The Dirac and Newton-Wigner Velocity Operators and the Speed of Light Fermion Perplexity
Jay R. Yablon
With some good advice along the way from Nuropulp in some recent threads
over at sci.physics.foundations, I have been studying Foldy-Wouthuysen
transformations and the velocity operator in both the Dirac and the
Newton-Wigner representations.
From what I can see it has known since the time of Dirac, that the
eigenvalues of the velocity operator are equal to the speed of light,
and have therefore been taken as suggestive that fermions must travel at
the speed of light (subject then, to Zitterbewegung and the like).
Notwithstabnding other "fixes" to this problem which, e.g., Fock
(Principles of Quantum Theory, 2nd ed.) saw it as a deficience, i.e.,
unsolved problem, I believe I have found a solution to this
"perplexity," which I have outlined a a brief paper posted below:
http://jayryablon.files.wordpress.com/2008/07/correcting-newton-wigner.pdf
I believe I have demonstrated that there is a long-standing error in the
Newton-Wigner velocity operator, and have corrected this error.
Especially, since the Foldy-Wouthuysen transformation devolves into +/-
a unit matrix for a fermion "at rest," the velocity operators in the
Dirac and the Newton-Wigner representations should become identical for
an at-rest fermion, yet, they do not.
Correcting this error, in turn, allows one to recognize that the
eigenvalues of the velocity operator (which must be invariant under the
choice of representation, e.g. no different as between the Pauli-Dirac
or Weyl, and the Newton-Wigner representations) are 1) equal to the
speed of light as is known and 2) invariant under Lorentz boosts applied
to a fermion. There is no problem with fermions moving at subliminous
velocities: all that happens is that the components of the velocity
operator are mixed, but the eigenvalues stay constant and remain equal
in magnitude to the speed of light.
I would appreciate any comments as to whether this looks to be accurate
and any other pertinent discussion .
Thanks,
Jay.
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm
Igor Khavkine
> To all:
>
> With some good advice along the way from Nuropulp in some recent threads
> over at sci.physics.foundations, I have been studying Foldy-Wouthuysen
> transformations and the velocity operator in both the Dirac and the
> Newton-Wigner representations.
Correction. There is only one velocity operator, just like there is
only one momentum and only one energy operator. The same operator may
have different matrix elements for different choices of basis in the
Hilbert space of states.
> From what I can see it has known since the time of Dirac, that the
> eigenvalues of the velocity operator are equal to the speed of light,
> and have therefore been taken as suggestive that fermions must travel at
> the speed of light (subject then, to Zitterbewegung and the like).
If you you found that the "velocity" operator has eigenvalues equal to
c, then you've looked at the wrong operator. For a free particle, the
velocity operator is p/m, where p is the 3-momentum operator. Its
components commute and their spectra consist of the [-c,c] interval.
BTW, Zitterbewegung has never been observed. In fact, it is highly
unlikely, as it requires the superposition of states of different
charge (which are generally believed to be excluded, aka
superselection).
> I believe I have demonstrated that there is a long-standing error in the
> Newton-Wigner velocity operator, and have corrected this error.
> Especially, since the Foldy-Wouthuysen transformation devolves into +/-
> a unit matrix for a fermion "at rest," the velocity operators in the
> Dirac and the Newton-Wigner representations should become identical for
> an at-rest fermion, yet, they do not.
I highly doubt this claim. There is only one FW transformation (again,
for a free particle) not separate ones for when it is at rest and for
when it is not. So, the above paragraph makes very little sense.
Similarly, there aren't different velocity operators for when a
particle is at rest and for when it is not. There is only one velocity
operator (with three components, of course).
Igor
Juan R.
> On Jul 21, 1:43 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> To all:
>>
>> With some good advice along the way from Nuropulp in some recent
>> threads over at sci.physics.foundations, I have been studying
>> Foldy-Wouthuysen transformations and the velocity operator in both the
>> Dirac and the Newton-Wigner representations.
>
> Correction. There is only one velocity operator, just like there is only
> one momentum and only one energy operator.
Foldy-Wouthuysen found different velocity operators, different momentum
operators and different energy operators *and* the relationships between
them.
