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.