What is the velocity of a relativistic electron?

Juan R. González-Álvarez

I find different answers:

((Classical electrodynamics))

v < c

((Relativistic Quantum Mechanics))

v = c using Dirac equation for |PSI>.

((Space-time approach to QED))

v = c because the spacetime kernel K_+ is derived from the Dirac equation
for |PSI>.

Feynman even writes in his well-known textbook on QED [1] that

{BLOCKQUOTE
This result is sometimes made pausible 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, 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. }

Then in a footnote, Feynman recognizes that argument is invalid since v
commutes with p. Thus, an incorrect value is justified using an incorrect
argument!

((Field approach to QED))

I have computed the value of v using also the field approach to QED. I.e
the value for the v corresponding to a Dirac field PHI(x,t) is again c.

Conventional thinking says that position is not an observable in field QED
(just a parameter) and thus v cannot be indentified with the velocity of a
particle.

This argument is difficult to accept because is if x and v are not
observables at some fundamental level of theory, then,

i) why there exist observables x and v at level of non-relativistic
quantum mechanics?

ii) How does one obtain classical observables x and v?

Already Dirac warned about QED but his quote is almost ignored (only his
related quote on renormalization is usually cited). The quote on
incompatibility of QED with previous theories is

{BLOCKQUOTE
Most physicists are very satisfied with this situation. They argue that if
one has rules for doing calculations and the results agree with
observation, that is all that one requires. But it is not all that one
requires. One requires a single comprehensive theory applying to all
physical phenomena. Not one theory for dealing with non relativistic
effects and a separate disjoint theory for dealing with certain
relativistic effects. [...] For these reasons I find the present quantum
electrodynamics quite unsatisfactory. }

Igor Khavkine

On 2008-02-13, Juan R. Gonzàlez-Àlvarez <Juan@canonicalscience.com> wrote:
> I find different answers:
>
> ((Classical electrodynamics))
>
> v < c
>
> ((Relativistic Quantum Mechanics))
>
> v = c using Dirac equation for |PSI>.

Eh, no.

You are confused by an archaic paradox. The paradox consists of
identifying the operator c*alpha (in the usual 3+1 Dirac equation
notation) with the physical velocity, which quite obviously doesn't
work. For a modern discussion of the Dirac electron's velocity operator,
see [1] where the author discusses the Foldy-Wouthuysen transformation
and how to use it to get the correct operator obeservable. See also [2]
for some more discussion and references.

The velocity operator thus obtained, satisfies the Heisenberg equations
of motion v = dx/dt and the well known relation v = p/m in the
non-relativistic limit. Note, here x is not the spatial argument of the
usual Dirac wave function. It is also obtained through the
Foldy-Wouthuysen transformation, or from the formula given by Newton and
Wigner (again, see [2]). Once a particular time axis has been chosen,
this operator x is unique (up to translations of the origin of
coordinates).

[...]

> i) why there exist observables x and v at level of non-relativistic
> quantum mechanics?

They exist alright, as evidenced by any elementary treatment of QM. What
you are perhaps wondering is how are the x and v observables generalized
to relativistic single particle QM and to QFT.

The answer to the first part of this question is contained in the
references I gave above. The generalization from relativistic
single-particle QM to QFT follows standard methods of second
quantization [3].

> ii) How does one obtain classical observables x and v?

Given that x and v exist and QM operators, the corresponding classical
observables are obtained as expectation values of these operators with
respect to coherent sates, which embed the classical phase space in the
QM Hilbert space of states.

It is important to note that the cascade QFT -> rel QM -> non-rel QM -> CM
is only a limiting procedure. There are quantum effects that become
non-negligible when a particle is no longer described by a coherent
state. There are relativistic effects that become non-negligible when
the particles speed aproaches the speed of light. There are
multi-particle effects that become non-negligible when the
single-particle sector cannot be sufficiently decoupled from the rest of
the QFT Fock space.

> Already Dirac warned about QED but his quote is almost ignored (only his
> related quote on renormalization is usually cited). The quote on
> incompatibility of QED with previous theories is
>
> {BLOCKQUOTE
> Most physicists are very satisfied with this situation. They argue that if
> one has rules for doing calculations and the results agree with
> observation, that is all that one requires. But it is not all that one
> requires. One requires a single comprehensive theory applying to all
> physical phenomena. Not one theory for dealing with non relativistic
> effects and a separate disjoint theory for dealing with certain
> relativistic effects. [...] For these reasons I find the present quantum
> electrodynamics quite unsatisfactory. }

I think Dirac would have been happy today, since his dream has been
achieved.

[1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
Chapter 7.

[2] J.P. Costellay and B.H.J. McKellarz
The Foldy-Wouthuysen transformation
arXiv:hep-ph/9503416

[3] Paul A. M. Dirac, _Principles of Quantum Mechanics_.

Igor

Juan R.

Igor Khavkine wrote on Fri, 15 Feb 2008 20:47:23 +0000:

> On 2008-02-13, Juan R. Gonzlez-lvarez <Juan@canonicalscience.com> wrote:
>> I find different answers:
>>
>> ((Classical electrodynamics))
>>
>> v < c
>>
>> ((Relativistic Quantum Mechanics))
>>
>> v = c using Dirac equation for |PSI>.
>
> Eh, no.
>
> You are confused by an archaic paradox. The paradox consists of
> identifying the operator c*alpha (in the usual 3+1 Dirac equation
> notation) with the physical velocity, which quite obviously doesn't
> work. For a modern discussion of the Dirac electron's velocity operator,
> see [1] where the author discusses the Foldy-Wouthuysen transformation
> and how to use it to get the correct operator obeservable. See also [2]
> for some more discussion and references.
>
> The velocity operator thus obtained, satisfies the Heisenberg equations
> of motion v = dx/dt and the well known relation v = p/m in the
> non-relativistic limit. Note, here x is not the spatial argument of the
> usual Dirac wave function. It is also obtained through the
> Foldy-Wouthuysen transformation, or from the formula given by Newton and
> Wigner (again, see [2]). Once a particular time axis has been chosen,
> this operator x is unique (up to translations of the origin of
> coordinates).

Actually the author of reference [1] does not maintain your own
distinction between "correct" and "incorrect" velocity operators. He
is
saying something *different* and, in fact, he is supporting my point.

By velocity operator i did mean its standard definition, obviously.
>From [1] one reads:

{BLOCKQUOTE [emphasis mine]
We have seen that the *standard* velocity operator gives the electron
speed as equal to the velocity of the light.
}

Next he discusses a *different* velocity operator (his eq. 7.41) and
adds

{BLOCKQUOTE [emphasis mine]
These two can be *reconciled* if we identify (7.41) with the *average*
velocity of the particle.
}

Then the *full* motion of the electron can be divided into two parts

{BLOCKQUOTE [emphasis mine]
Firstly there is the *average* velocity (7.41) and secondly there is
very
rapid oscillatory motion which ensures that if an instantaneous
measurement of the *velocity* of the electron could be done it would
give c.
}

It seems that the author of [1] think one cannot measure velocities
when
says "if an instantaneous measurement [...] could be done".