For instance they found (p_NW = p_D) but (v_NW /= p_D)
>> From what I can see it has known since the time of Dirac, that the
>> eigenvalues of the velocity operator are equal to the speed of light,
>> and have therefore been taken as suggestive that fermions must travel
>> at the speed of light (subject then, to Zitterbewegung and the like).
>
> If you you found that the "velocity" operator has eigenvalues equal to
> c, then you've looked at the wrong operator.
That is wrong. It is well-known that the velocity operator in Dirac theory
is (c alpha) and eigenvalues are +-c. Even Dirac was aware of that.
Today the Breit Hamiltonian uses alpha as velocity operators.
> For a free particle, the
> velocity operator is p/m, where p is the 3-momentum operator. Its
> components commute and their spectra consist of the [-c,c] interval.
This is plain wrong also. Your (p/m) even has not correct classical
relativistic limit!
In special relativity v is not p/m.
(...)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Igor Khavkine
> > Correction. There is only one velocity operator, just like there is only
> > one momentum and only one energy operator.
>
> Foldy-Wouthuysen found different velocity operators, different momentum
> operators and different energy operators *and* the relationships between
> them.
>
> For instance they found (p_NW = p_D) but (v_NW /= p_D)
I presume you meant to write v_D in the last inequality. Let me
restate my point. This v_D is not the velocity operator. If it is
referred to as the "velocity" operator, then it is a misnomer. Yes,
even Dirac used this misnomer. However, better information is now
available and there is no reason to continue using it. Just give v_D a
different name.
> > For a free particle, the
> > velocity operator is p/m, where p is the 3-momentum operator. Its
> > components commute and their spectra consist of the [-c,c] interval.
>
> This is plain wrong also. Your (p/m) even has not correct classical
> relativistic limit!
>
> In special relativity v is not p/m.
You're right, this was plain wrong (but not "also"). The correct
velocity observable, both classical and quantum mechanically is v=c*p/
sqrt(m^2+p^2). This formula is now consistent with my statement about
its spectrum.
Igor
Oh No
>On Aug 7, 3:41 pm, "Juan R." González-Álvarez
><juanREM...@canonicalscience.com> wrote:
>> Igor Khavkine wrote on Wed, 23 Jul 2008 21:08:31 +0000:
>
>> > For a free particle, the
>> > velocity operator is p/m, where p is the 3-momentum operator.
>>
>> This is plain wrong also. Your (p/m) even has not correct classical
>> relativistic limit!
>>
>> In special relativity v is not p/m.
>
>You're right, this was plain wrong (but not "also"). The correct
>velocity observable, both classical and quantum mechanically is v=c*p/
>sqrt(m^2+p^2).
Surely which one of these is "right" depends on whether one wants to
define a 4-vector velocity, or whether one wants to define rate of
change of position with respect to coordinate time. These are different
quantities, and both used in special relativity. I would not say either
is plain wrong.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Juan R. González-Álvarez
> On Aug 7, 3:41 pm, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Igor Khavkine wrote on Wed, 23 Jul 2008 21:08:31 +0000:
>
>> > Correction. There is only one velocity operator, just like there is
>> > only one momentum and only one energy operator.
>>
>> Foldy-Wouthuysen found different velocity operators, different momentum
>> operators and different energy operators *and* the relationships
>> between them.
>>
>> For instance they found (p_NW = p_D) but (v_NW /= p_D)
>
> I presume you meant to write v_D in the last inequality.
You are right, I copied and pasted and forgot to rewrite to v_D
v_D = c \alpha
> Let me restate
> my point. This v_D is not the velocity operator.
It is, as follows from basic principles of QM.
> If it is referred to as
> the "velocity" operator, then it is a misnomer. Yes, even Dirac used
> this misnomer. However, better information is now available and there is
> no reason to continue using it. Just give v_D a different name.
There is no misnomer and I am using the standard name.
In [1] v_D is named the standard velocity operator
In [2] v_D is named simply the velocity operator.
See also
http://delta.cs.cinvestav.mx/~mcintosh/comun/symm/node11.html
>> > For a free particle, the
>> > velocity operator is p/m, where p is the 3-momentum operator. Its
>> > components commute and their spectra consist of the [-c,c] interval.
>>
>> This is plain wrong also. Your (p/m) even has not correct classical
>> relativistic limit!
>>
>> In special relativity v is not p/m.
