We will continue our discussion on the electroweak
interactions understanding it through the
gauge transformations. Wo, we said that we
had to consider gauge symmetry with the group
SU 2 L cross U 1 y.
Where L denotes the that fact that only the
left handed particles are charged under s
u 2 and y denotes that there is a charge called
hyper charge, which dictates the interaction
or transformation through U 1.
And psi L left handed particles transforms
like psi L prime equal to e power i theta
a I a plus i alpha y earlier we had denoted
this hyper charge parameter along with the
hyper charge as y. But, for reasons that will
be clear as we go on we will denote it by
alpha now. Reason is that later on we will
consider the scalar field where we will usually
denote that by the standard notation phi and
we do not want to confuse you with this thing.
So, you will change the notation here compared
to the earlier this one this L a is equal
to 1, 2, 3. y is whatever the hyper charged
is, and essentially for infinitesimal transformation
we write; This is a doublet, for doublets;
I mean for usually the all the particles leptons
and quarks that has a nonzero s u 2 charge
is isospin it is taken in a doublet of isospin
equal to 1 by 2.
So, let us say nu e L, e L corresponding fields
the primed version of this is equal to 1 plus
i theta a I a plus i alpha y; sorry here y
by 2 is there y by 2 again there is 1 over
2 or y is a convention. This times, acting
on nu e L. Here I a is 2 by 2 matrix the first
term and the last term there is no 2 by 2
matrix, but we assume that there is a unit
2 by 2 matrix.
And in particular I can be written as I 1
as the Pauli matrix by 2; 0 1 1 0; I 2 is
equal to the second Pauli matrix sigma 2 minus
i is i 0, again 1 over 2; and I 3 equal to
1 over 2 1 0 0 minus 1. And this factor of
1 over 2 on y is mainly due to this. So, this
is the way it is. And the right handed field
transforms as 1 plus I alpha y by 2 e R. No
isospin, isospin is equal to 0 for this.
Now, let us look at the Lagrangian, let us
only look at the electron part. So, we have
i nu e L e L bar gamma mu D mu acting on nu
e L; nu e L, e L. This is one and this is
a D mu which act on the left handed doublet.
For the right handed field I have i e R bar
gamma mu D mu R e R. D mu L is doh mu plus
i g I a W me mu a i g prime y by 2 B mu.
So, we have introduced 2 not 2, 3 W mu and
one B mu. So, total 4 bosons; vector particles;
vector fields here. And they are the ones
which help us have the gauge invariants for
the Lagrangian. For the right handed particles
and the transformation is with the covariant
derivative doh mu plus i g prime y by 2 B
mu. And this is because I is equal to 0 for
this particular case.
Now, we said how the psi’s transform here.
Then W transform like W mu a goes to W mu
a prime equal to W mu a 1 over g derivative
of theta a epsilon a b c theta b W mu c.
B mu goes to B mu prime equal to B mu minus
1 over g prime doh mu alpha. So, this is familiar
to us. SU 2; W; the gauge bosons that correspond
to the s u 2 transformation W mu is transformed
in a similar way compared to the g mu. We
introduced in the case of s u 3; the gluon
fields. Whereas, B mu which corresponds to
a u and gauge transformation; transform similar
to the A mu, which we had in the electromagnetic
transformation. So we will not spend too much
time dwelling on this, because we are familiar
with these things.
Now let me write down the Lagrangian i psi
L bar gamma mu D mu L psi L plus i psi bar
gamma mu D mu R psi R. This is what we had
written down earlier as well. This is gauge
invariant. Now, but now to make the new part
nearly introduced particle; physical particles
we need to give add the kinetic energy term.
So, it is the usual notation that we consider
one over; minus 1 over 4 W mu nu a W mu and
nu a minus 1 over 4 B mu nu B mu nu, where
W mu nu a is defined in an exactly similar
way as g mu nu. Doh mu W nu doh nu W mu and
a non-abelian term g epsilon a b c W mu b
W nu c. Whereas, B mu nu corresponds to the
U 1 and it will be exactly similar to f mu
nu doh mu B nu minus doh nu B mu.