This argument usually goes over discussion of sequential measurements,
thus position at two times and then taking the limit when time
interval
goes to zero. This is also the point on [1].

This argument however is not acceptable, which is also the remark done
by
Feynman in the footnote on the page i cited (see
sci.physics.foundations,
where my post is not truncated, for the reference).

Feynman (and myself) point is that v commutes with p, therefore,
instantaneous measurement of v are possible (at least in theory).

Now we would analyze the author [1] is really showing. He splits the
"velocity of the electron" into two parts, say

v = <v> + Dv

he identifies <v>, the *average* velocity, with

{BLOCKQUOTE
the average motion of the particle, i.e. that given by classical
relativistic formulae
}

Thus <v> does not describe the full velocity of the *quantum*
particle.

And the author gives a value c for the latter. A simple transformation
cannot change that fact.

>> i) why there exist observables x and v at level of non-relativistic
>> quantum mechanics?
>
> They exist alright, as evidenced by any elementary treatment of QM. What
> you are perhaps wondering is how are the x and v observables generalized
> to relativistic single particle QM and to QFT.
>
> The answer to the first part of this question is contained in the
> references I gave above. The generalization from relativistic
> single-particle QM to QFT follows standard methods of second
> quantization [3].

I was just asking an inverse question. We know x is a classical
parameter in
relativistic quantum field theory. Thus the (unsolved) question is:

How a non-observable transform into an observable in the non-
relativistic
limit?

This is one of inconsistencies troubled Dirac.

>> ii) How does one obtain classical observables x and v?
>
> Given that x and v exist and QM operators, the corresponding classical
> observables are obtained as expectation values of these operators with
> respect to coherent sates, which embed the classical phase space in the
> QM Hilbert space of states.
>
> It is important to note that the cascade QFT -> rel QM -> non-rel QM ->
> CM is only a limiting procedure. There are quantum effects that become
> non-negligible when a particle is no longer described by a coherent
> state. There are relativistic effects that become non-negligible when
> the particles speed aproaches the speed of light. There are
> multi-particle effects that become non-negligible when the
> single-particle sector cannot be sufficiently decoupled from the rest of
> the QFT Fock space.

You missed my point, again.

If relativistic quantum field theory says that position is not an
observable. How it is supposed that something cannot be observed at
some
supposed fundamental level turns into a observable in the classical
limit?

>> Already Dirac warned about QED but his quote is almost ignored (only
>> his related quote on renormalization is usually cited). The quote on
>> incompatibility of QED with previous theories is
>>
>> {BLOCKQUOTE
>> Most physicists are very satisfied with this situation. They argue that
>> if one has rules for doing calculations and the results agree with
>> observation, that is all that one requires. But it is not all that one
>> requires. One requires a single comprehensive theory applying to all
>> physical phenomena. Not one theory for dealing with non relativistic
>> effects and a separate disjoint theory for dealing with certain
>> relativistic effects. [...] For these reasons I find the present
>> quantum electrodynamics quite unsatisfactory. }
>
> I think Dirac would have been happy today, since his dream has been
> achieved.

Difficult to believe since Dirac issues remain unsolved at present
date.

In [hep-th0501222], It is stated that Dirac maintained his
criticism on QED in a talk on 1983.

Moreover the quote is rather large and finalizes with Dirac hope for
the
development of a full and consistent relativistic quantum mechanics.

Several groups in all the world are now searching one.

> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
> Chapter 7.
>
> [2] J.P. Costellay and B.H.J. McKellarz
> The Foldy-Wouthuysen transformation
> arXiv:hep-ph/9503416
>
> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_.
>
> Igor


--
I follow http://canonicalscience.org/en/misce...guidelines.txt

Igor Khavkine

On 2008-02-16, Juan R. <juanrgonzaleza@canonicalscience.com> wrote:
> Igor Khavkine wrote on Fri, 15 Feb 2008 20:47:23 +0000:
>
>> On 2008-02-13, Juan R. Gonzlez-lvarez <Juan@canonicalscience.com> wrote:
>>> I find different answers:
>>>
>>> ((Classical electrodynamics))
>>>
>>> v < c
>>>
>>> ((Relativistic Quantum Mechanics))
>>>
>>> v = c using Dirac equation for |PSI>.
>>
>> Eh, no.
>>
>> You are confused by an archaic paradox. The paradox consists of
>> identifying the operator c*alpha (in the usual 3+1 Dirac equation
>> notation) with the physical velocity, which quite obviously doesn't
>> work. For a modern discussion of the Dirac electron's velocity operator,
>> see [1] where the author discusses the Foldy-Wouthuysen transformation
>> and how to use it to get the correct operator obeservable. See also [2]
>> for some more discussion and references.
>>
>> The velocity operator thus obtained, satisfies the Heisenberg equations
>> of motion v = dx/dt and the well known relation v = p/m in the
>> non-relativistic limit. Note, here x is not the spatial argument of the
>> usual Dirac wave function. It is also obtained through the
>> Foldy-Wouthuysen transformation, or from the formula given by Newton and
>> Wigner (again, see [2]). Once a particular time axis has been chosen,
>> this operator x is unique (up to translations of the origin of
>> coordinates).
>
> Actually the author of reference [1] does not maintain your own
> distinction between "correct" and "incorrect" velocity operators. He
> is saying something *different* and, in fact, he is supporting my
> point.
>
> By velocity operator i did mean its standard definition, obviously.
>>From [1] one reads:
>
> {BLOCKQUOTE [emphasis mine]
> We have seen that the *standard* velocity operator gives the electron
> speed as equal to the velocity of the light.
> }

Velocity is defined by experiment. If the "standard" velocity operator
does not reproduce experimental measurements, then it is still not the
*correct* one. It is only unfortunate that it has somehow become the
"standard" one.

In any case, reference [1] describes how to obtain an operator that does
reproduce experimental observations of electron velocities. If you were
wondering if such an operator exists, then that's your answer. If you
were wondering why this is not the "standard" velocity operator, then
the answer has more to do with history than physics.

[...]
> Now we would analyze the author [1] is really showing. He splits the
> "velocity of the electron" into two parts, say
>
> v = <v> + Dv
>
> he identifies <v>, the *average* velocity, with
>
> {BLOCKQUOTE
> the average motion of the particle, i.e. that given by classical
> relativistic formulae
> }
>
> Thus <v> does not describe the full velocity of the *quantum*
> particle.
>
> And the author gives a value c for the latter. A simple transformation
> cannot change that fact.

