Standard Model Lagrangian
with Electro-Weak Unification
The Standard Model assumes that the mass of the neutrino is zero and that
it is “left handed” -- travelling with its spin pointing opposite to its direction
of motion.
Since in this case there would be no “right handed” neutrino, the “flavor”
partner of the neutrino must be a “left handed” electron. This changes
the structure of the Standard Model Lagrangian – which is assumed to treat
only left handed flavor doublets.
The spinor for the neutrino is the same as for the electron,
if one substitutes m = 0 :
neutrino spinor
The helicity operator, p, satisfies the following condition
2
So when acting on the spinor
It’s a projection!
=u
1
u
From this we can determine conditions on a and b
left hand side gives


1
(p) (p)/pp = 1 (2x2 unit matrix)


So, we have the following condition on a and b:
These are the only spinors allowed for a zero mass neutrino!
positive helicity
negative helicity
The neutrino, if it has a zero mass can only have its spin
pointing along (or opposite to) it’s momentum.
Non-conservation of parity: Wu 1957
J
number of
electrons
e

Helicity of the neutrino: Goldhaber 1958
The neutrino could have both values of helicity, and Wu’s experiment,
while confirming non conservation of parity (left-right symmetry
broken), did not conclusively determine the neutrino’s helicity.
In 1958 Goldhaber determined the helicity of the neutrino in the K
capture of an electron: 63Eu152 (J=0) 62Sm*152 (J=1) + . The Sm*
decays giving off a photon with the same helicity as the neutrino.
Goldhaber measured the helicity of the photon by passing it through
magnetized iron. If the photon has the same direction of spin as the
magnetized iron, it would pass through, otherwise it would produce a
spin flip. He reported a helicity of -1.
Since 1998 it has been accepted that the neutrino has a small mass.
This produces some corrections in the Standard Model. For this
description of the Standard Model, it is assumed that the neutrino
has no mass.
Each term in the SM Lagrangian density containing quarks and
the leptons can be rewritten using the following expression. For
the neutrinos, however, only the left handed term exists.
In the following slide we use the notation d R = dR
The following is the interaction Lagrangian density for the first generation
of particles with the left and right handed parts shown explicitly.
B
B
B
B
B
W1
W1
W2+
W2+
B
W3
W3
sum over a = 1,2,…8
  aGa
  aGa
Weinberg’s decomposition of the B and W:
W = Weinberg angle
-- to be determined experimentally!
sin2W  0.23
Next steps: rewriting interaction Lagrangian density so
that interactions with the photon are identified.
1.
2.
3.
The neutrino has zero charge and can’t interact with the photon.
After substituting the expressions for B and W0 (which takes some work),
one can identify factors which equal e, the electronic charge, or the up
quark charge, etc. This permits one to find relationships between sinW ,
cos W , e, g2 and g1.
One finds that:
g 2 = e / sinW
g 1 = e / cosW
Also one defines:
T 3f
YL = -1
YR = 2 YL
= + 1/2 for the uL
= - 1/2 for the dL
= 0 for uR
= 0 for dR
The Standard Model Interaction Lagrangian for the 1st generation
(E & M) QED interactions
weak neutral current interactions
+
weak flavor changing interactions
+
QCD color interactions
The U(1) and SU(2) interaction terms
g2
= e / sinW
e
A
(E & M) QED interactions
g2
Z+
Z
weak neutral current interactions
g2
W+
weak flavor changing interactions
W-