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PROFESSOR: OK, let's start.
So last time we started thinking about phase transitions.
We said that a simple example obtained
by taking a piece of magnet such as iron, nickel
and seeing what happens as a function of temperature.
And there's a phase transition between a paramagnet
at high temperature and a ferromagnet
at low temperatures.
This transition takes place at a characteristic Tc.
And a nice way to describe what was happening thermodynamically
was to also look at the space that included a magnetic field.
And then it was clear that various thermodynamic
properties had a singularity, had a discontinuity at the line
h equals to 0 for all T less than or equal to Tc.
Then we saw that if we looked at characteristic isotherms going
from high temperatures going to low temperatures
where there was a discontinuity, we more or less
had to conclude by continuity that the one that
goes along Tc, the magnetization as a function of field,
has to come and hug the axis at 90 degree angle
corresponding to having infinite susceptibility.
We also said that once you have infinite susceptibility,
you can pretty much conclude that you
are going to have long-range correlations across the sample.
So that if you make a fluctuation here,
the influence is going to be felt at large distances away.
OK?
So given with that piece of knowledge,
we said what we can do is to basically do
some kind of averaging.
I can take pieces of the sample.
And at each piece, I can find locally
what the magnetization is and have this field
m as a function of x that varies from one point
to another point.
Presumably, close to this point either below or above, this m
fluctuates across the sample over large distances
and is typically small.
So then we use those pieces of information
to proceed as follows.
We said all thermodynamic properties of the system
can in principle be obtained by looking at a kind of partition
function or Gibbs partition function that depends
on temperature, let's say, which is obtained by tracing
the Hamiltonian that governs all microscopic degrees of freedom
that describe this system.
The electrons, their spins, nuclei, all kinds of things.
Now naturally, this I cannot do.
But I can focus on this magnetization field close to Tc
and say that each configuration of magnetization
has some kind of a weight.
And in principle, what I can do is
I can subdivide all configurations
of microscopic degrees of freedom
that are consistent with a particular macroscopic weight.
Macroscopic field m of x.
And hence, in principle compute what that weight is.
The analog of tracing over all degrees of freedom
would now become integrating over
all configurations of this magnetization
that I indicate through this symbol
of functional integration over all configurations.
Now, clearly I can no more obtain this
than I can do the original trace.
So what did we do?
We said I can guess what this is going to look like.
Because in the absence of the field,
it's a function that has rotational symmetry.
So what I can do is I can write the log of that probability
as something.
So far I haven't done anything.
I made the assumption that I can write it
as an integral over space of some kind
of a density at each point.
So that was this kind of quasi-locality assumption.
And then I would write anything that
comes to my mind that is consistent
with rotational symmetry.
Now, since m is small in the vicinity of this point,
it makes sense that I should make something
like a Taylor expansion.
So the Taylor expansion will start not with a linear term,
which violates rotational symmetry, but something
that is quadratic.
And I can add any even power such as m
to the fourth and higher order terms.
And I can add all kinds of gradients that are consistent,
again, with rotational symmetry.
And the first of those terms is gradient of m squared.
And there are many more.
Of course, there can be an overall constant term.
Maybe I will write it out front.
Insert a Z regular, which means that you-- in the process,
you have all kinds of other degrees of freedom
that are not reflected in the magnetization.
You're going to have phonon degrees of freedom.
So there will be a contribution from the phonons
to the partition function of the system.
We are not interested in any of those things.
We are interested in what becomes singular over here.
And the reason I write that as Z regular
is because it's presumably some benign function of temperature
as I pass through this point.
So it is, indeed, a function of temperature.
It is also worth emphasizing that not only is
this parameter that appears outside representing
all kinds of other degrees of freedom
a function of temperature, that these phenomenological
parameters that I introduced here
are also functions of temperature.
Because the microscopic weight, the true Hamiltonian,
is the one that is scaled by 1 over kt.
Just because of analogy, I sometimes
call this combination of what is happening here beta H or minus
beta H as appearing in the exponent.
But that, by no means, indicates that the coefficient here
are scaling inversely with temperature
like the true microscopic coordinates.
Because in order to do this coarse graining,
I have to integrate over a lot of different configurations.
So there is energy associated with this.
There is entropy associated with this.
There is all kinds of complicated things
that go into this-- these parameters.
So that's important to remember.
Finally, if I slightly go away from here just
to explore the vicinity of having a finite magnetic field,
then I can work in the ensemble where
I have added the field here.
And the weight here will be modified
by an amount that is h dot m.
So what is occurring here is this Landau-Ginzburg model
that I right now introduced in the context
of magnetic systems.
But very shortly, I will introduce
this in the context of super-fluidity.
I can go back to the original example
that we started with the liquid gas phenomena,
replace m with some kind of a density difference,
and then it would indicate a phase transition
of the liquid gas system.
So it is supposed to be very general,
applicable to a lot of things because it is constructed
on the basis of nothing other than symmetry principles.
But at the cost of having no knowledge of how
these phenomenological parameters depend
on the microscopics of the system-- on temperature.
Clearly, for example, even in the context of magnet,
I would construct the same theory for iron, nickel,
cobalt, et cetera.
But presumably these coefficients
will be very different from one system as opposed to the other.
All right, so that's supposed to be, as far as we are concerned,
sufficient to figure out what is going
on in the vicinity of this transition.
