So, welcome to lecture number 6. So, in this
lecture now we are going to start putting
electrons in a crystal and I will talk about
Bloch’s electron. So, we are still building
our base how to represent e versus k diagram.
So, if you recall that we just did last lecture
we did reciprocal that lattices, last two
lectures we did reciprocal lattices. Idea
was that of course, that electron wave which
had a small cave that in their vector its
dimension was in a reciprocal space.
Then we generated capital k vectors, those
reciprocal lattice vectors, those are done
with the purpose, those vectors we are going
to use to represent the ban band diagram that
is e versus k diagrams. So, ultimately we
are going to be using that, but still we are
building towards it. So, what I am going to
do is remember, now I am going to start including
the crystal crystal under lying a.
So, far we did free electrons, that we started
with free electrons. In free electrons what
we did, what are the things we did, let us
think about the free electrons? When we did
with free electrons, in this case what we
did was two things we did. First we said this
all this electrons are non interacting, neither
their was a ion in electron interaction nor
their was electron electron interaction. Therefore,
we did this V is equal to 0 is first thing
we did in our solving our the equation which
was this h square by 2 m e. In this equation
we had substituted V is equal to 0, we had
made this go away, we had made this quantity
equal to 0 and we had made this go away.
Then we said this is free electron and what
did we get? We got energy as equal to h square
by 2 m e k dot k or k square is what we got,
that is what we have derived for a free electron
energy. Now, here comes problem ok fine, saw
that it does a improvement it gives us some
better result then Drude’s theory. But then
we also said that many things free electron
theory does not explain, it does not show
why should a material should be a insulator?
Why material should be semi conductor? It
does not show, for example, why an aluminum
the hall coefficient should be of opposite
sign than expected, etcetera, there are failures
of free electron theory which we discussed.
So, what we need to do we said, that we will
have to include these electrons, what we have
not done is that we have not we have ignored
the underlined lattice, its periodicity etc.
So, what we have to not do as a improvement
do is we will solve the problem in two in
two steps. In first step what we are going
to do is that we are going to keep keep relation
between e and k simple, we will just use the
result of free electron theory that is e is
equal to h square by 2 m e k small k dot small
k which is a vector. Of course, in this case
in this case the solution we got was phi was
of position was e to power minus j k dot r
that was in formula solution we got for this.
So, we will continue to use that as solution
in there.
Now, so first step we are going do that, second
step we will include, we will include the
potential, we will include the include the
potential which will or may be present. Now,
what is that mean, just look at it, what is
the potential mean? Well where is potential
come from? Now, imagine V comes from what?
It comes from you can see that many, many
electrons, there are many, many electrons
in the system. Let us divide electrons we
can think of valence electrons, we can think
of core electrons, we can think of nuclei.
So, in a crystal in a crystal we will have
interactions between valance electrons, we
will have interaction between all these entities
nuclei, so and we have many, many electrons.
All the electrons, all the ions, all the nuclei
which are present in their in the system,
all these are interacting with each other.
In dead we can write a potential for it, so
if were to solving the equation for it, then
we will have the Hamiltanian 
will include some of kinetic energy I should
say. Energy for all these particles, kinetic
energy for all these particles apart from
potential, that in dead would be of a really
formatable problem and to date there is no
way of solving such a problem. So, first the
approximation we make is we think of valance
electron 
and ions which means core electrons plus the
nuclei.
So, that all is bared, the core electrons
get bared with nuclei and we call them ions,
that is one thing first. But still we have
huge number of valence electrons and we have
huge number of ions all of them all of them
are present still the Hamiltanian include
kinetic energies, because all these entities
and again the potential. Essentially then
what is done is that this problem is still
remains formattable and impossible to solve
with the present day computers. So, what can
you do?
So, one way is certain ingenious formulation
made in which you consider a single electron
problem, you reduce the whole problem, if
you have equivalent problem which has single
electron in an effective potential. In an
effective potential due to ions and all other
valence electrons, that to all other valence
electrons. So, that is what we are going to
have all other valence electrons and ions
will then will reduce that will in effective
potential the engineers formulation for it.
So, that you treat the single electron you
solve the single electron showed showed in
the equation in a effective potential.