I am eagerly awaiting your publications on the measurement of the "full
velocity of the quantum particle". Until then, we only have the usual
velocity measurement experiments, which are already well reproduced by
the velocity operator obtained with the help of the Foldy-Wouthuysen
transformation.

>>> i) why there exist observables x and v at level of non-relativistic
>>> quantum mechanics?
>>
>> They exist alright, as evidenced by any elementary treatment of QM. What
>> you are perhaps wondering is how are the x and v observables generalized
>> to relativistic single particle QM and to QFT.
>>
>> The answer to the first part of this question is contained in the
>> references I gave above. The generalization from relativistic
>> single-particle QM to QFT follows standard methods of second
>> quantization [3].
>
> I was just asking an inverse question. We know x is a classical
> parameter in relativistic quantum field theory. Thus the (unsolved)
> question is:
>
> How a non-observable transform into an observable in the non-
> relativistic limit?

Let x - position coordinates (parameter)
X - position observable (operator)
|x> - an eigenstate of the X operator
chi(x) - position-space wave function
Psi(x) - field operator
|chi> - single-particle state
|0> - Fock vacuum

1. Project onto position eigenstates: chi(x) = <x|chi>.
2. Embed state in Fock space: |chi> = int dx chi(x) Psi(x)^* |0>.
3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x).

On the LHS of step 3, we have an operator X (an observable), while on
the RHS of the same equality we have integration over a coordinate x (a
field theoretic parameter). Step 3 is the answer to your question.

This is a generic method for promoting single-particle operators to Fock
operators. In any particular situation, there will be differences.
Again, the Foldy-Wouthuysen transformation helps you pick out the right
position operator. See sections 59-65 of [3], for the general
prescription.

>> I think Dirac would have been happy today, since his dream has been
>> achieved.
>
> Difficult to believe since Dirac issues remain unsolved at present
> date.
>
> In [hep-th0501222], It is stated that Dirac maintained his
> criticism on QED in a talk on 1983.

Yes, it is unfortunate that in 1983 theoretical physics lost one of its
great thinkers. It is also unfortunate that Dirac can no longer join us
in a discussion of how his dream of a consistent treatment of
relativistic quantum mechanics has been achieved. I guess all we can do
now is rely on existing evidence and our own judgement, rather than the
opinions of someone whose mind can no longer be altered.

> Moreover the quote is rather large and finalizes with Dirac hope for
> the development of a full and consistent relativistic quantum
> mechanics.
>
> Several groups in all the world are now searching one.

I wish them the best of luck.

>> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
>> Chapter 7.
>>
>> [2] J.P. Costellay and B.H.J. McKellarz
>> The Foldy-Wouthuysen transformation
>> arXiv:hep-ph/9503416
>>
>> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_.

Igor

Juan R.

On Feb 17, 9:01 am, Igor Khavkine <igor...@gmail.com> wrote:
> On 2008-02-16, Juan R. <juanrgonzal...@canonicalscience.com> wrote:

> > By velocity operator i did mean its standard definition, obviously.
> >>From [1] one reads:
>
> > {BLOCKQUOTE [emphasis mine]
> > We have seen that the *standard* velocity operator gives the electron
> > speed as equal to the velocity of the light.
> > }
>
> Velocity is defined by experiment.

Velocity is defined by theory. Experimental outcomes are always
interpreted *inside* the framework of some theory.

> If the "standard" velocity operator
> does not reproduce experimental measurements, then it is still not the
> *correct* one. It is only unfortunate that it has somehow become the
> "standard" one.

There is nothing "unfortunate" except your insistence on cataloging
operators into "correct" and "incorrect". The terminology used in [1]
is more accurate:

- velocity operator for v

- *average* velocity operator for <v>.

As explained in [1] the *average* velocity operator corresponds to
measurements of the average velocity. Author on [1] does also clear
that the average measurement correspond to

{BLOCKQUOTE [1]
the average motion of the particle
}

Moreover, your claim that the standard velocity operator disagrees
with measurements is plain false.

As explained in my previous message (and in quotations from [1]) the
velocity operator can be divided into two components (see also
relevant equations on [1])

{BLOCKQUOTE [emphasis mine]
Firstly there is the *average* velocity (7.41) and secondly there is
very rapid oscillatory motion
}

v = <v> + Dv

If you are measuring *only* the *average* motion of the particle, then
the Dv term may vanish and

v = <v>


> In any case, reference [1] describes how to obtain an operator that does
> reproduce experimental observations of electron velocities.

Yes, it describes how to define an *average* operator for observations
of *average* velocities. I already explained this before also. Just
search in my previous comments when i wrote about the equation (7.41)
on reference [1]

But how the title of this thread reminder us, my question was "what is
the velocity of an relativistic electron?"

This is also answered on reference [1]:

{BLOCKQUOTE
We have seen that the standard velocity operator gives the electron
speed as equal to the velocity of the light.
}

[1] also remarks that velocity operator corresponds to actual velocity
of the electron, whereas the average operator only corresponds to
measurements of average velocities.

> I am eagerly awaiting your publications on the measurement of the "full
> velocity of the quantum particle". Until then, we only have the usual
> velocity measurement experiments, which are already well reproduced by
> the velocity operator obtained with the help of the Foldy-Wouthuysen
> transformation.

As explained in [1] those "usual velocity measurement" correspond to

{BLOCKQUOTE [emphasis mine]
the *average* motion of the particle, i.e. that given by classical
relativistic formulae
}

I was asking for the actual velocity of the electron.

And i already said this before (but you ignored twice) Feynman (and
myself also) shows that two-times arguments are not acceptable because
of the commutation relations between v and p.

> > How a non-observable transform into an observable in the non-
> > relativistic limit?
>
> Let x - position coordinates (parameter)
> X - position observable (operator)
> |x> - an eigenstate of the X operator
> chi(x) - position-space wave function
> Psi(x) - field operator
> |chi> - single-particle state
> |0> - Fock vacuum
>
> 1. Project onto position eigenstates: chi(x) = <x|chi>.
> 2. Embed state in Fock space: |chi> = int dx chi(x) Psi(x)^* |0>.
> 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x).
>
> On the LHS of step 3, we have an operator X (an observable), while on
> the RHS of the same equality we have integration over a coordinate x (a
> field theoretic parameter). Step 3 is the answer to your question.
>
> This is a generic method for promoting single-particle operators to Fock
> operators. In any particular situation, there will be differences.
> Again, the Foldy-Wouthuysen transformation helps you pick out the right
> position operator. See sections 59-65 of [3], for the general
> prescription.

Unfortunately my complaints already start with your step 1.

It seems you do not understand difference between classical parameters
and quantum observables. It seems also you lack knowledge of basic
relativistic localization issues. But let us continue.

I am not sure if my question was understood. First I will repeat it by
third time.