Maybe I should emphasize one more thing,
which is that I said that after all what we are trying
to figure out is the nature of singularities in free energy,
in phase diagrams, et cetera.
Yet, when I wrote this, I insisted
on making an analytical expansion.
And the reason making an analytical expansion here
is justified is because to get this expansion,
I summed over degrees of freedom--
I averaged degrees of freedom over some finite piece
of my system.
Maybe I took a 100 by 100 by 100 Angstrom cubed
block of material and averaged the magnetization of spins
in that area, et cetera.
And the idea that we also encountered last semester
is that as long as you are dealing
with a finite system, all of the manipulations
that you are doing involve analytic functions such as e
to the minus beta H applied to a finite number of integrations.
And you simply cannot get a singularity out of such
a process.
So the averaging process that goes in this coarse graining
must give you analytical functions.
That's not the origin of the singularity.
The origin of the singularity must
come from taking the size of the system, the volume
of this piece of iron, essentially to infinity.
That's what gives us the singularity.
Also, because of that I expect that at the end of the day,
this Z of T and h that I calculate
will be something that is proportional
to volume or something like that in the exponent.
That is, I will have something like e
to the-- from this component minus V some beta f regular.
And actually, let me just forget about the beta
and just write it as some other function regular.
It's an extensive quantity.
And the result of this integration
of all of the field configurations
should give me another contribution
that is proportional to volume.
And hopefully, all of the singularities that I expect
will come from this piece.
And I am going to expect those singularities to arise only
in the limit where V becomes large.
Now, last semester we saw a trick
that allowed us to do integrations
when the final answer was extensive.
It was the saddle point result.
Basically, we said that as we span the integration range,
there is a part that corresponds to extremizing whatever you are
integrating that gives you overwhelmingly
larger weight than any other part of the integration.
So let's try to, without justification--
and we'll correct this-- apply that same principle
of a saddle point approximation to the functional integration
that we are doing over here.
That is, rather than integrating over all functions,
let's find the extremum.
Where is this function maximized,
the integrand is maximized?
So what I need to do is to find maximum of exponent.
So what is happening over here.
I am going to make one further statement here
before I do that, which is that this term, the thing that
is proportional to gradient of m,
has something to do with the way spins
know about their neighborhood.
And if I am looking at something that is ferromagnet,
the tendency is for neighbors to be in the same direction.
That's how ferromagnetism emerges.
And that corresponds to having a K that is positive.
So that helps me a lot in finding
the extremum of the function because it
means that the extremum will occur when m of x
is uniform across the system.
Any variations in m of x will give you a cost from that
and will reduce the probability.
So with K positive, I know that the optimal solution
is going to correspond to a uniform value
across the system.
What is this uniform value?
If I put a uniform value, first of all,
the integration will simply give me the volume.
That's good because that's what I expect over here.
So at the end of the day, what do I get?
I will get that F singular is, in fact, the minimum of psi
of m where psi of m is obtained by simply evaluating what
is in the integrand at a uniform value of m.
t over 2 m squared mu m to the fourth.
Potentially higher orders, but no K. The K disappeared.
Minus h m.
Actually, I kind of expect also that the uniform solution,
if I put a magnetic field, will point along the magnetic field.
If there's an up field, the uniform solution
will be along the field.
If it's 0, then I don't know.
If it is down field, it will be the opposite.
So it makes sense that if I have a field,
the uniform magnetization should point
along it, which means that over here this dot product is really
just h times m.
So I reduced the complexity of the problem subject
to this saddle point approximation
to just minimizing some function.
Now, it is important to note, since I
was talking about analyticities, and singularities, and things
like that, that whereas the function that is appearing here
as psi of m as we discussed is completely analytical,
the operation of finding the minimum of a function
is something that introduces non-analyticity.
So that's how we are going to get that structure that we
have over there.
Let's explicitly show how that occurs.
So I'm going to show you the shape of this function
in the space that is spanned by the parameters
that I have here, which are t and h.
We will come to u later on.
But for the time being, since I want to show things in a plane,
let's stick with t and h.
So first of all, let's look at the regime where t is positive.
So this is 0.
To the right is t is positive.
Right on the axis when h equals to 0,
the function is m squared plus m to the fourth, et cetera.
The coefficient of m squared is positive.
So basically, right on this axis the function
is like a parabola.
So what I am plotting here are different forms of psi of m.
So this is m squared is a parabola.
If I go further out, m to the fourth
is something to worry about.
But if I'm really looking at case where the emerging
magnetization is small, I can stick just
with the lowest order term in the expansion.
Yes?
AUDIENCE: Are we plotting h or psi?
The axis label was h.
PROFESSOR: OK.
So there is a two-parameter plane t h.
At each plant in this parameter plane, I can plot what psi of m
looks like.
Unfortunately, I can't bring it out of the plane,
so I am going to put it in the two dimensions.
So this is why I wrote psi of m with green
to distinguish it with the h that is in white.
So this is supposed to be right on the axis.
If I go up on the axis, what would psi of m look like?
So now I have a term that is positive h m.
So it starts linearly to go down,
and then the m squared term takes over.
So now the minimum is over here.
Whereas, right on the axis-- again,
by symmetry-- the minimum was at 0.
If I go and look at what happens at the other side
when h is negative, then the slope
is in the opposite direction.
