Today, we will discuss the Gauge Symmetry
which is the way of understanding the interactions
of particles.
As we go on to be clear what we mean by that.
So far we have been discussing, the interactions
of particles through their equations of motion
and we did motivate the Dirac equation; how
it arises and also the Klein Gordon equation,
and then basically use the perturbative technique
to compute the cross section. We went on to
describe the weak interaction also along the
line of the electromagnetic interactions.
Now, we will try to see how everything can
be understood from a symmetry point of view.
Or which then can be taken as the basic principle,
guiding principle and things can be built
from these guiding principles. For that we
will start with what is called the Lagrangian
of the system.
So, the basic thing here is the fundamental
thing is the Lagrangian of the system. And
the equations of motion can be derived from
the Lagrangian. For point particles; a discrete
system of point particles we can disc talk
about the Lagrangian as, for many cases it
is the kinetic energy minus the potential
energy. So, T here is the kinetic energy and
V is the potential energy.
For example, free particle kinetic energy
is half mass of the particle times the velocity
square. If I consider for simplicity one dimensional
system particle moving in one dimension, then
it is velocity is the position coordinate
time derivative of the position coordinate;
and kinetic energy is half m x dot square
and potential energy for every part of free
particle is 0 that gives us the Lagrangian
equal to half m x dot square.
Now, without deriving we will describe, what
is the equation of motion.
In this case, you take the derivative of the
Lagrangian with respect to x dot and that
will give you 
derivative with respect to x dot of half m
x dot square which is equal to m x dot. Now
you take the time derivative of this that
is x double dot and equation of motion for
any Lagrangian with one such coordinates is;
time derivative of derivative of the Lagrangian
with respect to velocity minus derivative
of the Lagrangian with respect to the coordinate
itself is equal to 0. That is the equation
of motion which is called Euler- Lagrange
equation of motion equations of motion.
So, this gives us in our present case x double
dot equal to 0, which is the statement that
acceleration of the particle which is freely
moving is equal to 0. That it will move with
constant speed. So, this particular Euler-Lagrange
equation can be generalized to more than one
coordinate, and in particular we are interested
in continuous system of particles. For continuous
system of particles; continuous system say
block of matter, solid block or things like
that, we can we have infinitely many degrees
of freedom.
So, there are infinitely many particles very
large number of particles; so, infinitely
many, loosely speaking. And this can be then
considered as for the position of the particular
point. So, we can assign some kind of a function
which describes the; describes the configuration
of the system at a particular time.
So, for that actually we rely on what is;
will define some function which is a function
of the spatial coordinates and the time coordinates
or in general we can write it as a function
of the 4 vector and we can consider the Lagrangian
now as a function of instead of the coordinates
and the velocities, the case; like in the
case of the particle mechanics.
We can now consider it as function as a function
of these phi’s the which are the functions
of the coordinates and the derivatives spatial
and time derivative of this and it can also
depend explicitly on the spatial coordinate,
and this is basically what we will consider
to start with is the Lagrangian density. In
this case 
and actually Lagrangian is obtained by integrating
this Lagrangian density over a volume; over
the volume of the system. And the equations
of motion 
generalized to derivative of the Lagrangian
density with respect to the doh phi; doh mu
phi 
and take the derivative of this whole thing
minus derivative of the Lagrangian density
with respect to phi equal to 0.
%%%
Again, we are not going to prove this we are
not going to derive this, we are not going
to discuss anything more about how this equation
is arrived at. But one can actually look at
any standard classical mechanics book including
Goldstein, classical mechanics by Goldstein
or any other field theory classical field
theory book then or quantum field theory book
where introduction to classical field theory
is given to understand what is the Lagrangian
mechanics and how do you get the equations
of motion in Lagrangian mechanics.
Now, we will, but illustrate this in our situations.
So, let us consider a particular Lagrangian
density, I will write it as 1 over 2 doh mu
phi doh mu phi minus m square over 2 phi square
for some field phi. Some function of x mu
phi is some function of x mu. So, in that
case, I can write doh L by doh phi I will
get it as m square phi and doh L over doh
derivative with respect to doh mu phi, how
do I write it second term there is no doh
mu for the first term has. So, what I will
do is, to write it in a more convenient way.
