5Electroweak theory

III The Standard Model



5.4 Neutrino oscillation and mass
Since around 2000, we know that the mass eigenstates and weak eigenstates for
neutrinos are not equivalent, as neutrinos were found to change from one flavour
to another. This implies there is some mixing between different generations of
leptons. The analogous mixing matrix is the Pontecorov–Maki–Nakagawa–Sakata
matrix , written U
P MN S
.
As of today, we do not really understand what neutrinos actually are, as
neutrinos don’t interact much, and we don’t have enough experimental data. If
neutrinos are Dirac fermions, then they behave just like quarks, and we expect
CP violation.
However, there is another possibility. Since neutrinos do not have a charge,
it is possible that they are their own anti-particles. In other words, they are
Majorana fermions. It turns out this implies that we cannot get rid of that many
phases, and we are left with 3 angles and 3 phases. Again, we get CP violation
in general.
We consider these cases briefly in turn.
Dirac fermions
If they are Dirac fermions, then we must also get some right-handed neutrinos,
which we write as
N
i
= ν
i
R
= (ν
eR
, ν
µR
, ν
τR
).
Then we modify the Dirac Lagrangian to say
L
lept
=
2(λ
ij
¯
L
i
φR
j
+ λ
ij
ν
¯
L
i
φ
c
N
j
+ h.c.).
This is exactly like for quarks. As in quarks, we obtain a mass term.
X
i
m
i
ν
(¯ν
i
R
ν
i
L
+ ¯ν
i
L
ν
i
R
).
Majorana neutrinos
If neutrinos are their own anti-particles, then, in the language we had at the
beginning of the course, we have
d
s
(p) = b
s
(p).
Then ν(x) = ν
L
(x) + ν
R
(x) must satisfy
ν
c
(x) = C ¯ν
T
L
= ν(x).
Then we see that we must have
ν
R
(x) = ν
c
L
(x),
and vice versa. So the right-handed neutrino field is not independent of the
left-handed field. In this case, the mass term would look like
1
2
X
i
m
i
ν
(¯ν
ic
L
ν
i
L
+ ¯ν
i
L
ν
ic
L
).
As in the case of leptons, postulating a mass term directly like this breaks gauge
invariance. Again, we solve this problem by coupling with the Higgs field. It
takes some work to find a working gauge coupling, and it turns out it the simplest
thing that works is
L
L,φ
=
Y
ij
M
(L
iT
φ
c
)C(φ
cT
L
j
) + h.c..
This is weird, because it is a dimension 5 operator. This dimension 5 operator is
non-renormalizable. This is actually okay, as along as we think of the standard
model as an effective field theory, describing physics at some low energy scale.