So the function kind of looks like this.
And the minimum has shifted over here.
So if I were to plot now what the minimum m
bar is as a function of the field h,
as I scan from some positive t-- from h positive to h negative,
what we have is that this sequence then
corresponds to a curve in which, essentially, m
is proportional to h.
Something like this.
So this is for t that is positive.
Now, what happens if I try to draw
these same curves for t that is negative?
Let's stick first again to the analog of this curve,
but now for h equal to 0.
So for h equals to 0, in plotting psi
if m, whereas previously the parabola had
the positive coefficient, now the parabola
has a negative coefficient.
So it kind of looks like this.
Now, one of the things that I know for sure in my system--
again, mathematics does not tell us that, physics tells us that.
That I have a piece of iron and I
know that the typical configurations of magnetization
are some small number.
So if the mathematics says that I have a function such as this,
it is wrong because it would say that the extremum would
go towards infinity.
But that's where the u-term comes into play.
So I need now to have a u-term with a positive sign
to ensure that the function does not have an extremum that
goes to minus plus infinity.
Physics tells us that the function
that I get if t is negative should have a positive u.
Now, what happens if we take this curve
and go to h positive?
Well, now again, like we did over here,
I start to shift it in one direction.
In particular, what I find is that the minimum to the left
goes up and the minimum to the right goes down.
What did I do?
Drew this poorly.
I start to go like this.
So now there is a well-defined minimum over here.
If I go to negative fields, I will get the opposite
Where I have this.
So now if I follow a path that corresponds
to these t negative structures, what do I get?
What I get is that the extremum is at some positive value.
So this is for t that is negative.
As I scan h from positive values,
I am tracking this minimum.
But that minimum ends up over here at 0.
So basically, I come to here.
Whereas, if I come from the negative side,
I am basically following the opposite curve.
And then I have a discontinuity exactly at the h
equals [INAUDIBLE].
Again, by continuity if I am exactly at t equals to 0,
the function is kind of-- rather than the parabola
is m to the fourth, and a little bit of thought
convinces you that the shape of the curve kind of looks
like this.
We will quantify what that shape is shortly for t equals to 0.
So what did we do?
We were able to reproduce exactly the structure
of the isotherms that we are getting
for the case of the ferromagnet from this simple theory.
In order to just match exactly this with what is going on
with iron or nickel, what do I have to do?
I just have to ensure that this parameter t goes
to 0 at Tc of whatever the material is.
Now, remember that I said that my t is really
a function of temperature.
So I can certainly make an expansion of it
around any point.
Let's see the Tc of the material.
So what I require is that the first term in that expansion
has to be 0.
Then I will have something that is linear in T, and then
quadratic in T minus Tc squared and so forth.
So there is one condition that I have to impose,
that that function of t-- which again,
by all of the arguments that I mentioned,
completely has to be an analytic function of temperature.
Hence, expandable in a Taylor series.
The 0 to order term in that Taylor series has to be 0.
What else?
That I can expand any other function.
U of t is u0 plus u1 T minus Tc and so forth.
K of t is k0 plus k1 T minus Tc and so forth.
And I don't really care about any of these coefficients.
The only things that I know are the signs.
a has to be positive because the high temperature
side corresponds to paramagnet.
u0 has to be positive because I require the stability
types of things that I mentioned.
And k0 has to be positive because I
want to have this kind of ferromagnetic behavior.
Apart from that, I really don't know much.
So the Landau-Ginzburg Hamiltonian with one condition
reproduces the phenomenology of the magnet
and all the other phase transitions,
which as one of you-- I think it was David-- was pointing out
to me, is very much like dealing with a branch cut singularity
in the mathematical sense along this axis
terminating at that point.
So this branch cut singularity is a consequence
of this minimization procedure of a purely
analytical function.
And the statement of universality at this level
is that pick any analytical function
and do this minimization.
You will always get the same mathematical branch cut
that we will now explain in terms of the exponents.
So we said that experimentally these phase transitions,
their universality was characterized
by looking at singularities of various quantities
and looking at the exponent, the functional forms.
So let's first look at the magnetization.
So if I do the extremization of this, what do I do?
I have to have d psi by dm equals to 0. d psi by dm
is tm bar plus 4u m bar cubed minus h.
For h equals to 0 along the symmetry axis,
then I have two solutions.
Because that I can write as t plus 4u m bar squared
times m bar equals to 0.
If I state at that equation, I immediately
see that the possible solutions are m bar is 0.
And that's really the only solution
that I have for any t that is positive.
While for t negative, in addition to the solution at 0,
which is clearly unphysical because it corresponds
to a maximum and not a minimum, I have solutions at minus
plus square root of minus t over 4u.
And so if I were to plot this magnetization
as a function of t, essentially I
have a kind of coexistence curve because I have nothing above
and below I have a square root singularity.
So this corresponds to the exponent beta being 1/2.
So that's the prediction [? from each. ?]
What about the shape of this green curve, the isotherm
that you have at t equals to Tc?
Well, t equals to Tc, in our language,
corresponds to small t equals to 0,
which means that I have to look at the equation 4u m bar cubed
equal to h or m bar is proportional to--
well, let's write it.
h over 4u to the power of 1/3.
So this green curve that comes with infinite slope
corresponds to a 1/3 singularity.
The exponent delta was defined to be the inverse of this,
so this corresponds to having an exponent delta that is 3.