It is there is a factor of 1 over 2 there
and I will write it as doh alpha phi doh alpha
phi. So, that there is no confusion with the
mu that we are taking alpha is a dummy index
and mu written in the Lagrangian density is
a dummy index. So, I can write it in this
fashion. So, half derivative with respect
doh mu phi or now since I am taking derivative
with respect to a covariant derivative of
phi I will write everything in the square
bracket here in terms of the covariant derivative.
I can do that by using g mu nu or g alpha
beta or the metric tensor g.
%%%%
So, I will write it as d alpha phi as it is
and this one is g alpha beta doh beta phi.
I hope everybody understands understand the
operation that we are doing. g alpha beta
is a metric which is independent of space
time coordinates for special theory of relativity.
So, we will take that out. In our formalism
we do not have any space time dependence for
the metric.
So, then we have doh alpha phi doh beta phi
what is this, derivative of; derivative with
respect to mu doh mu phi of doh alpha phi
and doh beta phi. I have just rewritten this
here; is equal to. First take the derivative
I will go slow here. So, that we understand
the steps and then we will go a little faster
later on.
First let me take the derivative with respect
of the first factor here doh alpha phi and
the other one is a spectator plus, do the
other way around doh alpha phi is a spectator
and the derivative acts on doh beta phi, which
is equal to; these two are the same as long
as you have same index on both the one which
you are differentiating and the one which
you are differentiating with respect to. So,
this essentially gives you a delta mu alpha.
The kronecker delta which says that this is
equal to one if mu is equal to alpha is equal
to 0 otherwise doh beta phi plus similarly
doh alpha phi delta mu beta. In case you get
lost with the indices, what you have to check
at every stage is that the number of free
indices on either side and in all the terms
are the same. And wherever you have a repeated
index that is summed over and that you want
that to be summed over.
So, here for example, each of this term on
the right hand side has a mu as a contra variant
upper index and an alpha and a beta as the
covariant indices for both the terms and on
the left hand side we have a covariant alpha
and beta indeed and a covariant mu in the
denominator which is equivalent to a contra
variant mu in the numerator. So, that is fine.
This is what you have look back at what you
have in dual derivative of L with respect
to doh mu phi. g alpha beta. So, let me write
that down derivative of L with respect to
doh mu phi is then equal to g alpha beta by
2 g alpha beta contra variant by 2 into whatever
else that we have here in the this thing.
So, that is delta mu alpha doh beta phi plus
doh alpha phi delta mu beta; taking all these
contractions.
Now, again there is only one free index alpha
beta are dummy indices, contracted and on
the left hand side we have only one free index.
So, it all agrees. Now let us put this g alpha
beta and multiply the first term with that,
what is happening is; it will; Firstly, the
delta mu alpha will change the alpha 2 alpha
n g alpha beta 2 mu. Maybe let us do that
slowly in the next slide.
So, this therefore equal to g alpha beta I
will use the delta function first to change
mu 2 alpha sorry alpha 2 mu. So, I have a
g mu beta over 2 this is only the first term
into doh beta phi and for the second one we
have a delta beta mu. So, mu will be; sorry
beta will be changed to mu. So, we have doh
alpha phi as it is g alpha beta; alpha mu.
beta is changed to mu; again by 2.
So, this is equal to 1 over 2 the role of
g mu beta acting on doh mu, doh beta phi is
to raise the index beta to mu; make it contravariant.
So, we have first term doh mu phi, second
term instead of beta you have an alpha no
other difference. So, it is doh mu phi which
is equal to doh mu phi without a half factor
that is it all right. So, that is what you
have here. And you have one part this one
which will go into the equation of motion
and the other part here. So, that will give
you equation of motion as derivative of doh
mu phi minus m square of phi. Sorry, actually
there is a minus sign here already. So, it
should be plus m square phi this is equal
to 0, what is this? This is the Klein Gordon
equation.
So, summary is that if I use the Lagrangian
density, half doh mu phi doh mu phi; mu summed
over minus m square by 2 phi square for a
function phi, that will give us doh mu doh
mu phi plus m square phi equal to 0, the Klein
Gordon equation. So, we can say that the Lagrangian
that we have written down is the Klein Gordon
Lagrangian or the Lagrangian corresponding
to a Klein Gordon field or a field that satisfied
for a function that satisfies the Klein Gordon
equation.
