who pretended linear Hamiltonians to be really non-linear
>and thus giving a speed different from c.
That is silly, so let us ignore it.
>
>4)
>Another think that v = [H,x] follows from assuming certain QM picture and
>that for other QM picture the result would be (v = 0).
Let us ignore that also.
>
>5)
>The Newton Wigner operator was claimed to be wrong and then a long mistake
>corrected, but the author who was doing those strong claims not even
>understood what the operator was measuring.
I am not sure what this claim is, but I observe that operators do not
measure. Apparatus measures.
>I am curious, how many more funny attempts and revolutionary claims I will
>be reading before people starts to accept that Dirac theory predicts |v|
>to be +-c?
There is no dispute that the eigenstates of v = [H,x] give |v| = +-c.
The point is that this clearly does not correspond to a measurement of
velocity. Hans has shown, rather neatly imv, that the reason is that
this v gives chiral eigenstates. But measurement of velocity should not
create such an eigenstate. The superposition of states with velocity +-c
gives the right group velocity, resolving the apparent paradox.
Cl.Massé
news:N+mJGoEM...@charlesfrancis.wanadoo.co.uk...
> 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).
The velocity is given by the Hamilton equation q' = @H/@p and nothing else.
It is equal to p/m only for H = p^2/2m + f(q). With the Dirac Hamiltonian,
@H/@p is precisely equal to alpha. Igor isn't Jesus Christ.
> The use of alpha introduces spin indices which are not properly part of a
> measurement of velocity.
In the second term of: O psi = v psi, there is also a spin indice, which is
unavoidable. If spin is to be loosely interpreted as intrinsic rotation, it
is not surprising that it intervene in the velocity.
> This superposition is not directly related to zitterbewegung (I am not
> sure if anyone has drawn the inference that it is). Zitterbewegung can
> be understood in quantum field theory, where it arises from the
> commutator of the field operators, leading to the appearance of
> "virtual" particle/antiparticle pairs in the propagator.
The Zitterbewegung has been studied in first quantization, where the field
operators commute. Fashionable ideas aren't always necessary.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
>news:N+mJGoEM...@charlesfrancis.wanadoo.co.uk...
>
>> 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).
>
>The velocity is given by the Hamilton equation q' = @H/@p and nothing else.
>It is equal to p/m only for H = p^2/2m + f(q). With the Dirac Hamiltonian,
>@H/@p is precisely equal to alpha.
Generally speaking, that leads to meaningless results, so cannot be true
(viz, v=+-c).
>
>> The use of alpha introduces spin indices which are not properly part of a
>> measurement of velocity.
>
>In the second term of: O psi = v psi, there is also a spin indice, which is
>unavoidable. If spin is to be loosely interpreted as intrinsic rotation, it
>is not surprising that it intervene in the velocity.
It means that alpha does not correspond directly to classical velocity,
for which their is no spin, and results in meaningless eigenstates.
>
>> This superposition is not directly related to zitterbewegung (I am not
>> sure if anyone has drawn the inference that it is). Zitterbewegung can
>> be understood in quantum field theory, where it arises from the
>> commutator of the field operators, leading to the appearance of
>> "virtual" particle/antiparticle pairs in the propagator.
>
>The Zitterbewegung has been studied in first quantization, where the field
>operators commute. Fashionable ideas aren't always necessary.
The field theoretic analysis corresponds quite neatly to the first
quantised interpretation,.
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> Hans de Vries wrote on Wed, 06 Aug 2008 14:05:12 -0600:
> > The velocity operator [H,x] applied on a Dirac plane wave simply gives
> > the speed v.
>
> For instance, the eigenvalues of the velocity operator (14.25) are well-
> known to be c.
>
The information about the speed of the particle is contained within
its field psi.
The operator [H,x] acts *on* the field psi to extract the speed.
v = psi* [H,x] psi
The operator itself doesn't have any information about the speed.
The field psi varies from particle to particle but [H,x] is always
the
same.
Your statement is sort of assuming that [H,x] should have
eigen-values of v... How can that be....
> One can also compute <v^2> using an analogous to your (14.26) and the
> result is
>
> <v^2> = |v|^2 = c^2
>
Again, here you assume that the speed v is contained within
the operator [H,x] and that [H,x]^2 should therefor produce v^2.
The speed information is in the field psi, not in the operator.
> >http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
> > see sections 14.8 and 14.9
Regards, Hans
http://www.physics-quest.org
======================================= MODERATOR'S COMMENT:
Common textbooks state, that the measurable values of an observable equal the eigenvalues of the corresponding operator - this is, obviously, inaccurate
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> Oh No wrote on Wed, 06 Aug 2008 04:24:21 -0600:
>
> The Dirac theory of electrons predicts particles to travel at c. Is this a
> right prediction? Is this a wrong prediction? I will not discuss this
> point here, because by obvious motives...
>
> However, I find really interesting all kind of fantastic dancing that
> people is trying to avoid that prediction.
> 2)
> Other claim that v = [H,x] is the right operator and that this gives v for
> free particles.
>
Yes, the operator is correct. It's a 4x4 matrix which acts on
the (Lorentz transformed) spin of the particle.
The Lorentz transformation of the spinor depends on the velocity
and this is what the operator [ H,X ] extracts via
v = psi* [ H,X ] psi
v here being the "velocity density" which just means that integrated
over space gives the average speed v of the field.
You can check any QFT book that psi* [ H,X ] psi is just the
usual vector current density \bar{psi} \gamma^\mu \psi
The integral of the current density is the total current. The total
current is: The charge times the velocity
Regards, Hans.
Oh No
><juanREM...@canonicalscience.com> wrote:
>> Hans de Vries wrote on Wed, 06 Aug 2008 14:05:12 -0600:
>> > The velocity operator [H,x] applied on a Dirac plane wave simply gives
>> > the speed v.
>>
>> For instance, the eigenvalues of the velocity operator (14.25) are well-
>> known to be c.
>>
>
>The information about the speed of the particle is contained within
>its field psi.
>
>The operator [H,x] acts *on* the field psi to extract the speed.
>
>v = psi* [H,x] psi
>
>The operator itself doesn't have any information about the speed.
>The field psi varies from particle to particle but [H,x] is always
>the
>same.
>
>Your statement is sort of assuming that [H,x] should have
>eigen-values of v... How can that be....
It is what one would normally require of an observable.
>
>> One can also compute <v^2> using an analogous to your (14.26) and the
>> result is
>>
>> <v^2> = |v|^2 = c^2
>>
>
>Again, here you assume that the speed v is contained within
>the operator [H,x] and that [H,x]^2 should therefor produce v^2.
>
>The speed information is in the field psi, not in the operator.
>
>
>> >http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
>> > see sections 14.8 and 14.9
>
>Regards, Hans
>http://www.physics-quest.org
>
>
>======================================= MODERATOR'S COMMENT:
> Common textbooks state, that the measurable values of an observable
>equal the eigenvalues of the corresponding operator - this is,
>obviously, inaccurate
>
Yes, this is the usual definition. Something has to give. If we take
alpha as the velocity observable, we have to lose this postulate. I am
not too comfortable with that, which is why I prefer p/m as the velocity
observable. On the other hand, I am very happy if acting with alpha on a
plane wave state gives p/m (but not an eigenstate) since this makes
sense of the Ehrenfest theorem and resolves the paradox that Dirac
theory appears to allow only particles travelling at the speed of light.
It then remains that, neglecting spin, alpha=p/m.
Hans de Vries
> Common textbooks state, that the measurable values of an
> observable equal the eigenvalues of the corresponding operator -
Note that we still may say that:
psi* [ H,X ] psi = psi* v psi
Which holds locally, where,
psi* v psi = v psi*psi
is the velocity time the charge density, which is
the current density.
But the information about v is contained in psi.
It is extracted from psi by letting the operator
[ H,x ] acting on the field.
[ H,X ] is always the same. It does not contain
velocity information. It *extracts* the velocity from
the field psi.
======== The way it works =====================
[ H,X ] is alpha and it operates on the spinor part of the field.
The spinor is Lorentz transformed depending on the velocity.
The spinor thus contains the velocity information.
psi* [ H,X ] psi is just the current density (the vector current
as described in every QFT book)
The integral of the current density is the total electric current.
The electric current of the particle is vq (velocity times charge)
============================================
Regards, Hans
http://www.physics-quest.org
Juan R.
> On Aug 7, 12:36 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Hans de Vries wrote on Wed, 06 Aug 2008 14:05:12 -0600:
>> > The velocity operator [H,x] applied on a Dirac plane wave simply
>> > gives the speed v.
>>
>> For instance, the eigenvalues of the velocity operator (14.25) are
>> well- known to be c.
>>
>>
> The information about the speed of the particle is contained within its
> field psi.
I find this comment misleading, for instance there is not fields psi in
Feynman formulation of QED. The concept of field psi only applies to field
formulations of QED but not to particle formulations. See previous posts
titled "different flavors of QED".
> The operator [H,x] acts *on* the field psi to extract the speed.
>
> v = psi* [H,x] psi
>
> The operator itself doesn't have any information about the speed. The
> field psi varies from particle to particle but [H,x] is always the same.
I find this also misleading. How you think that on general form
O = psi* o psi
o may be just [H,x]? Do you tried different expressions for o: (H+x), (p/
E), [x,p], [H,x]... Or just derived it? If you derived [H,x] then you may
know why rates of x are of kind
d/dt x = [H,x]
and *not*
d/dt x = p/m
for instance.
> Your statement is sort of assuming that [H,x] should have eigen-values
> of v... How can that be....
>
>> One can also compute <v^2> using an analogous to your (14.26) and the
>> result is
>>
>> <v^2> = |v|^2 = c^2
>>
>>
> Again, here you assume that the speed v is contained within the operator
> [H,x] and that [H,x]^2 should therefor produce v^2.
This is more of the same. If you know from where the expression
d/dt x = v = [H,x]
becomes then you may know why is also
v^2 = [H,x]^2
> The speed information is in the field psi, not in the operator.