We had behavior of susceptibility characterized
by another set of exponent.
Now, the susceptibility, quite generally,
is the response of the magnetization
if you change the field.
And typically, we were interested in the limit where
we measured a response if you are just at h equals to 0
and then you put a little bit more.
The equation that we have that relates h and m bar
is simply that h equals to tm bar plus 4u m bar cubed.
Rather than taking dm by dh, let me
evaluate its inverse, which is dh by dm.
dh by dm is t plus 12u m bar squared.
And so this inverse susceptibility,
if I am for t positive, m bar is also 0.
So the inverse susceptibility is t.
If I am for t negative, m bar squared is minus t over 4u.
I put a minus t over 4u here.
And it becomes t minus 3t.
So it becomes minus 2t.
Again, nicely positive.
Response functions have to be positive.
If I were to plot this susceptibility, therefore
as a function of temperature, or t
that is related to T minus Tc, what do I get?
I will get a divergence that is inverse of 1 over T minus Tc--
on both sides.
So basically, I get something like this.
And we said that the divergence of the susceptibility we
characterize by exponent gamma.
And as I had promised, we explicitly
see that, in this case, the gammas
on both sides of the transition are the same and equal
to unity.
The inverse vanishes linearly, so the susceptibility
diverges with unit exponent.
But actually, we are making-- we can
make an additional statement here
that experimentalists can go and check
that I hadn't told you before.
Now, you see in all of these other cases
the only thing that I can say is universal side is
the functional form.
This is where the exponent beta came from.
But the amplitude of what is happening there,
or the amplitude of what is happening here,
these are things that depend on you and all of these things
that I have no idea about.
Similarly here, because I don't know
what the relationship between t minus Tc and the parameter t
is, it involves this number a that I don't know of.
But one thing that I notice is that the ratio of these two
things is a pure number.
So I say, OK, what you have is if you measure
the susceptibility on the two sides of the transition,
you will see amplitudes.
And then T minus Tc absolute value
to minus gamma plus or gamma minus.
I have told you that the gammas are the same.
I don't know what the amplitudes are,
but I can tell for sure that the ratio of amplitudes--
if this is the theory that describes things--
is a pure number of 2.
So that's another thing that you can go and say
the experimentalists can check.
They can check the divergence, and then
see that the amplitude ratio is a universal object.
OK, there's one other response function that I had mentioned.
There was the exponent alpha that
came from the heat capacity.
So how do I calculate heat capacity?
So the heat capacity, which is a function of temperature,
let's focus only in the case where h equals to 0,
is obtained by taking a temperature derivative
of the internal energy of the system.
Now, the energy, on the other hand,
is obtained by taking a d by d beta of log of the partition
function.
Now, I have all of my answers in terms of these parameters t, u,
et cetera.
But I know that to lowest order, there
is a linear relationship between small t
and the real temperature that I have
to put into these expressions.
So in particular, I can do the following.
I can say that something like d by dT d
by d beta, where beta is 1 over kt,
is something approximately when you
look at the linear regime that is close to Tc of the order--
1 over beta is going to be kb t squared.
And then I would have d by dT, and then another d by dT.
I have to evaluate all of these things
eventually in the vicinity of the critical point.
Everything else is going to be a correction.
So to lowest order, I will do that.
And then I note that this is related-- derivatives
between temperature and small t are
related through a factor of a.
So I do this and I put a squared up here.
Doesn't matter.
The only reason I do that is because through the process
that we have described here, I get an idea for what log of Z
is.
So in particular, log of Z has a part
that comes from all of the regular degrees of freedom
and a part that comes from this additional minimization
that we are doing.
So we have a minus V times the minimum of the function, which
is t over 2 m bar squared plus u m bar to the fourth
when evaluated at h equals to 0.
So this is log Z, which is some regular function
of temperature, and hence t.
Why is this part singular?
Because for t that is positive, m is 0.
So this is going to give me 0 contribution for t that
is positive.
Whereas, for t that is negative, I
have to substitute the value of m bar
squared as I found above, which is minus t over 4u.
So I will get t over 2 times minus t over 4u.
So that's going to give me minus t squared over 8u.
Here I will have-- once I substitute that formula,
plus t squared over 16u.
The overall thing would be minus t squared over 16u.
So I have to take two derivatives of this function.
You can see that the function will give me
0 above for t positive and it will give me a constant for t
negative.
So if I were to plot two derivatives of this function
as T or T minus Tc is varied.
Well, there is a background part that
comes from all the other degrees of freedom.
So there is basically some kind of behavior you would have had
normally.
What we find is that above t positive,
that normal behavior is the only thing that you have.
And when you are below, taking two derivatives of this,
something is added to this.
So basically, the prediction is that the heat capacity
of the system as a function of temperature
should have a jump discontinuity.
Now, I said that in a number of cases
we see that the heat capacity actually diverges
and we introduce then exponent alpha
to parametrize that divergence.
Since we don't have this divergence,
people have resorted to indicating this
with alpha equals to 0.
But since alpha equals to 0 is ambiguous,
putting a discontinuity in addition
to be precise about what is happening.
So the predictions of the saddle point method
applied to this field theory are the set
of exponents and functional forms
beta equals to 1/2, gamma as being 1, et cetera.
A nice set of predictions.
And of course, the test is, do they agree?
