In this video, I will solve the problem of classical crystal, or an infinite
chain of classical oscillators. And I will determine the normal modes in
this system. and as I mentioned, I'll do it actually
in the chain, which means it's in 1D. And I should mention that actually I
could have done in two-dimensions or three-dimensions.
It would have been a little bit more complicated.
But otherwise, actually the difference, the key features of the solution are
pretty much the same in one-dimension. And higher dimensions, apart from some
polarization effects. and for this reason, I will just
concentrate on the simpler case of 1D. But, you could have done it in any
dimension. So, the specific model I'm going to
consider is presented here. So, we will consider one dimensional
chain. So, we should imagine that this infinite
chain, actually more specifically, I will consider the so-called Periodic Boundary
Conditions for simplicity. But in an infinite system, it doesn't
really matter. And so, in this chain, as you can see,
there are two types of atoms. Well, these guides the brown, the red
circles, they correspond, well in my model to an atom of such 1D crystal.
And the positions of atoms will be labelled by x sub n for lighter atoms and
Y sub n for heavier atoms. And the index n here corresponds to, in
some sense, to an elementary cell. So here, at this elementary cell would be
you know, one lighter atom and one heavier atom.
And, let's say the size of this elementary cell will be two times the
distance between the atoms. And so n can be, you know, 1, 2, 3, 4, so
have an infinite series of these elementary cells.
And also for the sake of simplicity I will be measuring the distances in the
beginning of my solution in the units of this elementary cell size.
So, I will essentially set 2a equals to 1.
So, a is basically the equilibrium length of a spring.
So, 2a is essentially represents symmetry of my, this great symmetry in my
problems. So, if I translate everything by 2a, I
get my crystal. What I'm saying here is I'm going to be
using the units work a set to 1. But basically, this is my problem.
And so, let me write down first the energy of this.
And so, the energy is actually very simple to write down.
So the energy, as usual, will contain the kinetic energy part and the potential
energy part. And so, the kinetic energy part includes
the kinetic energy of all light atoms. So this letter case p sub n over squared,
or root 2m. And the kinetic energy of the heavy
atoms, so which actually have a color code.
The sort of brownish forms corresponds to terms that are associated with lighter
atoms, and the red forms naturally corresponds to terms which are associated
with heavy atoms. And then, there will be potential energy,
which is just the usual Hook's Law if you want.
But I now have Hook's Law for every single spring in this oscillator chain.
So for example, in this nth elementary cell, I'm going to have a term here
associated with the spring between this lighter atom and this heavier atom, M.
And so this term essentially as is corresponds to this this spring.
But also, there will be an interaction let's say, of this heavy atom with the
lighter atom from the next elementary cell, the nth pluth, plus 1, ele,
elementary cell. And so there is a term also here in this
Hamiltonian, this classical Hamiltonian, associated with this, in nearest neighbor
interaction. Now, and of course, I have to sum over
all these n's, and this will be will give me the total energy of my classical
crystal. Now, the next step is structurally to
solve this problem is to write down the equations of motion for all the particles
involved. And there's an infinite number of such
particles. And in our case, these equations have
motion, and nothing but Newton's equations.
well, associated with each of these particles.
And, as usual, while the Newton equation is m, m times a, my, mass times
acceleration is equal to the sum of all the forces.
And the force can be found by, as a derivative of radiant, in this case is
just one-dimensional radiant of the corresponding potential energy.
With the respect to the coordinate of a particle where we're in, investigating.
And so, for instance for the light rate, I'm for the nth light rate of this, will
be the Newton law. And for the nth heavy atom, this will be
the Newton Law. And so, if we just formally differentiate
our potential energy, we get these equations of motion for the lighter atoms
and the low, heavy atoms. And it's very nature of that, there are
two terms in each of these equations. Because this essentially corresponds to
having two springs attached to each of the atoms.
So, for example, here in this equation, we can understand the first term as being
associated with force exerted on this heavy atom by this spring.
And the second term is going to be the force exerted on the same atom by this
other spring to the right. So and the goal now is to solve these
equations. Of course, it's an infinite chain of
equations because n goes from minus infinity to plus infinity.
So, this is an integer index which labels our pair of heavy and the light atoms.
So here, I'm showing the same Newton's Law that we derived in the previous
slide, but now just slightly reconstructed.
So, I just grouped together this 2x sub n and 2Y sub n.
So, of course it's a very complicated looking system of equations.
We actually have an infinite system of coupled differential equations and well
naively, we would you could just say that it's hopeless.
You know, there's no way we could solve such a complicated model.
But it turns out that using combination of a powerful mathematical method which
we'll actually use the rate of transform and the reasonable physical guess.
We can solve the problem in a few relatively straight-forward steps.
So, the method involved essentially a free transform where we represent our
time dependent coordinates of the light and heavy atoms as linear combination.
Or an integral over a wave vector of a wave like excitation.
So here, I have essentially a plain wave in one dimension.
And so, remember that I introduced this unions of 2a equals to 1.