For the original Dirac wave function theory and Feynman QED, the
'instantaneous' velocity (I am using Paul Strange notation) is +-c.
For the field formulation of QED, it the 'instantaneous' velocity
associated to the field is again +-c.
Both Dirac wave function and QED present Zbw effect.
In field formulation of QED it can be shown that the velocity of the field
may be splinted into a 'classical' term (I am using Paul Strange notation)
more a Zbw velocity term.
For both cases relativistic wave function and field formulation the
velocity operator for an electron is *not* (c alpha).
Regards.
>> >http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
>> > see sections 14.8 and 14.9
>
> Regards, Hans
> http://www.physics-quest.org
>
>
> ======================================= MODERATOR'S COMMENT:
> Common textbooks state, that the measurable values of an observable
> equal the eigenvalues of the corresponding operator - this is,
> obviously, inaccurate
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
>>One pretends that v = [H,x] is a wrong operator and that right operator
>>is v = p/m
>
> There is no pretence. Clearly v = p/m is correct as it gives the correct
> results for measurement of velocity.
As pointed before your v = (p/m) is plain wrong.
>>I am curious, how many more funny attempts and revolutionary claims I
>>will be reading before people starts to accept that Dirac theory
>>predicts |v| to be +-c?
>
> There is no dispute that the eigenstates of v = [H,x] give |v| = +-c.
> The point is that this clearly does not correspond to a measurement of
> velocity.
This is not true. I also remember someone here critizing Feynman
derivation of the eigenvalues. Exact words from that poster were
(\blockquote
Feynman only uses alpha^2=1 as a quick trick to obtain the magnitude of
the eigen values of alpha (=+/- 1) which is not quite the same (link 2,
page 47)
He then concludes that the eigenvelocities are +/- c without actually
applying the operator.
)
But Feynman was right and the poster wrong. The eigenvalues are +-c with
square c^2.
This is not a trick but follow from direct mathematical calculus. See
[1,2] for instance
[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
Chapter 7.
[2] http://www.amazon.com/Quantum-Electrodynamics-Advanced-Book-Classics/
dp/0201360756
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Oh No
>
>>>One pretends that v = [H,x] is a wrong operator and that right operator
>>>is v = p/m
>>
>> There is no pretence. Clearly v = p/m is correct as it gives the correct
>> results for measurement of velocity.
>
>As pointed before your v = (p/m) is plain wrong.
use p/E if you want rate of change with respect to coordinate time,
rather than a velocity four vector.
>
>>>I am curious, how many more funny attempts and revolutionary claims I
>>>will be reading before people starts to accept that Dirac theory
>>>predicts |v| to be +-c?
>>
>> There is no dispute that the eigenstates of v = [H,x] give |v| = +-c.
>> The point is that this clearly does not correspond to a measurement of
>> velocity.
>
>This is not true.
If it is not true, you should be able to show a measurement in which the
velocity of an electron is found to be +-c. You cannot. All you have is
eigenvalues of an operator giving a non-physical result.
> I also remember someone here critizing Feynman
>derivation of the eigenvalues. Exact words from that poster were
>
>(\blockquote
> Feynman only uses alpha^2=1 as a quick trick to obtain the magnitude of
> the eigen values of alpha (=+/- 1) which is not quite the same (link 2,
> page 47)
>
> He then concludes that the eigenvelocities are +/- c without actually
> applying the operator.
>)
>
>But Feynman was right and the poster wrong. The eigenvalues are +-c with
>square c^2.
There is no dispute about that. The poster was not wrong. The
eigenstates of alpha are particular spin states, but measurement of
velocity does not lead to these spin states.
Juan R.
Yes please sen me to my email of above. I will study and will comment
here.
> P.
>
> PS: I agree that v=p/m is a special case only (body subject to velocity-
> *in*dependent potential)
Yes, then one find dependence on the potential. But Peter even for a free
particle in absence of potentials the velocity operator
v = p/m
that Oh no wrote above is plain wrong. It is pure nonsense for high
velocities and also when one takes into account the virtual clouds.
The correct operator is v = [H,x]. This may be the only part where Hans
and me agree.
This for for the velocity operator is not invented /ad hoc/ or postulated
but follows as a theorem from first principles of QM. Any textbook in QM
explains how one can obtain the rate of any operator O like
d/dt O = [H,O] + @O/@t
For position this is just
d/dt x = v = [H,x]
*Iff* the Hamiltonian is H = p^2/2m
then
v = [H,x] = p/m
But if the Hamiltonian is that of Dirac then
v = [H,x] = c \alpha
And if the Hamiltonian is that of FW then
v = [H,x] = \beta pc^2 / |H|
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> "Oh No" <No...@charlesfrancis.wanadoo.co.uk> a �crit dans le message de
> news:N+mJGoEM...@charlesfrancis.wanadoo.co.uk...
>
>> 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).
>
> The velocity is given by the Hamilton equation q' = @H/@p and nothing
> else. It is equal to p/m only for H = p^2/2m + f(q). With the Dirac
> Hamiltonian, @H/@p is precisely equal to alpha. Igor isn't Jesus
> Christ.
ha ha ha that was funny. And even if Igor was, what he said continues
being wrong even at undergrad eyes...
Your
q' = @H/@p
follows from
q' = [H,x]
when working in the momentum representation
q' = [H,x] = [H, -@/@p] = @H/@p
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> Thus spake Juan R. González-Álvarez <juanR...@canonicalscience.com>
>>Oh No wrote on Thu, 07 Aug 2008 05:28:21 -0600:
>>
>>>>One pretends that v = [H,x] is a wrong operator and that right
>>>>operator is v = p/m
>>>
>>> There is no pretence. Clearly v = p/m is correct as it gives the
>>> correct results for measurement of velocity.
>>
>>As pointed before your v = (p/m) is plain wrong.
>
> use p/E if you want rate of change with respect to coordinate time,
> rather than a velocity four vector.
No thanks I will *not* use your new v operator, because it continues
being so wrong for certain cases as your former operator was.
I just apply QM and in QM the operator is obtained from v = [H,x]
>From here one finds specific forms of operator v for specific
problems.
>>>>I am curious, how many more funny attempts and revolutionary claims I
>>>>will be reading before people starts to accept that Dirac theory
>>>>predicts |v| to be +-c?
>>>
>>> There is no dispute that the eigenstates of v = [H,x] give |v| = +-c.
>>> The point is that this clearly does not correspond to a measurement of
>>> velocity.
>>
>>This is not true.
>
> If it is not true, you should be able to show a measurement in which the
> velocity of an electron is found to be +-c. You cannot. All you have is
> eigenvalues of an operator giving a non-physical result.
You split my message in so a way that next you misread it. According to
guidelines (below) I will not comment on stuff I did not say.
>> I also remember someone here critizing Feynman
>>derivation of the eigenvalues. Exact words from that poster were
>>
>>(\blockquote
>> Feynman only uses alpha^2=1 as a quick trick to obtain the magnitude of
>> the eigen values of alpha (=+/- 1) which is not quite the same (link 2,
>> page 47)
>>
>> He then concludes that the eigenvelocities are +/- c without actually
>> applying the operator.
>>)
>>
>>But Feynman was right and the poster wrong. The eigenvalues are +-c with
>>square c^2.
>
> There is no dispute about that. The poster was not wrong. The
> eigenstates of alpha are particular spin states, but measurement of
> velocity does not lead to these spin states.
More of the same. You would read that the poster said and I have said
before reply.
Regards.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> Thus spake Cl.Massé <daniell...@gmail.com>
>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
>>news:N+mJGoEM...@charlesfrancis.wanadoo.co.uk...
>>
>>> 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).
>>
>>The velocity is given by the Hamilton equation q' = @H/@p and nothing
>>else. It is equal to p/m only for H = p^2/2m + f(q). With the Dirac
>>Hamiltonian, @H/@p is precisely equal to alpha.
>
> Generally speaking, that leads to meaningless results, so cannot be true
> (viz, v=+-c).
Not a single experiment contradicts that result. At contrary the velocity
operator one may use is just alpha (e.g. in Breit Hamiltonian or in
Feynman many electron kernel for path integrals...)
You reject the right operator v = [H,x] and substitute by another is
completely wrong: your v = p/m
Your operator is even wrong at non-relativistic limit (e.g. it is clearly
wrong for an electron in an external field)
I will not emphasize this again.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
======================================= MODERATOR'S COMMENT:
I guess that there are just some misunderstandings (by several sides)
Oh No
>
>> Thus spake Cl.Massé <daniell...@gmail.com>
>>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
>>>news:N+mJGoEM...@charlesfrancis.wanadoo.co.uk...
>>>
>>>> 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).
>>>
>>>The velocity is given by the Hamilton equation q' = @H/@p and nothing
>>>else. It is equal to p/m only for H = p^2/2m + f(q). With the Dirac
>>>Hamiltonian, @H/@p is precisely equal to alpha.
>>
>> Generally speaking, that leads to meaningless results, so cannot be true
>> (viz, v=+-c).
>
>Not a single experiment contradicts that result.
Untrue. Not only does not a single experiment support it, but it
contradicts the fundamental physical principle of relativity which
prevents a massive particle from travelling at c. The result is
experimentally impossible as it would require infinite energy to
obtain..
>I will not emphasize this again.
>
Good. Your posts are repetitive and should be blocked for that reason
alone. Blanket assertions are not the same as reasoned argument.
Oh No
>postulated but follows as a theorem from first principles of QM. Any
>textbook in QM explains how one can obtain the rate of any operator O
>like
>
>d/dt O = [H,O] + @O/@t
In fact the theorem you have in mind (Ehrenfest) states (ignoring
factors of ih)
d/dt <O> = <[H,O]> + <@O/@t>
>
>For position this is just
>
>d/dt x = v = [H,x]
by which you mean
d/dt <x> = <[H,x]>
Hans has shown at
http://www.physicsforums.com/showthread.php?t=219205
the operator [H,x] does give the correct expectation value,. p/E for
plane wave states. But this means it contradicts a fundamental
requirement of an observable operator, that measurement should put the
state into an eigenstate of the operator. IOW the Ehrenfest theorem is
not sufficient to define [H,x] as the velocity observable, and doing so
is inconsistent with a basic principle.