>
> You're right, this was plain wrong (but not "also"). The correct
> velocity observable, both classical and quantum mechanically is v=c*p/
> sqrt(m^2+p^2). This formula is now consistent with my statement about
> its spectrum.
Hum, I recommend to you to check dimensional analysis *also* :-)
But even if one amends yours to
v = pc^2 / sqrt(m^2c^4 + p^2c^2)
v = pc / sqrt(m^2c^2 + p^2)
your new velocity operator is plain wrong *again*. It is not the correct
velocity observable as you believe.
As follows from QM, the definition is
v = [H,x]
For instance your velocity operator is plain wrong for an electron in an
external field, neither it accounts for virtual cloud effects in a free
electron.
Just to illustrate how relativistic physics works, let me revise two
electron relativistic interaction in the zero frequency limit.
Here one usually starts from classical action with the well-known
magnetic term proportional to (v v').
And next one substitutes the velocities by operators
v_classical --> v_operator
For Dirac theory the operator is [1,2] see also
http://delta.cs.cinvestav.mx/~mcintosh/comun/symm/node11.html
v = [H,x] = c \alpha
substitution gives the well-known result
U = - e e' alpha alpha' / r
sometimes named the Gaunt potential.
I will not emphasize the same points again and will be not correcting
your, Jay, Oh no repetitive mistakes again.
I think available literature is clear and precise.
[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
Hans de Vries
Hi, Jay.
To recapitulate my remarks on this here:
======= The velocity operator ========
There is nothing wrong with [ H,X ] = c.alpha.
It correctly leads to the normal velocity v.
psi* [ H,X ] psi = velocity / per-unit-of-volume.
This operator acts on the spinor part of the field psi.
The bi-spinor is Lorentz transformed depending on
the velocity v. The bi-spinor contains the velocity
information independent of exp(-ip.x)
psi* [ H,X ] psi is just the vector-current-density as
defined in every QFT introduction.
The vector-current density transforms like a 4-momentum.
"velocity / per-unit-of-volume" also transforms like
a 4-momentum due to the Lorentz contraction of
the volume by a factor gamma in the direction of v.
Integrated over space it gives the velocity v.
See: Chapter 14 of my QFT book, section 14.8
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
======= zitter - bewegung =======
The applied acceleration operator psi* [ H,V ] psi
gives zero when psi is a Dirac plane-wave. There
is *no* zitter bewegung.
An arbitrary Dirac field psi contains the momentum
p at three different places: In exp(-ip.x) and twice
in the Lorentz transformed bi-spinor.
All these p need to be the same. The bi-spinor
needs to be transformed in the same way as
the phase part exp(-ip.x).
When however the p are defined differently then one
has defined a super position of multiple momentum
eigenstates which results in interference.
This interference shows up in the math as the
infamous "zitter-bewegung"
See: Chapter 14 of my QFT book, section 14.9
http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
Regards, Hans
--------------------------------------------------------
Hans de Vries.
Science Advisor since 2005 at http://www.physicsforums.com
Web Site: http://www.physics-quest.org/
Rock Brentwood
>
> --http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Indeed, and the thread, itself (as well as its origination) is a
demonstration that there's nothing more insiduously difficult to deal
with someone who doesn't even know what they don't know.
The issues surrounding the localization of position are, in fact,
quite well-explored and well-known.
I've, in fact, worked out the details of the representations for ALL
the groups -- Euclidean, Galilean, Poincare' and all the sectors
(Tardion, Luxon, Tachyon, "Synchon", "Vacuuon") in some detail
The Wigner Classification for Galilei Poincare' and Euclid
http://federation.g3z.com/Physics/index.htm#GeneralizedWigner
(though the mass-energy-momentum-velocity relations for the tachyon
and synchron sectors need to be refined a bit -- they're more general
than what's shown in the article)
A few articles I've been looking at lately:
(1) Simple Derivation of the Newton-Wigner Position Operator
Thomas F. Jordan
Journal of Mathematical Physics 21(8), 2028-2031
(2) Simple Proof of No Position Operator for Quanta with Zero Mass and
Non-Zero Helicity
Thomas F. Jordan
Journal of Mathematical Physics 19 (6), 1382-1385, 1978 June
(3) The Foldy-Wouthuysen transformation
John P. Costella, Bruce H. J. McKellar
hep-ph/9503416
(4) Poincare' invariance of relativistic quantum position
Farkas, Kurucz, Weiner
quant-ph/0009102
(5) A Physical Realization of the Generalized PT-, C- and CPT-
Symmetries and the Position Operator for Klein-Gordon Fields.
