GERBRAND CEDER: What we're going to do
is work through several case studies
so you'll see it get more and more applied.
I'll start with doing a little more of alloy theory
and cluster expansions, which is really
a follow up on the Monte Carlo work
we did and you'll see how that fits in with course grading.
It's essentially the part we didn't get to when I finished.
And then I'll do a variety of things on Tuesday.
One of them is [INAUDIBLE] accelerated molecular dynamics
technique, which is sort of a specialty topic
but it gives you an idea of where people are
trying to push the boundaries.
And then on Thursday, we'll actually have a guest lecture.
We'll have Chris Wolverton from Ford Motor Company lecture.
It's kind of cool, he actually lectures from Ford
so you're going to see him project on the screen.
He does that every year and people usually find
that a really good lecture.
He essentially talks about how they
use modeling in general but in particular, first principles
methodology at Ford Motor Company.
He has a few really cool stories.
He won't tell you how much money they save the company,
but it's actually in the tens of millions of dollars
so that's why they're still there.
He won't tell you that, but I can tell you
that before his talk show.
That's on Thursday, and then we go into the last week
and I think we have only one more lecture left there.
So one of the things I want to do today
is be a little more systematic about some of the models we've
introduced for your lab.
We introduced this lattice model, for example,
to model hydrogen absorption on palladium
and several people have questions, you know?
How do you know that this is a good model?
Those lattice models actually occur
all the time in a solid state structure equilibration
thermodynamics and I'll go through them a little more
systematic today and then towards the end
prove to you that they actually are
a very accurate representation of the thermodynamics
of real systems.
So let me show you again where they will tend to show up.
I introduced the lattice model too simply
as a magnetic spin model, and that's how we started.
You model a magnet and you have up and down spins.
But as soon as you realize that those spins are really
just general occupation variables,
they tell you the result of a binary outcome,
you can also use them to represent
a whole lot of other things.
For example, if you look at superstructures
in the copper-gold system--
copper is FCC, gold is FCC, all the structures in between
are in an FCC lattice.
So if you really want to know how they look, the question is,
how do you distribute copper and gold on an FCC lattice?
So really all you need to know is on every lattice site,
is there a copper there or is there a gold?
Again, that's a binary outcome, so you can represent
that with spin variables.
So essentially the energetics and thermodynamics of this
will be represented by a spin model, also by a lattice model.
It doesn't necessarily have to be atoms,
it can be atoms and vacancies.
This is an example I'll come back to later,
this is lithium cobalt oxide.
The purple things up there are lithium, red is oxygen,
and the green is cobalt. You can actually extract lithium
from these materials, and so start
removing some of the lithiums.
And so essentially what you have then
is a distribution problem of lithiums and vacancies
on a given set of sites.
In this case, those planes that you're looking head on
are actually triangular lattices, OK?
And I'll come back to that as well.
So here the spin variable would indicate whether the site is
occupied by a lithium or not.
You can do surfaces.
Lattice models are classics for surfaces.
Here's a typical 111 surface with stuff absorbed on it.
So if you look at an FCC metal like platinum
and you look at the 111 surface, the absorption sites
have the same symmetry.
For example, the hollows, which I've indicated here.
So the hollows form a triangular lattice, OK?
And so if you look at absorption,
again, all you need to know is, is there something
at the absorption site or not?
So again, you would use a binary variable.
So what I first want to show you is
that an Ising-like model or a lattice Hamiltonian,
as you've seen before, is actually
a very natural way of writing the energy
and actually, I extend that later
to any property of configuration.
So I'm going to do that a little more formally
than we've done before.
Essentially if you want to study all
of these models that are lattice models,
you need some Hamiltonian for them.
And so I'm going to use this sigma
between these squiggly brackets as a variable
to indicate the configuration on the whole lattice.
So the variable sigma is essentially a combination
of sigma 1, sigma 2, sigma 3.
So you can think of as a vector in the configurational space
of all the occupation variables.
So what you need is a practical way
of writing that configurational Hamiltonian,
and the way you make you write almost
any practical Hamiltonian or any practical energy
function that you want as accurate as possible
is that you write it as some expansion in a suitable basis
set of functions.
OK?
Think of how you explain wave functions
in plane waves, how you sometimes
expend in sines or cosine.
The typical basis that's used here
is what's called cluster functions.
A cluster function, which I'm going to label by phi alpha,
is essentially the spin product of all the sites
of that cluster alpha.
So if I make a pair cluster, let's say this one, i and j,
the cluster function phi alpha then is sigma i times sigma j.
But I can make many other ones.
I could make a four body one, they don't even
have to be connected.
And so for the red, phi alpha would be--
let's call it i, j, k, l, would be sigma
i, sigma j, sigma k, sigma l.
OK?
I think the computer gets more tired as well as we proceed.
OK.
The first thing, so when you expend in a basis,
you want to make sure that the basis is complete to describe
your full space and then sometimes it's
useful that your basis set is orthogonal,
although that's not really necessary,
but it can be practical.
It's easy to see here that it's complete.
I mean, if you have n lattice sites,
what's the size of the space?
What's the size of the configurational space?
Well, it's 2 to the n.
On each lattice point you can put two things, essentially,
so the number of combinations is 2 to the n.
Well, how many basis functions are there?
Well, if you think of what's the choice of basis function,
it's essentially picking a combination of sites.
So that's also 2 to the n.
There are 2 to the n combinations of sites
because it's also a binary problem.
For every site, you include it or you don't include it
in the basis function.
So there are 2 to the n basis functions.
I'm going to show you in a second that they're orthogonal.
The basis that is complete, in a very trivial sense,
there are as many basis functions as states.
OK?
So you may ask yourself, what have you won?
We're actually going to expend things in this basis set,
but there's as many basic functions as state.
Well, the hope is that we can formulate these basis functions
so that whatever quantity we expend
is more rapidly convergent.
So let me prove you that the basis set is orthogonal if you
take a suitable scalar product.
The way I'm going to define the scalar product between, say,
any two functions of configuration, which
I'll call f sigma and g sigma.
OK?
I'm going to define the scalar product simply
as the sum over the configuration space
properly normalized.
OK?
So this is equivalent to if you're
working in a continuous space defining your scalar
product as the integral of the product of the two
functions over the configuration space,
and you might be more familiar because we have
a discrete configuration space.
That integral is essentially a sum
over all the possible configurations.
Let me prove you that the functions are orthogonal.
First you can note that for any basis
function, if you sum its value over the space it gives you 0.
And you can easily see that--
let's say that function would look something
like sigma i, sigma j, sigma k, maybe even more terms,
essentially these terms will alternate
as having plus and minus 1 and so they will sum to 1.
