Re: fermions at c, still confussed

Eric Gisse:
>Bilge wrote:

>> You are trying to think classically here. There is nothing strange
>> about neutrino oscillations from the standpoint of quantum mechanics.
>> Operators which don't commute are diagnolized by the same states, so
>> regardless of what you choose for a set of basis states to diagonolize
>> one operator, the other will be a linear combination of those states.
>
>What I am trying to understand is what it would mean, if it wouldn't
>break physics, for a neutrino to oscillate between having mass and no
>mass.

Don't think of it that way. The mass eigenstates have a fixed mass.
The oscillations between flavor eigenstates determine the relative
probability of the type of neutrino it will be in a weak interaction.

By the way, the neutrinos in this model are but one of several
possibilities. There is also the possibility that a neutrino can
be its own anti-particle, in which case it's a majorana neutrino.
Searches for those are conducted primarily by looking for neutrinoless
double-beta decay (so far without success).

>This relates to what I wrote below because I feel it would provide some
>insight into the nature of mass if we had an example of something that
>had mass only sometimes. I'm not sure what insight would be provided,
>but I'm certain something would be learned.

Actually, the ``nature of mass'' is that mass is the m^2 in
E^2 - p^2 = m^2. For example, consider two photons produced
by anihilation. In the center of momentum frame, the photons
are back-to-back with equal and opposite momenta, so the
total four momentum is (\hbar = c = 1),


(q + q')^2 = q^2 + q'^2 + 2 q.q'

Since q = (w,k) and q' = (w, -k), we get,

q^2 + q'^2 + 2qq' = 0 + 0 + 2(w^2 - k(-k)) = 4w^2 = m^2

m = 2w

Which is just the result that e+ + e- -> \gamma\gamma gives you two
photons with the energy (m_pos + m_elec). Well, this is not really the
same thing, but it illustrates a couple of points when dealing with
relativistic theories. First, if you are going to use a hamiltonian, then
the mass is conserved, even if you end up with two massless particles from
anihilation of two massive particles. The hamiltonian give you the total
energy, which at a point in spacetime, happens to the mass of two massive
particles. Second, the mass is conserved, but is not locally constant. Now
this isn't really what we're talking about with neutrino mixing, since
obviously, the mass is locally constant for the mass eigenstates (at least
to the extent that one can take the classical limit for the velocity).

However, this is a fairly involved question, so I located an
article on neutrino phenomenology for you: hep-ph/9812360. The
article might be rather dense, material wise, but you can safely
skip a number of sections, in particular the sections on the
see-saw mechanism and effective lagrangians. Most of the stuff
that pertains to your question will be in the first half of
section 2, section 3 and the appendix on majorana neutrinos.
It would probably be easier to ask questions where you get stuck
(or since you previously mentioned that you had already looked
at several articles, which didn't work for you, you could give
me those references and post questions from those articles, which
I will try to answer).

[...]
>> The flipping back and forth between velocity states is called
>> zitterbewegung. If we include the spin and note that fermions are
>> left-handed, then fermion must be entirely left-handed througout the
>> zitterbewegung. Constructing the correct eigenstates is beyond the scope
>> of this response, but the result is that the mass of the fermion is
>> obtained from a superposition of the longitudinal spin states. In
>> general, longitudinal polarizations may be associated with particle
>> masses.
>
>It sounds like the neutrino would oscillate back and forth along its'
>direction of travel, but with a preferred direction.

Precisely, except the same thing applies to any massive particle.

>My first guess before doing a little research though, was that
>it travels in both directions at once.

Not really. It's an artifact of the uncertainty principle.
A precise measurement of the velocity requires measuring an
interval \Delta t in the limit that t -> 0, in which case the
energy is infinite and so any precise measurement of the
velocity must give the total energy, \gamma mc^2 = infinity,
which implies that v = +/-c.

[...]
>Do we know if there is a trigger for neutrino oscillation or if it
>happens "just because" ? (I like "just because" better than "random".)

None of the above. This is just a feature of quantum mechanics.
Operators that are diagonal in some set of basis states will
not be diagonal in a different set of basis states. If two operators
do not commute, there is _no_ set of basis states which diagonalizes
both operators (e.g., S_x and S_z). By choosing S_z, you leave S_x
completely indeterminate and therefore if your particles are all
in an eigenstate of S_z, a measurement of S_x is going to give you
+x and -x randomly. Actually, there's a lot about random to like.
If you imagine that every variable really has infinitely precise values
and that the uncertainty principle is merely a difficulty in measurement,
then the information required (by nature) to specify any interaction
is infinite, etc. If instead you consider the possibility that there
really is only a finite number of measurements one can perform to
extract all of the information that exists, then the randomness is not
very surprising. If the spin can have only two values, then it doesn't
make any sense to try and measure 2 for each of S_x, S_y and S_z.
Finding the spin along z necessarily leaves S_x and S_y not well-defined.




.

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