as well as these more indirectly related items (which, by the way,
also begin to show that the naive mass-momentum-velocity formulas
posed earlier in the thread don't even begin to scratch the surface,
either)
(The Poincare' tachyon sector)
(6) Quantum Field Theory of Interacting Tachyons
Dhar, Sudarshan
Phys. Rev. 174(5) , 1808-1815
(7) Quantum Tachyon Dynamics
Fried, Gabellini
hep-th/07090414
(The Galilei tardion sector)
(8) Path-integral quantization of Galilean Fermi fields
hep-th/07064106
The correct representation for momentum vs. velocity comes straight
from Hamilton's equations, interpreting the kinetic energy as the
Hamiltonian.
In relativity, the mass shell relation reads: P^2 - E^2/c^2 + m^2 c^2
= 0 for ordinary massive particles. To provide a unified framework
that encompasses both Galilei and Poincare', the first thing you
notice is that this requires getting rid of the c^2 in the numerator.
Therefore, the quantities that emerge as actually being the more
natural ones are not the total energy E (= mc^2 + H), but the kinetic
energy (H); and not the "rest mass" m, but the "relativistic mass" M
(= m + H/c^2).
(In fact, even if you restrict yourself to the Poincare' sector, the
"rest mass" isn't a good variable to work with, because it only
applies to one of the Poincare' sectors listed below and not the other
3).
In place of the quadratic invariant is, then, a linear invariant:
mu = M - alpha H
and a quadratic invariant
P^2 - 2MH + alpha H^2.
The parameter alpha then selects between the different alternatives
according to the following:
alpha > 0 = 1/c^2 for Poincare'
alpha = 0 for Galilei
and even
alpha < 0 for Euclid.
First, one can find a relation with velocity by treating H as the
Hamiltonian. Then, taking derivatives with respect to the momentum P
of the two invariants, one gets:
dH/dP = v
dM/dP - alpha dH/dP = 0 --> dM/dP = alpha v
P - Mv - H dM/dv + alpha Hv = 0 --> P = Mv.
Hence, the correct relation is always P = Mv (with an asterisk for the
exceptional case M = 0 and P != 0 -- the "synchrons").
One can start by asking what the most general transformation is that
leaves the two invariants unchanged. Applying an infinitesimal
transformation to the linear and quadratic invariant results in the
relations:
delta(M) = alpha delta(H), P.delta(P) = M delta(H)
whose general solution involves 6 parameters:
delta(P) = w x P + u M, delta(H) = u.P, delta(M) = alpha u.P
where ().() denotes dot products (P is a 3-vector, here).
Ignoring the rotation generators (the 3-vector w) and concentrating
only on the boost generators (the 3-vector u), one can integrate these
relations out to finite form as the transformation:
P1 -> (P1 + M V)/z
P2 -> P2
M -> (M + alpha P.V)/z
H -> H + P.v/z + (M/z) (V^2/(1 + z))
where
z^2 = 1 - alpha V^2.
Here P1 is the component of P parallel to V, P2 the component
perpendicular.
This transformation applies for all boosts where alpha V^2 < 1.
One can then distinguish the following cases:
(TARDIONS) There exists a frame in which P = 0, M != 0.
Taking v = V as the velocity relative to this frame, H = U as the
energy in this frame, and M = m as the mass in this frame (i.e., the
"rest mass"), one gets
P = mV/z, M = m/z, H = U + (m/z) (v^2/(1 + z))
The last expression generalizes H = U + 1/2 m v^2. For Poincare', it's
just H = (U - mc^2) + mc^2/z.
(TACHYONS & SYNCHRONS) There exists a frame in which P = Pi != 0 (i.e.
the "synchron impulse") and M = 0.
Taking V as the velocity relative to this frame, H = U as the energy
in this frame, one gets the following relations
P1 = Pi_1/z, P2 = Pi_2
M = alpha Pi.V/z
H = U + Pi.V/z
where Pi_1, and Pi_2 are respectively the components of the impulse Pi
parallel and perpendicular to V.
In the Euclidean case (alpha < 0), it turns out that this can be
transformed to a rest frame; so that Euclidean tachyons are already
included as part of the tardion sector.