Rather than carry on a silly semantic argument over the meaning of
velocity, it might be better to discuss whether one really wants to
change the fundamental principles of qm in order to discuss a non-
physical measurement of velocity with result +-c, particularly when we
can already write down an observable which does give a sensible result.
Oh No
>>>> The point is that this clearly does not correspond to a measurement of
>>>> velocity.
>>>
>>>This is not true.
>>
>> If it is not true, you should be able to show a measurement in which the
>> velocity of an electron is found to be +-c. You cannot. All you have is
>> eigenvalues of an operator giving a non-physical result.
>
>You split my message in so a way that next you misread it. According to
>guidelines (below) I will not comment on stuff I did not say.
I don't think there is much to misread there, unless you were already
commenting on something I did not say..
Cl.Massé
news:guest.20080807102211$3f...@news.killfile.org...
> PS: I agree that v=p/m is a special case only (body subject to velocity-
> *in*dependent potential)
If v = p/m, integrating v = [H,x] gives H = -@_i@^i/2m + f(x), the quantum
mechanical equivalent of H = p^2/2m + f(x) from the integration of
v = @H/@p. p/m is then valid for the Schrödinger equation, not the Dirac
one. Indeed, it is the non relativistic approximation, which deprives it of
all universality claim. v = p/m appears at some place in the textbooks, but
the reader must be aware (and intelligent enough) that it is only about the
Schrödinger equation.
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> Hans de Vries wrote:
> > The information about the speed of the particle is contained within its
> > field psi.
>
> I find this comment misleading, for instance there is not fields psi in
> Feynman formulation of QED. The concept of field psi only applies to field
> formulations of QED but not to particle formulations. See previous posts
> titled "different flavors of QED".
Apparently you don't do the calculations otherwise you wouldn't
make these (rather loud) claims.
> > The operator [H,x] acts *on* the field psi to extract the speed.
> > v = psi* [H,x] psi
> > The operator itself doesn't have any information about the speed. The
> > field psi varies from particle to particle but [H,x] is always the same.
>
> I find this also misleading. How you think that on general form
>
> O = psi* o psi
>
> o may be just [H,x]? Do you tried different expressions for o: (H+x), (p/
> E), [x,p], [H,x]... Or just derived it? If you derived [H,x] then you may
> know why rates of x are of kind
>
The form psi^ [ H,X ] psi is identical to way other operators act on
psi:
The spin, The orbital angular momentum, et-cetera.
> d/dt x = [H,x]
>
The first x here represents a coordinate, while the second x
represents the position operator. The velocity can be derived
from a coordinate, but x as a position operator doesn't contain
any information about the position of the particle.
So the correct interpretations are:
d/dt and [ H,X ] are the operators, and
dx/dt and psi* [ H,X ] psi are the applied operators.
> and *not*
>
> d/dt x = p/m
>
For the classical particle you have
dx/dt = p/E
and for the Dirac particle one has
psi* [ H,X ] psi = p/m
Where the factor gamma = p/m : p/E is due to the
fact that the second expression is a density which
increases due to the Lorentz contraction of the volume
containing the field.
alternatively one can derive the same from the
Lorentz transform of the phase rates instead
of the Spinor transform.
-i\hbar \bar{psi} ( \partial_i - ieA ) \psi = p/m
> This is more of the same. If you know from where the expression
> d/dt x = v = [H,x]
> becomes then you may know why is also>
> v^2 = [H,x]^2
[ H,X ] is not the velocity. It is an operator like d/dt is an
operator
Its evident that
v^2 =/= psi* [ H,X ]^2 psi
like classically
v^2 =/= (d/dt)^2 x
Regards, Hans
http://www.physics-quest.org
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> Hans de Vries wrote on Thu, 07 Aug 2008 11:15:27 -0600:
> > The operator itself doesn't have any information about the speed. The
> > field psi varies from particle to particle but [H,x] is always the same.
>
> I find this also misleading. How you think that on general form O = psi* o psi
> o may be just [H,x]
Note that this is exactly how Paul Strange defines it in
your link (eq: 7.36) where he uses:
psi* [ H,X ] psi
just ike I do. Unfortunately he doesn't complete the rather
simple calculation.
> >> One can also compute <v^2> using an analogous to your (14.26) and the
> >> result is
> >> <v^2> = |v|^2 = c^2
I guess that what leads to these mathematical errors
is the treatment in momentum space. While in general.
( d/dt )^ 2 x =/= ( dx/dt )^2
second-order-derivative =/= square-of-first-order-derivative
This is true for sinusoidal functions in momentum space.
But, the operator c.alpha works on the bi-spinor and *not*
on the exp(-ipx) part of the wave-function.
Regards, Hans
http://www.physics-quest.org
Juan R.
> On Aug 8, 1:07 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Hans de Vries wrote:
>> > The information about the speed of the particle is contained within
>> > its field psi.
>>
>> I find this comment misleading, for instance there is not fields psi in
>> Feynman formulation of QED. The concept of field psi only applies to
>> field formulations of QED but not to particle formulations. See
>> previous posts titled "different flavors of QED".
>
> Apparently you don't do the calculations otherwise you wouldn't make
> these (rather loud) claims.
Ha ha ha in Feynmam original derivation there is no fields, neither
second quantization. All his QED formulation works in *first*
quantization :-)
His original work has been posted in a cosmological framework recently
http://prola.aps.org/abstract/RMP/v67/i1/p113_1
Bye bye fields...
>> > The operator [H,x] acts *on* the field psi to extract the speed. v =
>> > psi* [H,x] psi
>> > The operator itself doesn't have any information about the speed. The
>> > field psi varies from particle to particle but [H,x] is always the
>> > same.
>>
>> I find this also misleading. How you think that on general form
>>
>> O = psi* o psi
>>
>> o may be just [H,x]? Do you tried different expressions for o: (H+x),
>> (p/ E), [x,p], [H,x]... Or just derived it? If you derived [H,x] then
>> you may know why rates of x are of kind
>>
>>
> The form psi^ [ H,X ] psi is identical to way other operators act on
> psi:
> The spin, The orbital angular momentum, et-cetera.
Once more you avoid to reply questions...
>> d/dt x = [H,x]
>>
>>
> The first x here represents a coordinate,
No in Dirac theory and Feynman QED neither in nonrelativistic quantum
mechanics.
> while the second x represents
> the position operator. The velocity can be derived from a coordinate,
> but x as a position operator doesn't contain any information about the
> position of the particle.
The information is contained in its spectral decomposition.
> So the correct interpretations are:
>
> d/dt and [ H,X ] are the operators, and
>
> dx/dt and psi* [ H,X ] psi are the applied operators.
Well interpretations depend on the theoretical formalism. That is
'correct' in one formalism may be 'incorrect' in other.
In relativistic quantum mechanics and Feynman QED psi is not an operator
and *both* x and (d/dt x) are operators.
However, psi is an operator and x is not an operator in the *field*
formulation of QED using second quantization.
But unlike the two formers, the *field* formulation of QED avoids
localization issues. It seems you pretend to mix parts of both... but I
am not sure.
>> and *not*
>>
>> d/dt x = p/m
>>
> For the classical particle you have
>
> dx/dt = p/E
Sure that is wrong for instance for a particle in an external magnetic
field.
> and for the Dirac particle one has
>
> psi* [ H,X ] psi = p/m
If one does not consider the cloud of virtual particles (aka the Zb
effect)?
But the velocity operator that gives the conventional result v is not (c
\alpha), which contains the Zb and gives eigenvalues +-c.
>> This is more of the same. If you know from where the expression d/dt x
>> = v = [H,x]
>> becomes then you may know why is also> v^2 = [H,x]^2
>
> [ H,X ] is not the velocity. It is an operator like d/dt is an operator
v is an operator, obviously, as follow from his definition stated above
v = [H,x]
and its square is
v^2 = [H,x]^2
> Its evident that
>
> v^2 =/= psi* [ H,X ]^2 psi
>
> like classically
>
> v^2 =/= (d/dt)^2 x
I am really impressed after reading this Hans!
Do you know that classically also
v^2 = {H,x}^2
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> On Aug 8, 1:07 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Hans de Vries wrote on Thu, 07 Aug 2008 11:15:27 -0600:
>> > The operator itself doesn't have any information about the speed. The
>> > field psi varies from particle to particle but [H,x] is always the
>> > same.
>>
>> I find this also misleading. How you think that on general form O =
>> psi* o psi o may be just [H,x]
>
> Note that this is exactly how Paul Strange defines it in your link (eq:
> 7.36) where he uses:
>
> psi* [ H,X ] psi
>
> just ike I do.
It is *not* the same.
He is working with Dirac relativistic quantum theory. You are working with
field theory. There is subtle differences
Moreover, the question was he derived the result
d/dt x = v = [H,x]
and
v^2 = [H,x]^2
You seems to offer no rationale on why is [H,x] and then fail to
understand what is v^2 and what are the eigenvalues.
> Unfortunately he doesn't complete the rather simple calculation.
Dont true. Unlike you, Strange notices in next page that v_x, v_y, and v_z
don't commute. It follows in a trivial way that
v = v_x + v_y + v_z
is not an vector observable. and this is reason that computation of the
observable using the vector operator v makes no sense. This is the last
reason which Feynman never applied v. You called that a "trick" but it is
pure QM.
However the magnitude of the vector is observable and Strange computes the
magnitude (as Feynman makes also)
v^2 = c^2
from which follows the well-known result that
|v| = +-c
that Shrödinger, Dirac, Feynman, Strange and everyone else has obtained.
It is also trivially obtained that the magnitude of the velocity
associated to the Dirac field in *second* quantization is +-c and *NOT*
that (|v| < c) you pretend.