Relativistic quantum field theory (supposedly a fundamental theory)
treat position x as a classical parameter (i.e. *not* an observable)
whereas non-relativistic quantum mechanics treat position x as an
*observable*.

The difficulty arises in a natural way. How does something (x) is not
observable (not even being quantum!) transform into something is
quantum and observable by making approximations?

This makes no sense for me.

That position is not a quantum observable in relativistic quantum
field theory is noticed in several places. E.g. [I]

{BLOCKQUOTE [emphasis in the original]
The symmetry between space and time is restored by regarding position,
as well, as time as a /classical/ parameter, so that the classical
four-vector (x, t) is the space-time point at or around which we make
our /field/ measurement.
}

That position observables are mathematically incompatible with
relativistic requirements is also known. See, for example, section
"1.6 The Position operator" on [II]. The section finalizes with

{BLOCKQUOTE
The outcome, then, is that a position operator is inconsistent with
relativity. [...] In field theory, we model localization by making the
observables dependent on position in space-time.
}

But that spacetime is a classical construct. This introduces
foundational issues [I]

{BLOCKQUOTE [emphasis in the original]
Field-theorists do not solve the non-locality of the one-particle
[quantum mechanical] theory. Rather, they avoid the problem [...]
}

Thus Dirac complaints still hold.

{BLOCKQUOTE [emphasis mine]
Most physicists are very satisfied with this situation. They argue
that if
one has rules for doing calculations and the results agree with
observation, that is all that one requires. But it is not all that one
requires. One requires *a single comprehensive theory* applying to
*all*
physical phenomena. Not *one theory for* dealing with *non
relativistic
effects* and *a separate disjoint theory for* dealing with *certain
relativistic effects*. [...] For these reasons I find the present
quantum
electrodynamics quite unsatisfactory. }

The solution? Dirac already advanced which one would be acceptable.

We may develop a full and consistent relativistic quantum mechanics.
Relativistic quantum field theory would be a *special* case derived
from a really fundamental relativistic quantum mechanics theory. The
scheme wold be

Relativistic quantum mechanics ---> Relativistic quantum field theory
|
|
v
'Wave' relativistic quantum mechanics
|
|
v
Non-relativistic quantum mechanics

There are several recent relativistic quantum mechanics theory
proposals on literature. I have illustrated the basis for some in a
recent posting in this newsgroup. To be remarked all of them address
questions are *avoided* by relativistic quantum field theory.

In my opinion the current search for a unification of General
Relativity and quantum theory is a waste of time (and it has been
indeed). One wold unify special relativity and the quantum first.

> Yes, it is unfortunate that in 1983 theoretical physics lost one of its
> great thinkers. It is also unfortunate that Dirac can no longer join us
> in a discussion of how his dream of a consistent treatment of
> relativistic quantum mechanics has been achieved.

First, QED has not changed in any fundamental way since 1983. And,
second, no single comprehensive theory applying to both relativistic
and non-relativistic phenomena" is available today (QED continues to
work only for infinite time scattering processes between free fields
over a classical space-time).

I believe that Dirac very much would agree with me today.

> >> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
> >> Chapter 7.
>
> >> [2] J.P. Costellay and B.H.J. McKellarz
> >> The Foldy-Wouthuysen transformation
> >> arXiv:hep-ph/9503416
>
> >> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_.
>
> Igor

[I] Why axiomatic field Theory? Ray F. Streater. In Philosophical
foundations of Quantum field Theory. 1990. Oxford University Press,
USA. Harvey R. Brown; Rom Harre (Editors). p.143

[II] Quantum Field Theory for Mathematicians. 1999. Cambridge
University Press. Robin Ticciati


--
I follow http://canonicalscience.org/en/misce...guidelines.txt

Arnold Neumaier

Juan R. wrote;

> (QED continues to
> work only for infinite time scattering processes between free fields
> over a classical space-time).

This is not true. One can compute - nonrigorously, in renormalized
perturbation theory - many time-dependent things, namely via the
Schwinger-Keldysh (or closed time path = CPT) formalism; see, e.g.,
http://theory.gsi.de/~vanhees/publ/green.pdf


See also the entry
S9c. What about relativistic QFT at finite times?
in my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt


Arnold Neumaier

Igor Khavkine

On 2008-02-26, Juan R. <juanrgonzaleza@canonicalscience.com> wrote:
> On Feb 17, 9:01 am, Igor Khavkine <igor...@gmail.com> wrote:
>> On 2008-02-16, Juan R. <juanrgonzal...@canonicalscience.com> wrote:

>> Velocity is defined by experiment.
>
> Velocity is defined by theory. Experimental outcomes are always
> interpreted *inside* the framework of some theory.

Well, then I guess that settles the question. I'm sure that people who
do electron transport measurements will be happy to hear that they need
to interpret their data so that electron velocities always come out
equal to the speed of light (to be consistent with the theoretical
definition given in your original post), instead of the centimeters per
second that their instruments have been telling them all along.

>> I am eagerly awaiting your publications on the measurement of the "full
>> velocity of the quantum particle". Until then, we only have the usual
>> velocity measurement experiments, which are already well reproduced by
>> the velocity operator obtained with the help of the Foldy-Wouthuysen
>> transformation.
>
> As explained in [1] those "usual velocity measurement" correspond to
>
> {BLOCKQUOTE [emphasis mine]
> the *average* motion of the particle, i.e. that given by classical
> relativistic formulae
> }
>
> I was asking for the actual velocity of the electron.

I see, so you want to talk about a quantity that *has not* been
measured and then compare it to a quantity that *has* been measured?
A rather strange endeavour.

> And i already said this before (but you ignored twice) Feynman (and
> myself also) shows that two-times arguments are not acceptable because
> of the commutation relations between v and p.

Juan R. originally wrote:
> Feynman even writes in his well-known textbook on QED [1] that
>
> {BLOCKQUOTE
> This result is sometimes made pausible 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, 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. }
>
> Then in a footnote, Feynman recognizes that argument is invalid since v
> commutes with p. Thus, an incorrect value is justified using an
> incorrect
> argument!

Let me get this straight. You are saying that, for an electron, v is not
equal to c. Also, both you and Feynman are saying that the argument
presented above does not actually show that v = c. I completely agree
with both these statements, as long as v is the same velocity that
measures at centimeters per second in typical copper wire.

>> > How a non-observable transform into an observable in the non-
>> > relativistic limit?
>>
>> Let x - position coordinates (parameter)
>> X - position observable (operator)
>> |x> - an eigenstate of the X operator
>> chi(x) - position-space wave function
>> Psi(x) - field operator
>> |chi> - single-particle state
>> |0> - Fock vacuum
>>
>> 1. Project onto position eigenstates: chi(x) = <x|chi>.
>> 2. Embed state in Fock space: |chi> = int dx chi(x) Psi(x)^* |0>.
>> 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x).
>>
>> On the LHS of step 3, we have an operator X (an observable), while on
>> the RHS of the same equality we have integration over a coordinate x (a
>> field theoretic parameter). Step 3 is the answer to your question.
>>
>> This is a generic method for promoting single-particle operators to Fock
>> operators. In any particular situation, there will be differences.
>> Again, the Foldy-Wouthuysen transformation helps you pick out the right
>> position operator. See sections 59-65 of [3], for the general
>> prescription.
>
> Unfortunately my complaints already start with your step 1.
>
> It seems you do not understand difference between classical parameters
> and quantum observables. It seems also you lack knowledge of basic
> relativistic localization issues. But let us continue.

