So in this video, we're going to complete the derivation of the spectrum or
dispersion relation of the quantum phonons.
And well, this part as well as the previous part is a bit technical, so I
will actually skip some of the calculations and we'll just show you sort
of the method, illustrate the method. We just called, by the way, in the
Bogoliubov transformation that is used to solve such a problem.
And as a side remark, let me mention that well this material that I'm going to be
covering is actually a pretty advanced, gradient material.
So if you've got to this point, you should be proud of yourself.
This is very complicated, very advanced material of quantum antibody physics
you're actually studying. So in this slide, I would like to review
the calculations we have done so far, sort of tell you where we have been and
also where we're going with this. And as we know, we started with this
classic oscillator chain which is characterized by the coordinates and
momentum for the atoms in this one-dimensional crystal.
And the first step in putting together a quantum problem was to quantize this,
which means basically to replace the variables x and t with the operators, and
so, we just put hats on top of x, p, and h.
So the following step involved expressing this momenta and coordinates with the
corresponding creation and emulation operators, which led to this complicated
looking Hamiltonian. And here, this an and am dagger
correspond to creating an oscillation in a particular site with a particular atom
labeled by m. So the next step that we took in the
previous segment was essentially to Fourier transform this creation and
anti-annihilation operators and to express them in the q space in the wave
vector space if you want. And the corresponding Hamiltonian is
shown here, so it has, as you can see, two types of terms, a dagger and a and
this aq a minus q. And so, now, we have to work with these
operators, these new creation, and annihilation operators, and this
Hamiltonian. So and now I'm coming to where are we
going with this? So if we did not have, let's imagine for
a second that we do not have this term. Let's say we have just this term.
So if this were the case, it would have meant that we already solved the problem.
The reason for this being that this linear combination, the inevitable altura
is a linear combination of this A dagger a's.
independent a dagger a's, they sort of correspond to a collection of independent
harmonic oscillators, the spectrum of which we know perfectly well.
So for each q, we have a, basically a simple harmonic oscillator such as in the
previous lecture. And we know its spectrum and now we just
collect these oscillators and write them as an integral, but all in all, we know
the results. So this is essentially already a
solution. And therefore, this A of q or more
precisely actually do to the omission conjugate part here twice A of q, would
have been the spectrum of our problem, the spectrum of this quantum force.
But, of course, we do have this true. We do have this true.
And we don't know what to do with this, right?
So this, so the spectrum of this Hamiltonian is not at all obvious.
So a way to solve the problem now is to actually to construct a new set of
creation and annihilation process. So here, I have an operator B which is an
annihilation operator, can you type? And there's also a B dagger of q, and so
lets assume that we can construct this B and B dagger as a linear combination of a
and a dagger in such a way that the new Hamiltonian won't have the bad terms.
By bad terms, I mean the terms that we don't know how to deal with and instead
we'll have only this standard combination of b dagger and b for which the spectrum
is obvious. And so this essentially would mean to
solve the problem. The ability to write the Hamiltonian in
this form, in the form of a liner combination of some creation and
annihilation operators of this sort, so would mean to solve the problem in this
why's this a tilde would give us the dispersion relation of these bs.
Which actually are the quantum phonons. So as we will see, so this b dagger hat
and b ha, hat, then correspond to creation, and anagolation operators of
quantum phonons. Essentially, the normal modes, the
quantized normal modes, in this oscillator chain.
But together is well, as you can see, is rather nontrivial for us, we have to do
this quantization and then, well, which was simple, but then we have to do with
this 3 Linear transformations to get to this point.
And this last linear transformation is what is actually known as the Bugoliubov
of the transfrmation. So here's a picture of a young Nikolai
Bogoliubov who was a very influential and talentented mathermatical physicist.
A ration mathermatical physicist. And he was an author of many very
important works and also he came up with a number of very important and practical
mathematical methods that are used up to these days.
And one of these methods is this Bogoliubov transformation that we are
going to be studying. And let me mention that this Bogoliubov
transfomation is still used very actively in research papers.
I myself have used it in a number of of my published works.
And this is basically in some sense an everyday tool of theories working In
quantum physics and then condensed matter physics as well as perhaps high energy
physics. So this is an important and important
method but well in the example that we're going to be considering to be sort of
method is rather straight forward and I want to get you advertised, sort of a
motivation for using this method so lets lets look again at this Hamiltonian that
we have for the Original the creation and annihilation operators a sub q dagger and
a sub q. So and again, as I said, so our sort of
strategy would be to correct in some sense, the definition of the creation and
anogolation operators, so that we can get rid of this unwanted true.