Using that, you can easily prove that the functions
are orthogonal.
So if you [? adopt ?] the two basic functions
into each other, so you're essentially
summing their product, OK?
So let's say we have a cluster alpha and a cluster beta,
maybe this is the cluster alpha and maybe
this is the cluster beta.
OK, so they have a partial overlap.
So you can write the product of alpha and beta
then as first as the product of the spins
that are in both alpha and beta.
And for those, you get sigma i squared,
and then the spins that are only in alpha
and the spins that are only in beta.
Of course, sigma i squared is always one
because sigma is plus or minus 1 your occupation variable
in a spin model, so essentially this goes away.
OK?
And this is simply a function.
You could write this as the basis
function of alpha plus theta minus alpha intersection beta.
OK?
You're essentially here multiplying
all the spins that are only in alpha or only
in beta, but not in both.
So that's trivially 0.
So the only time that this product is not 0
is when this part doesn't exist, so that
means when all things are in both alpha and beta.
OK?
So this product is equal to the Kronecker delta.
It's actually not this product, I'm sorry.
It's this product summed over all configuration,
so I should correct this.
It's 1 over 2 to the n some stigma
and then that should go in front of here.
So the basis set is orthogonal, OK?
So we have a very practical expansion now,
we have a complete and orthogonal expansion.
So one of the first things we're going to do
is expend the energy in this basis set,
because the energy is a quantity we'll need in things
like Monte Carlo simulation.
But you should right away see that this
is much more powerful, huh?
You can expend any property of configuration
in this basis set.
They may have different rates of conversions,
but essentially this is a basis set
in which you can expend anything that depends on how you arrange
the items on a lattice.
So people have done things like expend
the band gap of materials.
And if that converges fast, you have
a model that you can rapidly search then.
People have expended things like the vibrational entropy in it,
people have expended things like--
even tried to expend things like connectivity,
and so which goes not as well.
But essentially anything that becomes
a property of the arrangement of atoms on that lattice,
so in that substitution space can be expended.
For now we're going to stick to the energy.
So if we expend the energy as a function of configuration,
so we get an expansion over all the possible clusters alpha,
the basis function, and then the expansion coefficient,
which we'll call v alpha, and so of course the expansion
coefficient that's going to contain
all the energetic information.
And I'll show you in a second how it's defined,
but first I want to write out what this cluster expansion,
as it's called, looks like.
I mean, if you pull out what these cluster functions are you
essentially get something that looks like this.
There'll be an empty cluster function, so that's this term.
When alpha is just a point you'll have this term.
So this is almost like point site energies or chemical
potential like.
If alphas are pairs, you'll get the standard pair term
that we see in lattice models, but this keeps on going.
You'll have triple terms, quadruplet terms, and so on.
So essentially the cluster expansion
is the same as what's called a generalized Ising
model or "Easing" model, OK?
So where you go away from relying only on two body terms
to higher body terms as well, and what it essentially is
is a generalization of that approach.
I'm sorry, I'm going back and forth.
I'm interchanging and using e and h, h for the Hamiltonian,
e for the energy, and one of the reasons
is that I cut-and-paste equations
from various PowerPoint slides and wasn't
quite able to modify them due to the great Microsoft software.
OK.
But I think what's the really nice thing
is that this formal expansion will allow you to define
exactly what these v's are because you may have asked
yourself when you were doing in the lab
and are probably still doing the hydrogen on palladium,
what those interactions actually are?
Are they sort of physical potentials
between the hydrogen or not?
You'll see in a second that they're not,
they're much more complex than that.
OK.
So whenever you write an expansion in orthogonal basis
functions, how do you get the expansion coefficient?
Well, the expansion coefficient is essentially
the projection of your function onto the corresponding basis
function, and that's what you'll see here
and you'll get something really interesting come out.
So the way to know how a v alpha is defined,
you can essentially take the scalar product
of this expression on both sides with phi beta.
So you multiply with phi beta and then
some overall configurations.
So you're essentially taking the scalar product
here, phi beta, phi beta.
On this n, we know that this is the Kronecker delta,
so this is v beta.
OK?
So we've projected out v beta out of this expansion,
and then we have to work out essentially
what this is to know exactly what v beta means.
And you'll see this is the projection of the energy
onto the basis function [? phi a, ?]
but it's interesting when you work that out
what it physically looks like.
So that's what I'm going to do here,
so I've written out the scalar product here.
OK.
What I'm going to do is I'm going to--
this is the sum over all possible configurations
of the lattice.
Since I have a basic function phi beta here,
I'm going to separate that into a sum over the configurations
on the spins that belong to beta.
So let's say this is beta, sum triplet.
So I'm going to sum over those spins
and then separate that and sum over all the other spins
on the rest of the lattice.
So I've written that here as a sum over all the i's--
sorry, this should be sigma i--
that are in beta and summing over all the configurations,
you get that by just making all the sigma
i's plus or minus 1, OK?
And then summing over the configuration space
minus what lives on beta, and this is a bit shorthand
notation.
It's a little hard to write formally what you're actually
doing here.
But those of you who are familiar with quantum mechanics
and so may recognize this as a reduced trace.
You're essentially tracing over all of the configuration space
except what lives on beta, OK?
OK, so what I want to do is work out
the example for pair interaction since the math is about just
the right level to do in class.
The three body gets a little harder.
So we're going to work this out when phi--
sorry, when beta consists of sites i and j,
so it's a pair interaction.
So you see that in this first term
I'm going to get four terms because I have two sides.
I can do a or b, let's say.
Let me call plus 1 is [? ac ?] and minus 1 is [? b. ?] So
let's say I do the term i equals plus 1, j equals plus 1.
OK?
So the basis function is going to be 1
because I have plus 1 times plus 1,
and then what is this really?
Think about this.
So when i and j are plus 1, that means I essentially
decorate this cluster of sites i and k
with an a atom and an a atom.
So it's an a, a pair.
But what is this term?
Essentially I'm summing over all the configurations outside
of this cluster, all the spins.
So this is essentially the average energy
of the configuration with a, a pairs at i,
j, because look at what I'm summing over.
So I'm summing it over the energy of all
the possible configurations, but it's
configurations that have a fixed theta occupancy because I'm not
summing over beta.
So this first term is essentially
what I'll call EAA averaged So you see what it is?
It's I put a at i, a at j, and then
I essentially trace over the remaining configuration space.
So I put all possible configurations
on the rest of the system and calculate that average energy.
OK?
So you can do that for the other terms.