Exact solution of this problem is beyond the
scope of this course, so what we are going
to do interested in is we will assume what
the result will come out of this. But we are
going to do interested in is that notice because
because our lattice is periodic, it is natural
to assume potential is also periodic with
same periodicity. Since, the period in the
lattice is capital R therefore potential at
any position must be same as potential, it
should also have periodicity of lattice which
is capital R. So, that should always be true.
Now what we are going to do is solve this
whole problem in two parts.
First we are going to look at the first part
which is going to start now is not really
solve the problem. I am going to use this
energy relation is the same e is equal to
h square by 2 m k square, that means whatever
we got from free electron theory, let us keep
the energy relation same. Then what are we
going to say is that what is the consequence
of the fact that underline potential is periodicity
been shown here. So, if this is the periodicity
then what will be the consequence on. And
how can we use that to represent the band
diagram and energy will be used in as electron
energy. But using this idea of periodicity
of lattice and therefore that of potential,
how can you represent energy versus k diagram,
that is really we want to do. So, let me try
this one more time.
So, energy we are going to assume as h square
by 2 m k dot k, now you can already appreciate
what the problem is. Problem is my paper is
two dimensional, if I want to choose right
this energy versus k diagram is what I want
to make. What since k is vector, so it has
infinite possibilities it could be pointing
any possible direction, which direction should
I represent here, direction should be what,
what vector should I take? That is my problem.
You see we can pick only one particular direction,
let us say pick only one direction called
k x for example, if you pick this k x direction
then that is only one equi diagram.
Such diagram should be infinite number of
diagrams, because infinite number I will have
to plot to the represent this e versus k,
that obviously is not reasonable. So, what
the method by which I can represent the e
k diagram is really what we are proceeding
towards. So, what we are going to do is we
will use the idea of the fact, that underline
potential is periodic meaning there by V is
equal to r plus R and therefore, use this
to represent these energies in a ingenious
way, in a interesting way, that would be a
first step which I will do in next, today
and next couple of next one more lecture.
Following that only after that what I am going
to do is, then I going to approximately sure
approximate solution for a approximately assume
what value V is, right now we are not interested
in what the what V itself is. Such that in
today’s lecture we are going to use the
fact that it is periodic that is all, but
in subsequently we will introduce we use a
approximate value of what v r is. How about
the behavior of v r is. Use that to find a
analytical solution, using that I will show
you that because of presence of v r and its
value and its nature, how a band gap can emerge
and therefore material can become a semi conductor
or insulator.
Having done that to understand conceptually,
then of course, then we will we will assume
that there is a some residual potential, which
we will not we will not really do in this
class. But there are real potential based
on that people have done the calculations
of the band diagrams in the first step you
would have learn how to represent e versus
k using this relationship right here which
is given here. Once you learn the representation
and second part we will assume a potential
and I will show you the consequence of the
potential.
But then the real thing I will show in then
e k diagrams, what the e k diagram would look
like if the potential is real, but that final
one calculation will not be a part of the
course because that itself that a course by
itself and that is how do u need overview
of next 3, 4 lectures how we are going to
proceed. So, at this point of time let us
assume these two thing that energy relationship
is just this. So, you need to represent this
energy relationship. Second e versus k diagram,
second thing is that my potential is periodic
that is what only result we are going to use
right now and see what the consequence of
this is, all right? So, you may wonder then
why did you do reciprocal lattice, very soon
you will see that, just today you will see
why did we do reciprocal lattice.
They are special vectors and they will do
something for us and that is what we are going
to do. One more thing we can write down here
is solution for this is of form e to power
j k dot r that is plain reaction involved.
So, using this background let us start this
lecture and we will start with what is called
as Bloch’s electrons. So, let us start with
this what is called as Bloch’s electron.
What is this Bloch’s electron? One that
follows one electron Schrodinger equation
in a periodic potential, hopefully you see
a difference. In free electron that was a
only difference, in free electron this potential
was 0, otherwise that was also a 1 electron
problem. So, here also I am still dealing
with a 1 electron problem and I have already
shown you the ingenious ways by which a real
problem is reduce to 1 electron problem, 1
electron problem. Where the interactions of
the potential the potential really represented
represented by interactions of all valence
electrons, electron pron interaction and also
a electron and ion interactions.
