Welcome everybody. Today, we will discuss
an important aspect of a Gauge theory; that
is the Spontaneous Braking of Gauge Symmetry.
We learned that charged fermions like electron
interacting with an electromagnetic field
or photons can be understood with the Lagrangian
like this; where, we have the kinetic energy
term and the interaction term together here
of the fermion particle represented by psi.
Then, we can have the term which corresponds
to the mass of the fermions.
Then, the purely electromagnetic part. Here
the second term here, this along with the
psi bar gives the information about the interaction
of the current due to the electron or any
charge fermion with the photon, which represents
the electromagnetic field in this fashion.
Here F mu nu we had discussed earlier is a
compact way of writing derivative of the photon
field, or the electromagnetic potential A
mu.
And we saw that such a Lagrangian is invariant
under the gauge transformation, where if we
change A mu to some A prime mu which is equal
to A mu minus derivative of some scalar function
lambda; psi the fermionic field going to some
psi prime which differs by the; differs from
the original fermionic field by a phase factor
like this and this is not independent, but
it is easy to write it or it is more convenient
to just write it here.
So, 1 e e power i e lambda psi bar. Psi bar,
we are now familiar with is the conjugate
field. Lambda is some Scalar field. So, it
is any differentiable scalar field that can
give you this thing ok.
So, this is something which we know and here
we look at this Lagrangian we won’t go into
the details, but we had done it earlier and
you can check that explicitly putting all
these things here, taking the derivatives
appropriately and then, see that this is invariant
under the this first term is invariant under
the transformation given here.
Psi bar psi is trivially invariant under this
transformation, because you have an exponential
minus i e lambda here coming from this and
then, plus i e lambda coming from psi bar
and they will balance each other. This F mu
nu is actually invariant in itself. That is,
if you take this to F mu nu prime; then, doh
mu A prime nu minus doh mu A prime mu.
As you can see, this will give doh mu A nu
and doh mu doh nu lambda which will cancel
from a similar term coming from the other
term here. So, this is invariant under that.
And we will get the equations of motion from
here the Dirac equation with the interaction
term and again, when you look at the equation
of motion of the photon you will get the Maxwell’s
equation from here.
Now, let me look at the electromagnetic part
of this in a little detail.
L electromagnetic is I will write only this
part. This means that we are not considering
the electron there. So, it is a free electromagnetic
field only this one; no interactions. So,
that gives us the Maxwell’s equations F
mu nu equal to 0; no current, it is in free
space.
Now, this we had; we could write as in terms
of A mu, the vector potential as derivative
of doh mu A nu minus doh nu A mu. This is
doh mu doh mu A nu minus doh mu doh nu A mu
which is equal to 0. Now, we had earlier mentioned
that any electromagnetic field in general
electromagnetic field can be represented by
an A mu with and we can always choose there
is a freedom to choose this. So, that this
is satisfied. That is, the Lorentz condition
is satisfied.
In that case here, this is nothing but, we
can swap this derivatives easily and you will
get this equal to 0, because this is basically
the divergence for divergence of A mu, Lorentz
condition says that this is satisfied. So,
we can always choose an a mu so that this
is satisfied and then, in that case we have
now; So, let me represent this doh mu doh
mu as we had done earlier by the box A nu
equal to 0.
Sometimes it is represented as box and then,
you can also write it as box square; there
are different convention, notations. But we
will I do not know what exactly we had chosen
earlier. But whatever it is this is going
to represent this particular thing here. Look
at that and this we recognize as the Klein
Gordon equation ok. So, this is the Klein
Gordon equation of equation representing the
dynamics of A mu and from here, we can note
that it is a mass less field. Ok.
So, this mass term is absent here. That is
all right; that is what we have in case of
photon. Photon is mass less. So, all these
are consistent as long as the as far as electromagnetic
interactions are concerned. But then, we generalize
a similar study or we generalize this to the
case of weak interactions.
So, for weak interactions, we again can write
the Lagrangian; where, electron is interacting
with the weak field now, weak nuclear interaction.
So, we have this doh mu here and then, we
have plus i g over 2 tau dot W. We will not
go into the details of this thing. We have
mentioned this in the previous lectures. There
is the field corresponding to the weak interaction
like the photon here we have a W to represent
the weak interaction, weak interaction field
and the field corresponding to the weak interaction.
This then, you also have similar to the pure
electromagnetic field case, here again, we
have a field tensor W corresponding now to
the weak field and the Lagrangian term; which
talks about the weak interaction fields evolution
in the absence of anything else. So, nothing
is there and then, this is purely it is going
to talk about the motion or evolution of the
electromagnetic field. Experimentally it is
seen that W boson corresponding to the quantum
of W mu similar to the photon; this is a massive
particle, unlike the photon.