Therefore, this wave vector multiplies just an integer label n, which labels my
importance of my light atoms. And, I represent the heavy atoms as a
linear combination of their own wave is some new amplitudes Y sub q.
And here I use q times n plus 1 half because this guy is shifted by 1 half in
my units of length relative to the light atom.
And importantly, here I have the time dependence again which is consistent with
the wave like behavior. So, I assume essentially that whatever
solutions there are, this solution represent waves running through this
crystal. And now, the next step in the solution is
to plug in this trial functions into the Newton equations.
And to see if we can make, if we can find a self consistent solution, which is
consistent both with the Newton equations, and with this representation.
I'm going a bit ahead myself. I would say that, of course, the reason
I'm presenting the solution is because we will indeed find that it's self
consistent. But the self consistency will include in
a very essential way a relation between Omega and q, between the frequency of the
wave. And the wave vector.
So and this is the,this is going to be the dispersion of the waves, the possible
dispersion of the waves. And basically, to solve the problem in
this context means to find the this recursion basically to find Omega as a
function. So now, let me consider every single
term, let's say in the first Newton equation, corresponding to the light
atoms. So, the first term on the left-hand side
the second derivative of the coordinate with respect to the time.
The acceleration here is going to involve the integral over all wave vectors.
and that will just label this integral as as integral sub q, not to write the
differentials. and here I'm going to have x sub q.
And the derivative of the exponential with respect to time is going to produce
minus i Omega squared exponential of iqn minus i Omega t.
So, this is just minus Omega squared. Now, I also have here this term of Y sub
n plus Y n plus 1, which essentially corresponds here in this picture.
to having this red atom to the left and the red atom to the right.
And so, if I now calculate this term, Y sub n plus T n minus 1, I simply can
write this expression. So I have an integral sub q, Y sub q.
And here I have a sum of two terms, e to the power iq n plus 1 half plus e to the
power iqn minus 1 half. Because now this corresponds to n minus
1, e to the bar minus i Omega t. Or I can factor out this e to the bar
iqn, and here I will have e to the bar iq over 2, plus e to the bar minus iq over
2. And this is equal to the cosine,
proportional to the cosine. So, I can write it identically as an
integral over q, Y sub q twice, cosine of q over 2 multiplying the same exponential
as in this term here. And if I put everything together.
So what I'm going to get is and basically if I focus only on the coefficients,
which multiply the same plane waves in each of these integrals, I can ride the
essentially the same Newton's Law. But now, in the [UNKNOWN] space as minus
m Omega squared x of q from this guy is equal to k twice Y sub q cosine q over 2.
This term, which we just derived minus 2x sub q.
And this is essentially the Newton Law corresponding to the light] atoms written
in the Fourier space. Of course, we can repeat the same kind of
procedure for the second Newton Law for the heavy atoms.
And if we do so we are going to get the following system of equations.
So, the first one we just derived, and the second one I didn't explicitly
derive, but it's derived in a very similar fashion.
So and well the good thing about these equations first of all that instead of an
infinite chain of coupled equations. Actually just have two coupled equations,
okay? You see that of course, I have all
different q's involved. So, I can change my q, but there is no
coupling between different q's. Unlike in real space where we need to
have coupling between neighboring size n, and another thing which is good is that
its no longer a differential equation. Instead of the derivative, I now have I
mean I got squared here. So, it's much easier now to deal with
this essentially linear system of equations.
And to solve the remaining part, let me just rewrite the same equation in a
slightly different form. I will write it as symmetrics.
Well, it's the same equation I just had before, but now I wrote in a slightly
different way. I wrote it asymmetric, 2x2 metrics acting
on a vector if you want, which contains the Fourier component of the coordinates
of the light atoms and the Fourier components of the coordinates of the
heavy atoms. And the, the right-hand side is zero.
Because if we go back, we see that all the terms here are proportional to one or
the other coordinate, there is not free terms here which is just constants.
So I can write it like that and it's equal to zero.
And maybe most of your, or some of, you know, from linear algebra.
That in order for me to be able to have a known trivial solution to this equation,
which is not identically equal to zero. Basically, a solution where x and q are
not identically equal to zero. I have to ensure that the determinant of
this matrix, so the determinate of this matrix vanishes.
And well this is determinate of a 2x2 matrix, which I can calculate.
And well, if you do so, you will get the following equation, which is the
algebraic equation. And essentially, it reduces to a
quadratic equation. Well, it's, formally, it's an equation of
the 4th order. But there is only Omega squared involved.
So all, all in all, it's just a quadratic equation that I hope most of you know how
to deal with. And well, what is this equation for?
Okay, this is again, an equation for the frequency.
Or more precisely, for the dependence of the frequency on the wave vector, which
is known as a dispersion relation. Well, we have seen actually many
dispersion relations already in this course in the context of Quantum Physics
and also classical Physics. For instance, when we talk about the
energy of a particle, the free particle being proportional to p squared.