And, if you look at, at this Hamiltonian, you will see that this Hamiltonian
involves essentially, so if we didn't have this term again, so we would have
had just a linear combination of independent q operators.
But this term, the appearance of this term couples the q mode with minus q
mode. So whatever transformation we come up
with, it should probably involve a linear combination of at least a q and a minus
q. So let me just preset a form, sort of
knowing the result of course is easy but it will save us some time.
So let me just present a form Well if this generic Bogoliubov transform that
accomplishes what I want. Namely that it introduces new operators
in this case these are these operators b sub q and b daggers of q which are linear
combination of a and a dagger or vice versa a and a dagger are linear
combinations of b and b-dagger is here. And so here I will assume that this
assumption will be verified, sort of a posteriori.
And that, let me assume that this, these coefficients can be chosen to be real
numbers well, functions better to say. And that there even functions of q.
So, u of q is equal to u of minus q. So in this case actually this second
definition here is redundant because the second definition of the creation
operator can be obtained from the Definition of the annihilation operator,
simply by applying the permission conjugation.
But in any case, so, let us assume that we can construct such operator.
So the first thing we need to do, and this is, I'm basically describing the
strategy of constructing Bogoliubov transforms Is to make sure that this b
and b hat ID creation and anti-annihilation operators.
So, I can call them such, but actually, this should have a precise mathematical
meaning. And the precise mathematical meaning
would be that this guy. So, let's see if I have b of sum q and b
of b, let's say dagger. So these guys should be the commutator of
this guy should be zero if q is not equal to b and it should be a delta function
other wise. So infinite otherwise in some sense.
So this what I'm looking for. So my a and a dagger, so the reason I
call the creation and annihilation operators was that they satisfied this
relation. Now I want the new operators which are
actually a linear combination of those to satisfy the same commutation relation so
and this would involve a calculation which would result in a constraint on
these coefficients u and v. So if I calculate this commutation
relation you already know. So this is a the commutator of a sub q
and a dagger sub b. So i simply plug in these definition to
here and this guy to here with replacing q with p and I just calculate the
commutators term by term. So what I can write, I can just write so
we'll have one two three four different terms.
So the first term lets say will be the commutator of these two guys so such is
here and using this constraint I will require that the commutator of these two
guys will be two pi delta function of q minus p Now these two commutators which
result in this term should give me zero because these both are action and
emulation operators. So likewise these two guys resulting in
this term should give me zero as well because these are commutators of the
creation operator. And so finally I have a term which
couples b dagger of minus q and b of minus b and well since the commutator of
b b dagger is equal to well this delta function the commutator of b dagger and b
is basically minus the commuator of b and b dagger so it should give me this
function. And let me remind you that, the delta
function volt here, is actually an even function, so it doesn't really matter
that I have here minus signs, etcetera. So, and if I put everything together, so
these guys basically go away, right. So these guys go away, and I have these
two terms, which essentially all involve the same Delta functions and also since
there are delta functions here, Q is actually pint to piece so the delta
function is zero if Q is not equal to P and so there are only terms, here I can
simply write it as let's say UQ U of Q squared and this guy will V of two
squared Now with a minus sign. So if I put everything together from this
expression so this is what I get, okay. And now I have to required that the well
the right hand side reproduce what is supposed to be in the left hand side.
That is this two pi delta function of q minus p.
So which means that this coefficient here.
So this u squared minus v squared must be identically equal to one and this
introduces a constraint, an important constraint on this Bogoliubov
transformation. So basically what we are doing here again
so let me sort of go back to where we are going with this, so we want to put
together this new operators. And so we want to find explicitly the
expressions for these u sub q and v sub q, that first of all wouldn't re-, will,
would ensure that these b and b dagger nd that annihilation and creation operators
of something. And second another constraint on this
this guise would be that the resulting Hamiltonian now expressed in terms of b
and b dagger would be simple enough so that we know its solution.
And this basically the essence of the Bogoliubov transformation.
In this case this Bogoliubov transformation is relatively simple, and
actually most cases you will see in the literature, even the research literature,
it is relatively simple, but in general, it may be more complicated.