>> >> One can also compute <v^2> using an analogous to your (14.26) and
>> >> the result is
>> >> <v^2> = |v|^2 = c^2
>
> I guess that what leads to these mathematical errors is the treatment in
> momentum space. While in general.
>
> ( d/dt )^ 2 x =/= ( dx/dt )^2
>
> second-order-derivative =/= square-of-first-order-derivative
Quantum mechanics
v = [H,x]
v^2 = [H,x]^2
Classical mechanics
v = {H,x}
v^2 = {H,x}^2
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
Cl\.Massé wrote on Fri, 08 Aug 2008 08:13:22 -0600:
> "Peter" <end...@dekasges.de> a écrit dans le message de
> news:guest.20080807102211$3f...@news.killfile.org...
>
>> PS: I agree that v=p/m is a special case only (body subject to
>> velocity- *in*dependent potential)
>
> If v = p/m, integrating v = [H,x] gives H = -@_i@^i/2m + f(x), the
> quantum mechanical equivalent of H = p^2/2m + f(x) from the integration
> of v = @H/@p. p/m is then valid for the Schrödinger equation, not the
> Dirac one. Indeed, it is the non relativistic approximation, which
> deprives it of all universality claim. v = p/m appears at some place in
> the textbooks, but the reader must be aware (and intelligent enough)
> that it is only about the Schrödinger equation.
Correction the velocity operator
(v = p/m) is only valid for the the Schrödinger equation in *absence* of
vector potentials as Pete correctly pointed.
If the Hamiltonian is that of a nonrelativistic electron in an external
field
H = ((p - eA)^2 / 2m) + V
then v = [H, x] gives the well-known textbook result
v = (p - eA) / m
--
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Hans de Vries
> Ha ha ha in Feynmam original derivation there is no fields, neither
> second quantization. All his QED formulation works in *first*
> quantization :-)
>
> Bye bye fields...
>
This is all totally irrelevant....
I used the general word field in the general sense. Nowhere
I did refer to 2nd order quantization as you seem to imply.
> In relativistic quantum mechanics and Feynman QED psi is not an operator
> and *both* x and (d/dt x) are operators.
>
> However, psi is an operator and x is not an operator in the *field*
> formulation of QED using second quantization.
I didn't talk about second quantization anywhere. I never mentioned
psi
as an operator, on the contrary.
> But unlike the two formers, the *field* formulation of QED avoids
> localization issues. It seems you pretend to mix parts of both... but I
> am not sure.
That's the problem you get when double using a general word like
"field" also for a specific usage, Use "2nd order quantized field"
or
"operator field" for the specific case I would say.
>> For the classical particle you have dx/dt = p/E
> Sure that is wrong for instance for a particle in an external magnetic
> field.
It's evident that we are discussing a free particle here.
> But the velocity operator that gives the conventional result v is not (c
> \alpha), which contains the Zb and gives eigenvalues +-c.
No, no no
You are forgetting that the operator alpha acts on the bi-spinor and
not on momentum eigen states in momentum space.
The eigen values of +/-c are only relevant if the operator would be
working on just the momentum eigenstates. We are not working
with the Schroedinger equation here!
> Do you know that classically also v^2 = {H,x}^2
Doesn't mean that you can be mathematically sloppy and assume
that an operator acting on a function squared is the same as the
operator squared acting on the function.
( Operator . function )^2 =/= ( Operator )^2 . function
That's the all important difference between {A,B} and [A,B]
Regards, Hans.
Juan R.
>>> Generally speaking, that leads to meaningless results, so cannot be
>>> true (viz, v=+-c).
>>
>>Not a single experiment contradicts that result.
>
> Untrue. Not only does not a single experiment support it, but it
> contradicts the fundamental physical principle of relativity which
> prevents a massive particle from travelling at c. The result is
> experimentally impossible as it would require infinite energy to
> obtain..
This is all plain wrong Charles:
1)
The result is compatible with *indirect* experimental data. This was
already noticed by Dirac, who wrote in his Nobel lecture
(\blockquote
the velocity of the electron at any time equals the velocity of light.
This is a prediction which cannot be directly verified by experiment,
since the frequency of the oscillatory motion is so high and its
amplitude is so small. But one must believe in this consequence of the
theory, since other consequences of the theory which are inseparably
bound up with this one, such as the law of scattering of light by an
electron, are confirmed by experiment.
)
For example for the Gaunt and Breit potentials broadly utilized to explain
atomic and molecular experimental data, one uses
v_op = c \alpha
and *not* your wrong operators.
2)
Not a single known *direct* measurement of velocity contradicts the
result. Because as remarked in [1] page 209, current direct measurements
cannot access to the scale beyond 10^-13 seconds. Over that scale the Zb
averages to zero and then you are not really measuring instantaneous
velocity of the quantum particle but an *averaged* velocity given by a FW
operator (7.41). [1]
3)
Your statement about needing infinite energy (mass) for an electron to
move at c is another plain wrong statement from you. Since I want not
engage in another unuseful discussion with you, I will simply cite Strange
in his page 214:
(\blockquote
One final point to make in this section is on the mass of our particle.
At it is moving at the speed of light one may expect it to have infinite
mass. This is not true because as we have seen, the motion is oscillatory
and hence the particle does not remain in the same inertial frame.
)
The *quantum* electron moving at c does not requires infinite energy. Only
the *classical* electron moving at c would require infinite energy.
>>I will not emphasize this again.
>>
> Good. Your posts are repetitive and should be blocked for that reason
> alone. Blanket assertions are not the same as reasoned argument.
It would be interesting to heard other moderators think about your
pretense I would remain silent each time you say the nonsense that
v = p/m
*is* the velocity operator.
Moderator Peter already noticed that your p/m is not valid in presence of
external fields.
Cl Masse and Hans also pointed your velocity operator is completely wrong
for Dirac theory.
Etc.
[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
Chapter 7.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> Thus spake Juan R. González-Álvarez <juanR...@canonicalscience.com>
>>This for for the velocity operator is not invented /ad hoc/ or
>>postulated but follows as a theorem from first principles of QM. Any
>>textbook in QM explains how one can obtain the rate of any operator O
>>like
>>
>>d/dt O = [H,O] + @O/@t
>
> In fact the theorem you have in mind (Ehrenfest) states (ignoring
> factors of ih)
>
> d/dt <O> = <[H,O]> + <@O/@t>
You would stop from continuously misreading me.
The theorem I wrote above is *not* Ehrenfest theorem. I *never* said it
was, only *you* did.
The Eherenfest theorem is a less general theorem for average values,
whereas the operator identity
d/dt O = [H,O] + @O/@t
is *more* general. This is also remarked in pages 313-314 of Cohen-
Tannoudji volume 1.
I will not comment in rest of your overly-repetitive mistakes and must
sniped them.
I will just remark that in Paul Strange textbook page 206 and 207 is
explained what is wrong with Hans computation
http://www.physicsforums.com/showthread.php?t=219205
Strange *computes* the order for v_x (1 dimension) in page 206 and obtains
<v_x> ~ |v| < c
That Hans did is nothing new but you seem to think the contrary.
The part interesting is when Strange explains in page 207 how
*noncommutativity* of matrices alpha *modifies the result* in three
dimensions.
Hans just extrapolate its 2d 'chiral' equation to the 4d 'chiral' equation
without any rigor and pretend that operator (c \alpha) with eigenvalues
+-1 will give a velocity vector with magnitude less than c, which is
plain false as showed in the literature cited.
The 3D vector velocity in Dirac theory is not observable. Only its
magnitude is observable and this is computed in page 207.
The magnitude of vector velocity is (see page 207, see also Feynman
monograph on QED cited in other posts)
|v| = +-c
with
|v|^2 = c^2
Oh No
>Oh No wrote on Fri, 08 Aug 2008 07:50:16 -0600:
>
>>>This for for the velocity operator is not invented /ad hoc/ or
>>>postulated but follows as a theorem from first principles of QM. Any
>>>textbook in QM explains how one can obtain the rate of any operator O
>>>like
>>>
>>>d/dt O = [H,O] + @O/@t
>>
>> In fact the theorem you have in mind (Ehrenfest) states (ignoring
>> factors of ih)
>>
>> d/dt <O> = <[H,O]> + <@O/@t>
>
>You would stop from continuously misreading me.
>
>The theorem I wrote above is *not* Ehrenfest theorem. I *never* said it
>was, only *you* did.
>
>The Eherenfest theorem is a less general theorem for average values,
>whereas the operator identity
>
>d/dt O = [H,O] + @O/@t
>
>is *more* general.
This may be made as a definition, but it is not otherwise an identity.
It is only justified as a definition in so far as it has physical
relevance, i.e. in so far as the Ehrenfest theorem holds. Otherwise we
would do better to ignore it as worthless speculation.
>
>The part interesting is when Strange explains in page 207 how
>*noncommutativity* of matrices alpha *modifies the result* in three
>dimensions.
>
>Hans just extrapolate its 2d 'chiral' equation to the 4d 'chiral' equation
>without any rigor and pretend that operator (c \alpha) with eigenvalues
>+-1 will give a velocity vector with magnitude less than c, which is
>plain false as showed in the literature cited.
>
>The 3D vector velocity in Dirac theory is not observable.
If it is not observable, it has no relevance to physics, and again we
should ignore it.
>Only its
>magnitude is observable and this is computed in page 207.
>
>The magnitude of vector velocity is (see page 207, see also Feynman
>monograph on QED cited in other posts)
>
>|v| = +-c
>
>with
>
>|v|^2 = c^2
This is also non-observable, and again we should ignore it.
Oh No
>
>>>> Generally speaking, that leads to meaningless results, so cannot be
>>>> true (viz, v=+-c).
>>>
>>>Not a single experiment contradicts that result.
>>
>> Untrue. Not only does not a single experiment support it, but it
>> contradicts the fundamental physical principle of relativity which
>> prevents a massive particle from travelling at c. The result is
>> experimentally impossible as it would require infinite energy to
>> obtain..
>
>This is all plain wrong Charles:
>
>1)
>The result is compatible with *indirect* experimental data. This was
>already noticed by Dirac, who wrote in his Nobel lecture
>
>(\blockquote
> the velocity of the electron at any time equals the velocity of light.