Now, what is a mass term we need in the Lagrangian?
Some quantity like W mu W mu; if you go back
to the Klein Gordon equation in one of the
early lectures when we discussed the scalar
fields, you will see that there was m square
phi square there; instead of phi square we
have a W prime, W mu W nu. Why am I writing
u prime? It is not there here yet.
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So, there is W mu W nu here. Now, when we
take the transformations similar to the field
transformation without going into the details
let me say this thing. We can when we look
at a W mu prime W mu prime which is basically
the transformation of W W or W mu W nu goes
to W mu prime W nu prime which is now original
term plus some non-zero coordination which
means that such a term is not 
Gauge invariant under this.
Again, the gauge transformation corresponding
to W going into W nu we had written earlier
we can look at it, but that is not really
the focus of today’s discussion. So, we
will not spend a lot of time on that.
You can ask the question; that is fine, gauge
symmetry is broken or violated; this is the
Lagrangian is no more gauge invariant, but
we need mass therefore, let us add that. That
is how physics is done; you understand the
physical phenomena around you. If something
is not explained according to our assumptions
or something goes against as our assumptions,
we will change it. Change the assumptions,
so that the new setup is in agreement with
experimental results or the observations.
Since, W boson is found to be massive, we
need this. So, we can say that we don’t
want to insist on gauge symmetry.
Then, there is a little bit of uncomfortable
situation, because there are a lot of nice
features of this gauge symmetry. One of that
is, it says when you do the Quantum theory
of this interactions and the quantum theory
of these particle’s evolution involved,
I mean including their interactions with different
fields and they try to compute different observables
like the probability computations in the scattering
or the scattering cross section calculations;
we see that gauge symmetry helps there, in
terms of getting reliable computation, reliable
quantities. Or I would I should say the other
way round, in the absence of this case symmetry
or in the presence of symmetry breaking terms
like the mass square, mass term in the Lagrangian,
physical quantities like scattering cross
section etcetera will become un-computable;
meaning that when you try to compute them
there will be, there are difficulties there.
Again, we will not go into the details of
this and just mention the name that there
is something called Renormalization which
actually guarantees that there won’t be
any difficulties or any practical problems
that comes up with simple the computations
of these observables can be handled in a mathematically
consistent way. Theories which respect gauge
symmetry, which are called the Gauge theories
are renormalizable or are guaranteed to be
well behaved. That is one reason where we
think that we should somehow try to maintain
the gauge symmetry.
Now, question is how do you get that? How
do you get the mass at the same time keeping
the good features of the gauge symmetry, the
invariance of the Lagrangian under the gauge
transformation. Now, this is the topic of
today’s discussion. So, there is a way of
breaking this symmetry. Lagrangian will still
be invariant under the symmetry; but then,
we will also be able to generate mass. This
is called the Spontaneous Symmetry Breaking.
I will towards the end mention why is it called
Symmetry Breaking still where Lagrangian is
invariant under the symmetry. Is something
else which actually breaks the symmetry? We
will come to that. And that will also explain
why in the sense, why it is called Spontaneous
Symmetry Breaking.
So, let us consider a Lagrangian of some field,
which we denote by B mu and the corresponding
field tensor as B mu nu. We do not want to
take the photon because we know photon is
mass less and then we do not want to add confusion
with the using the same symbol here. But this
is a kind of a toy picture or a just a toy
model and just to understand what is spontaneous
symmetry breaking.
Then so, here this is with a field B mu, we
can have B mu nu is equal to similar to the
electromagnetic case. It can define the field
tensor, and then you will see you see immediately
like exactly like in the earlier case that
B mu going to B mu prime B mu minus. Minus
or plus it does not matter. So, let us take
plus doh mu lambda. One can check the gauge
invariance of this explicitly. We will not
go into that here.
But, we will add a 
part to the Lagrangian; another part of the
Lagrangian, L scalar. Where, L scalar is equal
to D mu phi star D mu phi. Ok, that is the
kinetic part; and potential, V which depends
on phi star phi the product of phi star phi.
And D mu is doh mu that the partial derivative
plus rather minus i g B mu. Now, in the gauge
transformation here, we will consider transformation
of phi as well, with phi going to a phi prime
which is exponential i lambda phi; the same
lambda. So, these two together represents
the gauge transformation here. Now, you can
see; I will leave that as an exercise; that
D mu phi along with this B transformation
and phi transformation. So, you have a D prime
mu and phi prime which is equal to exponential
i lambda D mu phi. That part I will leave
as an exercise for you to complete.