So it doesn't necessarily have to include just two, two, two, two terms, it may
include more terms. But in all these cases, so the strategy
will be the same. So basically to ensure that the operators
you introduce are indeed creation annihilation operators and second then
they simply our Hamiltonian. So now we can sort keep this constraint
in mind or we can resolve it once and for all by just using new barometrization if
you want of this u and v. So and this can be achieved by using the
following mathematical identities. So if we have, if we use, the if we
introduce u of q being a hyperbolic cosine.
So let me remind you that the hyperbolic cosine x is equal to e to the power x
plus e to the power minus x divided by two and the hyperbolic sign of x is equal
to e super x, minus e super minus x, divided by 2.
So, this is a real and sometimes real version of the sign and cosine in, in
which case I would have had And the metric[UNKNOWN] but the main case of this
guy's so satisfy the following identity which is sort of an analogue of the
identity that you most of you know that sin squared plus cosine squared is equal
to one. So here for the hyperbolic functions I
have the hyperbolic cosine squared minus hyperbolic sin squared is equal to one
and this is exactly the identity that I want For my ???
if I use this ??? we should resolve this identity once and
for all. Which basically reduces the problem with
2 parameter, to a problem with a single parameter.
In this case this is [UNKNOWN]. So my sort of expression for the creation
and angulation operators. A and a dagger now becomes this ok?
So I have a of q is equal to cosine hyperbolic of Lambda b q plus sin
hyperbolic of Lambda b of minus q dagger and the corresponding a dagger.
Alright so now what needs to be done is that we ahve to just do a brute force
analysis. We have just go ahead and plug in this
expression this expression sent to the Hamiltonian, and I will leave it for you
for those of you who are interested. For those of you for instance who are
interested in becoming theorists, or who are theorists, or who are mathematicians
or are just curious about the technical part so you can go ahead and do the
simple, relatively straight, at least I'm not going to call it simple, its a
straightforward exercise of just plugging in lets say this and this guy.
Here we're going to put this linear combination, instead of this guy we're
going to put this linear combonation, etc and so if you do so well you will see
some unpleasant algebra but this is an exercise I'm sure you can do it.
So here I'm showing the result of the calculation, if you did everything
correctly this is what you should get and the result here is a Hamiltonian which
now is expressed in terms of these new operators.
B of q and b dagger of q. And as you can see it has the form which
is pretty much the same as we had before just with some new coefficients which I
call now i a tilde and b tilde and here is the expressions for these coefficients
involving hyperbolic cosines of two Lamba and hyperbolic sins of two Lambda and
these are the regional coefficients. And the Hamiltonian for A's.
And so, you may ask me what, you know, why did we bother to you know, introduce
new operators, if we got exactly the same thing?
So, well, the answer to this, and actually this brings me to the second
step in this Bogoliubov transformation is that well lambda here is a free
parameter, so, better to say a free fun, function.
We can choose this function Anyway we see fit.
In particular we can choose it such that this unwanted term, the bad term that we
don't want to see is identically equal to zero.
So we can esssentially can set by hand this b tilde of q equal to zero which
leads to an equation for lambdance. So this equation as you can see from here
if we just move this guy to the, to the right-hand side and divide by cosine, so
sine hyperbolic over cosine hyperbolic is tangent hyperbolic.
So the tangent hyperbolic of 2 lambda of Q, is equal to minus B of Q versus over A
of Q. And so this essentially gives us a
specific result for this Lambda of q which ensures that our Hamiltonian in
terms of this new creation, annihilation operators looks simple.
So now let me clean it up a little bit and sort of summarize things again.
So this is the result that we got and also let me remind you that this a of q
and b of q, the regional coefficients we already know.
So we derived them in the previous part of this lecture and so let me remind you
of their explicit forms. So here is their explicit form.
So this explicitly b of q and this is explicitly a of q and well I wrote it
sort of explicitly this way but as you can see lot of things here cancel out
including this H omega and there loads of other simplifications involved.
And if you play around with this, well, with this The coefficients if you do a
little bit of algaebra you will see that well first of all Lambda of q actually
determines as you can see from here. So this guy is zero as we wanted it to be
but now this guy is determined by Lambda and we do know Lambda.
So we can simply plug it Y'know in here and if you, once everything is said and
done you actually has an expilict expression for this a tilde of q.
Which becomes in the end of the day surprisingly simple.