> This is a prediction which cannot be directly verified by experiment,
> since the frequency of the oscillatory motion is so high and its
> amplitude is so small. But one must believe in this consequence of the
> theory, since other consequences of the theory which are inseparably
> bound up with this one, such as the law of scattering of light by an
> electron, are confirmed by experiment.
>)
Dirac was writing in the early days of qm and before qed. I am
comfortable disagreeing with him because if it cannot be verified by
experiment it is clearly not a prediction. We can define a group
velocity for a wave, and we can define a velocity as a rate of change of
measured position, but here he is talking of something which does not
have a place in physics.
>2)
>Not a single known *direct* measurement of velocity contradicts the
>result.
I said the result is meaningless, not that it is contradicted. It is
also not supported. You claim what I say is plain wrong, but then you
say something which does not support your claim. .
>
>(\blockquote
> One final point to make in this section is on the mass of our particle.
> At it is moving at the speed of light one may expect it to have infinite
> mass. This is not true because as we have seen, the motion is oscillatory
> and hence the particle does not remain in the same inertial frame.
>)
One cannot say the motion is oscillatory, since one is working with a
physically undefined quantity.
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> Hans de Vries wrote on Sat, 09 Aug 2008 04:34:52 -0600:
> > Note that this is exactly how Paul Strange defines it in your link (eq:
> > 7.36) where he uses psi* [ H,X ] psi just ike I do.
>
> It is *not* the same.
>
> He is working with Dirac relativistic quantum theory. You are working with
> field theory. There is subtle differences
It *is* the same. We are both working with relativistic quantum
theory.
psi is never considered as an operator here.
> You seems to offer no rationale on why is [H,x] and then fail to
> understand what is v^2 and what are the eigenvalues.
> Dont true. Unlike you, Strange notices in next page that v_x, v_y, and v_z
> don't commute. It follows in a trivial way that
You are forgetting that the Dirac field psi has a bi-spinor part
on which [ H,X ] acts and continue to reason from Schroedinger
equation background automatism's
You need to understand that some automatism's valid when working
with the Schroedinger equation don't work with Dirac and why they
don't work.
c.alpha is an *operator* working on the bi-spinor in either the
position-
or momentum space. c or c^2 are *constants* working on only on the
the momentum eigenstates in momentum space.
[ alpha_x, alpha_y ] =/= 0
but
[ psi* alpha_x psi, psi* alpha_y psi ] = 0
That is: The components of v do commute like they should !
even though the components of the velocity operator do not
commute. Matrix operators do not behave the same as constants.
Regards, Hans
-----------------------------------------------------------
http://www.physics-quest.org/
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Pmb
"Hans de Vries" <Hans.de....@gmail.com> wrote in message
news:ada9553d-1864-4ab5...@d77g2000hsb.googlegroups.com...
Hans - Juan is referring to the "velocity operator" that is used in quantum
mechanics and is defined as V = P/m (This is defined in "Quantum Mechanics,"
by Cohen-Tannoudji, Diu, Laloe). In the Heisenberg representation where X =>
X' = U^XU in which case
dX'/dt = [X', H]/i h-bar
> [ H,X ] is not the velocity. It is an operator like d/dt is an
> operator
By this I assume that you mean dX'/dt is not the same as V = P/m. Is that
correct?
Thanks
Pete
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> I will just remark that in Paul Strange textbook page 206 and 207 is
> explained what is wrong with Hans computation
>
> http://www.physicsforums.com/showthread.php?t=219205
>
> Strange *computes* the order for v_x (1 dimension) in page 206 and obtains
>
> <v_x> ~ |v| < c
>
> That Hans did is nothing new but you seem to think the contrary.
>
> The part interesting is when Strange explains in page 207 how
> *noncommutativity* of matrices alpha *modifies the result* in three
> dimensions.
For a starter you could check that:
psi* alpha psi
Is just the vector-current-density as teached in
any QFT introduction. usually written as:
J_v = \bar{\psi} \gamma^\mu \psi
Note that the gamma matrices don't commute
but the components of J_v *do* commute.
Start thinking yourself instead of quoting once-
upon-a-time made remarks of others...
J_v transforms as a 4-momentum but the integral
over the (Lorentz contracted) volume transforms
like the total electric current which transforms like
the velocity since:
electric current = velocity * electric charge.
> The 3D vector velocity in Dirac theory is not observable.
> Only its > magnitude is observable and this is computed
> in page 207.
Wrong conclusion.
[ alpha_x, alpha_y ] =/= 0 but
[ psi* alpha_x psi , psi* alpha_y psi ] == 0
And therefor the components of v do commute
and v *is* an observable.
Regards, Hans
---------------------------------
http://www.physics-quest.org/
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Juan R.
> Hans - Juan is referring to the "velocity operator" that is used in
> quantum mechanics and is defined as V = P/m (This is defined in "Quantum
> Mechanics," by Cohen-Tannoudji, Diu, Laloe).
No Pete, I am not referring that you believe.
I am using the velocity operator (v = c \alpha)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> c.alpha is an *operator* working on the bi-spinor in either the
> position-
> or momentum space. c or c^2 are *constants* working on only on the
> the momentum eigenstates in momentum space.
>
> [ alpha_x, alpha_y ] =/= 0
Which implies that the 3D velocity operator
alpha_x + alpha_y + alpha_z
does not define an observable vector.
> but
>
> [ psi* alpha_x psi, psi* alpha_y psi ] = 0
>
> That is: The components of v do commute like they should ! even though
> the components of the velocity operator do not commute.
Which again clearly indicates you pretend to use 1D result and then
extrapolate them to 3D using tricks.
Instead doing inaccurate 2D 'chiral' analysis (without Zb) and
extrapolating the results to 'chiral' 4D without any rigor. Instead
playing games of that kind, why don't just compute
Int psi* c alpha psi d^3 x
If you take care in details you may obtain a non observable result with
total eigenvalues splinted as
c = v + v_Zb
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Cl.Massé
>>The Eherenfest theorem is a less general theorem for average values,
>>whereas the operator identity
>>
>>d/dt O = [H,O] + @O/@t
>>
>>is *more* general.
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
news:+93iFoDI...@charlesfrancis.wanadoo.co.uk...
> This may be made as a definition, but it is not otherwise an identity.
> It is only justified as a definition in so far as it has physical
> relevance, i.e. in so far as the Ehrenfest theorem holds. Otherwise we
> would do better to ignore it as worthless speculation.
It is an identity, called the Heisenberg equation of motion, consequence of
the "definition" of the Hamilton operator: H psi = -i@t psi. It has a
classical mechanical equivalent:
d/dt f = {f,H} + @f/@t (f is a function of p and q)
>>The 3D vector velocity in Dirac theory is not observable.
> If it is not observable, it has no relevance to physics, and again we
> should ignore it.
If is is not observable, it is meaningless to say that it is meaningless.
>From it, the observable smoothed velocity can be derived.
Cl.Massé
>> d/dt x = [H,x]
"Hans de Vries" <Hans.de....@gmail.com> a écrit dans le message de
news:ada9553d-1864-4ab5...@d77g2000hsb.googlegroups.com...
> The first x here represents a coordinate, while the second x
> represents the position operator.
No, both are operators.
> The velocity can be derived
> from a coordinate, but x as a position operator doesn't contain
> any information about the position of the particle.
Not in general, for instance in the Heisenberg and interaction pictures.
> So the correct interpretations are:
>
> d/dt and [ H,X ] are the operators, and
If d/dt is the operator, d/dt x is not since the operator have been applied
to a function, then the equality isn't homogeneous and can't hold.
Peter
> Hans de Vries wrote on Sat, 09 Aug 2008 09:21:11 -0600:
>
> > c.alpha is an *operator* working on the bi-spinor in either the
> > position-
> > or momentum space. c or c^2 are *constants* working on only on the
> > the momentum eigenstates in momentum space.
> >
> > [ alpha_x, alpha_y ] =/= 0
>
> Which implies that the 3D velocity operator
>
> alpha_x + alpha_y + alpha_z
>
> does not define an observable vector.
I don't understand this conclusion, because for the angular momentum, L,
[L_x, L_y] =/= 0,
too, please explain.
Thank you,
Peter
Juan R.
> Thus spake Juan R. González-à lvarez
> <juanR...@canonicalscience.com>
>>Oh No wrote on Fri, 08 Aug 2008 07:15:11 -0600:
>>
>>>>> Generally speaking, that leads to meaningless results, so cannot be
>>>>> true (viz, v=+-c).
>>>>
>>>>Not a single experiment contradicts that result.
>>>
>>> Untrue. Not only does not a single experiment support it, but it
>>> contradicts the fundamental physical principle of relativity which
>>> prevents a massive particle from travelling at c. The result is
>>> experimentally impossible as it would require infinite energy to
>>> obtain..
>>
>>This is all plain wrong Charles:
>>
>>1)
>>The result is compatible with *indirect* experimental data. This was
>>already noticed by Dirac, who wrote in his Nobel lecture
>>
>>(\blockquote
>> the velocity of the electron at any time equals the velocity of light.
>> This is a prediction which cannot be directly verified by experiment,
>> since the frequency of the oscillatory motion is so high and its
>> amplitude is so small. But one must believe in this consequence of the
>> theory, since other consequences of the theory which are inseparably
>> bound up with this one, such as the law of scattering of light by an
>> electron, are confirmed by experiment.
>>)
> Dirac was writing in the early days of qm and before qed.
1)
According to Weinberg volume 1 (Chapter in history of QFT) QFT was
formulated about 1930. Weinberg put the start in Pauli and Heisenberg
articles of 1929. By 1933 (When Dirac wrote that of above) Fock, Furry and
Oppenheimer had showed the equivalence between QFT and the hole theory of
Dirac.
ii)
But the point is not historical.
The relevant point is that QED has not changed Dirac statements.
If you had taken a look to modern textbook on RQM by Paul Strange and QED
monograph by Feynman cited in this thread.