So, all those are barred in one effective
potential, so that part can be done. So, we
have one electron problem in a periodic potential
of form V r and has its periodicity of the
direct lattice. Lattice periodicity is shown
by R remember, what is R? R of course, is
my primitive lattice vector defined by n 1
n i a i summation where i of course is 1,
2 and 3. So, in other words I expanded from
R is n 1 a 1 plus n 2 a 2 plus n 3 a 3 and
this is a vector where n 1 n 2 and n 3 are
all integers, in this what happens?
So, Bloch’s theorem says that solution to
such a problem solution to such a problem
remember free electron solution was this,
this was free electron solution. Now, question
is if you solve one electron problem Schrodinger
equation in a periodic potential, whatever
that potential is we do not know the potential
is, just that it is periodic a help periodicity
of direct lattice.
Then if that is the case then the solution
is solution of this equation is of form that
is of form of e to power j k dot r which we
already knew in free electron, but also in
a digital term u k r. What is this u k, this
u k all we know about this u k is that it
is also periodic. So, Bloch’s theorem says
that whatever the solution, exact solution
they are interested in at this point of time,
but that if the potential of a periodic then
solution will be of this form where u k will
be periodic that is a Bloch’s theorem. So,
that is a solution we are going to use and
see which of solution will be of this form,
the fact that solution of this form itself
gives us a huge amount of information and
that is what about this lecture will be about
this part only, so let us look at that.
So, what happens to wave function 
and notice I am using a subscript k, because
this corresponds to some value of k, k can
be there are many many k values will any k
value will satisfied this Schrodinger equation,
real solution will therefore summation of
all those k values. Hence, I am putting a
index k for a for a k value. So, with that
let us write this solution, what will happen
to phi k at r plus R that is the question
we are asking. Wave function at k this we
have derived in the previous equation, I have
written it here by it here, this is what we
have written here.
Now, what we are doing is we are asking what
will this quantity will be equal to? This
quantity let us just substitute in therefore,
so that we are going to write as e to power
j k dot now r plus R and u of k which will
be r plus R. So, that is what this quantity
will be is equal to I just substitute in there.
So, now, what happens? What is that mean?
That mean maybe we should keep labeling the
equations also, let us call this as equation
as number one, let us this as equation as
number 2, let us this as equation as number
3.
So, now look at equation number 2, equation
number 2 says, so let us write this first
down first. So, this quantity will be equal
to e to power e to power j small k capital
R times, what e to power j small k dot small
r. Now, I am going to use equation number
2 which says, let us use equation number 2
which says u of k r plus R is same thing as
u of k at r that is periodic, so I am going
to use that. So, I am going to replace this
by u k just r. Now, what is this quantity
equal to? By equation one this quantity equal
to simply phi of r, that is quantity simply
phi of r.
So, therefore phi of k r plus R is equal to
e to power j k dot r phi of k at r and all
these are vectors and that is just this quantity,
what is that mean? Wave function in one primitive
cell differs from another primitive cell by
a quantity by this, by a quantity by this,
its differs by. So, this a phase difference
in wave equation between, lets limit this
equation between one cell and another cell,
there is a phase difference in the wave function
that is what we have figured out.
But for a special values values of k meaning
for k equal to capital K, remember now here
comes the reciprocal lattice vector, this
is a special K is remember that is reciprocal
lattice vector. Now, for this values what
will happen? If you substitute capital K for
it, then phi of capital K r plus r will become
equal too, if you substitute capital K in
there, in this in this equation number four
if you substitute capital K. Then do you remember
the definition of the reciprocal lattice itself
was that e to power j capital K dot capital
R was always 1, that is how it was chosen.
Now, you can see why chose we that that way,
in that case phi of capital K would be equal
that is equal to 1, therefore this will happen.
Certain lattice points reciprocal lattice
points when small k or sorry not like this
point, a small k was becomes capital K for
those vectors, in that case vectors which
are pointing towards lattice cell versus another
cell would be identical, all right? So, that
is important result. Now, let us once proceed
further, now let us do a trick, if that is
capital K is special vector, then if I have
a small vector k let me do this, let me do
little trick and express this as another any
small vector k k prime another another plus
capital K vector I can always split any vector
in to sum of two vectors.
But just that I want to make sure that one
of those vectors is capital k, so if I take
the small k to be like this, then what happens?
Let us look at this. So, from four then from
equation four what happens? Just look at this
phi of k r plus r will become equal to just
lets substitute in their, so that will become
equal to e to power j k prime dot r in to
power j capital K dot R. Since, this quantity
is equal to 1, therefore phi of k r plus R,
this will be equal to j k prime vector R phi
of k r and let us call this equation five.