Now, you can see that this Lagrangian, this
part of a Lagrangian is trivially invariant,
because this doh mu phi star will pick up
a minus sign here, in the same phase and you
will get cancellation of the face phase these
two together. V again is taken as the potential
part is taken as function of the product of
these two. So, from psi and psi star these
phases will cancel. And therefore, the whole
thing is invariant under this and then this
part has nothing to do with phi, but it is
invariant under this transformation as we
know.
So, this is invariant. So, the Lagrangian
is invariant under such a gauge transformations.
And we do not have a mass term which is purely
B mu B mu term. So, notice that no mass for
B mu. Now, let me do certain re-parameterizations.
We need the scalar part of the Lagrangian.
So, we will keep it here or rather yeah, let
me keep it is it as it is will take it. Let
me take phi which is a complex field as we
see here; phi star is already denoted along
with this and then, we also said in the gauge
transformation it can pick up a phase, etcetera.
So, we are talking about a scalar field. So,
let me define this or parameterize it as a
phase and another real a length scale real
part. Or the magnitude and the phase.
So, that you can always do for example, you
can consider a function x plus i y and write
x as r cos theta and y as r sin theta that
will give you r e power i theta and r is nothing
but square root of x square plus y square.
And theta is when you divide this by this
is tan inverse of y by x. This is very familiar
to you.
So, you can always write a complex number
or a complex quantity, complex field phi in
terms of the magnitude and then, the phase
factor, all right. Why are we doing it?
And then, I will write this xi as in a particular
fashion. So, that is equal to e power i theta
or minus i theta some v plus h x over square
root of 2.
So, here I will introduce the position dependence
explicitly. All of these parameters depend
on x. And here, we cannot have too many functions
independent. So, we have theta which still
depends on the x and the xi we have written
in a particular faction fashion a constant
plus the function h x. Why have we done this?
It will be clear in a moment.
Once you do this, now we can actually consider
some Gauge transformations. I will take a
specific Gauge transformation; where, phi
goes to phi prime which is equal to e power
i lambda now, but lambda is taken to be theta.
And 
B mu will go to e prime mu which is B mu plus
1 over say g, we have a g here. So, we will
take that g, derivative of theta.
So, this is invariant under such a transformation;
the Lagrangian L scalar is invariant under
such a transformation, which means that I
can very well write; we do not need this.
So, I will remove this. We can write the same
a scalar L scalar as D mu prime. So, I will
write a D in the explicit form. So, we have
a doh mu plus i g B mu prime. Plus, because
I am taking here this star acting on phi star.
Then, we have the doh mu minus i g B prime
mu. I wanted it to be written in terms of
the phi prime and phi.
So, here a phi prime. Plus or minus V phi
prime star phi prime. Since, V is a function
of the product of the complex conjugate phi
star and phi that is automatically invariant
under such a transformation and there is no
B there. And this term is guaranteed to be
invariant under such a simultaneous transformation
of phi and B. We have this, all right.
Let me take a look at this. That is I just
rub this off.
But it is basically L scalar doh mu plus i
g B mu prime, I will keep it as it is; phi
star is e power i theta phi and star of this.
Then, you have doh mu minus i g B prime mu
e power i theta phi minus V. V now, I am not
considering any changes in V. I will just
keep it is as it keep it as it is. This is
equal to doh mu plus i g B prime mu.
What do you have here is e power i theta and
phi which is e power i theta; e power here
minus i theta into psi that is equal to 1
over root 2 v plus h. Remember that h is a
field which is a function of space time co-ordinates;
whereas, v is a constant ok. So, we have that.
So, I can write this. This is nothing but
v plus h by square root of 2 and you have
doh mu 
minus i g B prime mu; same thing v plus h
over root. That is a real quantity now. Therefore,
star will not affect it. So, this is what
we have; I will rub these things.
So, this gives us L scalar is equal to 
you have a 1 over root 2 doh mu h.
So, that is from here, doh mu doh mu acting
on v is 0. It is a constant doh mu acting
on h, we have there, there is a square root
2. So, this is there. Then, you have i g B
mu prime. So, that is a plus i g B mu acting
on v; then, you have i g B mu prime acting
on interacting with h byte; that is coming
from this here. Similarly, from that; now
therefore, this.
So, there is a minus V here that potential.
So, here we have 1 over root 2 doh mu h minus
i g B prime mu v by 2 minus i g B prime mu
h by root 2; it is all square root of 2 here.