You would see both authors make the same statements than Dirac about the
prediction of velocity of an electron.
Charles, do you will argue now that Strange or Feynman were writing in
the early days of QM and before QED? :-)
>>2)
>>Not a single known *direct* measurement of velocity contradicts the
>>result.
>
> I said the result is meaningless
You already decided that before going to the lab to measure...
>>(\blockquote
>> One final point to make in this section is on the mass of our particle.
>> At it is moving at the speed of light one may expect it to have
>> infinite mass. This is not true because as we have seen, the motion is
>> oscillatory and hence the particle does not remain in the same inertial
>> frame.
>>)
>
> One cannot say the motion is oscillatory, since one is working with a
> physically undefined quantity.
I am not sure if you understand the difference between the observability
of a component (say z) and the observability of the 3D vector.
But in any case above picturesque description helps to understand why your
claim about infinite energy was plain wrong. This is QM, doesn't special
relativity.
Regards.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Cl.Massé
> Wrong conclusion.
>
> [ alpha_x, alpha_y ] =/= 0 but
>
> [ psi* alpha_x psi , psi* alpha_y psi ] == 0
>
> And therefor the components of v do commute
> and v *is* an observable.
Ah! I had all wrong:
[x,p] != 0.
But... tada...
[<x>,<p>] = 0,
And qm is all of a sudden easy.
Juan R.
In quantum theory two observables A and B are measurable (or are defined)
at same instant t, when [A,B] = 0
If commutator is not zero, then state |Psi> either defines values for A or
B but not both simultaneously, either
A|Psi> = a|Psi>
B|Psi> /= b|Psi>
or
A|Psi> /= a|Psi>
B|Psi> = b|Psi>
This is the case for momentum and position because [p,x] /= 0.
This is also the case for components of angular momentum (and also for
spin)
Eigenvalues for L_x, L_y, and L_z are not defined at same instant t. This
implies that vector
L = L_x + L_y + L_z
is not observable. However,
L = |L| u
and |L| is observable. Indeed, you may check that
[L^2, L_i] = 0
That is, the magnitude of the vector and one of components are
simultaneously observable. It is common to choose component z. Then one
measures L_z and |L|
L_z |Psi> = l_z |Psi>
L^2 |Psi> = l^2 |Psi>
Note that L^2 is not a vector but only the square of the length |L| of the
vector L.
http://en.wikipedia.org/wiki/Angular_momentum_operator
http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/
AngularMomentum.htm
Since only one component and magnitude may be known at once, instead a
vector we only know a cone. Next is common visual representation
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/vecmod.html
Notice that you know the cone but you cannot know the vector. A vector is
here draw by educative purposes, but it is not specified because you don't
know L_x and L_y.
This is also valid for spin. Where one measures S_z and S^2:
S_z |Spin> = s_z |Spin>
S^2 |Spin> = s^2 |Spin>
but not S_x and S_y.
Similar stuff applies to velocity vector in *Dirac* theory. The components
v_x, v_y, and v_z don't commute. This mean you cannot built the velocity
vector
v = v_x + v_y + v_z
but its magnitude |v| is still well-defined . As in spin and angular
momentum case, we can obtain the magnitude of vector v from the operator
v^2 which, remember, is not a vector but a scalar. This gives
<v^2> = Int Psi (c alpha)^2 Psi d^3x = c^2
This implies that magnitude of the velocity in Dirac theory is
|v| = +-c
as stated in many textbooks. See Strange or Feynman cited before.
Someone worked component alpha_z and then simply extrapolate the result to
3D. Assuming that
v = v_x + v_y + v_z
would be an observable vector of magnitude |v| < c.
But if we compute the 3D case directly
v = Int Psi (c alpha) Psi d^3x
We obtain a non-observable vector
v = v_class + v_Zb
with non-observability *contained in* the v_Zb term.
One can also split the Hamiltonian
H = H_class + H_Zb
with
v = [H,x] = c alpha
v_class = [H_class,x] = beta pc^2/E
etc.
It is also easy to verify that
|v_class| < c
v_class^2 = [H_class,x]^2 = beta^2 (p^2c^4)/E^2
= (p^2c^2) / m^2c^2 + p^2
< (p^2c^2) / p^2
= c^2
The magnitudes of the vectors v verify the identity
c = |v_class| + |v_Zb|
It may be remarked that the structure of the v_Zb is different in
relativistic quantum mechanics than in QFT.
In the original Dirac wave theory, the v_Zb arises from interference
between positive and negative energy branches. This interference has
origin in the fact that alpha matrix is odd.
In QFT, there is not negative energy (positrons are not holes in negative
energy spectrum) and the v_Zb arises from virtual particle antiparticle
clouds around an electron.
There is also subtle relations between the term v_Zb and the spin that I
have not discussed here.
P.S: You 'promised' send me a paper, stating in its abstract no existence
of Zb effect in QFT for further analysis, but I never received.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> This implies that magnitude of the velocity in Dirac theory is
>
> |v| = +-c
>
> as stated in many textbooks. See Strange or Feynman cited before.
OK, let's see what Strange and Feynman you cited say:
Paul Strange calls v =+/-c *Ludicrous* and *Awkward* !
(Relativistic Quantum Mechanics, page 207)
Feynman says he does not accept the arguments of people
who sometimes want to make v = +/-c look plausible.
(Quantum Electrodynamics, page 47, footnote)
Yet you use them over and over again as the authorities
who back your claim that the electron's velocity is c...
Regards, Hans
http://www.physics-quest.org/
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Oh No
>Oh No wrote on Sat, 09 Aug 2008 09:12:44 -0600:
>
>
>If you had taken a look to modern textbook on RQM by Paul Strange and QED
>monograph by Feynman cited in this thread.
>
>You would see both authors make the same statements than Dirac about the
>prediction of velocity of an electron.
>
>Charles, do you will argue now that Strange or Feynman were writing in
>the early days of QM and before QED? :-)
No, but I do observe that Hans has posted from both authors showing that
they agree with my view.
Paul Strange calls v =+/-c *Ludicrous* and *Awkward* ! (Relativistic
Quantum Mechanics, page 207)
Feynman says he does not accept the arguments of people who sometimes
want to make v = +/-c look plausible. (Quantum Electrodynamics, page
47, footnote)
(thank you Hans)
>
>>>2)
>>>Not a single known *direct* measurement of velocity contradicts the
>>>result.
>>
>> I said the result is meaningless
>
>You already decided that before going to the lab to measure...
If such a measurement became possible, it would cease to be meaningless.
As things are, this is not so. Since you have merely defined v formally,
it does not seem likely that you could even show that alpha was the
correct operator to describe such a measurement.
>
>>>(\blockquote
>>> One final point to make in this section is on the mass of our particle.
>>> At it is moving at the speed of light one may expect it to have
>>> infinite mass. This is not true because as we have seen, the motion is
>>> oscillatory and hence the particle does not remain in the same inertial
>>> frame.
>>>)
This is meaningless too. To say a particle does not remain in an
inertial frame is akin to saying it ceases to exist. I think you mean
that the particle does not exhibit inertial motion - the very
explanation I gave you which you rejected so forcefully when you first
introduced this topic.
>>
>> One cannot say the motion is oscillatory, since one is working with a
>> physically undefined quantity.
>
>I am not sure if you understand the difference between the observability
>of a component (say z) and the observability of the 3D vector.
I am not sure if you understand that physics is an empirical science.
>But in any case above picturesque description helps to understand why your
>claim about infinite energy was plain wrong. This is QM, doesn't special
>relativity.
This is relativistic qm, and one must still give classical special
relativity when classical quantities are described. If you find v=+-c as
an eigenstate following the measurement, then a second measurement will
give the same value. That is the point of the projection postulate
(unless, of course, the projection postulate is not obeyed, in which
case Hans result is perfectly good and consistent, and there is no
issue).
Assuming the projection postulate is obeyed, then an eigenstate is
established, the motion is not oscillatory, and the particle acquires
infinite energy, as originally stated. This is also consistent with the
point that a precise measurement of instantaneous velocity would require
two precise measurements of position over a vanishing time period.
First, however, one should respect the fact that physics is an empirical
science, and that measured numerical quantities are given meaning
through the manner of their measurement. One cannot simply formally
define an operator and say it is the same thing as a measured quantity
(or, as in this case, an unmeasured one).
Oh No
>Thus spake Juan R. González-Álvarez <juanR...@canonicalscience.com>
>
>>>The Eherenfest theorem is a less general theorem for average values,
>>>whereas the operator identity
>>>
>>>d/dt O = [H,O] + @O/@t
>>>
>>>is *more* general.
>
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
>news:+93iFoDI...@charlesfrancis.wanadoo.co.uk...
>
>> This may be made as a definition, but it is not otherwise an identity.
>> It is only justified as a definition in so far as it has physical
>> relevance, i.e. in so far as the Ehrenfest theorem holds. Otherwise we
>> would do better to ignore it as worthless speculation.
>
>It is an identity, called the Heisenberg equation of motion,
>consequence of the "definition" of the Hamilton operator: H psi = -i@t
>psi. It has a classical mechanical equivalent:
Please explain how the operator d/dt O is defined, if not by the
relation, for all |f> |g>
<f| d/dt O |g> = d/dt <f| O |g>
Peter
> Peter wrote on Mon, 11 Aug 2008 04:15:08 -0600:
> > <juanR...@canonicalscience.com> writes:
> >> Hans de Vries wrote on Sat, 09 Aug 2008 09:21:11 -0600:
> >> > c.alpha is an *operator* working on the bi-spinor in either the
> >> > position-
> >> > or momentum space. c or c^2 are *constants* working on only on the
> >> > the momentum eigenstates in momentum space.
> >> >
> >> > [ alpha_x, alpha_y ] =/= 0
> >> Which implies that the 3D velocity operator
> >>
> >> alpha_x + alpha_y + alpha_z
> >>
> >> does not define an observable vector.
