From: mcirvin@husc8.harvard.edu (Mcirvin)
Newsgroups: sci.physics
Subject: Re: FTL
Date: 29 Aug 92 01:50:36 GMT
Article-I.D.: husc8.mcirvin.715053036

aephraim@physics.Berkeley.EDU (Aephraim M. Steinberg) writes:

>If you try, in the context of the Dirac equation, to measure velocity,
>you find its only eigenvalues along any given axis are +/- c, oddly
>enough, so not only does nothing travel faster than light, nothing travels
>slower either!  (Hence "Zitterbewegung," the supposed jitterbug electrons
>are constantly performing within... oh, perhaps John Baez will remind me
>whether it's the Compton wavelength or the classical radius, though I'm
>always embarassed to have an uncertainty of more than two orders of
>magnitude... of its approximate (nonrelativistic) trajectory)

I'll save John the trouble:  It's the Compton wavelength.  Zitterbewegung
shows up when you interpret the Dirac equation as the wave equation for
a particle theory of "positive-energy" and "negative-energy" electrons;
it results from the fact that the positive-energy solutions by themselves
do not span the Hilbert space.  When you interpret the Dirac equation
correctly, as a field equation for a field whose "positive-energy modes"
annihilate particles and whose "negative-energy modes" really create 
antiparticles (so that it's no longer necessary to have negative energies),
what you get instead of a Compton-wavelength-sized jitterbug is an
impossible-to-remove Compton-wavelength-sized uncertainty in particle
position, which is kind of cute.  All this is done in Sakurai's red book,
_Advanced Quantum Mechanics._

What is also cute is that, in 1+1 dimensions at least, you can exploit
the +/- c eigenvalues of the particle-velocity operator when deriving
the Dirac propagator using path integrals.  You assume that the particle
can only travel to the left or to the right at a speed c, and the path
integral is just a sum over numbers and times of path reversals.  You
obtain the propagator for a Dirac particle.  This is an exercise in 
Feynman and Hibbs. 

>I think more in the spirit of the question, however, is an experiment
>where you measure a position at two different times, and divide distance
>by time to get velocity.  In field theory, you might say you "create"
>(creation operator) a particle at one point in space-time and then
>see what happens when you try to "annihilate" (annihilation operator) it
>at another.  Sarfatti's postings aside, the fact that spacelike-separated
>(i.e., separated by d > ct) operators commute means that the amplitude
>for the above event is zero-- it never happens.

I don't think this is quite the case.  What you've described is a
two-point Green's function, basically the propagator in position
space.  This extends outside the light cone, even at first order.
The tail exponentially decays, but it's there.  The question of why
this disallows superluminal information transfer is a subtle one.
I don't really understand it myself.  Clearly the equal-time commutation
relations disallow such information transfer, but how it follows from
perturbation theory is less than clear.  In _Theory of Fundamental
Processes_, I think, Feynman gives a justification for why the tail
exists; he says that such a measurement outside the light cone corresponds
to events in which the "annihilation" actually creates an antiparticle
which annihilates the original particle.  (I see now that you've 
described exactly this situation.  It does happen at zero temperature;
the propagator is a zero-temperature thing.) I suspect that when one goes
to non-covariant perturbation theory this is actually what it looks like,
but that still doesn't answer the question of why tails don't transmit
any information.  I suspect it is related to the fact that the "pancake"
of electromagnetic fields around a charged particle in straight-line
motion doesn't transmit superluminal information either; this may
be, at least, a helpful analogy.  Past boundary conditions are important. 

>Does someone know anything about this?
>Isn't the commutation an exact relation, though, meaning this process could
>not possibly happen?

The commutation is certainly exact.  Keep in mind that it only refers
to bosonic operators, ones that actually have number eigenvalues.  All
*observables* are of this nature, and so are boson fields, but fermion
fields are not; they *anticommute* at spacelike separation, which is
one reason why you can't measure them.  I don't think this is the source
of the trouble that's causing us both cognitive dissonance, though, 
since one could obtain the same results with, say, scalar boson fields.
Commutation does not imply that the Green's functions disappear at
spacelike separation, even for the uncorrected propagator at zero
temperature.

>Happily making a rather simple question even more confusing than before,
>aephraim

Gleefully muddying the waters still further,
Matt

-- 
Matt McIrvin, Cambridge, Massachusetts, USA 
Newsgroups: sci.physics,alt.folklore.science,alt.cascade,sci.skeptic,alt.religion.kibology