Now, these functions are called fields; and
we can actually say that this represents or
correspond to a particle. So, this is; field
is the one which will represent a particle.
So, this is what is the quantum field theory
does. In relativistic quantum mechanics when
we discussed the Klein Gordon equation we
considered the phi as wave function and there
is a subtle difference here when we actually
consider this phi now as a field. Again, we
will not go into the details of what is what.
It can be taken as a kind of a technical difference
here, but the bottom line for us today here
is that we represent particles now by fields
or rather the fields acting on some vector
space; or the state vector. And corresponding
to each; different particles will have different
fields representing them.
So, one such field is this phi. There is a
way to actually relate the spins of the particles
to understand the spins of the particles also.
And we will see that usually the scalar particles
are the ones which are described by; or the
spin 0 particles are the ones which are described
by Klein Gordon kind of Lagrangian that we
have just now written down which we are staring
at. Now another type of particle that we had
encountered earlier is this Dirac; the ones
which satisfy the Dirac equation. And in passing
we had mentioned that this is basically the
type of particles which are described by Dirac
equation are spin half or fermionic particles.
For example, electron.
So, let us consider that kind of a Lagrangian.
Consider the Lagrangian i times psi bar gamma
mu doh mu psi minus m psi bar psi. If you
take this Lagrangian and then consider psi
and psi bar as independent 
degrees of freedom independent wave functions.
So, Lagrangian is a function of psi psi bar
and doh mu psi in this case.
Now, let us look at the equation of motion
with respect to; there are 2 such degrees
of freedom. So, with respect to each of these
we had to consider the equation of motion.
When we consider it with respect to psi bar
we have no derivative term for psi bar. So,
this is equal to 0. And when we consider doh
L by doh psi bar, we have I gamma mu doh mu
psi minus m psi and equation of motion will
tell you derivative or doh mu of the first
line here plus or minus doh L by doh psi bar
second line is equal to 0. Since the first
one is 0 we have i gamma mu doh mu psi minus
m psi equal to 0. Indeed we recognize this
as the Dirac equation.
So, this Lagrangian gives us the Dirac equation.
What about the other degree of freedom psi.
So, when we take derivative with respect to
doh mu psi, we have i psi bar gamma mu. And
doh L by doh psi is equal to minus m psi bar.
Together this will give you i doh mu psi bar
gamma mu plus m psi bar equal to 0. Indeed,
this is the conjugate equation corresponding
to the above Dirac equation. So, we have the
Dirac equation, and the conjugate equation
from this Lagrangian given here.
Now, this; it can therefore, be considered
as the Dirac Lagrangian, if you want to name
it. So, psi here is the field corresponding
to the 
fermionic particle. And it is a function of
the space time coordinates and momentum as
we had written down earlier, all right.
So, then let us go on to another Lagrangian,
which I will write as minus 1 over 4 F mu
nu F mu nu the field tensor corresponding
to the electromagnetic potential minus some
current J mu A mu; potential A mu. F mu nu
for us is equal to doh mu A nu minus doh nu
A mu. So, if I expand the Lagrangian F mu
nu I have minus 1 over 4 doh mu A nu minus
doh nu A mu contracted with doh mu A nu minus
doh nu A mu minus J mu A mu. Remember also
the Lagrangian density should not have any
free index. So, all the indices are summed
over. So, then only you will have an invariant
quantity Lagrangian density is a Lorentz invariant
called Lorentz scalar.
So, this is when I open the bracket doh mu
A nu doh mu A nu minus doh mu A nu doh nu
A mu minus doh nu A mu, second term in the
first bracket times the first term in the
second bracket plus doh nu A mu doh nu A mu.
So, that closes the bracket there minus J
mu A nu. So, this is equal to first term and
the last term in the curly bracket if you
look at they are the same excepting that mu
and nu are interchange since they are dummy
indices if you interchange, nothing happens.
So, I mean they are dummy indices.