Now, let us call this equation six lets call
this equation six. Now, can you compare equation
number four and equation number six, compare
equation number four with equation number
six, what do you see? Something interesting
isn’t it? Notice that they look the same
if you place k by k prime, if you replace
k by k prime it makes no difference, it makes
no difference is means the same.
So, when you compare equations four and six,
if compare equation four and six implies wave
function has no unique k, that is k and k
prime yield the same wave function, unique
k, but a set of k values which which differ
by capital K vector. Now, you can see why
is that reciprocal lattice vector, reciprocal
lattice vector is special. Remember k was
equal to k prime plus K, so these two vector
differ by k prime.
And k vectors differed from each other by
some k vectors by a capital k vector, as long
that is true then the wave nature of wave
function remains the same. Wave function does
not change; that means or that gives us a
handle, that gives us a handle. So, now this
this conclusion which comes out of a Bloch’s
electrons from the fact that single electron
periodic potential we can now begin to use
them for our purposes.
So, what I am going to do is show you what
it means how does one reduce this problem.
What I am now going to do is, will now do
is for sake of representation of e k diagram,
make sure that k prime lies in first. Do you
follow that? Why I want to do is, since it
does not matter k you can choose as long as
if I have a small k vector. I can always find
a small k I can find a small k prime vector
which and the these two vectors different
are different by a capital k vector and it
does not matter, if I find a small k vector
wave function will remain the same.
So, that is the logic I will use and I will
make sure that the vector I find that is k
prime vector always lies in the first and
then instead of plotting e versus k diagram,
I will plot e versus k prime diagram. Now,
since k prime is always in first, therefore
the k axis does not need to go to infinity
instead it is only confined up to the boundary
of the first. You see that is the first thing
we want to do to limit our size of the band
diagram. So, in order to do that let me show
you a little bit how we can do that.
So, let us take this is as example, this is
the same example which I had shown you for
a reciprocal lattice where I showing you brag
refraction condition, construction of belanzone.
I had shown you construction of belanzone
in a two dimension and three dimension the
same construction which you do. So, I am showing
you again going back to the same same diagram
which I have drawn in the previous lecture
in that direction were my b 1 and b 2 were
my lattice vectors and therefore this k was
equal to m 1 times b 1 plus m 2 times b 2
in general of course, k is equal to m i b
i where m is the summation where m summation
is over i is equal to 1 2 3 write 1 2 3 in
genral.
But Since this is a two dimension problem
instead I am going to represent here, so k
is of that and this is lattice which I had
shown you. Blue line I had shown construction
I had shown you that was a belanzone. So,
now, what I am going to do is I want to show
you is that choose whatever k you want choose
whatever k you want will always represent,
will find a k prime which lies in the first
belazone, let us do that, let us take there
is an example. In order to do that lets do
it like this, that let me remain consistent
with this example I have chosen, so that the
construction is easier.
So, let us go somewhere here, let us go somewhere
for example, let us go somewhere here. Suppose,
this is my k vector, suppose I choose a k
vector which looks like this, suppose I have
k vector which is like this, this is my k
vector, this is my k vector. If this is the
k vector I can always find a capital K vector
which is reduce my k vector in to the first
belanzone, let me show you that is the example.
So, what we do is we choose another capital
K vector which is like this, remember capital
K vectors are those vectors which are drawn
by or I will write it more explicitly here.
So, this capital K vector then is equal to
lets choose the value of m 1 and m 2. So,
this is 1 and 2, so 2 b 1 2 b 1 plus this
is 2 b 2 this is the capital K vector I have
chosen, if so then what is k prime vector?
This is k prime vector. So, this is k prime
vector, it is k prime vector it is k prime
vector I can represent the same k prime vector
by here. Draw the same parallel vector from
this origin because origin must remain the
same because the k prime vector.
Now, notice I will show you that it does not
matter it does not matter this k prime is
not unique vector, many such k vectors lead
to same k prime vectors that is what I want
to show you from this diagram, it does not
matter. This is second conclusion I want to
show you, first I have shown you that if you
give me this k vector, if you give me this
k vector I will find you it is always possible
by taking a neighborhood lattice point. That
means, capital K vector it is always possible
to reduce to another k prime this is within
the first belanzone, you can see this is within
the belanzone.