So I'll write out explicitly now V ij,
because beta is a cluster consisting of i and j.
We're going to find this n beta is the number of sites in beta,
so we get 1 over 4 because the number of sites is 2.
So the first term I got was EAA.
OK.
If I do i equals minus 1 and j equals minus 1 I'll get,
again, plus 4 phi beta.
So I get plus EBB.
And then if I do i and j a different spin, so plus 1
and minus 1, I'm going to get minus EAB minus EBA.
OK?
Does everybody roughly see where that comes from?
So I essentially explicitly evaluate
the four terms in that sum and every time, I essentially
get the average energy of that configuration.
If your system is symmetric, if i and j are equivalent,
then these are the same.
OK?
If you have translational symmetry between i and j,
so then you'll get something like minus 2EAB.
And those of you who've seen regular solution thermodynamics
may recognize this very much as a sort of effective
pre-factor in the entropy of mixing.
You'll get something like the AA interaction plus the BB
interaction minus 2 times the AB interaction,
just sort of pointing that out.
But it's sort of interesting, this
is the formal definition of the expansion coefficients.
First of all, note that they are not potentials,
they are energies.
OK?
And they are energies of configurations.
So this is not the potential between the a atom at i
and the a atom at j, OK?
The reason is that the configuration around it
modifies that interaction substantially, OK?
Typically what you'll find is that these v's are much,
much smaller than direct interaction energies,
and the reason is they contain enormous amount
of cancellation.
If you think about it, this is AA energy, BB energy,
and minus 2 times the AB energy.
So they're often several orders of magnitude
smaller than direct interactions if such a thing were to exist.
The other thing is that indirect effects
can be included, like if you truncate your lattice
model at pair terms.
OK?
Because it's an infinite expansion, at some point
you'll have the truncate it.
You are still including many body quantum mechanical effects
because the many body quantum mechanical effects enter
in the calculation of the energy of these, OK?
So what you should understand is that the level of physics
or the level at which you truncate the cluster
expansion, the lattice Hamiltonian,
has no correspondence to the level of physics for, say,
your electron physics.
OK?
Let me give you an example just to think
through what these things are.
Let's say you had a model on a lattice where things just
interacted with pair potentials, OK?
So a very simplistic energetic scheme.
If the atoms don't relax away from the lattice points,
then these energies can really be written as a pairwise sum
because then the total energy of a system
is really the pairwise sum between all the pairs.
Then now what you're going to find
is that essentially, all of the pairwise sum
cancels in this difference except the AA
interaction, the BB interaction, and the AB interaction.
So if you do just a pairwise energy expression--
and should try this one, it's very pedagogical--
and the atoms don't relax away from the lattice,
then you'll find exactly that these are pair energies, pair
potentials.
Now as soon as you let the atoms relax away from the lattice
point, like maybe the a one is bigger than the b one,
even in a pairwise energy Hamiltonian these will not be
the pair interactions because you cannot remove from these
energies.
You cannot cancel the energetic effects between a and the rest
of the material and b and the rest of the material, OK?
So the really cool thing about these expansion coefficients
is that they will include all that extra physics,
like relaxation away from the equilibrium lattice site,
and I'll come back to that later.
Come on.
Shouldn't write so much.
I think class is over, no?
Ah, there we go.
So I'll go back and forth between writing the cluster
expansion in terms of the phi alpha, so the basis function,
or being more explicit and writing
the basic functions out as the products of spins right away.
So all your physics of your system,
all the energetics sits in those coefficients v of course.
So the question is, how do you get them?
Well, there's a lot of different ways
that people have tried to get them
and it depends a little bit on the problem.
For simple Hamiltonians, you can often really
just calculate them directly by their definition,
essentially as the projection of the energy
onto that basis function.
You can do that for pairwise Hamiltonians even
including forced relaxation.
So even when atoms don't sit on a lattice,
you can do it by Fourier transforming,
and that's in one of the reference that I give you.
But as soon as your physics is very multibody,
it becomes very hard to do it analytically.
There are people who do it numerically very much
according to the definition.
So you literally take a system and put an AA atom here
and then calculate a lot of different configurations
around that.
And so literally start to approximate those averages,
EAA average, EBB average.
The practical way that people pretty much do it now most
of the time, which is also the least elegant way,
is that essentially, you fit your whole expansion at once
to a large number of energy calculations.
So the first thing you have to do, of course,
is to truncate your Hamiltonian because this
is an infinite expansion.
You should really see this, this is a trivial expansion.
You're expending 2 to the n quantities and 2
to the n basis function, OK?
So it's only practical if you truncate it,
but once you truncate you have a finite set of coefficients
to determine.
So what you can do is literally calculate the Hamiltonian,
or in this case, the energy for a lot of these configurations.
Let's say there's 20 of these v's you have to determine, then
essentially if you have the energy at 20
different configurations you could determine the interaction
coefficients.
Typically people use very over-determined sets.
So if you want 20 of these, you may do 50 of 50 energies
and do a squares fitting just to make
the determination of interactions much more stable.
So that's essentially how it's done.
You have an energy method.
I wrote it as a first principles method
here, could be any energy method.
You calculate a lot of energies of different configurations
and then you fit the Hamiltonian and you plot the interactions
out.
It's really that simple.
And once you have your lattice model Hamiltonian,
you now know that it's fairly easy to equilibrate that.
So you can get ground states out of that,
you can do a Monte Carlo simulation on it
just like you're doing in the lab,
and then get phase diagrams thermodynamic properties out.
So essentially you can get both the structures and all
the thermodynamic information and more important,
as I'll show you in a second, all
the information you can derive from
the thermodynamic information, which is something people
often forget.
So what I was going to do is show you
a few examples of how this is used and then in the end,
essentially come back to some of the theory of that.
So here's one that shows how elegant it can be.
What I'm going to show you is a calculation
to try to get the solid part of the calcium oxide,
magnesium oxide phase diagram.
This is perfect for a cluster expansion.
Calcium oxide and magnesium oxide
are both rock salt structures.
So a rock salt is essentially an FCC lattice
of oxygens, which I think are in this picture, the purple ones.
And then the cations, either calcium or magnesium,
sit in the octahedral sites, octahedral interstitials.
And the octahedral interstitials of an FCC lattice
also form an FCC lattice.
So you can think of rock salts as two interpenetrating FCC
lattices, one of cations, one of anions.
So when you mix the two, you're essentially
asking the question, at each interstitial
is there a calcium or a magnesium sitting there?
So again, you have a perfect binary problem.
It's a lattice model.
But this already illustrates a nice thing of this approach,
in your whole lattice model, the oxygens won't appear.