I will let my record attest to my understanding of the matter (or lack
thereof). As to your own record, you've yet to say much concrete on the
subject at all. While I may not be an expert in relativistic
localization, I understand the topic well enough to point out that the
question below has nothing in particular to do with relativity.

> Relativistic quantum field theory (supposedly a fundamental theory)
> treat position x as a classical parameter (i.e. *not* an observable)
> whereas non-relativistic quantum mechanics treat position x as an
> *observable*.
>
> The difficulty arises in a natural way. How does something (x) is not
> observable (not even being quantum!) transform into something is
> quantum and observable by making approximations?

The answer, as I've already stated, is in the formula of step 3 above.
There is no approximation, and this method works for non-relativistic as
well as relativistic theories. It describes the transition from particle
theory to field theory. The relativistic -> non-relativistic
approximation may be made at will at either end of the transition.

> Thus Dirac complaints still hold.

Unfortunately, as I've pointed out, Dirac is not here to be convinced
otherwise. Neither are any of the other authors whose objections you
have also quoted.

I've outlined a number of steps that, through a sequence of well defined
transformations and approximations, convert operators defined on the
relativistic field-theoretic Fock space to the usual position and
velocity operators on the non-relativistic particle Hilbert space, as
well as the reverse steps. In my understanding of your question, there
should be enough information here to give you an answer. While you do
not seem to be satisfied with this answer, I've yet to see you raise
concrete objections to any of these steps.

>> >> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998).
>> >> Chapter 7.

>> >> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_.

Igor

Juan R. González-Álvarez

Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000:

> Juan R. wrote;
>
>> (QED continues to
>> work only for infinite time scattering processes between free fields
>> over a classical space-time).
>
> This is not true. One can compute - nonrigorously, in renormalized
> perturbation theory - many time-dependent things, namely via the
> Schwinger-Keldysh (or closed time path = CPT) formalism; see, e.g.,
> http://theory.gsi.de/~vanhees/publ/green.pdf

{BLOCKQUOTE [1]
The closed time path formalism proposed by Schwinger and refined by many
others, notable Keldysh, has been frequently used to study nonequilibrium
situations. But a close inspection of the formalism reveals that it is
more of a steady state theory than a truly time-dependent theory.
}

"steady state theory" = "time-independent theory"

See also the prose before and after (eq 13) and after (eq 36) in the pfd
you are citing.

Moreover, the SK formalism is based in inconsistent mixture of elements
from contradictory theories, e.g. Feynman diagrams for S-matrix (only
defined for infinite times) are mixed with a LvN equation for finite
times. See also for extra criticism on SK [1].

>
> See also the entry
> S9c. What about relativistic QFT at finite times?
> in my theoretical physics FAQ at
> http://www.mat.univie.ac.at/~neum/physics-faq.txt

I agree when says:

{BLOCKQUOTE
Current 4D QFT is based on perturbation theory for free
(i.e., asymptotic in- and out-) states; therefore it gives only
predictions that relate the in- and out-states.

[...]

In nonrelativistic QM, one has a well-defined dynamics at finite
times, given by the Schroedinger equation. This dynamics can be recast
in terms of Feynman path integrals. Unfortunately, this does not
extend to the relativistic case.

[...]

But I haven't seen a single article that gives meaning (i.e.,
infrared and ultraviolet finite, renormalization scheme independent
properties) to, say, quantum electrodynamics states at finite t and
their propagation in time.

People don't even know what an initial state should be in a
relativistic QFT (i.e., from which space to take the states at
finite t); so how can they know how to propagate it...

Thus the standard theory gives an S-matrix (or rather an asymptotic
series for it) but not a dynamics at finite times.
}

and disagree with other claims.

I can see you do not cite any of the classical results over which
relativistic QFT was developed. For instance, the Peirls-Landau
inequality is the reason S = U(-infinity, +infinity) and that QFT lacks
dynamical description at finite times.

It is interesting you cite the recent work of my colleague Eugene V.
Stefanovich. He agrees with me on the non-existence of finite time
relativistic QFT. E.g. see section 9.1.2 on [2].

You also write in the FAQ

{BLOCKQUOTE
The missing consistent dynamical theory in 4D relativistic QFT
may also have consequences for the foundations of quantum mechanics.
Clearly, measurements happen in finite time, hence cannot
be described at present in a fundamental way (i.e., beyond the
nonrelativistic QM approximation). Thus foundational
studies based on nonrelativistic QM are naturally incomplete.
This implies that it is quite possible that a solution of the
unresolved issues in relativistic QFT are related to the unresolved
issues in quantum measurement theory.
}

One of main points of Eugene (and others, including me) affirms there is
*not* quantum dynamical relativistic theory based in Minkowski space-time
unification: "The Einstein-Minkowski 4-dimensional spacetime is an
approximate concept as well."

Do not forget that relativistic QFT breaks with quantum mechanics and
treats spacetime as a *classical* object. See references and quotations
in my previous message.


[1] A unified formalism of thermal quantum field theory. 1994. Int. J.
of Mod. Phys A 9(14), 2363. Chu, H; Umezawa, H.

[2] http://arxiv.org/abs/physics/0504062


--
I apply http://canonicalscience.org/en/misce...guidelines.txt

Igor Khavkine

On 2008-03-07, Juan R. González-�lvarez <juan@canonicalscience.com> wrote:
> Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000:
>> Juan R. wrote;
>>
>>> (QED continues to
>>> work only for infinite time scattering processes between free fields
>>> over a classical space-time).
>>
>> This is not true. One can compute - nonrigorously, in renormalized
>> perturbation theory - many time-dependent things, namely via the
>> Schwinger-Keldysh (or closed time path = CPT) formalism; see, e.g.,
>> http://theory.gsi.de/~vanhees/publ/green.pdf
>
> {BLOCKQUOTE [1]
> The closed time path formalism proposed by Schwinger and refined by many
> others, notable Keldysh, has been frequently used to study nonequilibrium
> situations. But a close inspection of the formalism reveals that it is
> more of a steady state theory than a truly time-dependent theory.
> }
>
> "steady state theory" = "time-independent theory"

> [1] A unified formalism of thermal quantum field theory. 1994. Int. J.
> of Mod. Phys A 9(14), 2363. Chu, H; Umezawa, H.

