Now, welcome to lecture number two, we will
start from here what we learned in the first
lecture, I gave you the Drude’s theory.
The Drude’s theory elegant theory in it
is time more than 100 years ago when Drude’s
proposed this theory it was really elegant
theory it was really remarkable that he and
the kind assumptions he made. Then, kind of
predictions which could be made appeared that
the Drude’s theory seemed to work except
that it was a good fortune that there was
any matching with the experimental results
with Drude’s theory.
That is just because two has cancelled out
then wanted to show that if you do not have
those two errors cancelling out, then in cases
of ball effect or magnitude resistance. Also,
for example, this plasma frequency for transparency
of metal and there are many other examples
also where you see that group theory seems
to work well. For group one element such as
lithium, potassium, cesium etcetera, but it
begins to fail little for many other metals
and many other situations.
So, clearly something is wrong with that theory,
so first improvement came by what was done
was that you see what Drude assumed based
on kinetic theory of gases that when electron
collides with the ion it comes out with random
velocities. Those velocities are taken through
kinetic theory of gases which means they follow
Maxwell Boltzmann distribution. Now, we know
that electron is electron follows from the
direct distribution not Maxwell Boltzmann
distribution. Therefore, first improvement
which was made was by what is called as Somerfield’s
theory of metal in which essentially he took
Drude’s model, but substituted in there
a Fermi direct distribution for electron velocities.
Then, he applied that theory also which gave
some improvement, but there is a objection
to it that in otherwise it what is a classical
theory Drude theory that is how could you
put a distribution which is whose nature is
quantum mechanics. So, that is the second
which is the objection for Somerfield theory
also which therefore, I have done Drude’s
theory because of it is historical importance.
So, I will not take care of this, I will not
cover this Somerfield theory in here, but
move on to perfectly quantum mechanical theory
based on first attempt. We will make is free
electron theory; I assume that the students
are familiar with quantum mechanics.
So, what I will do in this lecture is do a
very quick 2 minute, 5 minute review of quantum
mechanics the relevant portion for free electron
theory thus to just to jog your memory, but
I expect you to know already. Then, I will
introduce this idea of free electron theory
and then in that free electron theory we will
see quantum states emerging and we will try
to understand the meaning of k vector which
is very important. You will see that eventually
when we derive properties of materials, we
will derive them from an energy versus k type
of curve from which all the properties of
these metals and semiconductors.
We derive and hence it is important understand
to meaning of k vector which will emerge out
of this free electron theory. Then, we will
start putting since these are; I am saying
already k represents a quantum mechanical
state in this state. We will start putting
electrons which are there in the system, so
this is where we will end this lecture midway
in the free electron theory and then in third
lecture we will continue over this. Let us
start with the quick review of quantum mechanics,
essentially what I will do is that that just
basic thing.
Remember, recall that we deal with this wave
function xi and I am using a notation where
I put hat on top of all them this quantity
contains all the relevant information about
the electron it is a function of position
and of time. This is the wave function which
has all, which contains all the information
of electron, you can derive anything you want
through this through this electron through
this wave function. Then, of course he says
that f corresponding to classical operator,
there is a quantum mechanical operator. This
operators is let me write these operators
are given here that for position, there is
a classical operator r corresponding quantum
mechanical operator is r.
So, if this is a function f, the classical
variable is f there is a quantum operator
f for example, the momentum if the momentum
is p in classical variable. Then, in case
of there is an equivalent quantum operator
in quantum mechanics, there is an operator
where as in the variable this quantum momentum
operator is like here. If there is a quantity
called energy in classical variable, then
there is a energy operator in here in quantum
mechanics. There by, if I take this operator,
if we apply this operator on to this operator
which ever operator we are talking about here
operators applied to this wave function which
I have written.
Now, here then it gives you an Eigen value
of this particular operators if I apply momentum
operator on this wave function. Then, gives
me it gives me momentum Eigen values multiplied
by this wave function itself that is the basic
idea of these operators in quantum mechanics,
you will recall. So, what I will do here is
that I will sort of derive if you wish derive
for Schrodinger’s equation wave equation,
obviously you can understand that I am writing
derive in inverted commas. What that means
is obviously you cannot derive, it is basically
a postulate, but I will pretend to derive
by saying that in classical, you think of
1 over 2 m that means the kinetic energy is
the total kinetic energy plus the potential
energy is equal to total energy.
Thus, what you think of in a classical way
what I will do is, I will take this momentum
operator apply on the wave function to derive
this kinetic energy the potential energy and
the total energy well as thus use this these
operators which I have written down here.