So the variables you use are only
going to be defined on the cation lattices.
So the cations are the smaller ones.
So on this one, you'll have sigma i say plus 1
and on this one, you'll have sigma j equals minus 1.
So you're essentially going to express the energy solely
in terms of spin variables on the cation sites,
and that makes sense.
There's nothing unknown about the anion sites.
They're always fully occupied, OK?
So the anions will show up when you calculate
the energies of these configurations, of course,
but there's really no need to expend the energy
in their occupation variables because they're always there.
They're always minus 1.
So you see, we're going to end up
with an effective Hamiltonian between the cations that
includes the energetic intermediary of the anions
without explicitly including them.
So how do you do something like that?
Well, you truncate your Hamiltonian [? number, ?]
it's always going to be the same procedure.
You calculate the energy, a lot of different configurations.
This is old work, so it wasn't fully quantum mechanical.
There are two sets of points-- so this is actually energy--
I'm sorry, this is in electron volt per cation.
The blue points are done with empirical potentials.
Remember, we did that all the way in the beginning of class?
These are essentially Buckingham potentials
with electrostatics and shell models attached to them.
And the open red circles is a self-consistent PIB,
which is a really poor man's quantum mechanics.
It's a sort of intermediate between Thomas Fermi theory
and better quantum mechanics, but think of it
as a poor man's quantum mechanics,
and these are mixing energies.
So they're referenced to 0 here and here, that's
why they have that parabolic shape because you force them
to 0 at 0 and 1.
First of all, they're positive so
that means the two things don't want
to mix, which is in accordance to the phase diagram,
but you may notice something already
that's kind of interesting.
The quantum mechanical values actually
tend to be lower than the potential ones,
and this is very common.
Typically a better energy model will
lower the energy of low symmetry states
more than a bad energy model.
And the reason is that usually a good energy model has
more degrees of freedom, OK?
If you think of your potential model has no electrons in it.
There can be no electronic relaxation
by definition in a model with empirical potentials
because you don't have electrons.
A quantum mechanical model has electrons in it
so you have more ways for the system to relax.
And the lower the symmetry, the more a system
can actually use those methods.
If your system is fully constrained by symmetry,
then there's barely any way the system can relax.
OK?
So this is something you very typically see.
If you have energy methods that are fitted to perfect bulk
properties, so they're fitted in high symmetry environments,
if you then use them on defect properties,
the better the physics of the model
or the more degrees of freedom usually
the lower the defect energy will be.
But anyway, so you use this.
You'd fit your Hamiltonians to it, your cluster expansion
to it, and you'd get a bunch of pair interactions.
So the first six are pair interactions.
Again, blue-- oh, dear.
Changed the color scheme here.
Blue is the quantum mechanics, now the open circles
are the potentials, and then there
are some triplets interactions and
some quadruplets interactions.
And notice that the first pair interaction is negative,
so what does that mean?
That means the nearest neighbor's spin product
wants to be positive, get your energy down.
So that means like things want to be together.
Plus ones want to be with plus ones,
minus ones want to be with minus ones,
so that's phase separation.
You can already see it from the interaction,
but notice the scale of the interaction.
So this is now about 40 milli electron volt per cation.
So this is a far cry.
This is much lower than any direct electrostatic
interaction or even any direct evaluation of a Buckingham
potential because you have effective interactions.
Actually, if you think of the definition
of the effective pair interaction in this case--
so let me write it out.
So Vij would be 1/4.
Let me actually put calcium and magnesium on here.
So it would be E magnesium, magnesium averaged plus E
calcium, calcium minus 2e calcium, magnesium on the i,
j sites.
So you should think about where that effective interaction
comes from because look at the electrostatics.
Does the electrostatics contribute?
Magnesium and calcium are both 2 plus.
Essentially if you froze those two on their lattice sites,
you'd have in the first term 2 plus 2 plus plus the 2 plus 2
plus minus 2 times the 2 plus 2 plus.
So the electrostatics completely vanishes
from the effective cluster interaction.
So it's interesting, the fact that you
do this projection essentially filters out the largest
part of the energy here in the system,
but because it's the same for both magnesium, magnesium
and calcium, calcium and magnesium, calcium interactions
it's irrelevant for the configurational thermodynamics
and the projection essentially filters that out.
So where does the remaining Vij come from?
Where does the balance come from?
If you remember what we taught you empirical potentials,
the way you set up an empirical potential model for rock sites
is that you usually have, of course, the electrostatics,
then you usually have an oxygen, oxygen potential--
because the oxygen is the biggest ion, they touch--
and then you have a cation, oxygen potential.
Rarely do you have a cation, cation potential.
So there's not even a direct potential between the two.
So if you actually froze these things on their lattice site,
that Vij is 0 because we just said that the electrostatics
disappears.
There is no direct cation, cation term,
so Vij is 0 in the case.
So I sort of give the answer away
and I said if you freeze these things on their lattice site,
where does the interaction come from?
Well, if you freeze them, there's no interaction.
So all the interaction comes from relaxation.
OK?
The interaction comes from the fact
that calcium is bigger than magnesium
and so that sets up an effective interaction.
So it's kind of cool because with these interactions
you have a Hamiltonian and a lattice model,
which people always tend to associate
with frozen positions, but the interaction comes essentially
from the positions not being frozen,
from the fact that the bigger atom pushes the smaller
one away.
OK?
And so this is maybe the key point
to remember that once you parameterize your Hamiltonian
with relaxed calculations, that Ising model represents
the energy of those relaxed configuration.
Not the ones where you're frozen at lattice sites.
So when you do hydrogen on palladium,
a lot of that interaction that you get between hydrogen
on one spot on the surface and the other one
has nothing to do with hydrogen, hydrogen interaction.
It's actually all has to do with relaxation of the surface sites
around the hydrogen.
You put a hydrogen here, the palladium around it relaxes.
That's how it sees the other hydrogen, not because
of a direct interaction.
And with hydrogen that's not so dramatic.
If you that with other absorbents like oxygen,
it's very dramatic.
They largely interact through a strain induced in the surface.
OK.
So then you do Monte Carlo, and here's the phase diagram.
It's a bit too many things on this slide,
but essentially this is the line you should be looking at.
This is the one with potentials, so these
are the solubility limits at low temperature.
So it's a two phase region in between,
but you have some solubility of MgO and CaO and of CaO and MgO,
and this is the quantum mechanical one.
So of course this becomes a miscibility gap
because there's no liquid in the calculation
so if you somehow could include the free energy of the liquid
it would cut off that miscibility gap
and form a [INAUDIBLE].