It turns out that there is one and only one case where
you do the experiments and you get these precise exponents.
And that's something like a superconductor.
And the picture that I showed you last time
for the gas, et cetera, corresponds
to totally different set of exponents.
So at this point, we have to face one of two alternatives.
One, the starting point is wrong.
We put everything we could think of in the starting point.
Maybe we forget something, but it seems OK.
The other is, maybe we didn't do the analysis right when
we did the saddle point approximation.
And we'll gradually build the case
that that is, indeed, the case and that we should treat
the problem in a slightly better fashion.
Any questions?
OK.
Yes?
AUDIENCE: Just to remind me, the saddle point approximation
was saying m was continuous, a continuous number
across the substrate?
PROFESSOR: The saddle point approximation
is to evaluate the functional integral, which corresponds
to looking at all configurations,
replacing that integration with the value of the integrand
at the point that is most probable.
In this case, the most probable point was the uniform case.
But maybe in some other case, the most probable configuration
would be something else.
The saddle point is to replace the entire functional integral
with just one value of the integral.
Yes?
AUDIENCE: So your second claims that analysis is probably wrong
somewhere.
It is most likely when we are trying
to compute the energy of the system from all of field m
and we just assumed something incorrectly,
and that's why we get incorrect exponents?
PROFESSOR: No.
My claim is that up to the place that I say Landau-Ginzburg,
I have been extremely general.
It may be that I missed something,
but I will convince you that that's not the case.
Then the line below that says saddle point approximation.
My claim is that that's where the error came.
AUDIENCE: Also, we can do a similar kind of analysis
for liquid gas transition in critical [INAUDIBLE]?
PROFESSOR: Yes.
AUDIENCE: And then, how would it be
reasonable to assume uniform density?
Because I guess the whole point of behavior--
PROFESSOR: We did exactly that approximation in 8.3.3.3.
I wrote down some theory for the liquid gas transition
out of which came the van der Waals equation.
AUDIENCE: Yes.
PROFESSOR: And the assumption for that
was that the density in the grand canonical ensemble,
in the grand canonical ensemble was uniform.
So that you either got the density
for the liquid or the density for the gas.
But I made there the saddle point approximation also.
I assumed that there was a uniform density that was--
AUDIENCE: OK.
So was just likely to be the point where--
PROFESSOR: Yes.
AUDIENCE: --something breaks up.
PROFESSOR: Yes.
So before doing that, let me point out
to something interesting that happened.
And it's just a matter of terminology.
Note that we constructed the Landau-Ginzburg Hamiltonian
for h equals to 0 on the basis that we
should have rotational symmetry.
Nonetheless, even for h equals to 0, what we find
are solutions where the magnetization is pointing
in one direction or the other.
So it is possible to have the state that
emerges as a result of a weight that has some symmetry
to not have that symmetry.
So the symmetry is spontaneously broken
and the direction in space is selected.
Now, of course, what that means is
that if you apply the rotation operation to one
of these ground states, then you will
generate another equally good ground state.
You can take everything that is pointed along the z-axis
and make them all point along the x-axis.
That's an equally good ground state.
So essentially, you have a manifold of possible states.
And making a change from one state deforming
to another ground state does not cost you any energy.
So one consequence of that is that slow deformations
should cost little energy.
What do I mean by that?
So let's imagine that I start with a state
where after I minimize, I find that all of my magnetizations
are pointing up.
Now, as I said, I could rotate everybody into this direction
and the formation of my state would cost no energy.
That's a uniform deformation.
What if I took a deformation that is very slow?
So I gradually rotate from one to the other state.
Then in the limit where the wavelength of this deformation
becomes of the order of the size of your system,
you should have no energy cost.
And it kind of makes sense that in the limit where
you have long wavelengths, you should have little energy cost.
So you should have slow wavelength, no energy
distortions or modes, called Goldstone modes.
But you can only have this for a broken continuous symmetry
such as what I have depicted there,
where all orientations are equally likely.
But if I had the liquid gas system,
the density was either above average or below average.
If I had uniaxial magnet, the spin
would be either pointing up or down.
Then I can't deform slowly from one to the other.
So for discrete symmetries, you don't have these modes.
For continuous symmetries, you have these modes.
And actually, we've already seen one set of those modes.
These were the phonons.
When in the first lecture I was constructing
this theory of elasticity, I said
if we take the whole deformation and move it uniformity,
there is no cost.
And then we were able, based on that,
to conclude that long wavelength phonons have little cost.
And we wrote their dispersion, relation, et cetera.
So phonons are an example of Goldstone modes.
These kinds of rotations of spins in a magnet-- magnons
are another example of these modes.
But something else that we said, therefore in the first lecture,
is something that we should start to think about.
Which is that we said that because these modes exist
and they have so little energy cost, if I
am at some finite temperature, I will be able to excite them.
So I know for sure that if I'm at finite temperature,
there are at least these fluctuations that
are going on in my system.
And maybe in lieu of that, I should
be wary of assuming that only the state where everything
is uniform is the thing that is contributing.
What about the fluctuations?
So let's think about these fluctuations that
are easiest and most easily generated
and look at their thermal excitations and consequences
for the phase in phase transition.
And let's do that in the context of superfluid.
So we saw the problem of super fluidity
towards the end of 8.3.3.3.
You had helium that was an ordinary liquid.