> > I don't understand this conclusion, because for the angular momentum, L,
> >
> > [L_x, L_y] =/= 0,
> >
> > too, please explain.
> >
> > Thank you,
> > Peter
> In quantum theory two observables A and B are measurable (or are defined)
> at same instant t, when [A,B] = 0
>
> If commutator is not zero, then state |Psi>
can
> either defines values for A or
> B but not both simultaneously...
> Eigenvalues for L_x, L_y, and L_z are not defined at same instant t. This
> implies that vector
>
> L = L_x + L_y + L_z
>
> is not observable.
I was confused by your rather uncommon use of 'observable' for vectors
...
> Similar stuff applies to velocity vector in *Dirac* theory. The components
> v_x, v_y, and v_z don't commute. This mean you cannot built
??
> the velocity
> vector
>
> v = v_x + v_y + v_z
>
> but its magnitude |v| is still well-defined. As in spin and angular
> momentum case, we can obtain the magnitude of vector v from the operator
> v^2 which, remember, is not a vector but a scalar. This gives
>
> <v^2> = Int Psi (c alpha)^2 Psi d^3x = c^2
>
> This implies that magnitude of the velocity in Dirac theory is
>
> |v| = +-c
>
> as stated in many textbooks. See Strange or Feynman cited before.
My impression is that the fact that v is an operator in spinor state (4x4
matrix) is not suffiently accounted for. What is the meaning of the 16 matrix
elements? Which are the eigenvalues of v_z and v?
> Someone worked component alpha_z and then simply extrapolate the result to
> 3D. Assuming that
>
> v = v_x + v_y + v_z
>
> would be an observable vector of magnitude |v| < c.
>
> But if we compute the 3D case directly
>
> v = Int Psi (c alpha) Psi d^3x
>
> We obtain a non-observable vector
>
> v = v_class + v_Zb
>
> with non-observability *contained in* the v_Zb term.
>
> One can also split the Hamiltonian
>
> H = H_class + H_Zb
>
> with
>
> v = [H,x] = c alpha
>
> v_class = [H_class,x] = beta pc^2/E
>
> etc.
>
> It is also easy to verify that
>
> |v_class| < c
>
> v_class^2 = [H_class,x]^2 = beta^2 (p^2c^4)/E^2
> = (p^2c^2) / m^2c^2 + p^2
> < (p^2c^2) / p^2
> = c^2
>
> The magnitudes of the vectors v verify the identity
>
> c = |v_class| + |v_Zb|
Again, my impression is that the fact that v is an operator in spinor state
(4x4 matrix) is not suffiently accounted for. If beta is a matrix, is |beta|
its determinant? What about the eigenvalues of v_class?
...
> P.S: You 'promised' send me a paper, stating in its abstract no existence
> of Zb effect in QFT for further analysis, but I never received.
Sorry, I have lost the reference, please rewrite the bibliographical data
Thank you,
Peter
FrediFizzx
news:guest.20080812095250$40...@news.killfile.org...
> <juanR...@canonicalscience.com> writes:
>> P.S: You 'promised' send me a paper, stating in its abstract no
>> existence
>> of Zb effect in QFT for further analysis, but I never received.
>
> Sorry, I have lost the reference, please rewrite the bibliographical
> data
Here is the counter paper to the one Peter mentioned.
http://arxiv.org/abs/0712.0491
I believe this is the paper Peter mentioned at the link below.
http://prola.aps.org/abstract/PRL/v93/i4/e043004
Peter, can you please email to me also?
Best,
Fred Diether
Juan R.
> On Aug 11, 5:12 pm, "Juan R." González-
àlvarez
> <juanREM...@canonicalscience.com> wrote:
>
>> This implies that magnitude of the velocity in Dirac theory is
>>
>> |v| = +-c
>>
>> as stated in many textbooks. See Strange or Feynman cited before.
>
>
> OK, let's see what Strange and Feynman you cited say:
Let us see then!
> Paul Strange calls v =+/-c *Ludicrous* and *Awkward* ! (Relativistic
> Quantum Mechanics, page 207)
What one gross misquoting and misunderstanding!
In that section Strange is not saying that v =+/-c was. Indeed, several
times Strange claims that v =+/-c and even explain how that result can be
measured. That Strange call ludicrous and awkward is that when *comparing*
the Dirac velocity operator (7.35) to the *non-relativistic* operator
(7.34) both seem in contradiction in a *first* look.
However, finally Strange agrees that inconsistency was *only apparent* and
that a FW transformation helps to understand the relationship between both
operators, making the theory internally consistent. I.e. that (7.35)
really reduces to (7.34) in the non-relativistic limit when one does
right.
Now let us in what context Strange used words "ludicrous" and "awkward".
And we will see it has nothing to see with your misunderstanding.
>From page 207:
(\blockquote
Clearly this is a ludicrous result. It gives us the unlikely sounding
information that the electron always has at least the velocity of light
(even in its rest frame). As if this wasn't awkward enough the x-, y- and
z-components of velocity commute with each other in the non- relativistic
theory, but don't in the relativistic case. So in Dirac theory they are
not simultaneously measurable to arbitrary precision, but in the
non-relativistic theory they are. What has gone wrong? It is impossible
to believe that two such different operators representing the same
physical quantity can both be correct. Indeed, such a contradiction is
good reason to suspect that the whole theory is up to creek. However,
those clever fellows Foldy and Wouthuysen come to our rescue again. The
essential point is that these two apparently contradictory operators do
not actually represent the same observable
)
Above Strange is saying that *apparently* the velocity operator in Dirac
theory was incompatible with the non-relativistic operator, which would be
both ludicrous and awkward. But this was only apparent because the
observables are different
v_Dirac = v_NR + correction terms
Once the non-relativistic limit has been solved, Strange writes about the
velocity of a Dirac electron.
>From page 208
(\blockquote
We have seen that the standard velocity operator gives the electron speed
as equal to the speed of light. However, the inverse F-W transform of the
position operator given by (7.38) leads to a velocity operator that has a
sensible non-relativistic limit. These two can be reconciled if we
identify (7.41) with the average velocity of the particle. Then the
motion of the electron can be divided into two parts as was done in
chapter 3 for Klein-Gordon particles. First 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 would give c.
)
Strange also remarks technological difficulties to measure the
instantaneous component and how the theoretical prediction (|v| = c) is
not in contradiction with current experiments when says
>From page 209
(\blockquote
From (7.47) we see again that the x-component of velocity has two parts.
The second term is the F-W terms and is associated with the average
motion of the particle, i.e. that given by classical relativistic
formulae. The first term is the /Zitterbewegung/ which oscillates
extremely rapidly. In any macroscopic experimental time there will be a
vast number of oscillations because of the hbar in the denominator of the
exponent. [...] 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
the inverse of this, i.e. in about 3x10^-13 seconds! Clearly this is not
a feasible experiment.
)
And of course Paul Strange again remarks that the speed of the Dirac
electron is c in other pages. For example in page 214 again writes
(\blockquote
As it is moving at the speed of light.
)
And in figure 7.1 writes
(\blockquote
The electron executes oscillatory motion within the sphere to maintain a
speed c at all times.
)
> Feynman says he does not accept the arguments of people who sometimes
> want to make v = +/-c look plausible. (Quantum Electrodynamics, page
> 47, footnote)
Another far misquoting.
Feynman says
(\blockquote
But alpha^2 = 1 so the eigenvalues of alpha are +-1. Hence the
eigenvelocities of dotx are +- speed of light. This result is sometimes
made plausible by the argument that a precise determination of velocity
implies precise determinations of position at two times. Then, by the
uncertainty principle, the momentum is completely uncertain and all
values are equally likely. With the relativistic relation between
velocity and momentum, this is seen to imply that velocities near the
speed of light are more probable, so that in the limit the expected value
of the velocity is the speed of light.*
)
And in a footnote * adds
(\blockquote
This argument is not completely acceptable, for dotx commutes with p;
that is, one should be able to measure the two quantities simultaneously
)
Feynman somewhat criticize the argument, not the result v = +-c. Which
Feynman uses during development of QED.
For instance, Feynman obtain the quantum kernel for electron-electron
scattering from the classical one doing substitution
v_classical --> v_operator = c alpha
as waited.
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
>> Eigenvalues for L_x, L_y, and L_z are not defined at same instant t.
>> This implies that vector
>>
>> L = L_x + L_y + L_z
>>
>> is not observable.
>
> I was confused by your rather uncommon use of 'observable' for vectors
Any observable <O> = <Psi| O |Psi>
may be scalar <E>, vector <p>, tensor <g_ab>
>> <v^2> = Int Psi (c alpha)^2 Psi d^3x = c^2
>>
>> This implies that magnitude of the velocity in Dirac theory is
>>
>> |v| = +-c
>>
>> as stated in many textbooks. See Strange or Feynman cited before.
>
> My impression is that the fact that v is an operator in spinor state
> (4x4 matrix) is not suffiently accounted for. What is the meaning of the
> 16 matrix elements?
What 16?
Remember beta is not vector, alpha is.
> What about the eigenvalues of v_class?
are the |p|c^2/E because beta is the unit matrix.
> ...
>
>> P.S: You 'promised' send me a paper, stating in its abstract no
>> existence of Zb effect in QFT for further analysis, but I never
>> received.
>
> Sorry, I have lost the reference, please rewrite the bibliographical
> data
Was this one
http://prola.aps.org/abstract/PRL/v93/i4/e043004
> Thank you,
> Peter
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Oh No
>
>
>Once the non-relativistic limit has been solved, Strange writes about the
>velocity of a Dirac electron.
>
>>From page 208
>
>(\blockquote
> We have seen that the standard velocity operator gives the electron speed
> as equal to the speed of light. However, the inverse F-W transform of the
> position operator given by (7.38) leads to a velocity operator that has a
> sensible non-relativistic limit. These two can be reconciled if we
> identify (7.41) with the average velocity of the particle. Then the
> motion of the electron can be divided into two parts as was done in
> chapter 3 for Klein-Gordon particles. First 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 would give c.