OK?
But because you don't have a liquid,
you're essentially looking at only solid state transitions.
And again, you see the same thing.
The quantum mechanics gives you a higher solubility limit, so
a lower temperature which you can get full solution.
Because it's sometimes easier it can deal better
with the lower symmetry state.
OK.
So this is the classic technique now
to deal with any substitutional problem and I would say almost
any crystalline problem because even if you
work on systems where there are multiple lattice
types in play--
I mean, if you think of simple metals you could think,
well, maybe there's some that ordered
on a BCC lattice and some that order on FCC lattice,
you can cluster expend them separately and obtain
free energies from Monte Carlo and then put them together.
That's why this is essentially the method by which
any multicomponent crystalline problem is equilibrated,
but I'll show you some of the older work
just because historically it's kind of cool.
This is the lithium aluminum system,
which was I think only the second or third system that
was done at [INAUDIBLE] by this method.
This was done in the early '80s.
So the reason they did this in the early '80s
because everybody thought we were going to make airplanes
out of lithium aluminum because lithium
is the lightest solid element in the periodic table.
So any lithium you can put in an alloy
makes it really considerably lighter,
and the interesting piece occurs here, which you can see.
There's a delta phase here, which is metastable,
sits about here.
And what typically people do is they solution
to heat up high temperature then [? quench ?] down so that you
precipitate out delta phase to strengthen the alloy.
It's difficult to do experimentally
because this is a metastable phase,
so it's a little hard to study it.
But of course computationally, all metastability means
is that there's something else that's more stable, so as soon
as you remove that from the picture and the simulation
you can stabilize it.
And so this was one of those early calculations where
they did exactly what we showed here,
build a cluster expansion, parameterize
it, [INAUDIBLE] functional theory.
In this case, did Monte Carlo do the statistical mechanics
to find temperature and essentially get
this calculated phase diagram?
And this is the equilibrium phase diagram,
but then you can also do the metastable.
This is the delta phase precipitated by essentially
removing the free energy curves of the stable phases
and look at them.
The liquid was kind of empirically added, I think,
to this calculation.
This was my own PhD thesis, so I have to show it.
Most of my PhD thesis were just what is 1,
2, 3, 4, lines, so I got off easily.
But this is the phasing [INAUDIBLE] copper oxide,
which probably most of you familiar
is a fairly well known high temperature superconductor.
And the critical issue, or one of the critical issues
was, how the material behaves as a function of oxygen content.
Essentially that z up in the formula
can go anywhere from 6 to 7 and at 7 it's a high temperature
superconductor, at 6 and 1/2 it's
a low temperature superconductor,
and at 6 it's an anti ferromagnet.
And so figuring out what the phase ion looked like was,
I guess, worth the PhD thesis.
And essentially it's, again, a lattice model problem.
You have a bunch of sites where the oxygen can sit.
When you start taking out the oxygen
it's now the site can be vacant or filled with oxygen.
So it's a binary problem.
So you cluster expended the same thing.
In those days, the quantum mechanics wasn't all that good
and you did Monte Carlo and you get a phase diagram.
And what's neat about this O2, to my knowledge, this phase--
I think there's just some history with it--
was probably the first structure I
think that ever was predicted with quantum mechanics
before it was seen experimentally.
Now there have been a lot of cases of that,
but this was actually in the late '80s,
and it may have been the first case where
the experimental evidence came afterwards.
OK.
Here's a lot more complicated systems,
so this is the lithium cobalt oxide
I showed you in the beginning.
There's cobalt layers, there's oxygen layers,
but the key component is the lithium layer.
As you take lithium out you have a vacancy lithium problem.
So again, you expend in variables
that only describe the occupation on this layer, OK?
So you don't really have to deal with the cobalt,
with the oxygen, you just deal with that
when you do the energy calculations
and you get a bunch of effective interactions.
And again, let me stress that these ECI are
effective interactions.
If you look at the first neighbor,
it's only 30 milli electron volt, OK?
To give you an idea, those first neighbor distances
in the lithium plane are about 3 Angstrom.
If you evaluated the electrostatic source of a plus
1 charge and a plus 1 charge over 3 Angstrom,
you'd get something that's close to an electron volt. OK?
But you see, because you're taking
effective interactions you essentially
remove most of that.
Well, first some of it is screened--
because you're not in a vacuum doing your electrostatic,
some of it is screened by the oxygens that
are close by-- but you're also canceling
a lot of energetic effects and really seeing
the only part that's relevant for the thermodynamics.
And then you can, again, do Monte Carlo
just like we've done in the lab, get a phase diagram.
This is the kind of phase diagram
that you would not get out of straight Monte Carlo.
The hydrogen and palladium problem we gave you,
you can do that literally by looking
at the output of the Monte Carlo,
by looking at the heat capacity, looking at the energy,
because the system equilibrates very well.
When it doesn't equilibrate you really
have to do free energy integrations
and look at where they intersect.
This space also was predicted three years before it
was seen experimentally, so this approach has a really very
powerful predictive capability because you control
almost all the approximations.
So essentially, you have to live with the approximations
in density functional theory.
So whatever the accuracy of your electronic Hamiltonian,
you can't do any better.
OK?
But from there, you control all the approximations.
So if you don't like the convergence of your cluster
expansion, well, you add more terms to it.
OK?
So essentially you can make as accurate a representation
of your quantum mechanics as you want.
And that's why you can make Hamiltonians
that are very accurate, but they live on a lattice model
so they're extremely fast to search.
I mean, you've done the hydrogen on palladium.
Hydrogen on palladium, that trivial job [INAUDIBLE],,
this 4, 5 million spin flips per second on a stupid, old IBM
computer.
So on a new machine, you can run 10,
15 million spin flips per second,
so you can search configuration space extremely fast.
OK?
And so that's why it's so good at finding new structures.
OK, I was going to show you a surface problem.
So here's platinum 111.
Reason I took platinum, it's sort
of an important material for all kinds of catalysts and so.
And every time you have these FCC surfaces,
the possible absorption sites essentially
form a triangular lattice.
I should go to a different color here.
So you have these really old form of triangular lattice.
There we go.
Triangular lattice.
So if you want to study the absorption of something
like hydrogen or oxygen, again, you build a lattice model.
Then you have to do density functional theory calculations
to parameterize it and I think you've
done that now in the lab.
So because this is surface you would set up a slab geometry.
So you essentially would take a few layers of platinum
and then put stuff on the surface to absorb it,
and you arrange a vacuum above it and do the DFT calculations.
OK.