Second statement I am making here is that
this k prime is not unique, this k prime is
result of many different k’s it is possible
that is from very small k’s. This k prime
could be same and that is example I will show
you now. So, for example, let us take it like
this, suppose you want to take somewhere here
like this, suppose this is your k vector,
another k vector which is like this, this
is your small k vector. So, in order to bring
this into first belanzone, let me choose this
as my capital K vector, this is my capital
k vector this capital k vector you can see
now is my minus b 2 vector, it is just a minus
b 2 vector, remember this is b 1 and this
is b 2, this is b 2 and this is b 1, it is
just minus b vector. So, therefore what does
it mean?
This means that this is k prime vector, if
that is the case then you see this k prime
vector is same as this red colored k prime
vector I have shown if I transfer to. So,
whether it is this k or whether it is this
k, both leads to same k prime. So, now what
are the conclusion we draw, next page I am
going to write down write down the conclusion
which which I draw from this. So, what, but
let me say in it words now one more and I
will repeat it again. So, as a consequence
of a periodic potential Bloch’s therom said
that, first I should told you that is always
possible, not always possible.
It is possible to reduce the problem to 1
electron problem with an effective potential,
number 1, number 2, once I have one electron
problem, then a 1 electron problem is being
solved by in periodic potential which is natural
to assume, since the lattice has a periodicity
so the potential should also have the same
periodicity of r. If there is a periodic potential,
then there is Bloch’s therom which gives
you formal solution, given that form of Bloch’s
that form of a solution.
What we are able to tell is that wave function
in one premitive cell to another premitive
cell is different by a only a phase. But for
a special values of k there is no phase difference
that is number 1, number 2 conclusion we maid
was that wave function, there are many many
different values of small k is which is lead
to a same wave function which is lead to a
same wave function, we have use this as a
consequence of Bloch’s theorem. We will
use this result to show that if you have.
Remember I saw it seeing at when I want to
plot e k diagram, I have infinite values of
k possible and you extend of course, how far
it can go with a infinite extend.
Now, since any k value which you have I can
always represent by a small k, I can represent
by a k prime vector which has within in the
first at least what we have achieved is that
we have confined that infinite extent of this
k vector within a narrow fun fine of a first
only. That means, that much we are able to
do. So, let me write down and second thing
I have shown you is there is no uniqueness,
whatever k prime vector you find the same
prime k vector represents many different many
manymanymany infinite number of not infinite
very large number of small k vectors.
I have shown you two examples I have taken
this k vector and I have taken this k vector
and I have shown to you that by using an appropriate
capital K vector, reciprocal lattice vector
you can reduce it that to a small k prime
vector which are both are same. Both are within
the first belanzone that is a red line which
I have shown you, so that is a reciprocal
lattice vector. So, let me write down the
conclusion here.
So, conclusion number 1 for a k value it is
always always possible to find an equivalent
k prime which differs from k by amount capital
K only. It is always possible to find an equivalent
k prime in first belanzone, that is point
number 1 I have conclusion, but still but
still energy is equal to h square by 2 m k
dot k knot h square by 2 m e k prime dot k
prime, it is not that. Energy is still, so
x axis of e k diagram I can find k prime,
but the energy that I represent it is still
going to be real energy which will use be
using k not k prime. So, that is the one point
that is one point I want to make. Second within
this conclusion set one is that many k values
can be represented by same k prime in first,
that is the conclusion I want to draw.
So, if these are the two conclusion m then
now you can see. Now, therefore since it is
same k prime in first belanzone I can, this
k vector small k vector represents, k vector
represents electron state. Remember in free
electron theory we found that this is discreet
and some state which electron can take, this
k vector represents electron state and this
can represents and many many this k states
k can be represented by same k prime within
the belanzone. So, question I want to ask
is that in one therefore, how many natural
question to ask is how many electrons can
I put? How many k states therefore, are available
only within the first belanzone itself, in
first belanzone how many case states are available?
So, that is when next logical question is
next logical question is, because k prime
k prime represents many k states a small k
states therefore, and this k prime is in first
belanzone. How many total amount of states
are there in the first belanzone? That is
the question we are trying to ask and since
once you known a number of states, then we
known how many electron we can put in the
first belanzone. Next logical question is
how many e states in first belanzone, that
is the question we are have to answer now.
In order to do that let me do this, consider
and I just do a simple algebraic calculation.