So, then they are the same terms, similarly
you will recognize that the second term and
the third term doh mu A nu doh nu A mu is
minus; doh nu A mu doh mu A nu are the same
for example, when you have dummy indices like
this you could change the upper and lower
indices without changing anything. You will
see that they are the same. So, I will leave
that as a small exercise for you to raise
everything and I recognize that they are the
same.
Essentially, this term that I will circle
with a blue line so, the one which I circled
with the blue line are the same and adds up
similarly the other 2 the left the other 2
which are left, so, the which I circle with
red are also the same. So, they also add up.
Together you have minus 1 over 2 doh mu A
nu doh mu A nu minus doh mu A nu; the switched
A nu doh nu A nu minus J mu A nu. So, that
is your Lagrangian density in terms of the
potential derivative of the potential.
Now let me take the derivative of the Lagrangian
with respect to doh alpha A beta. So, I have
taken alpha beta in order not to confuse with
this mus there in the Lagrangian. Let me take
the first term here. So, the first term is
going to give you; so, this is what we want
to look at. So, let me take derivative with
respect to doh alpha A beta or first term
of the Lagrangian apart from minus 1 over
2 is doh mu A nu doh mu A nu. So, doh mu A
nu doh mu A nu. So, this acts only on that.
This is equal to derivative with respect to
doh alpha A beta. Again, let me write it as
g rho mu g nu sigma or sigma nu doh rho A
sigma. So, idea is to raise all the indices
of doh mu and A mu A nu and you have doh mu
a nu.
Now, you can apply this take the derivative
here. So, we have rho mu sigma nu was a common
factor which is independent of space time
or A the vector potential. So, we do not have
this affected by the differentiation. But
the other one first it will give you a delta
of rho alpha and delta of sigma beta; and
doh mu A nu as it is. Plus, you have delta
of mu alpha delta of nu beta and doh rho A
sigma. Which is equal to; Now, delta rho alpha
will change the rho on the g rho mu to alpha.
g alpha mu g beta nu doh mu A nu plus similarly,
g alpha rho g sigma beta doh rho A sigma;
which is essentially equal to doh alpha A
beta. And the other one also give you rho
alpha A beta. So, this is twice this. so,
first term here will give you twice doh alpha
A beta and do a similar analysis.
And you will see that the second term will
give you doh L over doh of doh alpha A beta
as minus doh alpha doh beta A beta minus doh
beta doh alpha.
So, and this is the first part other term
in the Lagrangian is J mu A mu with A minus
sign. So, that will give you doh L over doh
A alpha equal to minus J alpha, here let me
take the same J beta and that will give you
J beta. So, together that will give you derivative
of the first line with respect to alpha in
the numerator of minus doh alpha A beta plus
doh beta A alpha plus J beta is equal to 0.
So, this is equal to essentially minus doh
alpha F alpha beta plus J beta equal to 0
or doh alpha F alpha beta equal to J beta.
You recognize this as the Maxwell’s equations.
So, therefore, the Lagrangian that we considered
here leads to Maxwell’s equation all right.
So, we started we have described 3 different
Lagrangian.
One L is equal to 1 over 2 doh mu phi doh
mu phi minus 1 over 2 m square phi square
and. So, that this gave us the Klein Gordon
equation, which is essentially box square
plus m square acting on phi equal to 0. Then
we looked at i psi bar gamma mu doh mu psi
minus m psi bar psi and that gave us the Dirac
equation, i gamma mu doh mu psi minus m psi
equal to 0. And then we considered the Maxwell’s
Lagrangian F mu nu F mu nu minus J mu A mu
and that gave Maxwell’s equation doh mu
F mu nu equal to J nu.
So, we have somewhat familiarize ourselves
with the Lagrangian and equations and then
we understand how we can write down the Lagrangian
corresponding to the Dirac equation, Klein
Gordon equation, Maxwell’s equation which
are the 3 familiar equations that we had discussed
in the earlier this ones. So, the task that
we have now next is how to write down the
Lagrangian with interactions of the particles
included.
So, not just the free particle is not a very
interesting thing that will not tell us what
the dynamics is. It tells us only how the
free particle is propagates that is fine.
But if you want to understand the dynamics
how they behave under various different situations.
Especially under the basic forces and we need
to include the interaction in that theory.
So, we will just now see how this interaction
can be taken into account this time.
