In this video we're going to consider the same problem of an oscillator chain and a
galactic moves in the chain. But now we're going to start the quantum
collective moves or quantum phonons in a quantum oscillator chain.
So this problem sounds very complicated because we are going to be dealing with
an infinite number of quantum degrees of freedom now.
But as we shall see actually at the end of the day the results are going to be
very similar to those in the classical cases, so namely, the spectrum of quantum
phonons apart from the zero point energy is going to be identical To the spectrum
of classical phonons. Now, to solve the problem, first we need
to define what we want. So let me consider slightly simpler
version of what we looked at in the previous video, and now I'm going to
study an oscillator chain, consisting of the same type of oscillators, and same
type of atoms. So I have here identical atoms with a
mass m and indentical osciallators, identical sort of springs.
Now its going to be quantum springs of sorts with a stiffness key.
So this the x corrdinate and the corresponding classical Hamiltonian
assocaited with this osciallator chain is Presented here.
So we have a sum going over all these atoms.
So n here is sort of lets say this is going to be n equals one, n equals two, n
equals three. So you just label the atoms and x sub n
is the position of the corresponding atom.
Now this is of course the kinetic energy and this the potential energy.
So its essentially the same Hamiltonian we had before but now just for a single
species of atoms if you want. Now to go to quantum problem we need to
quantize this Hamiltonian. We need to make it quantum.
So how do we do this? Well, it turns out that, that it's
actually very easy to do so, so it's as easy as it was in the case of a single
harmonic oscillator. So all we have to do is just to put hats
on top of H. P and X, so which means that we made this
variables in the classical case into operators in the quantum problem.
So, for example if we are working in a position space, so let's say P sub n.
Is simply[UNKNOWN] minus i h bar g over dx m.
So it adds only on the x coordinate of the oscialltor.
Now this[UNKNOWN] satisfied the uh[UNKNOWN] goal of commutation
relations. So if they correspond to the same atom so
x n and p n they commute to i h bar. If the correspond to a different atom,
lets say if n is not equal to, well lets say if we have x n and p n So then is
going to be 0 of n, is not equal to n. So and, we can sort of unify these
computation relations, by writing that delta symbol here.
So which means that it's equal to i h bar 1, if n equals to m, and 0 otherwise.
So this, this is what it means, to, to quantize this on a Hamiltonian.
So now the next step we're going to perform is we're going to use our
definition of the creation and annihilation operators that was that were
introduced in the solution of the single harmonic oscillator And we're going to
represent each coordinate xn and momentum pn through the corresponding creation and
annihilation operators a and a dagger. So here are the same, you can check,
going back to the first segment of the lecture today, that as a matter of fact
these are exactly the same definitions We used before but now we just have this
additional index end which refers to the atoms in this chain.
And of course this creation and annihilation operators, they satisfy
their own canonical commutation relations as here.
They commute with one another their commuator is equal to 0.
If they've responded to different atoms, and otherwise if they respond to the same
atom, a n commutation with a dagger n is equal to one.
So what we're going to do now, we're going to plug in this expression of x and
p through a and a dagger into this Hamiltonian.
And let me do that. So if we do so the Hamiltonian takes this
form. Which actually appears to be a more
complicated looking form since there are more terms here than we had before.
But we are on our way to solving the problem.
So the strategy for solving the problem would be actually different from the sort
of typical solution in single particle quantum mechanics where we're dealing
with the Schrodinger equation wave function.
So here of course we still would like to solve the Schrodinger equation.
We have in mind this I shrink your equation, but we don't worry too much at
this stage about the form of the wave functions.
We just want to know the energies that are possible in this in this system.
Just like in the previous video, we just worried about the spectrum of the
dispersion relation of classical wave. So here we just want to know the
dispersion relations. For quantum waves.
So how do we do this? a scenario to solve the problem here
instead of working with the wave functions, we're going to be working with
the operators themselves. And we're going to try to transform the
Hamiltonian into a new, simpler form so that basically we're going to be looking
for a new set of operators. We're going to go from the operators,
let's say a n. And a n dagger, with all possible n's, to
a new operator. So let me ahead, let's say b q, and b q
dagger. Such that the form of the[UNKNOWN], in
terms of this new creation, and annihilation operators, is going to be
much simpler that the form, which is written here.
So basically the challenge here is to find a linear transformation of the
original operators into new forms. Such that the Hamiltonian, which looks
simple, and that such that the solution to Hamiltonian this would be null.
So this is a bit of a technical challenge and a technical exercise.
So and the first stop in this exercise is just to write the Hamiltonian in terms of
this a and a dagger as is. So we have to basically expand these
brackets and calculate the square of these brackets, which leads to a rather
lengthy expression, as here. So and I have omitted here an overall
constant. There could be a constant which
corresponds essentially to an energy shift And also I have written it in a
slightly more[UNKNOWN] form. well its not as long as [UNKNOWN]
covering it because of this term which is corresponds to your mission conjugate.
So basically I have this guy so you can just work it out and we can work it out
as an exercise if you want. So and you will see that we have many
many terms and each term here but be compensate with the permission congregate
terms somewhere else because Hamiltonian in the end of the day must ask permission
of so, and in this case, it simplifies out life a little bit because we don't
have to write two lengthy expressions but still it looks pretty bad.
So there are a lot of terms we don't want to see.
So we don't know what the spectrum, what the solution of this Hamiltonian is.