In the Galilean case, one can -- without loss of generality -- take U
= 0 in the special frame. Then one has, simply P = Pi, H = Pi.V/z, M =
0.
A synchron is an instantaneous impulse across a distance in space
(i.e., an action-at-a-distance transfer of momentum).
In the Poincare' case, the tachyon is just an action-at-a-distance
synchron ... except now, since simultaneity is relative,
"instantaneous" generalizes to "faster than light". Still, the
characteristic quantity is the "synchron impulse" Pi. In the
literature, there is a lot of confusion on this matter, and it is
usually termed the "imaginary rest mass" m = iPi/c ... which doesn't
make sense, since neither synchrons nor tachyons have a rest frame.
The velocity of the tachyon comes out of the momentum mass relation: v
= P/M. This is the vector parallel to P (or anti-parallel if M < 0),
with magnitude v given by:
v^2 = (Pi^2 c^2 - (Pi.V)^2)/(Pi.V)^2.
(VACUONS) There exists a frame in which P = 0 and M = 0.
Then this relation holds in all frames and one has
P = 0, M = 0, H = U
Vacuons are translation-invariant states which may have non-trivial
representation under the boost and rotation group. The little group
for vacuons is homogeneous Lorentz (alpha > 0), homogeneous Galilei
(alpha = 0) or homogeneous Euclidean (alpha < 0) SO(4) = SU(2) x
SU(2).
Finally, one has
(LUXON) In all frames, P != 0 and M != 0.
Given the transformation behavior under boosts, this can only occur if
alpha > 0 and the mass and momentum are related by M^2 = P^2/c^2. In
that case, the invariants reduce to
mu = M - H/c^2, P^2 - 2MH + H^2/c^2 = (mu c)^2.
One can actually go further with al this by treating the quadratic
invariant as the line element for the dual metric:
P^2 - 2MH + alpha H^2 = g^{-1}((P,-H,M), (P,-H,M)).
Then it can be factored into a Dirac square root to yield the
generalized Dirac equation -- which can be written for all sectors.
Supplementing it with the equation (M - alpha H) = mu = constant, one
obtains (in the Galilean tardion case) the equivalent of the
Schroedinger equation or (in the Poincare' tardion case) the usual
Dirac equation.
In all cases, the underlying Clifford algebra is that associated with
the metric above, which always has signature (+,+,+,+,-), independent
of alpha. This Clifford algebra is just the complexified Dirac
algebra, itself: (gamma_0, gamma_1, gamma_2, gamma_3, gamma_5).
The dual metric is obtained by writing down the canonical 1-form
theta = P.dr - H ds + M du = P.dr - H dt + mu du
where t = s - alpha u. This yields the line elements
g = dr^2 + 2 ds du - alpha ds^2 = dr^2 + 2 dt du + alpha dt^2.
The infinitesimal transformations induced on the coordinates (which
may be obtained by requring that the canonical 1-form be invariant
under infinitesimal transformation) are
delta(r) = w x r + u t
delta(t) = alpha u.r
delta(u) = -u.r
delta(s) = 0.
Adding a translation component (5 generators here -- 3 for r, 1 for t
or s, and 1 for u) results in the 11-parameter unified Galilei/
Poincare'/Euclid group, itself -- the one and the same as was referred
to above.
In finite form, (again ignoring the rotation generator), the boost
becomes
r1 -> (r1 + vt)/z, r2 -> r2, t -> (t + alpha r.v)/z
and
s -> s, u -> u - r.v/z - (t/z) (v^2/(1 + z)).
The last two are completely novel for anyone who's only seen the
Lorentz (or Galilei) transformation before. But it's a central
ingredient in the representation theory for the Galilei group. In the
Galilei limit, the transformation for u becomes
u -> u - r.v - tv^2/2.
Basically that's where the v^2/2 comes from in the Kinetic energy.
What's remarkable about this is that one has an Absolute Time (s)
present in ALL THREE cases. This is the lingering vestige of the
absolute time of the Galilean case. At the same time, one has a
lingering vestige of the simultaneity shift even in the Galilean case
-- only now, it's locked up with the "mass phase" u coordinate. But
it's still there.
