# What is the velocity of a relativistic electron?

by Juan R. Gonzlez-lvarez
Tags: electron, relativistic, velocity
 P: n/a 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. }
 P: n/a On 2008-02-13, Juan R. Gonzlez-lvarez 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
P: n/a

## What is the velocity of a relativistic electron?

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:
>>>
>>> ((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.

>
> {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

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

 P: n/a 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