Consider a crystal of volume V and you see
a primitive cell has a volume V p is a primitive
cell volume, if n are the number off primitive
cells, then clearly V is equal to N times
V p. Now, assume that remember, so if volume
V is enclosed by vectors N 1 a 1, N 2 a 2
and N 3 a 3, so think of this like this, that
if in N 1 times in direction of a 1, if N
2 times in direction of a 2, N 3 times in
direction of a 3, that is the extent of my
total volume. That means, so I take a in terms
of primitive lattice vectors, if I take N
1 steps in a relation of a 1, N 2 steps in
direction of a 2, N 3 steps in direction of
a 3, then I construct my whole volume.
If that is so, then clearly N is equal to
N 1 multiplied by N 2 times N 3, so how many
primitive cells there are therefore, if I
took N 1 step in this direction, N 2 step
in this direction and N 3 step in this direction.
Then total number of primitive cells which
are defined by 1 a 1 a 2 a 3 will be N 1 into
N 2 into N 3, then that many primitive cell.
I have said N is that number of primitive,
total number of primitive cell in volume V.
Therefore, this N must be equal to N 1 N 2
and N 3, to the whole extent of the volume
is determined by the multiples of a 1 a 2
a 3 which I have N 1 N 2 and N 3.
Now, remember we apply again there are good
old Born Von Karman boundary condition, we
apply which was what you will recall in free
electron theory. I have written phi of x comma
y comma z being equal to phi of x plus l y
comma z you will recall, I have written it
something like this. In as a Born Von Karman
boundary condition namely there what goes
in it is like this, if something comes out
of here, it basically entered from that side,
it has entered from the other side of the
boundary, if this boundary then it is comes
out something comes out here, essentially
enters from this side.
As a poster saying let us make the wave function
0 on the both sides, so it does not leak.
In order to avoid the leak of wave function,
confined the wave function of material itself,
what we are used was Born Von Karman boundary
condition, a cyclic boundary condition we
said we will not allow the wave function to
leak out of the material, but the way we will
do it is that whatever coming out of here
it goes back in from the other side. So, that
is now we are going to write the same, so
now the material boundary is defined by the
multiples of N 1 N 2 N 3, so we will use it
in that context.
So, I am going to write therefore in that
case, phi of k at any position r will therefore
be equal to phi of any position r plus N 1
N i a i were i could be equal to 1 2 or 3
not summation 1 2 or 3 it is just like this.
If you where is boundary in this direction
after N 1 this is the a 1 direction, so after
N 1 times I have my boundary here, that is
why I have written N 1 multiplied by a 1 or
if it is in direction a 2 it is N 2 times
a 2 is in count of my boundary. So, I have
written it like that in general form where
i is the square quantity.
So, now by Bloch’s theorem what happens?
By Bloch’s theorem phi of k of r plus this
N i a i this is vector, a i whatever N i is
you know as 1 2 or 3 and you apply in equation
in 1 2 there, this at period is equal to e
to power j times, now I am going to write
this is n i k dot a i, I am going to write
like this now phi of k at r. Remember where
this come from? This whole stuff comes from
phi of r plus R being equal to e to power
j k dot r times phi of r, that was Bloch’s
theorem solution, that is what I have applied
here essentially. What is that mean? It implies
if you substitute in there.
So, what is that mean? Since, this quantity
should be equal to remember use this equation
here, since this quantity should be equal
to phi of r implies that e to power j N i
k dot a i should be equal to 1, that is what
it implies. So, if since a i is is real, it
is direct lattice vector it is lattice vector,
so it is a real quantity and if it is real
quantity then dot product. So, that means
k must also be real otherwise I would have
got in an imaginary implies that k must also
be real. If k was not real then I would have
got in an imaginary component also, so therefore
this implies that k is also real.
So, what I am going to do is therefore, since
k belongs to reciprocal lattice since k not
to reciprocal lattice, k belongs to belongs
to reciprocal space, I am just going to express
this k vector also as some multiples of l
1 times b 1 plus l 2 times b 2 plus l 3 times
b 3. And these are reciprocal lattice vectors
where l 1 l 2 l 3 need not be integers because
if they were integers then then the small
k will look like capital K vector. So, it
is just I am trying to do is let us express
we have b 1 b 2 b 3. So, let us express k
as some multiple some coefficient of that
l 1 l 2 and l 3 we are going to choose and
use that. Which I am going to suficly write
as summation over l i and bi, so that you
get the idea that i is 1 2 and 3. So, now
as I said these all are not necessarily integer.