This reference is from the authors of a formalism called "thermo field
dynamics" (many papers and at least one textbook has been written on the
subject). From your quote it appears that the authors perceive the
Schwinger-Keldysh formalism as a direct competitor and alternative,
distinct from theirs. The point of both, BTW, is to capture time
dependent phenomena in field theory. Unfortunately, the authors of [1]
are mistaken. A very clear side by side comparison of "thermo field
dynamics" and the Schwinger-Keldysh formalism reveals that the latter
completely reproduce and is more general than the former.

Lawrie, I. D.
On the relationship between thermo field dynamics and
quantum statistical mechanics
J. Phys. A: Math. Gen. 27 (1994) 1435-1452

> See also the prose before and after (eq 13) and after (eq 36) in the pfd
> you are citing.
>
> Moreover, the SK formalism is based in inconsistent mixture of elements
> from contradictory theories, e.g. Feynman diagrams for S-matrix (only
> defined for infinite times) are mixed with a LvN equation for finite
> times. See also for extra criticism on SK [1].

Perhaps you'd like to share some specific criticism of (eq 13) and
(eq 36) from van Hees's notes and why the Schwinger-Keldysh formalism
appear inconsistent to you.

>> See also the entry
>> S9c. What about relativistic QFT at finite times?
>> in my theoretical physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physics-faq.txt

> I can see you do not cite any of the classical results over which
> relativistic QFT was developed. For instance, the Peirls-Landau
> inequality is the reason S = U(-infinity, +infinity) and that QFT lacks
> dynamical description at finite times.

It's unclear what you mean here. Peierls and Landau only wrote two
papers together, neither of which dealt with this topic.

> It is interesting you cite the recent work of my colleague Eugene V.
> Stefanovich. He agrees with me on the non-existence of finite time
> relativistic QFT. E.g. see section 9.1.2 on [2].

Eugene's misconceptions regarding the content of QED have been pointed
out numerous times in this group.

Igor

Arnold Neumaier

Juan R. González-�lvarez wrote
(in the thread: What is the velocity of a relativistic electron?):
> Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000:
>> Juan R. wrote;
>>
>>> (QED continues to
>>> work only for infinite time scattering processes between free fields
>>> over a classical space-time).
>> This is not true. One can compute - nonrigorously, in renormalized
>> perturbation theory - many time-dependent things, namely via the
>> Schwinger-Keldysh (or closed time path = CPT) formalism; see, e.g.,
>> http://theory.gsi.de/~vanhees/publ/green.pdf
>
> {BLOCKQUOTE [1]
> The closed time path formalism proposed by Schwinger and refined by many
> others, notable Keldysh, has been frequently used to study nonequilibrium
> situations. But a close inspection of the formalism reveals that it is
> more of a steady state theory than a truly time-dependent theory.
> }

This ''close inspection'' cannot have a deep basis.
How would it then be possible that the paper
E. Calzetta and B. L. Hu,
Nonequilibrium quantum fields: Closed-time-path effective action,
Wigner function, and Boltzmann equation,
Phys. Rev. D 37 (1988), 2878-2900.
derives finite-time Boltzmann-type kinetic equations from quantum
field theory using the CTP formalism?


> Moreover, the SK formalism is based in inconsistent mixture of elements
> from contradictory theories, e.g. Feynman diagrams for S-matrix (only
> defined for infinite times) are mixed with a LvN equation for finite
> times. See also for extra criticism on SK [1].

Any current formalism for interacting quantum field theory is
inconsistent and ill-defined when viewed from a logical point of view.

Nevertheless, people derive from it predictions, often very accurate ones.


>> See also the entry
>> S9c. What about relativistic QFT at finite times?
>> in my theoretical physics FAQ at
>> http://www.mat.univie.ac.at/~neum/physics-faq.txt
>
> I agree when says:
>
> {BLOCKQUOTE
> Current 4D QFT is based on perturbation theory for free
> (i.e., asymptotic in- and out-) states; therefore it gives only
> predictions that relate the in- and out-states.

This referred to the standard textbook approach.
I improved the FAQ by providing clarifying words.


> But I haven't seen a single article that gives meaning (i.e.,
> infrared and ultraviolet finite, renormalization scheme independent
> properties) to, say, quantum electrodynamics states at finite t and
> their propagation in time.

Nevertheless, there are very useful approximations towards that goal,
which I had mentioned in the part that follows, but which you
didn't quote.


> I can see you do not cite any of the classical results over which
> relativistic QFT was developed. For instance, the Peirls-Landau
> inequality is the reason S = U(-infinity, +infinity) and that QFT lacks
> dynamical description at finite times.

There is little point in citing old results which no longer reflect
the current state of the art.


> It is interesting you cite the recent work of my colleague Eugene V.
> Stefanovich. He agrees with me on the non-existence of finite time
> relativistic QFT. E.g. see section 9.1.2 on [2].

Quoting a paper does not mean that I consent to every statement in it.
A few years ago we had extensive discussions here on s.p.r. where
you can find out what I think of his work. Many of his claims at that
time were immature, and what you write below about his current views
does not indicate that things have changed.


> You also write in the FAQ
>
> {BLOCKQUOTE
> The missing consistent dynamical theory in 4D relativistic QFT
> may also have consequences for the foundations of quantum mechanics.

The point here is ''consistent'', which means logically impeccable.
The theory of Stefanovich has the same logical flaws as all current
4D quantum field theory - no convergence theory which would prove
that the objects to which perturbative or nonperturbative
approximations are computed are logically well-defined.

I had read the first version of his online book (your reference [2]),
and it presents a view which completely ignores much of what has
been established in the field. His arguments are provably wrong
in 2D quantum field theory. So there is no reason to trust them in 4D.


> Clearly, measurements happen in finite time, hence cannot
> be described at present in a fundamental way (i.e., beyond the
> nonrelativistic QM approximation). Thus foundational
> studies based on nonrelativistic QM are naturally incomplete.
> This implies that it is quite possible that a solution of the
> unresolved issues in relativistic QFT are related to the unresolved
> issues in quantum measurement theory.
> }
>
> One of main points of Eugene (and others, including me) affirms there is
> *not* quantum dynamical relativistic theory based in Minkowski space-time
> unification: "The Einstein-Minkowski 4-dimensional spacetime is an
> approximate concept as well."

I find this view mistaken and not supported by the existing
research literature.


> Do not forget that relativistic QFT breaks with quantum mechanics

No. It is fully based on quantum mechanics, as my FAQ
amply demonstrates.


[2] http://arxiv.org/abs/physics/0504062


Arnold Neumaier

Juan R. González-Álvarez

Igor Khavkine wrote on Sun, 02 Mar 2008 23:24:23 +0000:

> Well, then I guess that settles the question. I'm sure that people who
> do electron transport measurements will be happy to hear that they need
> to interpret their data so that electron velocities always come out
> equal to the speed of light (to be consistent with the theoretical
> definition given in your original post), instead of the centimeters per
> second that their instruments have been telling them all along.