So here's that result. Again, this
shows both LD [? and GGA ?] but again, we plot them--
this is oxygen absorbed on platinum 111.
We reference the full coverage and the zero coverage to 0.
You may be a little surprised that you
can set two points to 0 on the energy scale,
but one, of course, is obvious.
So you can always reference one point to 0 on energy scale.
The reason you can set the second one
to 0, what you're essentially doing
is setting it 0 for the chemical potential
because the chemical potential is, in some sense,
the slope of that curve.
It's the energy changed just by adding oxygen in this case.
So you're essentially referencing
the chemical potential to 0 by setting two points to 0.
So you do that, same thing.
You fit interactions.
I've actually made some copies of real notes
so you see how this is really done.
So people sketching out on triangle lattice,
different interactions.
You find their value, again, they're very small.
The black points is for oxygen or platinum,
the other one is on ruthenium.
You know it's about 75 for a near neighbor interaction
and then it decays very quickly very fast.
And you know that they actually have to be in that range
because if you're going to see phase transitions in the range
of, say, 0 to 1,000 Kelvin, your energies have
to live on that energy scale and 1,000 Kelvin is what?
About 100,000-- well, 1,000 Celsius
is about 100 milli electron volt so your interactions
have to be in that range for stuff to happen.
So I thought I'd show you, rather than a nice picture
of a phasing-- hold on.
So how would you look for phase transitions?
You do the same stuff you do in the lab,
you track heat capacities, you track energies.
So this is for the same run, different oxygen chemical
potentials.
So with your experience, what's happening here at about 940k?
This clearly looks like a phase transition,
first or second order.
What's your guess?
Second?
It's hard to say, but it looks like second order.
I mean, it's maybe not refined enough to look,
but there's clearly a peak in the heat capacity.
Let's say we take, I don't know, the green curve,
which is at the particular oxygen there's
a peak in the heat capacity.
It's not very well defined, you shouldn't let yourself mislead.
If you didn't draw the lines here but just the points,
that peak is essentially defined by one point.
I mean, you should always be careful.
You may have very spurious behavior going on
in one similar Monte Carlo solution.
The other peaks have slightly more points,
but normally if you really wanted
to know which is first or second order
you would refine it a lot more near the phase transition.
Again, the energy seems to be continuous,
but don't let the lines fool you.
You've taken, say, this point and that point
and connected them, so of course they're continuous,
but maybe what this thing does is it continues and then shoots
down to the other vertically, so it could still be first order.
So you would normally refine this a lot more.
And the other thing, if you really
want to know that something's second order,
you look at the scaling of the heat capacity with system size
and that's the really accurate way
to know that something's second order.
You're normalized heat capacity in a second order transition,
so the heat capacity per side scales
like the square root of the number of lattice sites, OK?
You'll find that already in your simple example,
your simple homework problem of hydrogen on palladium,
you'll find it remarkably hard to identify in certain regions
whether the phase transition is first or second order.
That's why you put them in, they are
are both first and second order.
So [INAUDIBLE] beautiful phasing,
I thought I'd show you guys how it's really done.
When we do Monte Carlo, this is really like a puzzle, you know?
In papers we always show these nicely drawn things,
but this is how you get to it.
You do an enormous amount of scans in mu.
So you scan this way, then you scan in temperature,
and you slowly start piecing together
which phases exist where, where they disappear,
how they disappear, first or second order reactions.
So this is one that is near completion with a few question
marks whether these are like first or second order.
And then later you draw nice lines,
but this is one of the problems with Monte Carlo
because it's a stochastic method.
It really is a lot a lot of puzzling.
And what you'll find is that the last thing that's
useful, the least useful thing is actually
looking at the spins.
That's people's first intuition.
I'm going to actually look at the configuration in my system.
You know, any time systems are partially disordered,
just looking at them is the best way to fool you.
If you look at something partially disordered,
half of the class will say it's ordered and half of the class
will say it's disordered, and it's really
the thermodynamic behavior that tells you in which phase
it still is.
If it's in an ordered state, so in a compound state,
that means that you should be able to push it
to a disordered state.
OK?
That means you should see somewhere a phase transition
as you raise the temperature or drive the chemical potential
to another phase.
If it's already disordered, if you heat it further up
it won't go to a phase transition anymore.
OK.
Just to show you that you can push this a lot harder
and make it a lot more complicated.
I'll show you something that gets a little more abstract,
and it's from a problem that I worked on quite recently.
It's absorption induced segregation,
and here's the problem.
So the problem before was oxygen absorbing on platinum,
but if you have actually a platinum metal
alloy like a platinum metal catalyst, in this case,
platinum ruthenium, then you have
two configurational problems.
So the first question is, what's on the surface?
Is it oxygen or vacancy?
Is the site oxygen or vacancy?
But there's also segregation thermodynamics now.
On the surface, you now may not have pure platinum,
you may have either platinum or ruthenium and the two interact.
What segregates to the surface is influenced by what absorbs.
You see, you may think of that as a quaternary problem
because essentially, what's in play now?
Platinum is in play, ruthenium is in play, oxygen's in play,
and vacancies are in play--
vacancies being vacant absorption sites.
But it's actually a 2 plus 2 problem.
It's really a binary configuration problem
in the surface layer and then a binary configuration
problem on top of the surface layer,
and so you can still perfectly solve
this with a binary cluster expansion.
It's just that your spin means something different
whether you're in the surface layer
or whether you're in the absorbent layer.
So you could call them different,
but you don't have to.
You could say, well, there are variables that is,
say, this describes what's in the absorption layer
and I could just call a new set of variables delta,
this describes within the surface layer,
and then I could literally expend
in the products of these.
But to be honest, you could just all call them sigmas.
OK.
They just mean something different
whether they're defined on this site or on that site, OK?
And so you can do the same technique
and now essentially expend a quaternary problem.
And you do the same whole thing, you calculate
a lot of configurations.
Because your configuration space is essentially bigger,
you'll need more information to parameterize it.
And then you do Monte Carlo and equilibrate it
and here's an example.
This is, at a given temperature, driving up the oxygen chemical
potential so we expect more oxygen to absorb,
and that's what you see.
But you can actually track the amount of ruthenium segregating
to the surface and as you bring more oxygen to the surface,
you actually start increasing the amount of ruthenium
in the surface as well.
And so you can really solve the thermodynamics
of these problems by [INAUDIBLE] initial methods
by going through this cluster expansion,
but you can actually look at the sites.
Orange is the platinum surface, red is the oxygen absorbed,
and this picture here is the same
with the oxygen stripped off so you see underlying ruthenium.
