Re: fermions at c, still confussed

Eric Gisse:
>
>
>I see an interesting possibility though. What if one of the neutrino
>masses were zero?

They cannot be zero. It's easiest to see this using just two neutrinos,
but the same holds for three. It's just more complex.

The neutrino eigenstates for the weak hamiltonian are are \nu_e and
\nu_u. However, those eigenstates might not diagonalize a different
hamiltonian. The eigenstates which diagnolize some other hamiltonian
might be linear combinations of \nu_e and \nu_mu. Call those \nu_1
and \nu_2:


|\nu_1> = cos(M)|\nu_e> + sin(M)|\nu_u>

|\nu_2> = cos(M)|\nu_u> - sin(M)|\nu_e>

and the inverse,

|\nu_e> = cos(M)|\nu_1> - sin(M)|\nu_2>

|\nu_u> = cos(M)|\nu_2> + sin(M)|\nu_1>

Where M is the neutrino mixing angle. If the mass eigenstates are
defined by some hamiltonian H, then the time dependent states of
H are,

|\nu_1(t)> = \exp(-iHt/hbar)|\nu_1>
|\nu_2(t)> = \exp(-iHt/hbar)|\nu_2>

and since H|\nu_1> = E_1 |\nu_1>, in the \neutrino rest frame, you
have E_1 = m_1 c^2, E_2 = m_2 c^2. If you now invert the the relation
to obtain the time dependence for \nu_e and \nu_u, you have states
which depend on terms like, \exp(-i(m_1 - m_2)c^2/hbar).

Note that if H = H_weak (or if [H, H_weak] = 0) then m_1 = m_2,
and there is no oscillation, since you can define \nu_1 = \nu_e
and \nu_2 = \nu_u, unless some other observable exists which
doesn't commute with H_weak.

>I find the concept of a massless particle changing of
>its' own volition [I assume neutrino oscillation is causeless, if not
>please correct me] into a massive particle very interesting.

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.

>Personally, I find the standing of mass in physics to be not especially
>solid.

It's just more complex than you probably imagine. The only eigenvalues
for the velocity in the dirac and klein-gordon equations is +/-c, so
for example if we have a massive neutrino, the mass eigenstates cannot
be the eigenstates of the velocity. Instead, we must obtain the velocity
as the expectation value of the two velocity eigenstates, i.e., the
probability for fiding the velocity to be +/-c is given by,


P(+c) = |a|^2 and P(-c) = 1 - |a|^2

so that

v = cP(+c) + (-c)P(-c) = c |a|^2 - c (1 - |a|^2)

v = -c (1 - 2|a|^2)

which gives the amplitude for the motion in the +c direction as,
|a|^2 = (1/2)(1 + v/c) and in the -c direction as (1/2)(1 - v/c).

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.

>I feel something interesting as that might be able to teach us
>something new. Of course, as has happened many times before, a little
>more education will probably change that perception.

Personally, I think the neutrino mixing has the potential to
resolve a number of questions, however, I admit to being biased
due to my personal interest in weak interactions.


.

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