We cooled it below 2.8 degrees and suddenly it
became a new form of matter that has this ability
to flow through capillaries, et cetera.
And we pointed out that there was
some kind of quite likely quantum origin
to that because of the similarities
that it showed to Bose-Einstein condensation.
And basically, it was in this context
that Landau introduced something like this The theory
that we write down where he chose as order
parameter as the analog of the m of x that we have over there,
a complex function psi of x.
And very roughly, you can regard this--
and again, this is very rough-- as overlap of wave
function with the ground state at position
x in some coarse-grained sense.
Now, anything quantum mechanical we
saw has an amplitude and a phase.
So this is actually a number plus a phase.
Or if you like, it has a real part and an imaginary part.
And there is no way that we know anything about the phase.
The phase is not an unobservable.
So the probability that when we scan the system
we have identified some psi of x that the probability should
depend on the phase is meaningless.
It's not an observable.
So this functional should only depend
on things like absolute value of psi.
If, like Landau, we assume that it is a local form,
then the kinds of terms that we can write
are absolute value of psi squared, absolute value of psi
to the fourth power.
And the tendency for the order to expand across the system you
would put through a term such as gradient of psi squared.
Now, for the case of the superfluid,
there is no physical field that corresponds to the h.
That just you don't have that field.
You can convince yourself that if you write psi
to be psi 1 plus psi 2, i psi 2-- real and imaginary part.
And put that in this formula, that corresponds exactly
to the theory that we wrote over there as long
as we choose a two-component magnetization.
So these two theories are identical.
Now, if I look at this system for t negative
and try to find a minimum of the functional
that I have over there, then the shape
of the function functional psi-- poor choice of notation.
Psi both being the wave function as well as the function
that I have to extremize.
But let's stick with it.
It has a minimum that goes along a circuit.
So basically, take this picture that we
have over here that corresponds to essentially one direction
and rotate it.
And what you will get is what is sometimes
called the wine bottle type of shape, or the Mexican hat
potential, or whatever.
But essentially, it means that there
is a ring of possible ground states.
So the minimum occurs for psi of x
having some particular magnitude which corresponds
to the location of this ring-- how far
away it is from the center.
And that will be given by the square root of t formula
that we have up there.
But then there is a phase that is something
that you don't know.
Now, let's ask the question.
Suppose I allow this phase to vary
from one part of the sample to another part of the sample.
So that's the analog of this slow distortion
that I was making up there.
So essentially, as I go from one part of the sample
to another part of the sample, I slowly
move around this bottom of this with potential.
And I ask, what is the cost of this distortion that I
impose on the system?
If I calculate beta H for psi bar e the i theta x, what I get
is whatever I have put over here,
such as this function which minimizes the-- which
is the location of the minimum.
But because of the variation in theta that I have allowed,
there is a cost.
So let's write that as beta H0 plus this additional cost.
The additional cost comes from this term.
If I simply put psi bar in to the i theta over there,
I will get an integral d dx k psi
bar squared over 2 gradient of theta squared.
So there is an additional energy cost.
This is very similar to the energy cost
that we had for phonons.
Because if I Fourier transform, you
can see that I get a k squared just
like we got for the case of phonons.
And just like for the case of phonons,
I expect that at some finite temperature,
these kinds of modes are thermally excited.
So in reality, I expect that if I'm at some finite temperature,
this phase will fluctuate across my system.
And maybe I should note that whereas I'm here
thinking in terms of thermal fluctuations,
by appropriate boundary conditions one can establish
a gradient of theta that is uniform across the system.
And that actually corresponds to a superfluid flow.
So the case of a superfluid flow can
be regarded by having a gradient of theta
being proportional to velocity, and then this
is something like the kinetic energy.
But that's a different story.
Let's just stick with the fact that this
is the cost of making these distortions.
And I want to know, what's the probability of having
one of these distorted shapes?
Yes.
AUDIENCE: Question.
When you're introducing the psi as another parameter
and you call it overlap of [INAUDIBLE] what is boundaries
and what values this sort of parameter can take?
So I basically wonder if we have this Mexican hat-shaped
potential with minima on the ring of radius at 1,
can the value of potential-- of the other parameter principally
be further than that?
PROFESSOR: OK.
So again, here we are trying to phenomenologically explain
an observation that there is a transition between a case where
there is no super fluidity and right when
a certain small amount of super fluidity
has been established in the system.
The question that you asked over here is legitimate,
but you could have asked it also for the case
of the magnetization.
So you could have said, why not to have
that potential with the minimum somewhere else?
But that does not explain the phenomena
that we are trying to explain.
The phenomena that we are trying to explain
is the observation that I go from having nothing
to having a little bit of something.
And I choose the mathematical form
that is capable of describing that.
AUDIENCE: My question basically was,
when we were talking about magnetization, if you take
a piece of metal, you can magnetize it
from 0 to a pretty large value.
PROFESSOR: No.
AUDIENCE: If we are interested in something not too large--
PROFESSOR: No.
AUDIENCE: And as I say, [INAUDIBLE].
PROFESSOR: What is your scale?
What is very large?
For a magnet, there is a maximum magnetization--
AUDIENCE: In this case, I mean that spontaneous magnetization
for a magnet would be lower than saturation, no?
PROFESSOR: What is saturation?
AUDIENCE: When all spins are same direction.