And then you have a V which is not relevant
at the moment it is important, but we will
come that.
This is equal to now let me expand this. So,
I have a 1 over 2 doh mu h doh mu h together
this. Then, you have doh mu into this. So,
what is that? Doh mu h minus i g B prime mu
v by 2. But then, notice that you also have
1 over root 2 doh mu h plus i g B prime mu
v by root. So, they cancel each other. So,
this is square root, that cancels. Similarly,
this and this cancel. So, you do not have
that cross term. So, the only term here is
this.
Then, you have base times, it is taken care
of this into this is plus g square v square
by 2 B mu prime B mu prime, coming from the
product of these two. This times, this is
again you have plus g square h v by 2 B mu
prime B mu prime.
So, there is one coming from here this times
this and another coming from here. So, actually
there is no 2. Then, you have this times this,
which is plus g square h square by 2 B mu
prime. Now, this term, this Lagrangian along
with the potential there; minus potential
V. Look at this term, v is constant; g is
some constant like the charge; 2 is constant
and then this is it.
So, this is like m square, if we denote this
by m square; we have a mass term there in
terms of B prime mu. So, let me look at this
once again. Initially, we had a scalar field
which is doh mu minus doh mu plus i g B mu
phi; and potential V. This is an interaction
term with in terms of phi when you look at
this phi here.
So, this Lagrangian here written in terms
of phi has no term which is purely quadratic
in B; or there is no mass term. Whereas, once
we make the Gauge transformation that phi
goes to phi prime which is equal to e power
i theta phi which is equal to; So, finally,
we had e power i theta cancelling with e power
minus i theta and a 1 over square root of
2 v plus h. And simultaneously, we had B going
to B mu prime which is equal to the original
B mu plus 1 over g derivative of theta. Such
a B prime when we write in terms of this phi
prime; now, phi prime is without any theta
here is essentially this whole thing; we have
a mass term here.
There are few things; a couple of things we
can notice. We may notice that theta disappears
from phi. The original phi had a theta that
disappears in phi prime. So, this says that
theta disappears. Original B did not obviously,
have any theta, but after the gauge transformation,
theta appears here.
So, this theta is now part of B, and not part
of phi that is interesting. What has happened
is in this re parameterization, we have absorbed
theta, part of phi into B to become B prime
and that B prime, then gets a mass, ok. So,
that is interesting. Now, we have mass term
for B and this particular mechanism to generate
mass to the scalar (read gauge vector) field
is the celebrated Higgs Mechanism.
So, that tells you that mass of B is for the
square of that this and B prime is now considered
as the actual physical field. B mu alone will
not suffice, B mu has to be added with theta;
added with doh mu theta. Now, let us look
at, the potential part that we have not looked
at so far.
We may need this or let me leave it there.
Potential V is chosen so that we have a lambda
phi star phi minus v square by 2 whole square.
Now, this when we ask question about what
is the Minimum energy Configuration, any physical
system tries to have a Minimum energy Configuration.
So, that is one of the things and the lowest
energy configuration is basically called the
Vacuum which in particle physics ok.
So, we will say that the Minimum energy Configuration
or the vacuum. So, the minimum energy configuration
corresponds to the minimum of the potential,
because any system can have the potential
energy and kinetic energy, the other part.
So, potential energy can be positive or negative,
whereas, the kinetic energy cannot be negative.
This is the always positive. So, the kinetic
energy minimum of the kinetic energy is 0
and therefore, minimum of the potential will
basically correspond to the minimum of the;
minimum energy configuration.
So, here we can ask the question, what would
be the minimum of this; which means we can
go by the usual way we can get this by differentiating
this taking phi and phi stars two independently
degrees freedom. So, this we have with the
phi star here. Setting it to 0 ok, we get
the extrema minimum or maximum. So, that will
give you two ways; one is this is 0. So, phi
star phi is equal to v square by 2 or this
is 0. So, there are two configurations which
corresponds to the extrema, derivative vanishing.
If we take the second derivative, we will
see that it is basically the first one which
corresponds to the minimum. Or we can actually
draw this potential in phi and so, here; either
I write it as real and imaginary parts of
phi and you will see that ok. You can actually
think about a kind of circle in here. So,
basically you rotate this about this axis,
axis potential here; and you will get all
variations of real and imaginary part.
But, the point to note is that here there
are two extremas to the turning points; one
is when phi equal to 0 both imaginary and
real parts of this equal to 0, the other is
here which we can consider as the minimum
of the potential. So, that part mathematically
you can check what that this corresponds to
the minimum of the potential and that means,
that for minimum energy the expectation value
is v by root 2; of phi.