You can begin to get a bigger picture with what's going on with the
time coordinates, by writing out the metric in terms of s and t. This
applies, however, only away from alpha = 0, since it becomes singular
around alpha = 0. In fact, what one gets is:
g = dr^2 - dt^2/alpha + ds^2/alpha.
One sees here, that the s coordinate, in effect, is just the proper
time, itself (which is why it's invariant). But what one also sees is
that while for alpha > 0, one has
g = dr^2 - c^2 dt^2 + c^2 ds^2,
for alpha = -1/K^2 < 0, one has
g = dr^2 + K^2 dt^2 - K^2 ds^2.
There is a transition of the "real time" coordinate (t) from
Lorentzian time (when alpha > 0), through absolute time (t = s, when
alpha = 0), to Euclideam time (when alpha < 0). This is offset by s,
which shifts from "Euclidean" proper time to "Lorentzian" proper time.
One can, in fact, to yet further and write down the configuration
space variables for the gravity field by treating the (frame 1-form) +
(connection 1-form) as the gauge potentials of the 11-parameter
generalized Galilei/Euclid/Poincare' group. The 5 1-forms are those
which generalize dr, ds and du (or dr, dt and du).
The result is a generalization of Riemannian geometry that (1)
transcends signature (Lorentzian -> Newton-Cartan -> Euclidean) and
(2) incorporates an absolute time (the 1-form corresponding to ds
remains an exact 1-form, in the absence of torsion, and so can be used
to define a foliation by s).
It's tempting to exploit this feature to do a poor man's resolution on
the Relativity vs. Quantum Theory conflict in their handling of time.
One might try to take the (s) time as the one in which the
Schroedinger equation is evolved, while the (t) time is just the
coordinate time of the ordinary 4-dimensional space-time manifold,
treated the same as the coordinates (r = (x,y,z)).
The key features of what's just been discussed are outlined, here,
below, in the conclusion:
Defining the non-relativistic correspondence limit for the Poincare'
group is not a trivial exercise.
(1) Since the Galilei group has a central charge (mass), then it is
necessary to include an 11th generator into the Poincare' group and a
5th coordinate to the underlying geometry. The result is a unification
of Galilei and Poincare' into a 1-parameter family of gauge groups
that also happens to include the 4-dimensional Euclidean group. All
members of the family are embedded into the 4+1 Poincare' group.
(2) The extra coordinate gives us an "absolute time" while also
incorporating a vestige of "Lorentzian time" into the Galilei case.
For massive particles, the absolute time coordinate coincides with
proper time. Under a continuous change in the family parameter,
Lorentzian time transforms smoothly into Euclidean time.
(3) The irreducible representations are much richer, including notably
the "Synchrons" (0-mass, infinite speed, action-at-a-distance modes)
in the Galilei case, as well as "Vacuons" (0 mass, 0 momentum).
(4) Total energy no longer plays a fundamental role. Instead, it is
kinetic energy that generates time translations. Similarly, "rest
mass" no longer plays a fundamental role, but rather the "relativistic
mass". In place of 1 mass shell invariant are two: one linear, one
quadratic; thus making it possible to define more meaningful mass-
energy-momentum relations for the tachyons and "synchrons".
(5) A Dirac equation exists for all sectors. The underlying 4+1
Clifford algebra is just the complexified Dirac algebra <gamma^i, i =
0,1,2,3,5>. The Dirac equation for the Galilei tardions is equivalent
to the Schroedinger equation with extra spin degrees of freedom.
(I didn't go through this in any detail, but it's in the article I
cited above. The 3 cases for the rotation group are distinguished by
the sign of the derived invariant
M^2 - alpha P^2 = mu^2 - alpha (P^2 - 2MH + alpha H^2).
For the tardion sector, where this is positive, one may THEN define a
rest mass m as the square root of the invariant. For the 0 and
negative cases, there is neither a rest frame, nor a rest mass.
Cl.Massé
news:dbc66f91-f243-442c...@m36g2000hse.googlegroups.com...
> You're right, this was plain wrong (but not "also"). The correct velocity
> observable, both classical and quantum mechanically is
> v=c*p/sqrt(m^2+p^2). This formula is now consistent with my statement
> about its spectrum.
What gives that velocity when integrating the Hamilton equation v = @H/@p
or the quantum mechanical counterpart v = [H,x]? I highly doubt it be the
Dirac Hamiltonian.