So, this is actually dispersional relation, say of an electron or any other
free particle. When we talk about the dispersional
relation of a photon where Omega scales linearly with the wave vector.
This is well the expression relation of electromagnetic waves.
So now, we have here an equation which determines the dispersion relation of
elastic waves, other, otherwise known as phonons.
Which can exist in this sort of simple model of a one-dimensional crystal.
So, now before solving this equation, let me go back to the sort of physical,
regional physical units where the dimension of length is restored.
So, here we have this q, the wave vector which is measured in 1 over distance.
And remember that my so the convention was that 2a, a being basically the size
of the length of a spring was equal to 1. So, if we want to restore the right
physical dimension, I should multiply this q here by 2e and also of course, Y
minus cosine squared is equal to sine squared.
So, I can just rewrite in the usual physical union, sorry, I can write this
term as sine squared q times a. And now well again, this is just the a
quadratic equation Omega, for Omega. So, if I solve it I will get two branches
with a plus and minus sign. And well, I get they're actually
non-linear and non-trivial relation between m lambda and q.
There are two relations as a matter, two types of expectations, two types of
waves. A normal moves that I get out of this
solution. So, and if I plug this dispersional
relations, so this would be my Omega of q, and q here is defined.
I should have mentioned it before between minus pi over 80 up to pi over 80.
So, we'll get two terms. There will be your range corresponding to
plus sign, the plus range which will look like this.
And there will be a range corresponding to minus sign which will look
approximately as so. So, this is Omega minus and this is Omega
plus. And the Omega minus is called Acoustic
Phonons, and the Omega plus is called Optical Phonons.
So the, as you can see, there is a gap separating this optical phonons from zero
energy. So, if we hit our is, our oscillator
chain with a hammer and it's going to be a non-linear perturbation.
We're going to exact all kinds of moles. But if we just touch it very slightly,
very gently and apply just a weak perturbation, there is no way we can
excite an optical phonon. We're only going to excite these energy
excitations in the vicinity of zero energy.
So, this is basically the energy associated energy that we would need to
spend in order to excite in this case acoustic moles.
And the dispersion of the acoustic moles as q goes to zero, we'll also discuss it
in the next slide is linear dispersion. And this type of linear dispersion is
called sound, sound-like dispersion. And the coefficient of proportionality,
basically Omega as a function of q, in the q to 0 limit, for this acoustic
phonons. And this coefficient of proportionality
is the speed of sound. I actually forgot to mention a couple of
things here. So in this equation, there appears this
variable mu. So, mu here is simply the reduced mass
we, we actually have seen before, quite a few times.
Which is, in this case, the product of the mass of light atoms and heavy atoms
and divided by their sum. And the last thing I'm going to mention
on this slide is something about the optical phonons.
Namely that the existence of this optical branch is related to the fact that we
have two type of two types of atoms present in our crystal.
So, if we had just one type of atom, let's say these blue atoms were the only
type of atoms. Then, we would not of had the optical
branch and we only would of had the acoustic branch.
Which by the way, the acoustic branch response to what I was discussing in the
end of the previous lecture to the presence of this sort of broken symmetry
in the crystal. So finally, let me calculate the speed of
sound, that is the coefficient of proportionality in this this portion of
the acoustic wave as q goes to zero. So as we can see here so if we just focus
on the minus sign in this equation, okay? So and if we consider the limit of q very
small, then this guy can be expanded in Taylor series just up to first two order.
So, we're just going to have qa squared instead of the sine.
And also well, this whole thing, the square root of 1 minus this 4 mu squared
over m M qa squared in the limit of small q.
It can be written again using the Taylor series if we have 1 minus epsilon.
And the epsilon is small, we can approximately write it as 1 minus epsilon
over 2. So, in this case, epsilon is a useful
thing. And this thing is small because we are
near the zero momentum and zero energy. So, we can write that approximately is 1
minus 2 mu squared mM qa squared. So, if we plug it back in here, so we're
going to see that 1 minus 1 gives us zero, and we have just this guy
appearing. So, the spectrum of the acoustic phonons
in the q to zero limit is approximately equal to k over mu 2 mu squared and mM qa
squared. Or we can cancel these guys and we're
going to have simply recalling the definition of the reduced mass two times
the stiffness divided by the sum of two masses, qa squared.
Or, if we take finally the square root of this expression, we're going to get Omega
minus, in the q to zero limit, is approximately equal to the square root of
2k a squared, m plus M times q. And this is our speed of sound.
So in the next video, I'm going to redo the problem.
Actually, in the quantum limit. And, the, surprisingly, we're going to
see that the solution is not more complicated than the classical limit.
Well, I actually find that the quantum solution actually easier than the
classical solution. But in any case however you classify the
solution, easier or more complicated, the solution is actually going to be quite
similar to what we found here. And the spectrum of the quantum phonons
will turn out to be exactly equal to the spectrum of the classical formulas.
So which is another sort of striking manifestation of quantum to classical
correspondence, and actually a rather complicated system.
In this case, this system models a crystal.