So, phi I am just saying this has the vacuum
expectation value or whatever is the value
of phi corresponding to the vacuum is v by
root 2. So, that phi star phi is v square
by 2.
What does this mean? This means that when
we say vacuum, in fact, we call the minimum
energy configuration as vacuum assuming that
in such vacuum or minimum energy configuration,
we do not have presence of physical particles
or physical fields. Then, quantum fluctuations
or particles can be created giving energy
or with energy larger than the minimum energy
configuration, we can think of having particles
in it; or the presence of particles.
But here, what it says is that the Higgs field
cannot be set to 0 even in vacuum. So, there
is a presence of the Higgs field in the vacuum.
When we look at the scalar field, the interaction
of this field with other particles in the
vacuum say the gauge fields in the vacuum
gives its mass. So, we had, say for example,
we had written the mass of B equal to g square
v square over 2, the square of this thing.
So, when v equal to 0, B is mass less.
So, if v was 0. So, we had if we had started
with a potential just phi star phi whole square
without p square v; then, this would have
been 0. The presence of this gives the mass
through the interaction that we saw. It is
the scalar Lagrangian had the term doh mu
plus i g B mu phi. So, it is basically the
interaction of this B with phi that essentially
gave rise to the mass here. But there is no
phi; the physical field corresponding to phi;
which is essentially h is presence its value
in the vacuum that is what that is what matters
as far as the mass is concerned.
So, it is not the Higgs fields interaction
outside, I mean other ways in general that
is giving this mass, but its interaction which
is there even in vacuum, even in the minimum
energy configuration that gives this thing.
We will not really be able to resolve this
as the interaction of a particle like the
phi field with B, but we will see its manifestation
in terms of measurable mass of this particle.
And along with this we also have as a consequence
a field. So, we did introduce this scalar
field and then, there is a remnant scalar
particle corresponding to that which is denoted
here as h the field there. Now, this h is;
looked for such a scalar particle notice that
we did not have any need of such a particle
otherwise, when we study electrons nucleus,
atoms or any other elementary particles; we
do not really see this Higgs filed. And there
was no reason to actually think about it,
but for the generation of this mass.
And such a particle was discovered in 2012
at the LHC.
In 2012 at LHC. And that was a great success
of Higgs mechanism, which therefore, was subsequently
awarded the Nobel Prize. Now, one more point
that I wanted to mention. So, just a final
point in this discussion. Why are we calling
it the Spontaneous Symmetry Breaking? Symmetry
Breaking; where is the Symmetry Breaking?
We saw that the Lagrangian is invariant.
So, we said this Scalar Lagrangian goes to
itself when we do this transformation. So,
there under the Gauge trans transformation,
the Lagrangian is symmetric. Or even if you
add the gauge field part that B mu nu B mu
nu, Lagrangian is invariant under this thing.
But, when we look at the vacuum configuration,
if it is here phi equal to 0 or the vacuum
expectation value is 0; is invariant under
the gauge transformations.
As if we say vacuum expectation value is not
0 which is some value, it cannot keep changing
it is the vacuum we have to choose a particular
energy level, energy configuration and then,
that is what it is. If we say it is v, whatever
will be is that is the value of the field
in that. Incidentally this is the measurable
quantity; So, the L experimental because they
interact with particles, we can actually measure
this not from this. Ok sorry um.
So, here we have a toy model, but in actuality
we have the gauge bosons where we can actually
measure these things. So, in the standard
model here I do not want to really write the
value of this because this is not the realistic
model that we discussed. But in the real standard
model realistic Higgs mechanism explained
in the standard model, this v is measured
to be 246 GeV. So, it is a measurable quantity.
So, it is a constant, it is a fixed value.
Then, when we actually say we transform is
this is not invariant under the gauge transformation,
because gauge transformation takes phi to
e power i theta phi.
So, we would have gone picked up a phase.
So, that is actually trying to move around.
So, you are picking this thing here and then,
it is like moving around these things. So,
now, if you pick something there, it is a
particular value that we are fixing and this
particular choice that you are making here
for the vacuum a particular configuration,
particular value of the choice of v that we
are considering field phi that you are considering
the Vacuum is not invariant under the gauge
transformation; whereas, the Lagrangian is
invariant under the gauge transformation.
So, this is what is called the Higgs Mechanism
which rely on this Spontaneous Symmetry Breaking.
Ok.
We will stop here and we will continue with
our discussion or other topics; we would not
continue with this discussion here, but we
will discuss other things in the next class.