There are the operators, all these operators
which I have written down, let us use these
operators and derive the Schrodinger equation
quote and quote derive the Schrodinger equation
for the wave. We will write it is as same
thing, I will take the momentum operator,
so this will become and j indicates by the
way this j. I use symbol j, what is j j is
equal to square root of minus 1.
So, it is an imaginary quantity something
to use, so I am using, I am going to use j
in this in here. Therefore, this j is when
j square forces is equal to minus 1, so what
I am going to write this as minus h square
by 2 m and remember let us introduce another
symbol, this h bar. I hope you are familiar
with this meaning of h bar, h bar is equal
to applying constant divided by 2 pi. Therefore,
this is written as h bar, so h bar 2 m and
radian square and this thing is square of
this applied to this wave function xi of r
comma t.
This vector r comma t is what I have written
here plus this potential of position and times
you wish equal to this energy. Now, I am going
to write as times xi, sorry since this is
the function here and corresponding operator
is this of r comma t. This is then equal to
this energy this energy, now I am going to
use the energy operator minus h bar by j del
by del t of this wave function here.
So, here the Schrodinger’s wave equation
which we have just derived from the classical
sense, you noticed that essentially this operator
then becomes a postulate. So, I assume that
you already know quantum mechanics, this is
just a quick review of what we intend to do
here and so then then what I am going to do
is, I am going to assume separation of valuable
variables.
I assume that it is possible to separate this
wave functions spatial part and time dependent
part it is possible to separate them out.
That means write this as xi of position only,
the difference in symbol is that I draw these
hats, here in case of when it is r comma t,
but if it is only this is a different symbol
when I do not draw the hats here.
Here is only function of position versus phi
which is of time dependent I am assuming that
it is possible it may not always be possible,
but I am assuming that is possible that to
write the wave function as two separate functions
one of position and one of time. So, then
we can substitute this in into our Schrodinger
equation which changes what I wrote down.
So, in that case, this becomes h square by
2 m times phi which is time dependent, hence
need not take it is and this is only special
derivative. So, only phi appears here plus
v times phi times phi equal to and then of
course only the time dependent part.
We need to consider here by j and phi comes
out and del phi by del t, we separate it now
like this divided by 
what do you get divided by this quantity what
you get is minus h square this is the exercise
you have seen in past. Also, in point of time
if you do not recall today, then you need
to brush up whenever you going to introduce
to this, what do you notice? What you notice
is, this is only phi dependent which is a
position dependent if only phi r dependent
and this is time dependent and yet they are
equal, yet two sides spatial part and the
time dependent part are equal. That only means
that they both must be equal to same constant
they must be both equal to same constant and
that constant is that constant we are taking
as a energy and, therefore because this spatial
part.
We are going to write minus h square by 2
m, therefore we are going to write this is
as one over phi plus v as equal to this energy
e implies in more common form as we see it
2 m h into square of phi which is of position
only plus v phi equal to energy times phi.
This is a spatial part and time dependent
part of the wave function, then likewise since
this quantity e this energy should be equal
to the same quantity here in the same both
quantity should be equal. Therefore, minus
h by j write this as phi del phi by del t
equal to this, this implies that I am going
to write this as h by j del phi by del t is
equal to which is time dependent only.
We will assume we will assume, in fact in
doing, so maybe I have made a mistake, I go
back a little bit in order for this to be
true. I have make sure that this v is only
function of position, I should move this part
and should be only function of position, only
then it will be true that this part is only
spatial dependent. So, hence we will assume
that, so we will assume that time dependence
is harmonic, this implies that we will take
phi to be e to power some constant times e
to power j omega t.
That means rotating in time with the angular
speed omega you can clearly see you can substitute
this into this equation here and you will
see that this is a solution to this equation.
So, we are going to assume that whatever we
do subsequently, what we will assume is we
will solve for only the spatial part of phi
solve for only spatial part xi and assume
phi to be e to power to j omega t. Then, this
whole wave function r comma t to be equal
to phi of r times some constant times e to
power j omega t time dependent, so this point
and this is the part we going to solve.
So, this is all that we going to do in this
class, we going to put focus only on this
part, you assume that once we have calculated
this part of the wave function, it is always
possible to multiply this by e to power j
omega t and get the whole wave function. So,
essentially that was a quick review of that
was the quick review of the quantum mechanics
which we need to do here. These are operators,
we will require energy operator s, we will
require these different operators we will
require, now let us move on to this free electron
theory.
What do we mean and what are we doing here
essentially like I said this Drude’s theory
and this Drude’s theory which was a classical
theory Somerfield theory in which a distribution
was put in which was quantum mechanical. Otherwise,
a classical theory, now we are moving towards
something which is purely quantum mechanical.