More of the same.

What velocities are measuring? Instantaneous velocities or averaged
velocities?

Each one has its own operator. This was discussed four or five times
before.

It is ironic, you also seem completely unaware on how atomic and molecular
physicists, and quantum chemists work.

Take the classical (Darwin) expression

U_D = -1/2 (v v' + v r v' r) (e^2/r) 1/c^2

For a Dirac electron the velocity operator is

v_op = [H,x] = c alpha

substitution {v --> v_op} gives

U_B = -1/2 (alpha alpha' + alpha r alpha' r) (e^2/r)

This is the Dirac Breit operator broadly used in atomic and molecular
physics, and quantum chemistry.

Another example is the Gaunt potential, a standard in relativistic
molecular hamiltonians for bound states. Again its derivation involves

v --> c alpha.

Ironically, this is also the way Feynman derived QED -now considered the
most precise theory of physics-. Take a look to his historical paper [5].

Feynman starts from the classical interaction Lagrangian (he uses c = 1)

{ 1 - (v v') delta_+(s^2) }

and substitutes velocity by its quantum operator

{1 - (alpha alpha') delta_+(s^2) }

This gives [5] "our fundamental equation for electrodynamics."

In all those *standard* cases, the rule has been

v --> {v_op = c alpha}

Is not ironic physicists and chemists work with an velocity operator
"which quite obviously doesn't work" in your own words?

> I see, so you want to talk about a quantity that *has not* been measured
> and then compare it to a quantity that *has* been measured? A rather
> strange endeavour.

Everything there is due to your own confusion.

I have differentiated between velocity and average velocity, including
explicit formulae for both. And the title for this thread is: What is the
velocity of a relativistic electron?

Why do I need to clarify the same point again and again in each post?

> Let me get this straight. You are saying that, for an electron, v is not
> equal to c.

More confusion and superfitial reading of my posts. What v is not equal?

I wrote four velocities on my original post in their respective
theoretical contexts. And a fifth v was introduced during discussion!

> Also, both you and Feynman are saying that the argument presented above
> does not actually show that v = c.

You got all completely wrong once more.

Feynman first computes the velocity for a Dirac electron. As stated before
this *well-known* result is

v = c

Over the basis of that result Feynman writes:

{BLOCKQUOTE
This result is sometimes made pausible by the argument that a precise
determination of velocity implies precise determinations of position at
two times.
}

And just present a standard argument done in literature regarding the non
commutativity of x and p.

Then in a footnote Feynman recognizes that two-times argument is actually
invalid because v and p commute.

But of course he continues using the result

v --> {v_op = c alpha}

In fact, Feynman *got* QED in that way [5].

> While I may not be an expert in relativistic localization, I understand
> the topic well enough to point out that the question below has nothing
> in particular to do with relativity.

I provided an adequate quotation "1.6 The Position operator" in a previous
message. You deleted but i reintroduce here:

{BLOCKQUOTE
The outcome, then, is that a position operator is inconsistent with
relativity.
}

What part, do you think, has nothing in particular to do with relativity?

Maybe, do you thik¡nk, when he says "is inconsistent with relativity.."

> The answer, as I've already stated, is in the formula of step 3 above.
> There is no approximation, and this method works for non-relativistic as
> well as relativistic theories. It describes the transition from particle
> theory to field theory. The relativistic -> non-relativistic
> approximation may be made at will at either end of the transition.

As noticed before your answer is incorrect. And your 'derivation'
inconsistent already at step 1.

Also You seems to be not disturbed becaue i was explcitely asking for the
QM well-known result

x =/= parameter

whereas your 'answer' began with assumption

x = parameter

My previous reply to your 'answer' points some of your confussions. In
your new reply you are adding new misunderstandings.

Thanks because this remember me why i 'invented' guidelines

http://canonicalscience.org/en/misce...guidelines.txt


>> Thus Dirac complaints still hold.
>
> Unfortunately, as I've pointed out, Dirac is not here to be convinced
> otherwise. Neither are any of the other authors whose objections you
> have also quoted.

Yes, but pointing the obvious -Dirac is not here- would not hidde the
important: that his complaints remain in print.


[5] Space-Time Approach to Quantum Electrodynamics. 1949. Phys. Rev. 76,
769. Feynman, R. P.



--
I apply http://canonicalscience.org/en/misce...guidelines.txt

DRLunsford

On Mar 8, 10:27 am, "Juan R. González-Álvarez"
<j...@canonicalscience.com> wrote:

> More of the same.
>
> What velocities are measuring? Instantaneous velocities or averaged
> velocities?

The answer is simple, and is founded on the theory of invariants as
expressed in the Dirac algebra.

The alphas are spacetime bivectors - surface elements in spacetime,
thus

alpha_i = gamma_0 gamma_i

Thus, covariants formed from a spinor and the alphas cannot possibly
represent the velocity of a material body, which is vectorial, not
bivectorial (e.g. a free electromagnetic wave). The Dirac current, and
by implication the ponderable velocity of the Dirac particle itself,
is rather formed directly from the gammas,

J_mu = psibar gamma_mu psi

If you want to concoct a ponderable velocity from this, then divide
through by the (positive definite) fourth component to produce a
relativistic unit vector, which is the four-velocity hiding in this
current. That is all.

So what does psibar alpha psi represent? It's a group velocity
resulting from interference of the positive and negative energy
components of the spinor field. It is not vectorial, as a true
velocity must be, rather, the time-space part of a bivector, like the
electric field. That the expected value of it is always c is
physically interesting, but not in any way connected to the ponderable
velocity of the massive Dirac particle.

I would advise a study of the Foldy transformation.

-drl

Igor Khavkine

On 2008-03-08, DRLunsford <antimatter33@yahoo.com> wrote:

> I would advise a study of the Foldy transformation.

Incidentally, here's an interesting, short memoir written by Foldy,
concerning the origin of his work with Wouthuysen.

Physics at a Research University, Case Western Reserve 1830-1990
http://www.phys.cwru.edu/history/book%20page.htm

Appendix G, Foldy's Essay: Origins of Foldy-Wouthuysen Transformation
http://www.phys.cwru.edu/history/boo...p%20G%20FW.pdf

Igor

Igor Khavkine

On 2008-03-08, Juan R. <juan@canonicalscience.com> wrote:
> Igor Khavkine wrote on Sun, 02 Mar 2008 23:24:23 +0000:
>
>> Well, then I guess that settles the question. I'm sure that people who
>> do electron transport measurements will be happy to hear that they need
>> to interpret their data so that electron velocities always come out
>> equal to the speed of light (to be consistent with the theoretical
>> definition given in your original post), instead of the centimeters per
>> second that their instruments have been telling them all along.
>
> More of the same.
>
> What velocities are measuring? Instantaneous velocities or averaged
> velocities?