We see a bunch of this a and a daggers in various combinations and we see also
couplings between different sides, et cetra, et cetra, et cetra.
So in order to solve this Hamiltonian in the way I just described, that is to find
new set of operators that simplifies form so just like in the classical case it is
convenient to perform a for your transform but now its going to before
your transform for eh the creation And, annihilation operators themselves.
So what we have here is a representation of the annihilation here, and the
creation operators, in terms of some other operators a sub q, and a sub q
dagger. So, I use the same symbols, but, this is
really a different operators. And so this a sub q are in a sense
Fourier images of this a sub n. And so the two are related by this
Fourier transform, which now is an integral going from minus pi to pi.
So and it's not from minus infinity to the plus infinity because we have a
periodic. Array of these oscillators.
Now at the continuum of oscillators. So its more like actually very serious
than Fourier transform that we're using. In any case I'll be using this symbol
here. So itegral with the substrate q with two
sort of label all the the terms that I have here, and in this notations, for
example, the Fourier-transform of the creation operator is presented here.
So it can be obtained from this one simply by Conjugating, both sides.
That's why I have a minus sign here as opposed to a plus sign here.
So there are a couple of identities that I want to emphasize without deriving
them. So the first one is the sum over all
integers, basically over all the oscillators in our problem.
And, so n goes from zero, plus minus 1, plus minus 2, etc.
So and if I sum over all all these n's. These exponentials will give rise to two
pi times the delta function and sort of the inverse identity to it, if we
integrate the exponential from minus pi to pi so then the right hand side is
going to be delta Simple, so this integral is equal to zero.
If n is not equal to zero and it's equal to one, well, in a trivial way,
otherwise. I'm mentioning these identities because
they're very useful in deriving a form of the, Hamiltonian in the q-space.
From the Hamiltonian we derived in the previous slide in real space, which is
which was dependent on the index n. And this derivation, which takes a little
bit of time and effort, I will leave mostly to you, and this will be one of
problems and you'll Upcoming homework for self-assessment, and probably this is
going to be the most complicated problem there.
However, there is one important identity that I want to prove explicitly here, and
this will be an identity involving a commutator of two q space operators.
So again our goal would be to rewrite the annihilation in terms of this a sub q and
a sub q[INAUDIBLE] and I refer to this guys as annihilation and creation
operators but I haven't proven that they are indeed such operation.
So what does it mean for them to be creation annihilation operators?
So they , it would mean that they satisfy certain type of computational relations
similar to the economical computational relations we have discussed in the
previous lecture. So, and these commutation relations for
the real space operators a sub n, let's say a sub m dagger, we know that this is
delta symbol of n and m. So now we want to check what kind of
limitation relations these guys satisfy and in order for me to work it out, work
out these commutation relations I'm going to use an inverse Fourier
transform. The first term in this commutation
relation is sub q q. It can be written as a sum over n1 a sub
n1 e to the power minus iq1n1. So this is an inverse Fourier transform
sometimes opposite to this guy here and the second term here is going to involve
for some let's say over n2 a dagger' n2 E to the bar plus iq2m2.
And this corresponds to the Fourier-transform sort of reversing this
relation. So now the commutator only cares about
the operators. Oh, and this is a linear operation.
So we can factor out in some sense the sum, and we can factor out these
exponentials. So[UNKNOWN] commutation relation as so.
So its going to be a sum where n one and n two e to the power i q two n two minus
q one n one. In here I'm going to the commutator of a
sub n 1a dagger sub n 2 and so if we now compare this canonical commutation
relation with this guy we're going to see that this is nothing but delta symbol of
n one and n two which means that it sort of extracts just one term out of this sum
where n one is equal to n two and what we can do we can simply set m one equals to
m two and lets say We will set n 1 equals n 2, and equals to some other n.
And so, we're going to write it as follows.
Is going to be equal to the sum, over n now, e to the power i, q 2 minus q 1,
times n. And that's it, so the operators are gone.
And now at this stage we're going to use finally this identity, the first identity
that I just defined and we'll write the final results.
So let me actually write it in red and its going to be equal to two pi delta
function of q one minus q two, of q two minus q one.
So it needs to be compared with this conical commutation relation.
So and as you can see of course here this is in some since a continuum version of
this conical commutation relation and this means actually that these guys a sub
q one well actually a sub q and a sub dagger The two represent annihilation and
creation operators, but what do these creation and annihilation operators
create or destroy? So like a sub n and a sub n dagger, which
sort of create local oscillations of a particular side, of a particular
oscillator somewhere In the well defined position in space so these a's of q and
a's of q dagger they anagolate and create waves if you want.
Propogating through the oscillator chain. So these guys are not local in space They
instead create waves with a certain wave vector.
A cube basically is the wave vector. So here again now I'm using the units
where the lengths are measured in the units of the oscillator lengths.
So the distance between the neighboring. Equilibrium size but in any case so this
is really the interpretation of what we are doing and now if we plug in this
expression. This expression sent to Hamiltonian in
the previous slide and use these identities and other simple algeabraic
manipulations, we're going to arrive At the following expression for the same
Hamiltonian. So this is actually, this is the very
same Hamiltonian we started with but now rewritten in terms of this a sub q and
sub q dagger and well its not at all obvious that this is the result and well
this a and captial a and capital b, these coefficients here have this form and its
not obvious at all. So this actually To derive this guy would
be challenge for you in the bonus problem of your homework here.