So again this calculation is not obvious from here.
It requires a little bit of work and due to the format of the lectures And lack of
time I'm not presenting it but those of you who are interested can of course
check that this is indeed the result. So this is by construction but a of q has
this form. So if we plot it so this a of q has this
form which is actually a very familiar form so this is a of q which we have seen
before in the classical problem in the context of acoustic forms and this is
exactly what they are but now they are quantum formula.
So sort of to summarize the result and also to emphasize a very important point
so this is what we actually got. We have found explicitly a series of
linear transformations Of the regional Hamiltonian, of the regional operators
involved with the Hamiltionian. So I should have the Hamiltonian can be
sort of was casted in this form. Which is the so called diagonal form of
the Hamiltonian. Sometimes if you happen to read research
papers, the reasearchers call about diagonalization of a Hamiltonian.
So today in lies the Hamiltonian, and, and in sort of in the problem, in a
problem which involves mainly quantum particles, essentially means, the ability
to write it, as a linear combination of independent oscillators or some other
particles, the sort of behavior of which we know very well.
And so, this is exactly what we accomplished here.
And so here we have also this Hermitian conjugate, so if we couple it, it will be
simply the same term, so we can write it integral over q, well twice A tilde of q
b dagger q b q. So, and well, I should also mention one
thing that I sort of swept under the rug and this thing is that I haven't paid
attention to constants that appear in the Hamiltonian so, there could be constant
here. If I had done it, so the correct result
that I would have found is well, this is correct, but there also is a constant
term here, which corresponds to zero point motion.
And this term happens to be exactly one half, so if you remember the solution to
the simple harmonic oscillator, we have essentially h omega and plus one half and
so similarly it appears here. So in this two a tilde of q is exactly
the energy modes of the oscillator. So we can call it so this guy is equal to
h m n of q and this would be the spectrum of the quantum phonons, and this is as
presented here. So the spectrum of the quantum phonons
here is twice omega. Omega here, is the frequency of a single
oscillation in some sense. In the original classical picture it
would have been the frequency, the natural frequency of this spring
connecting the atoms in some sense. And so, the remarkable thing, that I want
to emphasize in the very end, and this will be the end of the, today's lecture
is that, it turns out that this spectrum of quantum phonons, is actually identical
to the spectrum of classical phonons. There is absolutely no difference
whatsoever. So if you go back to the problem that we
solved earlier today, in today's lecture, so namely the classical problem, so it
was slightly more complicated in the sense that it involved two types of
atoms, but if we have just one type of atom so this is the result which we would
of got. So an important conclusion of this sort
of discovery that quantum and classical phonons turn out to be exactly the same
is the following so we can say that we could have quantized our problem at any
stage. So in the solution I present it we did so
at the very beginning. So we essentially converted the original
coordinates and momento of the atoms into quantum operators, x hat sub n and p hat
sub n and then we went ahead with this quantum calculation with the newer
transform. But instead of doing that sort of there
is a second version of the solution. One could have solved the problem, the
classical problem first and which means finding the trajectories of particles and
the corresponding waves. And so one who could have quantized this
normal mode. So basically so the normal modes can be
converted then into this Essentially operators, h omega q, sum b dagger q, b q
plus one half. So we could have done that.
And the results, basically we have proven that the result are, are exactly the
same. And this is very important.
It shows us something about the quantum to classical correspondence, and this
quantum to classical correspondence sort of persists Also in more complicated
problem. Let's say problems in three dimensions
with more degrees of freedom. With different types of atoms.
And this is true as long as we have non-interacting theory, which in the
language of this creation and annihilation operators essentially means
a theory which is bi-linear in this b's and b daggers.
So as long as we have this bi-linear forms we're fine with this quantum two
plus equal correspondents but once we have y'know more complicated nonlinear
terms, so this situation becomes much more complicated.
So the last thing I'm going to mention is that the appearance of such terms, which
we're not going to discuss at all but you know if we had this quadratic terms, b
dagger b dagger b b for instance. So they would have meant that the photons
interact with one another. So in some sense it would have responded
to scattering of waves off of each other, waves running through the crystal.
And this is a very complicated challenge to understand the behavior of such an
interacting system. And there is still on going theoretical
challenges in describing it. So it's definitely beyond the scope of
this course, but this theory is well understood and I hope some of you got an
idea about phonons and collective modes and classical and quantum physics.