We will keep it simple, we will keep it very
simple and we will use the same similar assumptions
like Drude did.
For example, this called free electron, we
will assume that it is free electron when
we say free electron in the same sense as
Drude did. We will assume this free electron
here to mean as free electron and independent
electron meaning thereby neglect any interaction
between electron and electron and will neglect
interaction any potential arising of electron
and ions. That means, we assume that in a
metal the ion positive charges well shielded
by the valance electrons and sorry the inner
shells of the electrons inner shells and because
of the screening the valance electrons are
very loosely bound to the ions. Since, they
are very loosely bound, we will ignore this
interaction and we will say this interaction
is nearly 0 and that is what is really meant
by free electron.
So, then how would we solve for this problem
of conductivity, eventually how it is how
we would think of conductivity in in a material
in metals. If this is a situation in quantum
mechanical sense is essentially what we are
going to attempt to do. Now, recall so what
we going to do we recall that h square by
2 m phi, we derive this as to be equal to
this is the Schrodinger equation. The spatial
part of the Schrodinger equation what we had
derived the free electron theory implies free
electron theory, essentially implies v is
equal to 0 that means all the potential the
potential is 0.
There is no interaction, we have ignored electron
which is fairer approximation which is not
too terrible though there are systems were
electron interactions are very important,
but electron ion interaction. Some typically
presented material which you cannot ignore,
but in some metal if it is possible to ignore
them in that case potential is totally 0.
There is no interaction and essentially free
electron theory, therefore is about solving
this equation of position. Of course, let
me write it again this time is about solving
this Schrodinger equation with some boundary
conditions, what boundary conditions? Now,
then you are, you can, since if have a material,
you can make sure you can confine the electron
into the material after all do not allow this
electron to leak out of this material.
So, it stays in the material that is one way
and that you can do by not letting the wave
function leak out of leak out of this material
that is one way, but when you are and that
leads such a solution. If I try to solve it
this way, then you will end of it is standing
waves in the material. Now, you are interested
in when we are interested in transport of
energy then it is much better to deal with
running waves and in order to obtain a solution
which is running wave like solution.
So, not that the other solution is wrong,
just that for convenience, if you want to
a running wave like solution, then it is possible
to write what are called as cyclic boundary
conditions. So, I will write down the cyclic
boundary conditions mathematically because
it can only draw it, I am going do it in three
dimensions, but in one dimension, I will show
you what it means.
So, I am going to write down this Born Von
Karman boundary conditions, what are these
boundary conditions?
I am going to think of free electron, so imagine
this free electron 
confined in a cuboid of edge length l and
volume v. So, that means 1 by 3 power of volume
is in the edge length, so imagine that this
electron confined in all three directions
in imagine, this is confined in a cube.
In that case phi, you can write at position
x plus l comma y comma z is same as phi of
x comma y comma z. I am going to write phi
of x comma y plus l slowly y plus l comma
z is equal to phi x comma y comma z. Similarly,
x comma y comma z plus l is equal to phi x
comma y comma z this is a mathematical written
down in three dimension if you wish to see
it in one dimension you can think like this.
If this is starting point and in one dimension,
let us consider the dimension one dimension
to follow a circular path a closed loop imagine
this to be the starting point and imagine
this to be the terminating point.
So, imagine a wave function which goes something
maybe we use a different pen, so we draw a
closed loop and we using the color here and
we say draw the wave function. Then, it goes
something like this, let us say wave function
is like this this wave so that it terminates
like this, it is a starting point imagine
this that x equal to l is viewed back on 2
x. So, this is the starting point let us say
and that is this is the finishing point x
plus l is the total length here total length
here, this is the length this is the length
l we are talking about this is we are going
around this and this is of length l.
So, we going around x, we start from x and
go x plus l the whole start from x go round
the length l you end up the same point. Therefore,
wave function is the same essentially we are
going to saying is what comes in from this
boundary leaves from this boundary in case
running in case of this this kind of boundary
condition come in come out, energy comes in,
energy goes out. So, this point is considered
a basically collapsed onto that starting point
and this in order to write, I cannot one cannot
draw that in three dimension.
So, in three dimension, therefore we are writing
it mathematically, so these are the boundary
conditions we will apply for this running
wave like solution. So, then notice without
regard to this boundary condition solution
to this particular equation is of form you
can see that phi of k a symbol I am giving
this r is equal to 1 by square root V e to
power j k dot r this k is the wave number.
You can take this and substitute in there
substitute this into this equation which is
right here, up here into this equation you
substitute this solution.
