9 Feb 2006
A friend of mine (with a PhD in math but a strong interest in physics)
wrote to ask me some questions about neutrino mixing and neutrino
mass. He started out asking for intuition about why neutrino mixing
implies neutrino mass; my reply follows. For the record, he knows a
_lot_ of physics and he's interested in all sorts of details of the
subject, so this explanation may be hard to understand for a lot of
people.

My own understanding of this comes entirely from quantum field theory,
and in particular from the Feynmann diagrams that correspond to mass.
I'm pretty sure you're familiar with Feynmann diagrams; if not, take a
look at the relevant slides in my "Intro to String Theory" lecture
online. In terms of diagrams, mass shows up as a "vertex" during the
propagation of a single particle:
e^* x e
(As a side note, that's related to how the Higgs mechanism works for
producing apparent masses for particles: a particle's mass comes from
an interaction with the "vacuum expectation value" of the Higgs field.
The interaction is (H e^* e), and the diagram looks like this:


e^* o e
Because that is just the vacuum value rather than the fluctuating
particle itself, the vertical line just acts like a constant, so this
behaves like the simple mass "vertex" that I drew first. The strength
of that vertex, and hence the mass of the particle, depends on the
coupling strength of the Higgs to the electron and antielectron; it
also depends on the vacuum expectation value . For the record, the
very same diagram and coupling strength gives the interaction between
the electron and the Higgs itself, which hopefully we'll observe in the
laboratory before too terribly long.)
Anyway, back to mixing. What kind of Feynmann diagram would you draw
to show one particle (say, 'n') transforming into another (say, 'm') on
its own? I can only think of one possibility:
n x m
The 'x' here represents some sort of interaction vertex, with some
associated "coupling strength" that determines how likely this process
is. Look familiar? This acts precisely like a mass term. In fact,
the natural thing to do in a situation like this is to put all of the
mass terms "i x j" into a matrix (i,j) and diagonalize it. That
way, the eigenvectors of the matrix will correspond to fields that
don't change while they propagate, and its eigenvalues will be their
masses. But in any case, a matrix with nonzero elements can't have
all its eigenvalues be zero! So mixing implies mass.
> Is there some intuition for the reverse implication, "if neutrinos
> have mass, then they must mix"?
Not only is there no intuition for it, it isn't true. (Otherwise, you
could make the same argument for, say, quarks: "Up quarks and top
quarks both have mass, so they must mix." But they don't.) If the
mass matrix for neutrino flavors were perfectly diagonal, then there
wouldn't be mixing.
"But wait!" you cry, "why don't we just always use the eigenstates of
the mass matrix? Why even consider this 'mixing' business?" The issue
with neutrinos (and a variety of other particles, notably the K_short
and K_long) is that the mass eigenstates are not necessarily the same
as the eigenstates of other interactions. In particular, weak
interactions like "W^+ e^ \nu_e^*" pick out a specific neutrino state
as the electron's partner, whether it happens to be an eigenstate of
the mass matrix or not.
So that's the real issue: when neutrinos are created or destroyed, the
most reasonable way to think of them is \nu_e, \nu_\mu, and \nu_\tau.
When they propagate from one place to another, the reasonable basis is
the mass eigenstates \nu_1, \nu_2, and \nu_3. The trick is in figuring
out what the overlap is between those two bases of states. Experiments
are underway to figure that out (as well as figuring out the actual
masses of the mass eigenstates).
> Do the three neutrinos all have the same quantum numbers in the same
> way that the two neutral pions (upantiup and downantidown) have the
> same quantum numbers?
What two neutral pions? The \pi^0 is specifically the "vector"
combination "(u u^*  d d^*)/(\sqrt(2)), to combine with the \pi^\pm
into an "isospin 1" triplet. The other combination, "(u u^* + d
d^*)/(\sqrt(2)), is a component of the \eta, which is about four times
heavier than the \pi^0 (presumably because it contains some strange
quark components as well).
Having said that, yes, the neutrinos all have the same quantum numbers
as far as I'm aware. (The evidence for neutrino mixing seems to
indicate that not even quantities like "muonness" are conserved.)
Steuard

One other comment that I passed along later was that no mixing would
happen if the neutrinos' masses were all the same. In that case, the
mass matrix would be a multiple of the identity matrix, so _every_
linear combination would be an eigenstate. That's why neutrino mixing
can only tell you about the relative mass _differences_ between the
mass eigenstates. (In fact, I believe that it only tells you the
absolute values of the differences; based on current experiment,
there's still a chance that the "mostly electron type" mass eigenstate
is heavier than the "mostly muon type".)