So as you increase the oxidation,
you essentially get more ruthenium
to the surface and more oxidation.
This is sort of more and more, higher and higher
oxygen pressure.
OK.
Let me show you one more quickly before I continue
the theory and the reason I want to show you
this is because it's sort of a little different.
This is a structural problem from hydrogen
modified aluminum fracture.
OK.
The problem there is that impurity embrittlement
in metals is a big issue.
There's a lot of metals whose ductile behavior becomes very
poor when there's impurity present,
and hydrogen embrittlement in particular is a bad one.
And it's not that hydrogen's so much
worse than other impurities but first of all,
hydrogen tends to be a common impurity because it's available
everywhere.
Any time you have corrosion reactions going on,
you often produce hydrogen.
Any time you're working in acid environments,
you produce hydrogen, and it tends to be a fast diffuser.
So there's a lot of other impurities,
but a lot of other impurities sort of never
make it to the problem areas in the material
but hydrogen tends to be a fast diffuser
So the way people have dealt with this in quantum mechanics
is literally by looking at how planes decohere here
with and without hydrogen.
And of course, if you hydrogen in between it,
it decoheres more easily.
There's less work of decohesion.
But if you want to look at the structural problem of what
happens when you do this with hydrogen,
if you want to equilibrate the hydrogen
it becomes a lot harder.
So let me sort of first tell you how
would you do this without hydrogen.
You could make a slab, you would define a plane where
you open up the slab and you remove the two
slabs farther and farther apart so it's
like you're opening a crack.
You keep track of the energy and the derivative of that is,
of course, the force required to separate the slab.
But think of the problem now if you want
to do that with an impurity.
You have to now equilibrate the impurity at every separation
of the slab.
OK?
So you pull the crack open a little more, you now
have to figure out what happens to that impurity
there because as that impurity equilibrates,
not only its structure equilibrates by the way,
but also its composition because segregation to those surfaces
is an open system problem.
The concentration there is not fixed
because it comes from the bulk or from a gas phase.
So it's actually a grand canonical problem,
it's a constant chemical potential problem rather than
a constant composition problem.
So you see that the hard problem you have.
Not only do you have to arrange the impurities
in the most stable way because that's
going to determine the energy and therefore the force
it requires to open up those planes,
but you also need to equilibrate the amount of impurity
in that crack.
And so this is actually why it's so hard to include chemistry
in mechanical calculations, because you're really
working on two different time scales that
are very hard to combine.
And so the solution here was essentially
to calculate the relevant free energy of hydrogen
for any state of crack opening, for any state of separation
of the planes you tried to get the equilibrium hydrogen
content by a cluster expansion and then
from that envelope of free energies you get the force.
So it's again, the same thing.
You define a lattice model.
This was hydrogen on one side of the crack opening,
hydrogen sites on another side of the crack opening.
So now you have interactions within one plane,
but also between the two.
You do a bunch of first principles computations,
so it's, again, the same thing to fit those interactions.
Then you're going to do Monte Carlo.
You're going to, in this case, get your free energies
in the derivatives of the free energy with respect
to the separation is going to be the force.
So in this case, this is the force
as a function of slab separation for all
the different hydrogen coverages.
So I plotted the force.
I really should have plotted the energy because the energy is
the quantity that's expended.
And then you get interactions.
In this case, I'm only showing two.
So there's an in-plane hydrogen interaction,
that's the red one here, and there's an inter-plane.
The inter-plane is between the two openings of the crack
and of course, that one tends to go to 0
as you separate more and more.
OK?
So now the cool thing is at every state of the crack
opening, you can actually get the free energy of the system
fully equilibrated.
And so if you get the free energy,
you can actually find the force, and this
is the fully equilibrated thermodynamic trajectory,
that red line.
And what's really surprising is that because you fully
equilibrate the system, it suddenly
looks very different from what you thought it would be
and what you would have gotten in
an unequilibrated calculation.
These black lines, this is what you
got in an unequilibrated calculation.
It sort of makes sense, you know?
As you pull, the force goes up.
No surprise, I mean.
And then you reach a critical point,
a point of maximum force.
At that point, you're unstable because essentially the force,
as you pull further, goes down so you're unstable.
On the downslope you're essentially unstable.
But look at the red line.
So this as a chemically equilibrated system
behaves very different.
Essentially you pull a little bit,
you open the crack a little bit, and then
the crack opens with no increase in force.
OK?
And then you pull the saturated crack apart.
And essentially what this is is, because you've
done your thermodynamics right, you actually
have a first order transition in length space.
OK, this is a length discontinuity.
It's just like a volume discontinuity
in a regular first order transition, OK?
Here, essentially the crack is separating at constant force.
OK?
It's like when you do a first order transition.
Upon heating, you're putting an entropy
at constant temperature for example,
or if you do a pressure induced phase transition,
you get a volume change at constant pressure.
OK?
So it's the same thing.
But you see that by doing the chemistry
right and the equilibration, you get very different results.
OK.
So the last thing I want to do is come back to the idea
that we started with and I wrote down
the energy as a function of configuration
and and then showed you how you cluster expended.
What I want to do is now come back again
to what the definition should be of that energy,
and we did that at the back end of the Monte Carlo lecture
but because it's kind of important
I want to come back to it because it's not totally clear
what that energy or that Hamiltonian in the lattice
model should be.
So let me take a step back-- and I know you've seen this,
but I'm going to go through it.
If you want to get the free energy of a system
you normally want to include all excitations because essentially
you have to calculate the partition function,
you have to trace over all excitations,
and that will give you the full partition function
and the [INAUDIBLE] will give you the free energy.
So if you think of the most basic system,
we'll always have vibrational excitations, OK?
If we have anything with a binary problem like A's or B's
or something and vacancies, we will
have configurational excitations,
and then unless you're in an insulator
you'll have electronic excitations.
The idea we talked about last time
is that the way you can define a Hamiltonian that
lives in this space, which is the one with the slower time
scale, is by integrating over the faster timescale.
And I started to show you last time
I think when we finished the Monte Carlo lecture.
And let me sort of go through that formal description
again so we see exactly what our latest model Hamiltonian is.
So I'm going to do it here only with vibrations
of the fast time scale and substitutions
as the slow timescale, and after you'll
see how it would go with electronic excitations as well.
So I replace the coordinate ri, so this is a set of coordinates
that describe the atom, into a lattice site
position and a deviation from that lattice site.
OK?
The lattice site position, that is fully
defined by the set of configurational variable,
sigma, and the deviation from the ideal lattice site position
are called configuration nu.
So the combination of sigma and nu
is really the same as the set of vectors ri, OK?