Now, since we have since we have that exponential,
so now let us substitute this k, since we
let us go back to our equation here let us
go back to this equation here substitute in
here it called j N i k dot a i, so we will
use this e to power e to power j let us j
N i k dot a i, let us substitute in there,
so what we get here?
So, what we get is e to power j N i for k
we substitute l 1 b 1 plus l 2 b 2 plus l
3 b 3 b 3 dot a i equal to 1, what is that
mean? Remember whatever i value is i 1 2 3
then you will pick only that l 1 or l 2 or
l 3 according to that in the dot product.
So, therefore e to power j N i l i is what
we get picked out of this and b i dot a i
will it to 2 pi, so I should put a 2 pi that
should be equal to 1, that dot product of
a i and b i is simply 2 pi. You can see our
definition of b i, the reciprocal lattice
vectors, this implies that N i into l i should
be an integer, this quantity should be an
integer and let us give it a name let us give
it a name of let us say just call it q i,
just give it a name of q i, these is a integer.
So, this quantity should be integer quantity
what is that mean? That means k vector should
be equal to i by N 1. I am going to substitute
in here and here from here I am going to write
k as l 1 which to 1 times b 1 divided by 1
by N 1 plus a integer quantity q 2 times b
2 divided by 1 by N 2 plus 1 by N 3 q 3 times
b 3. Now, what is the volume enclosed by this
case state what is the volume as a consequence
of this state.
Remember a vector k which is due to b 1 plus
b 2 plus b 3 primitive cell leads to a primitive
cell volume. This primitive cell volume is
will be equal to this b 1 dot b 2 cross b
3 will be just this quantity, that is the
volume of the primitive cell in the reciprocal
space. Similarly, if you look at this k, what
will be the volume occupied by this case state
and this quantity is of course will be equal
to you can see 2 pi whole cube divided by
V p that you can carry out this as a exercise.
Likewise and reciprocal space imagine equalent
to that you think like this, these are integer,
these are integers quantities, these are integer
quantities.
So, similarly I am going to write what is
the volume volume occupied by this k state
which is defined by 1 by N 1 b 1 plus 1 by
N 2 b 2 plus 1 by N 3 b 3. What is the volume
occupied by this? This is a one case state
occupies this small k volume which will now
in this case will simply be, if I substitute
in their then instead of b 1 I will substitute
b 1 by N 1 instead of b 2 I will substitute
b 2 by N 2 and therefore, and b 3 b three
by N 3 I will substitute.
If I do that substitution then I will get
this as 2 pi whole cube divided by V p divided
by 1 by N 1 N 2 N 3 which is equal to 2 pi
whole cube V p times by by N. That implies
that volume of primitive cell should be equal
to volume associated N times what the volume
associated with a case state and what is the
implication of this? Implication of this I
will write in next page and finish the lecture
here.
That since V k is independent of how the primitive
cell was chosen, so this result must be true
as a general case. So, that means I should
have; that means we should this equation therefore
says, since this is a volume associated with
one case state, in first belanzone and this
the total, this is the volume of the primitive
cell. Since that means, I should have and
belanzone represents that means, a primitive
cell in the volume of the belanzone that means,
primitive cell in reciprocal space is really
a belanzone.
So, therefore this is the volume belanzone,
so this is a volume belanzone associated with
one case state; that means I should have N
case states in the first belanzone. We must
have N k states I wish again put N first belanzone,
this is about to end my lecture and shown
you now that means, if you had N primitive
cells in the real lattice, in the real material
which has given to you, if you have N primitive
cell, then you have N k states available in
the first belanzone. So, you know how many
electrons you can put in that, how many first
belanzone will contain how many electrons.
So, this is the conclusion in we have drawn,
we will use these two facts to now draw out
our e k diagram. So, from in next lecture
on what I am going to do is, I am going to
fist show you how e k diagram could be drawn
for a hypothetical one dimensional system.
Then what I will do is then I will take it
from metals and I will show you how e k diagram
is to be drawn for three dimensional system,
for say aluminum or copper or some material
like that and in from that point you will
start appreciating what e k diagram is. After
that we will introduce a actual potential
and see how the band gap origin originates.
Thank you.