PROFESSOR: OK.
So you have a microscopic picture in mind.
Now, the place that we are is far
from that saturation magnetization.
Similarly, in this case, presumably
if I go to 0 temperature, there is some uniformity.
And if I call this an overlap, the maximum of it will be 1.
AUDIENCE: Yeah.
But basically, I just don't understand--
what is overlap of wave function?
PROFESSOR: Well, that's why I didn't
want to go into that detail.
But basically, the overlap of two wave functions
would be the psi 1 star psi 2 of x.
And if you are thinking about the ground state,
let's say that I have normalized this function
to have a maximum of 1.
The point is that what that maximum is,
is folded into all of these parameters-- a, u, et cetera--
and is pretty irrelevant to the nature of the transition.
AUDIENCE: OK.
PROFESSOR: OK?
So the probability of a particular configuration
of theta across the system is given by this formula.
I can unpack that a little bit better just
like we did for the case of phonons
by writing theta in terms of Fourier modes.
e to the i q dot x theta of q, which for the time being,
I assume I have discretized appropriate values of q.
I choose this normalization root V in this context.
If I substitute that over here, what
I find is that beta H as a function of the collection
of theta q's is, again, some beta H0,
which is not important.
I put gradient of theta.
Once I do gradient of theta, you can
see that I get a factor of iq.
I will have two of them, so the answer
is going to be proportional to q squared.
Let's call this combination k psi bar squared k bar
so that I don't have to write it again and again.
k bar over 2.
And then theta of q squared.
OK?
So the probability of some particular combination
of these Fourier amplitudes is proportional to exponential
of this.
And therefore, a product of independent
Gaussian-distributed quantities.
q squared.
Yes.
AUDIENCE: When you plug in your summation of your-- well,
your Fourier series into the gradient
and then you square that, why don't you
get interactions between the different Fourier amplitudes?
PROFESSOR: OK, let's do it explicitly.
I have integral d dx gradient of theta gradient of theta.
Gradient of theta is iq sum-- OK.
Is i sum over q q e to the i q dot x theta tilde of q.
And I have to repeat that twice.
So I have i sum over q prime q prime into the i q prime dot
x theta tilde of q prime.
So basically, this went into that.
This went into that.
OK?
So I have a sum over q, sum over q prime,
and an integral over x.
What is the integral over x of e to the i q plus q prime dot x?
It is 0, unless q and q prime add up to 0.
This is delta function.
So you put it there.
Only one sum survives.
Actually, I had introduced the normalizations that
were root V. The normalizations get rid of this factor of V.
That's why I had normalized it with the root V.
And I will get one factor of q remaining.
Since q prime is minus q iq iq prime becomes q squared.
And then I have theta tilde of q theta tilde
of minus q, which gives me this.
So each mode is independently distributed according
to a Gaussian, which immediately tells me
the average of theta of q tilde is, of course, 0.
Let's be careful and put the tildes all over the place.
While the average of theta tilde of q squared
is 1 over k bar q squared.
Again, all that says is that as you go to long wavelength modes
and q goes to 0, the fluctuations
become larger because the energy cost is smaller.
OK, so that's understandable.
But now, let's look at what is happening in real space.
I pick two points, x and x prime.
And I ask, how do the fluctuations
vary from one point to another point?
So I'm interested in theta x.
Let's say minus theta.
Let's calculate the following first,
theta of x, theta of x prime just because the algebra
is slightly easier.
Now, theta of x I can write in terms of theta tilde of q.
So this becomes sum over q q prime e to the iq dot x
e to the iq prime dot x prime.
There is a factor of 1 over V that
comes from the normalization I chose.
And then the average of theta tilde
of q theta tilde of q prime.
But we just established that the different modes
are independent of each other.
So basically, this gives me a delta function
that forces q and q prime to add up to 0.
If they do add up to 0, the expectation that I get
is 1 over k bar q squared.
And so what I find is that this becomes related
to a sum over q 1 over V e to the iq dot x
minus x prime divided by k bar q squared.
If I go to the continuum limit where the sum over
q I replace with an integral over q,
then I have to introduce the density of states.
And so then I find that theta of x theta of x
prime is 1 over k bar integral d dq
2 pi to the d the Fourier transform of 1 over q squared.
Now, the Fourier transform of 1 over q squared
is something that appears all over physics.
So let's give it a name.
So we're going to call the integral d dq 2 pi to the d
e to the iq dot x divided by q squared.
And let's put a minus sign in front of that.
And I'll give it the name the Coulomb potential
in d dimensional space.
And for those of you who haven't seen this,
the reason this is the Coulomb potential
is because if I take two derivatives
and construct the Laplacian of that function--
so I take two derivative with respect to x here,
the two derivatives will go inside the integral.
And what they do is they bring down
two factors of iq divided by q squared.
The minus sign disappears, q squared over q squared
goes to 1.
Fourier transform of e to the iq x is simply the delta function.
So this Cd of x is the potential that would emerge from a unit
charge at the origin at a distance x.
So again, for those who have forgotten this or not seen it,
let's calculate it explicitly using Gauss' theorem.
The potential due to a unit charge
is going to be spherically symmetric, so it's only
a function of the magnitude of x.
It doesn't depend on the orientation.
And Gauss' law states that the integral of Laplacian
over volume is the same as the integral of the analog
of the electric field, which is the gradient over the surface.