>)
The idea that this represents oscillatory motion is a gross misstatement
of the principle of superposition.
Juan R.
>> What about the eigenvalues of v_class?
>
> are the |p|c^2/E because beta is the unit matrix.
>
Sorry, would say
are the |p|c^2/E because beta^2 is the unit matrix.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
>...
Conveniently, You did not include *what* Paul Strange
calls a ludicrous result in your quote from his book.
Let me add it here: Paul Strange, page 207
" This gives us that v_x is +/- c and similarly for the
y- and z-components, Clearly this is a ludicrous result"
So what he calls a ludicrous result is that the three
components x, y and z are all three +/-c
He would have obtained the correct speed v if only
he had only calculated the exact result of his equation
7.36 which is quite simple actually.
quote from Paul Strange:
> the motion of the electron can be divided into two parts as was done in
> chapter 3 for Klein-Gordon particles. First 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 would give c.
> )
OK. This U-turn is the origin of your beliefs
It is from Schrödinger’s anzatz to integrate
the acceleration operator. Strange’s equations (7.42)
through (7.48) are a direct copy of Dirac’s 1930 book
(The principles of Quantum mechanics, page 263)
quote from Paul Strange:
> 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
> the inverse of this, i.e. in about 3x10^-13 seconds! Clearly this is not
> a feasible experiment.
"in about 3x10^-13 seconds" ? Oops.....
How about 10^-21 seconds ? The electrons rest
frequency is 1.2355899729e20 Hertz. Hence he
is off here by a factor of c…
Clearly, the experiment could be done by measuring
the bath of infinitely intensive 1 MeV soft gamma
radiation produced By the “zittering” electrons…..
quote from Paul Strange (in figure 7.1)
> The electron executes oscillatory motion within
> the sphere to maintain a speed c at all times.
> As it is moving at the speed one might expect it
> to have infinite mass. This is not true because,
> as we have seen, the motion is oscillatory.
Relativity does not work because the motion
is “oscillatory” ??
Misconceptions like this apparently allows him
to accept what he previously called ludicrous.
Regards, Hans
http://www.physics-quest.org
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Cl.Massé
>> My impression is that the fact that v is an operator in spinor state
>> (4x4 matrix) is not suffiently accounted for. What is the meaning of the
>> 16 matrix elements?
In <psi|alpha|psi>, the spin indices are contracted, so that they disappear.
Note that the Dirac Hamiltonian has spin indices and 16 components too.
Juan R.
(my previous post vanished in the air. I am resubmitting it again)
> Conveniently, You did not include *what* Paul Strange calls a ludicrous
> result in your quote from his book.
I see you again cite out of context.
> Let me add it here: Paul Strange, page 207
>
> " This gives us that v_x is +/- c and similarly for the y- and
> z-components, Clearly this is a ludicrous result"
As explained before Strange is *comparing* the relativistic and non-
relativistic operators. At the start of that section Strange writes
(\blockquote
Let us consider the velocity operator. Up to now we have been working
with an implicit contradiction which we have deliberately failed to point
out. This concerns the non-relativistic limit of the velocity operator.
)
And rest of that section is devoted to study the relation between both
operators and show both are compatible.
> So what he calls a ludicrous result is that the three components x, y
> and z are all three +/-c
This misreading was already corrected before.
Strange is explaining that is an (apparently) ludicrous result for the
direct *comparison* with the non-relativistic operator. It would be
ludicrous that one operator gives +-c always and other gives < c always.
Is not?
But then Strange explains that there is not contradiction. One operator
measures the instantaneous velocity and other the averaged (beyond Zb)
with
c = v_class + v_Zb
Which trivially implies (v_class < c) and no contradiction...
Strange also discusses how we would measure electron instantaneous (i.e.
c) speed latter (see full quotations in previous message)
Paul Strange *remarks* that the speed of the Dirac electron is c in
several other pages. Where Strange *never* uses the words ludicrous or
awkward on those:
For example in page 214 writes
(\blockquote
As it *is* moving at the speed of light.
)
And in figure 7.1 Strange writes
(\blockquote
The electron executes oscillatory motion within the sphere to *maintain*
a speed c at *all* times.
)
If you read both of above and think that Strange is saying that speed of
an electron is not c or if you think that Strange is saying that that
result is ludicrous or wrong, then I can say little more...
> He would have obtained the correct speed v if only he had only
> calculated the exact result of his equation 7.36 which is quite simple
> actually.
Unlike you, he obtained the right magnitude c^2 for the velocity and right
eigenvalues |v| = +-c.
He is also aware on how one may correctly extrapolate the 1D result to 3D.
> quote from Paul Strange (in figure 7.1)
>
>> The electron executes oscillatory motion within the sphere to maintain
>> a speed c at all times. As it is moving at the speed one might expect
>> it
>> to have infinite mass. This is not true because, as we have seen, the
>> motion is oscillatory.
>
> Relativity does not work because the motion is “oscillatory” ??
This is another part where you get clearly confused.
Relativity does not 'work' because the special relativistic operator is
v_avg = pc^2 / E
which gives (v_avg = c) *only* when (m = 0). A massive particle would need
infinite energy for reach v_avg = c.
But this is NOT that Dirac theory and quantum mechanics say.
But this operator does not account for Zb effects. The operator which
account for 'oscillatory' motion is
v_ins = c \alpha
which is valid for finite mass.
Misconceptions between special relativity like formulae and Dirac quantum
formulae are the basis for amateurs rejections of the idea that Dirac
electrons travel at c. Unlike you think that result is not violating
special relativity or any known experiment.
Regards
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
>
> (my previous post vanished in the air. I am resubmitting it again)
>
This discussion is getting just as ugly and meaningless
as many of the other ugly fights you seem to have all
over the place:
Some titles of your posts for example in this thread:
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/be956bf14b9bff00#
"Re: new ad hominem and lie by "Dirk van de Mortel"
"Re: new lies by professional liar Tom Roberts"
"Re: new lie and ad hominem by crackpot Eric"
Well, I refuse to get dragged down into the dirt and
I guess the only way to do so is stop responding
to your post and let your insinuations and insults
for what they are.
That's made somewhat easier after seeing that I'm
in good company. Like Steve Carlip for instance
recently when he claims that speed of gravity is c
http://groups.google.com/group/sci.physics.relativity/msg/d647ae6731d38506
Steve Carlip is "completely wrong" according to
you and you claim the speed of gravity is infinite...
Steve, as one of the most respected scientist in
General Relativity is a "professional liar" according
to your blog:
http://canonicalscience.blogspot.com/
Which is one big angry rant against half the world.
Regards, Hans.
======================================= MODERATOR'S COMMENT:
Indeed, it is time to terminate this discussion
Juan R.
(snip nasty stuff)
> you claim the
> speed of gravity is infinite...
I have never claimed that and I will never do.
In fact, the word "infinite" is not to be found in the message you cited
less still your whole phrase.
Stop from attributing to me stuff I am not saying.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
======================================= MODERATOR'S COMMENT:
This mail is passed before closing this thread for denying a statement in the foregoing post
Hans de Vries
<juanREM...@canonicalscience.com> wrote:
> I have never claimed that and I will never do.
>
> In fact, the word "infinite" is not to be found in the message you cited
> less still your whole phrase.
>
> Stop from attributing to me stuff I am not saying.
Well. this is from your post:
http://groups.google.com/group/sci.physics.relativity/msg/d647ae6731d38506
" The conclusion is always the same:
STATEMENT
The speed of gravitational and electromagnetic influences is not
retarded by c but both contain an *instantaneous* component."
Regards, Hans
Peter
> Peter wrote on Tue, 12 Aug 2008 07:00:18 -0600:
>
> >> My impression is that the fact that v is an operator in spinor state
> >> (4x4 matrix) is not suffiently accounted for. What is the meaning of the
> >> 16 matrix elements?
> In <psi|alpha|psi>, the spin indices are contracted, so that they
> disappear.
I know - do wish to say that they don't have got a physical meaning? If yes,
they should be removed using Ockham's razor - and, possibly, Dirac himself
had done this eventually
> Note that the Dirac Hamiltonian has spin indices and 16 components too.
What is the meaning of these components?
Thank you,
Peter
FrediFizzx
You can see that perhaps Hestenes has given them more physical
meaning.
http://modelingnts.la.asu.edu/pdf/ZBW_I_QM.pdf
http://arxiv.org/abs/0802.2728
Fred
Hans de Vries
They do have a simple physical meaning.
In case of the velocity operator they are equal
to the boost generators of the spinors. See for
instance Peskin and Schroeder eq (3.26)
They can be used to transform the spinor
from one reference frame to another but
also to determine the rest-frame of the
spinor, which then gives you its speed.
The Dirac field has a spinor component which
denotes a spin density. You can extract the
3 spin-vector density components with the
use of the spinor rotation generators. See
Peskin and Schroeder (3.27)
The spinor rotation generator can be used
to rotate the spinor over an arbitrary angle
but they can also be used to determine
the direction of the spin vector of the spinor.
Regards, Hans
http://www.physics-quest.org
http://physics-quest.org/Book_Chapter_Dirac.pdf
http://www.physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Peter
> >> >> My impression is that the fact that v is an operator in spinor state
> >> >> (4x4 matrix) is not suffiently accounted for. What is the
> >> >> meaning of the 16 matrix elements?
> >> In <psi|alpha|psi>, the spin indices are contracted, so that they
> >> disappear.
> > I know - do wish to say that they don't have got a physical meaning?
> > If yes,
> > they should be removed using Ockham's razor - and, possibly, Dirac
> > himself had done this eventually
> >> Note that the Dirac Hamiltonian has spin indices and 16 components
> >> too.
> > What is the meaning of these components?
> You can see that perhaps Hestenes has given them more physical
> meaning.
>
> http://modelingnts.la.asu.edu/pdf/ZBW_I_QM.pdf
> http://arxiv.org/abs/0802.2728
>
> Fred
Thank you very much!
Will print-out and read them and hopefully come back ;-)
Peter
Peter
This sounds plausible, thank you!
Peter