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There is nothing; no third term and we have
only one B. So, we cannot make an anti symmetric
combination with one particular component
object. So, this is the Lagrangian which is
invariant under s u 2 cross u 1. And everything
seems to be fine what gives what is missing,
there is something missing here. The kinetic
energy term is there no problem and there
are interaction terms through this D mu; D
mu L has both B mu and W mu in it.
So, there are interactions of the left handed
particles with both Ws and B right hand side
right handed particles in the second term
of the Lagrangian has D mu R which was only
B mu in it apart from the normal derivative.
So, right handed particles interact with B
not with the W’s. But what is missing is
the mass term. Where are the mass terms. We
had written down the mass term, both in the
case of electromagnetic interaction and QCD
the strong interaction, quantum chromodynamics,
whereas, in this case we do not write down
that.
Nature of s u 2 group leads to or cannot be
answered either way I a I b equal to i epsilon
a b c I c. So, these are again called the
structure constants that that is what we had
used and looking at the W mu nu very similar
to the g mu nu a, we have here again self
interaction of the Ws. So, this leads to self
interaction of the W gauge bosons both triple
interaction and quadratic interactions this
is proportional to do g the coupling constant
and this is proportional to g square. In addition
to the 
interaction with fermions Ws or half of these
interactions whereas, the B do not have any
self interactions it is similar to the photon.
Let us look at the Mass term, we know that
electron has mass, so are the quarks, and
muons and tau particles. Neutrinos do not
have mass, within the standard model and then
it has a tiny mass experimental determined
which we will come to in a different way later
on. We will have this in a completely different
way. Let us focus on the masses of the other
particles the electrons, all the charged leptons:
electron, muon and tau electron and all the
quarks. They are found to be massive and then
how do we generate mass. Firstly, why could
not we write a mass term in the earlier Lagrangian?
Let us see how to write a mass term. What
do I do? I have to have a term psi L m psi
L why should I club this with this psi L with
psi L it is because a that is the only way
to do it. With SU 2 gauge invariance, because
psi L transform in a way which will be canceled
by psi L bar. So, this is fine. But the other
let me write down the other thing. We can
have let me call this m L psi bar R m R psi
R ms are some coefficients it could be metric
it could be a metric in the first it could
be a 2 by 2 matrix.
In the second it need not to be. It cannot
be a matrix anyway there is only one term.
Plus you can also have if you want psi bar
L m psi R and the Hermitian conjugate of this.
Plus Hermitian conjugate. Now problem with
this terms are the following. Let me start
with the last term. So, the third term psi
L bar m psi R is not or cannot be made s u
2 L cross u 1 y invariant. Actually problem
is with s u 2 L invariants.
That is because the doublet transforms in
a particular way including the s u 2 part
and right handed wave function you know field
transform without including any isospin part.
So, this will not be conserved. There is no
way to cancel that from other phase factors
of psi R and psi L bar. What about the other
term? Let us look at first psi R bar m R.
So, let me write it as m R psi R. This is
equal to whatever m is, psi bar 1 minus gamma
5 by 2 1 plus gamma 5 by 2 psi for psi R is
equal to 1 plus gamma 5 by 2 psi. And psi
L is equal to 1 minus gamma 5 by 2 psi.
So, this is equal to m R by 4 psi bar 1 minus
gamma 5 plus gamma 5 minus, gamma 5 square
psi. Gamma 5 square is equal to 1 that makes
this psi R bar psi R equal to 0. For any psi,
this is equal to 0. So, we cannot have an
interaction like this a term like this possible.
Similarly, psi L bar psi L is also equal to
0. So, that tells you that Lagrangian mass
term is equal to 0; we cannot have, if you
want consistent with s u 2 symmetry. What
does this mean? This means that only possibility
so, in this Lagrangian here only possibility
is to have third term, first 2 terms are not
possible technically I mean it is just a question
of I mean they are identically when it where
is nothing to do with gauge invariance or
anything they are gauge invariant of course,
but reason for that would be 0 is to do with
the way the projections are multiplied.