We can substitute there and check whether
this satisfies or not, if this you find that
this solution will satisfy there with definition
that k of course k vector is equal to k of
x, x hat k of y, y hat. Let us write this
explicitly and that this energy then in that
cases h square by two m k dot k this is the
same k, k dot k or equal to h square by 2
m k square. Thus, the energy, so this is a
solution and this square root v basically
appears for normalization 
that is you must make sure that should be
1.
So, essentially in to make sure that this
wave function is normalized this square root
of v is appearing there thus about it, but
the form of solution is this here, let us
use a different pen here now. So, this is
a form of solution essentially which you can
substitute into this equation right here into
this equation you can substitute in there.
So, these two together then constitute the
solution to this particular equation which
we are trying to solve as a Eigen value problem.
Now, we going to go and apply the boundary
conditions in there and see how the quantization
starts appearing in this except or even before
I do that notice we have introduced k which
clearly is this solution is plain wave like
solution. So, if you look at this vector k
then you can see I had mentioned that wave
k vector has a meaning. So, obviously it is
a wave vector first of all, first of all it
is a wave vector that is fairly obvious from
the solution itself, but what else it means
eventually we will see and this is you can
see that this k vector automatically started
becoming related to energy starts.
If already this starts, this k is related
to energy we will see that this k also has
implication has some meaning to it relates
to the momentum also. Therefore, meaning of
k becomes very important which we will try
to understand subsequently as we go along
in here.
So, let us apply the boundary conditions now
the boundary conditions one of the boundary
conditions is that we call let me repeat this
again let us apply this one that x plus l
comma y comma z is equal to phi x comma y
comma z, let us apply this. so, what we will
do let us substitute that into the solution,
so we have one over square root of v which
is the solution here e to power j. We are
going to write this k x times x plus l plus
k y times y plus k z times z as the equal
to one over square root of v which is basically
this wave function.
This we know over j k x x plus x k y y plus
k z z k dot r basically e to power remember
the solution is of form j k dot r. Essentially,
that is what we are writing here essentially
that is what we are writing 1 over square
root of v was our solution form for substituting
in there what does that imply?
That implies that from this two equations
implies e to power j k x l is equal to 1 that
is what it means boundary condition. This
essentially that is what implies with this
boundary condition this forces the k x cannot
be any value, but k x can only be k x can
only be 2 pi multiple of l n x where n x is
integer where n x can only be integer. So,
n x value can be essentially, now notice that
my value of k x is our value of k x is quantized.
So, n x is a quantum number in that sense
k x is a quantum number, since n x is only
be integer k can take only definite values
fix values and recall since k is related to
energy. So, energy can therefore only discrete
values essentially thus a common feature which
you see in quantum mechanics, so similarly,
by same token.
If I use the other two boundary conditions
you will find, then two by l n y would become
for the second boundary condition and a third
boundary condition we will find k z is equal
to two pi by l n z these are the only allowed
of course n y n z are all integers. Also,
these are also integers, so this is what we
get out of this that, let us summarize of
the more time that here is the wave function
which satisfies the Schrodinger equation together
with the Eigen value of energy right here,
let us use another color.
So, here is the wave function which satisfies
the Schrodinger equation and correspondingly
the Eigen value of energy which is dependent
on this quantity called k there this quantity
k here. Then, if you move on, then we find
that if you apply the boundary condition to
it than the k can only take definite values
and since k can take only definite values
than energy is also discrete. Therefore, also
what do you see, therefore we are now have
three quantum number n x, n y, n z.
Loosely speaking, we can think of three quantum
numbers to be k x, k y, k z along with than
the fourth one which is the spin of the electron.
We can now think of we call that I have written
the wave function as only that of spatial
and time. In addition to that, we should also
consider the spin state which I have ignored
we can assume that spin state is plus half
minus half up spin or down spin. So, two spins
are possible so along with these three quantum
numbers plus the spin quantum number then
completely describes my system.
Now, this k values 
are discrete let us see what it means let
us do some calculation h bar is equal to 6.63
into 10 to power ten to 34 minus joules second
divided by 2 pi which is equal to 1.055 into
10 to power minus 34 joules second. Mass of
electron of course is 9.11 into 10 to power
minus 31 kg, so let us than therefore calculate
this quantity h square h bar square by 2 m.
I have calculated this to be 6.11 into 10
to power minus 39 in appropriate units and
if you take l to be 0.01 meter that means
1 centimeter. If we take the l to be that
dimension then two pi by l is 628 meter inverse
and then in that case if you look at this
remember 2 pi by l of some integer multiple.