So what I'm going to do is--
you already see where I'm going probably with this.
I'm going to trace over the nus to get
a Hamiltonian in the sigmas.
OK.
So the full partition function is
the sum of the exponential of the energy, which
depends both on sigma and nu, and I
sum over the sigma states and the nu states.
OK?
What I'm going to do is I'm going to do the inner sum
so that I'm only left with the sum over sigmas of sum
f, which is the Hamiltonian in the configurational space.
And if you look to make these to the same, what
that f should be?
Well, f is the free energy that I get by tracing over the nus.
OK.
I mean, there's no physics here.
This is pure and simple substitution of variables.
So I call, essentially, this sum here
is the exponential of minus beta f,
which means that f is the log of that exponential.
So look at what I formally end up with.
I have a partition function that only sums
over substitutional states.
So this is a problem that lives in the same space as a lattice
model because I'm only coming over discrete configurational
states.
So what I'm really saying is that if I do this partition
function correctly, I have the full partition function
and therefore, the free energy.
But to do it correctly, my Hamiltonian in my lattice model
state needs to be f, not e.
And what is that f?
Well, f is essentially a free energy.
It's the log of a partition function.
And what partition function is it?
It's the partition function by summing over
all the faster states.
So the problem is, essentially, simple.
The Hamiltonian you should use in your slow time scale space
is the one you get by integrating
or by calculating the free energy of your fast time
scales.
So in this case, what should you do?
In everything I showed you earlier this morning,
I used the energy as the Hamiltonian.
OK?
I use the direct to DFT energy in my Hamiltonian.
What I really should have used is the vibrational free energy
to be correct.
If I use the vibrational free energy in that Hamiltonian,
I would have gotten the full partition function
and therefore, I would have gotten the total free energy
of the system.
So what did I do instead?
I didn't calculate this.
I approximated this by something.
I approximated this by the state in this ensemble
with the lowest energy.
I mean, if you think of this as a free energy,
this is going to be a free energy plus some entropy term,
which you could think of as a ground state
energy plus contributions from excited states.
OK?
I've basically said, I'm taking this just as the ground state
energy in this [? ensemble ?] because I'm
taking the lowest energy configuration that I
can get by varying over this.
OK?
And what is varying over this?
What are the nus?
The nus are the displacement from the lattice sites.
So essentially I'm relaxing my system,
letting it displace from the lattice site,
I get the lowest energy state in the ensemble,
and that's what I expend.
So when I'm done, what do I have?
When I'm done and I do Monte Carlo with that Hamiltonian,
I have essentially the right energetics,
I have configuration entropy, but I
don't have vibrational entropy.
To have vibrational entropy, I would have to do this sum,
not just take the lowest energy term.
OK.
Come on.
OK, let me sort of skip that.
OK.
People actually have done examples
where they do the full integration
over vibrational states rather than just approximating
that ensemble with its lowest energy term.
You see that it's a lot more work.
Remember that to parameterize the cluster expansion,
I need the Hamiltonian evaluated for a bunch of configurations.
Even in the simplest case that's easily 40, 50 configurations.
So if you want to include the vibrations,
you actually have to do, for 50 configurations, the full phonon
DOS or do molecular dynamics on them and integrate the phonon
and get the vibrational free energy through that.
So it's like two orders of magnitude more work.
So rather than just the energy, you
want to get the whole vibrational spectrum,
and that's why people usually don't do it.
But the nice thing is that in this approach,
you can do it if you really want to.
And I'll show you one example where people did it and looked
at the difference.
It was the very simple case of the cadmium magnesium phase
diagram.
It was done by Mark Asta, who's now at Northwestern.
So this is the experimental phase diagram actually.
So this is the calculation with no vibrational entropy
and with vibrational entropy.
So the first one is the one that I sort of
showed you as the standard one.
You just take the lowest energy for a lattice model
configuration and you do Monte Carlo then with that.
And this is if you literally do the whole phonon
DOS for every configuration.
And what you see is that there is some change in topology.
I mean, look at this point has come down considerably
and even has changed the topology somewhat,
and usually that's what you see.
You rarely see new phases new ground states come
in when you include vibrations, but you
will change the topology of phase diagrams, for example.
There are a few known cases.
Usually people write a physical review letter
whenever they find one where phrases
come in purely stabilized by their vibrational entropy.
It's a bit of a puzzle, actually, why
it doesn't happen more because the vibrational entropy is
large.
It's essentially 3r.
Remember the law of the [INAUDIBLE]??
The entropy is 3rt.
The [INAUDIBLE] is 3r, so the entropy is 3rt.
The configurational entropy at its maximum is only r log 2.
At 50-50, in a binary alloy, you have the maximum entropy
and it's actually the logarithm of 1/2,
so it's minus the logarithm of 2.
So you either have--
that's about, what, 0.67?
So the configurational entropy only about 0.67r,
the vibration one is 3r, so it's a lot bigger.
But of course, what matters is the difference
in that quality between different structures
and different states.
And even though the configurational entropy
is the smaller one, it tends to be the one that's
more different between different states than the vibration one,
and that's why we sort of all hope
that it's OK to often leave the vibrational entropy out
because often most of it tends to cancel
between different states.
OK.
So sometime in 1994, somebody compiled all the systems
done with these methods.
This was 1994.
We stopped compiling after '94 because that was just way too
many studies going on, but this kind of approach
has now been used for the study hundreds
and hundreds of [INAUDIBLE] structure
and thermodynamics problems.
People do higher component now, ternary's
it's still a lot more work.
This is a simple ternary.
And then I gave you a series of references.
I wanted to show you some of the older work, that's actually
the last references here.
Often the older work, even though it's not up to date,
but it's also not confused by a lot of complications
so it's often very easy to read, very elegant to read.
But the key review paper is essentially
this one, which is like 150 pages or something
like that which contains the ideas of the cluster expansion
and a lot more.
And then my own work is just a short review on how that works,
how you apply the [INAUDIBLE] sites, and the one by Alexander
Zunger is essentially the same thing
but how you apply that in semiconductors.
And the fourth one is the most recent one,
which my own grad student wrote, Axel Van de Walle,
because he's been kind of instrumental.
He's gone to making this into automated codes
and so on, has a whole code suite and the paper
is about that, how you train Monte Carlo
and so to automatically look for phase transitions and things
like that.
OK.
I think that's it.
So let's see, what day's today?
Today's Thursday, so Tuesday we'll meet again here.
I'll start talking about accelerated molecular dynamics
and find another topic to talk about as well,
and then Thursday is Chris Wolverton.
So I'll see you, have a good weekend.