Now the third term is the only hope for us,
but we do not have gauge invariance for that.
Actually that is something which we have to
live with. We need to break s u 2 symmetry;
or s u 2 plus c 1 symmetry. In order to have
mass term in the Lagrangian. This is for the
fermions; even for the gauge bosons,
we have added interaction term for the various
things including psi bar gamma mu D mu L psi
L. Interaction is added through this D. And
we have added 1 over 4 W mu nu a W mu nu.
Kinetic energy term for this and self interaction
term, but the self interaction term always
had 3 W’s or for W’s in it and wherever
you we had 2 W’s let us go back to that
here. Whenever you have W mu nu W mu nu 2
either you have 3 terms 3 W’s with 1 derivative
acting on one of those or we have 4 W’s.
Where we had 2 W’s we had 2 derivatives
I think which is actually the kinetic energy
term there is no mass term.
The mass term for gauge boson is essentially
one which is equal to something, some constant
into W mu W mu. This is the kind of term that
will give you the mass square. How do I justify
this? All right, let us look at the Lagrangian.
Doh mu W; minus 1 over 4 F mu nu F mu nu for
simplicity we will take a non abelian kind
of this thing. Minus; or let us take a B mu
nu, which is immediately familiar. m square;
m B square B mu B mu. What is the equation
of motion? So, what this part is purely W
derivative term the first term has only derivative
terms.
So, that will give you doh L over doh B mu
is equal to; because of the symmetry; minus
2 m B square B mu. And doh L by; you write
down the equation of motion for this that
will give you the equation of motion del square
the B mu box square mu B mu minus, minus of
minus plus m B square B mu equal to 0. So,
the mass term in the Lagrangian is similar
to the mass term for the phi the Klein Gordon
equation part, these are Klein Gordon. So,
this is for there is part of this thing is
there then there is also term minus doh mu
doh nu B nu.
So, this gives you this tells you that the
mass term is similar to this, but now if you
focus on that. On the mass term, you see that
this is not invariant.
under gauge transformation, where B mu goes
to B mu minus 1 over g doh mu alpha, and B
mu 1 over g doh mu alpha. This is not equal
to B mu B mu. It is not possible to have this.
This is the reason why we cannot have mass
term for the gauge boson as well. Therefore,
that leads to mass terms for B mu W mu or
g mu will not be gauge invariant.
So, there are 2 issues, one is to generate
the gauge boson, the Fermion, particle masses,
other is to introduce masses for the gauge
bosons. We will see later on that the gauge
why should we have the gauge boson mass. We
will see that the photon is mass less anyway
and then the gluons are also mass less, but
we will see that weak gauge bosons are massive.
Weak gauge bosons are massive we have determined
the mass of this experimentally we have discovered
these particles.
And then we have seen that they have mass
that is one thing, but the other thing is
that it was expected that they are massive
because they belong to, they correspond to
the weak nuclear interactions which are short
range, and for short range particle interactions
the exchange particle is expected to have
mass, otherwise the range of the interaction
will be infinite; like the electromagnetic
interaction. It is a different story why strong
interactions are short range, even with gluon
as the exchange particle without any mass.
That we will not consider for the time being.
And we saw also that s u 2 L cross u 1 y cannot
be an exact symmetry, because there are masses
for all these particles and then to generate
the mass for these particles, we need to break
the theory; break the gauge invariance. So,
we cannot add any mass term explicitly, to
the Lagrangian and break the symmetry. Breaking
the symmetry explicitly, by explicit we mean
by adding mass terms to the Lagrangian. This
is not a good idea. Not a good idea because
this leads to various problems. So, in order
to avoid all these problems, is there a way
out? Way out is what is called spontaneous
symmetry breaking.