So I have the surface integral of gradient.
Now, for the case that we are dealing with,
the left-hand side Laplacian is a delta function.
So when we integrate that over the sphere, I simply get 1.
So this gives me 1.
What do I get on the right-hand side?
It's just like the flux of the electric field
that you have calculated.
It is the magnitude of the electric field
times the surface area.
And I am doing this generally in d dimensions.
So the surface area in d dimension
grows like x to the d minus 1.
And then there is a factor such as 2 pi, 4 pi, et cetera,
which is the solid angle that you
would have to put in d dimensions.
And to remind you, the solid angle in d dimensions
is 2 pi to the d over 2 d over 2 minus 1 factorial.
So the magnitude of this derivative dC
by dx following from that is simply
1 divided by x to the d minus 1, or x to the 1
minus d divided by Sd.
So this is generalizing how you would calculate
Gauss' law in three dimensions.
So now I just integrate that and I find that the d dimensional
Coulomb potential is x to the 2 minus d
divided by 2 minus d Sd.
And of course, there could be some constant of integration.
So it reproduces the familiar 1 over x law in three dimensions.
But the thing that is important to note
is how much this Coulomb potential
depends on dimensions.
It determines these angle-- angle fluctuation correlations.
And typically, here in this context
we want to know something about large distances.
If I make a fluctuation here, how far away
is the influence of that fluctuation felt?
So I would be interested in the limit of this
when the separation is large-- goes to infinity.
And we can see that the answer very much
depends on the dimensions.
So we find that for d that is greater than 2,
like the Coulomb potential in three dimensions,
you basically go to a constant.
While in d less than 2, it is something
that grows as a function of distance
as x to the 2 minus d 2 minus d Sd.
And actually, write at the borderline dimension
of d equals to 2, it also grows at large distances.
If you do the integration correctly,
you will find that it is 1 over 2 pi log x over some distance
or [INAUDIBLE].
Now, what you are really interested-- actually,
this thing that I wrote down is not particularly meaningful.
The thing that you are interested in
is what I had originally written,
which is that if I look at the angle that I have at x and then
I go far away-- because the angel itself is not
an observable, but angle differences are.
So the average of this quantity will be 0.
But there will be some variance to it,
so I can look at this quantity.
And that quantity-- I can expand this--
is twice the average of theta at some particular location.
Presumably, it doesn't matter which location I look at.
So it's the variance locally that you have in the angles.
And then minus twice theta of x theta of x prime,
which is the quantity that I calculated for you above.
So all I need to do is to take the Coulomb potential that I
calculated, multiply it by a factor of 2,
divide by a factor of k bar that I basically
indicated as part of the definition.
So this object is going to be 2 x to the 2
minus d divided by k bar 2 minus d Sd.
And actually, the reason I do this is because now I
can indicate the overall constant as follows.
Remember that all of our statistical field theories
are obtained by averaging.
And I shouldn't believe any of these formulas
when I look at very short wavelengths.
So I shouldn't really believe any answer
that I got from those formulas when the points x prime and x
come too close to each other.
So there is something of the order of a lattice spacing,
averaging distance, et cetera, that I'll call a.
And by the time I get to a, I expect that my fluctuations
vanish because that's the distance over which
I'm doing the average.
So I manage to get rid of whatever this constant is
by substituting [INAUDIBLE] with the scale over which I expect
my theory to cease to be valid.
But again, what I find is that if I
look at the fluctuation between two points at large distances,
if I am in dimensions greater than 2,
this thing eventually goes to a constant.
Which means that if I'm in three dimensions,
the fluctuation in phase between one place and another place
are not necessarily small or large because I don't know what
the magnitude of this constant is,
but they are not getting bigger as I
go further and further along.
Whereas, no matter what I do in d that is less than
or equal to 2, this thing at large distances blows up.
So I thought that I had a system where
I had broken spontaneous symmetry and all of my spins,
all of my phases were pointing in one direction.
But I see that when I put these fluctuations, no matter
how small I make the amplitude, the amplitude doesn't matter.
If I go to far enough distances, fluctuations
will tell me that I don't know what the phase is from here
to here because it has gone over many multiples of 2 pi,
so that it has become divergent.
So what that really means is that because of fluctuations,
you cannot have long-range order.
So destroy continuous long-range order
in dimensions that are less than or equal to 2.
This is called the Mermin-Wagner theorem.
So you shouldn't have any, for example, super fluidity,
magnetization, anything in two dimensions of one dimension.
If you go to long enough, you will
see that fluctuations have destroyed your order.
So we can see already how important fluctuations are.
This d equal to 2 is called the lower critical dimension.
It is this phenomena of symmetry breaking, ordering,
phase transition, et cetera, that we
are discussing for continuous systems--
for continuous symmetry breaking can only
exist in three dimensions but not in two dimensions.
We'll see that for discrete symmetries,
you can have ordering in two dimensions, but not one
dimensions.
So there, the lower critical dimension is 1.
Yes.
AUDIENCE: And does this hold for any n?
This example you were doing--
PROFESSOR: Yes.
AUDIENCE: --it just has two components.
PROFESSOR: Any n. n equals to 2, 3, 4, anything.
We'll see later on, towards the end of the course,
that there is a slight proviso for the case of n equals to 2,
but that we'll leave for later.