These measurements correspond to what has been called the "averaged
velocity" observable. Since no other kind of velocity has ever been
measured, I am completely justified in calling it "the velocity".

> Each one has its own operator. This was discussed four or five times
> before.

I've challenged you to come up with an experiment that has measured what
you call the "instantaneous velocity" (so far without response). Until
one can be found, the "instantaneous velocity" is merely an operator
with units of m/s and with little physical relevance.

> It is ironic, you also seem completely unaware on how atomic and molecular
> physicists, and quantum chemists work.
>
> Take the classical (Darwin) expression
>
> U_D = -1/2 (v v' + v r v' r) (e^2/r) 1/c^2
>
> For a Dirac electron the velocity operator is
>
> v_op = [H,x] = c alpha
>
> substitution {v --> v_op} gives
>
> U_B = -1/2 (alpha alpha' + alpha r alpha' r) (e^2/r)
>
> This is the Dirac Breit operator broadly used in atomic and molecular
> physics, and quantum chemistry.
>
> Another example is the Gaunt potential, a standard in relativistic
> molecular hamiltonians for bound states. Again its derivation involves
>
> v --> c alpha.

These potentials are derived directly from the Dirac equation and remain
in the same 4-component spinor representation. These can also be
subjected to the Foldy-Wouthuysen transformation and put into
2-component Pauli-Schroedinger-like form (see [6]). Since the FW
transformation is unitary, spectral properties of the Hamiltonian remain
unchanged. However, the Hamiltonian does get rewritten in terms of the x
and p canonical variables (different from the ones in the Dirac
representation), where v = dx/dt = p/E, as expected.

> Ironically, this is also the way Feynman derived QED -now considered the
> most precise theory of physics-. Take a look to his historical paper [5].
>
> Feynman starts from the classical interaction Lagrangian (he uses c = 1)
>
> { 1 - (v v') delta_+(s^2) }
>
> and substitutes velocity by its quantum operator
>
> {1 - (alpha alpha') delta_+(s^2) }
>
> This gives [5] "our fundamental equation for electrodynamics."
>
> In all those *standard* cases, the rule has been
>
> v --> {v_op = c alpha}
>
> Is not ironic physicists and chemists work with an velocity operator
> "which quite obviously doesn't work" in your own words?

There are many equivalent formulations of quantum mechanics
(encompassing QED), including Feynmans. What is true of one formulation,
by equivalence, must be true of all other ones. The use of the Dirac
representation, including its alpha and x operators, is not in the least
ironic. Most of the time, only spectral properties of the Hamiltonian
are of interest. The expectation values of the x and alpha are not
measured, and hence need to be interpreted. However, whenever high order
relativistic corrections are desired (to be computed in the
Pauli-Schroedinger representation, where position and momentum
observables have the common interpretations), these corrections are
invariably written in the FW representation (again, see [6]).

>> While I may not be an expert in relativistic localization, I understand
>> the topic well enough to point out that the question below has nothing
>> in particular to do with relativity.
>
> I provided an adequate quotation "1.6 The Position operator" in a previous
> message. You deleted but i reintroduce here:
>
> {BLOCKQUOTE
> The outcome, then, is that a position operator is inconsistent with
> relativity.
> }

I do not have access to Ticciatti's book at the moment, so I cannot
judge his statement in context. However, whatever his position is, it
doesn't change the fact that massive relativistic particles (whether
formulated on their own or in the context of field theory) possess a
well defined, frame dependent position operator (granted that the
situation does become more complicated once multiparticle effects are
taken into account).

> What part, do you think, has nothing in particular to do with relativity?

The part where a field theoretic parameter (x) gets converted into a
single particle operator (X), or vice versa.

I wrote:
>> Let x - position coordinates (parameter)
>> X - position observable (operator)
>> |x> - an eigenstate of the X operator <-- X|x> = x|x>
>> chi(x) - position-space wave function
>> Psi(x) - field operator
>> |chi> - single-particle state
>> |0> - Fock vacuum
>>
>> 1. Project onto position eigenstates: chi(x) = <x|chi>.
>> 2. Embed state in Fock space: |chi> = int dx chi(x) Psi(x)^* |0>.
>> 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x).
>>
>> The answer, as I've already stated, is in the formula of step 3 above.
>> There is no approximation, and this method works for non-relativistic as
>> well as relativistic theories. It describes the transition from particle
>> theory to field theory. The relativistic -> non-relativistic
>> approximation may be made at will at either end of the transition.
>
> As noticed before your answer is incorrect. And your 'derivation'
> inconsistent already at step 1.
>
> Also You seems to be not disturbed becaue i was explcitely asking for the
> QM well-known result
>
> x =/= parameter
>
> whereas your 'answer' began with assumption
>
> x = parameter

Aha, perhaps you missed the second line of what I wrote above, where I
introduce X (capital X) as a single particle position operator
observable. Are you perchange denying the fact that a self-adjoint
operator has a real spectrum or that any state vector can be written as
a linear superposition of its eigenstates? You may want to read over
what I wrote above somewhat more carefully.

[6] W. A. Barker, F. N. Glover, Reduction of Relativistic Two-Particle
Wave Equations to Approximate Forms. III (1955) Phys.Rev. v.99, p.317

Igor

Oh No

Thus spake DRLunsford <antimatter33@yahoo.com>
>I would advise a study of the Foldy transformation.

Agreed. This link to the mathematical detail may also be helpful

http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf



Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

DRLunsford

On Mar 8, 2:59 pm, DRLunsford <antimatte...@yahoo.com> wrote:
> On Mar 8, 10:27 am, "Juan R. González-Álvarez"

> So what does psibar alpha psi represent? It's a group velocity
> resulting from interference of the positive and negative energy
> components of the spinor field. It is not vectorial, as a true
> velocity must be, rather, the time-space part of a bivector, like the
> electric field. That the expected value of it is always c is
> physically interesting, but not in any way connected to the ponderable
> velocity of the massive Dirac particle.

Oh I forgot to add, c is interesting in this context because it
amounts to an invariant expression of constraint upon the co-local
presence of matter and antimatter. These are tied together by the
conservation of charge, and also angular momentum. Hmm.

-drl

Hans de Vries

Igor Khavkine wrote:
>> v = c using Dirac equation for |PSI>.[/color]

>Eh, no.
>
>The paradox consists of identifying the operator c*alpha (in the
>usual 3+1 Dirac equation notation) with the physical velocity,



Actually, this IS the right operator and it DOES produce v when
applied on a plane wave solution of the Dirac equation in the
Chiral representation.

1/(2E) ( phi* c*alpha phi ) = v

http://www.physicsforums.com/showpos...50&postcount=6


Regards, Hans