So, let us look at this quantity h square
by 2 m 2 pi by L square this quantity, then
becomes 2.41 into 10 to power minus 33 joules
now notice that as n x changes or n x y changes
and n z changes by integer one integer k x
changes by this much amount.
So, imagine if one state to be at n x equal
to zero and then in that case we will see
the energy to be 0 k to be in x, x direction.
Let us assume this big assumption, so let
us assume n x is equal to 0 n y equal to 0
and n z to be equal to be 0 and then in next
case let us assume n x to be equal to 1 n
y to be 0 and n z to be 0. Then, next state
next accessible state for electron, so if
you look at this for this case energy is equal
to 0 as we see in k square depends on k square
h square by 2 m h square by 2 m k square.
That means energy is 0, in next case this
k value will be equal to 2 pi by l, so in
this case n x is equal to 1.
So, n x is equal to 1 that means k x will
be equal to 2 pi by l, so that is why I have
done h square by 2 m 2 pi by L whole square.
This gives me the idea of energy, next energy,
so energy will be 2.41 into 10 to power minus
33 joules, but this gap in terms of electron
volts you can divide this number by 1.6 into
10 to power minus 19 only minus 19 and minus
33. So, that means in electron volts also
the two energy levels there is one energy
level and then next level is only separated
by very small energy.
So, it is always possible though this energies
are discrete this because this n x because
this n x n y and n z are all discrete, yet
we can think them to be quasi continuous.
We can think of quasi continuous energies
meaning there by that as n x n y n z integers
change k x k y k z change very small amount.
Therefore, the energies though discrete are
changing by very small amount. So, it will
appear that as if it is a quasi continuous
energy, we can think that energy almost continuous
though in principle they are discrete as they
should be in quantum mechanical system.
So, let us go further now what is meaning
vector k what of course we know it is a wave
vector that we know second what we know momentum
operator 
is we will recall before the momentum operator.
So, let us find the momentum Eigen value,
so what happens let us do this, therefore
we will take this operator and apply on the
wave function. So, we will take this operator
apply on the wave function and should give
me my momentum Eigen value times wave function
since I already know my wave function.
So, I substitute in there what do we get we
are going to get is essentially you substitute
in there. You would you would get is, therefore
h k times phi as p phi implies this momentum
as be equal to h bar k very interesting which
says is that k vector is related to electron
momentum k vector is related to electron momentum.
What that means is that actually, but, I will
at this point of time I will also do a clarification
that indeed in free electron k vector is exactly
giving you electron momentum, but it was not
free electron theory. Then, in that case also
though not done, here k vector is related
to the momentum in sense that change of this
quantity h k.
This quantity h k rate of change of this quantity
h k is still gives you what is the external
force that if you apply the external electric
field to that force how does electron respond
that to that force change in this momentum
is still. This still appears as momentum fluid
in external force, what I mean is external
force when external force rate of change of
this quantity h k is still is effected by
this electric field which is applied from
outside. Hence, we think of this also as crystal
momentum that means that it is not actually
an electron momentum the electrons, if it
is not free electron in free electron case,
this quantity h k is precisely momentum of
the electron in case.
If it is not free electron still h k behaves
like momentum to the external source though
it is not actual momentum. This is then third
thing we can notice essentially having defined
the momentum and the crystal momentum though
in free electron theory. I mention again that
this is electron momentum, now let us see
what dimensions of k are k is k as dimension
of reciprocal length 
belongs to reciprocal as written down here.
It is dimension is reciprocal of length and
belongs to reciprocal space it is consequence,
we will see as we go along in this course,
so these are three points which you should
remember.
Now, so what do we have we have states so
let us look at occupation, next topic the
last topic for this lecture, so what have
we derived we have derived k x as equal to
2 pi by L n x k y equal to 2 pi by L n y k
z equal to 2 pi by L n z. We said that these
are the three quantum numbers n x n y n z
these, therefore define the k state that means
they together define a k state k they define
a k state 
and each k state can take 2 electrons 1 1
spin up and 1 spin down. That means the four
quantum numbers essentially k x k y k z in
a spin quantum number that means I take consider
one combination of k k x value k y k z value
corresponding to 1.
Then, that defines my k state and in this
case state, I can put two electrons, now I
have defined what the states are now. Suppose,
in volume v a volume v we have we have n electrons
and small n be equal, therefore N divided
by v which is number of electrons per unit
volume. So, I have n electrons in volume v
question we ask is how do we fill these electrons
in these k states remember n x n y n z can
take any integer values up to infinity in
that sense I have an great number of k states
never ends is continuous continuously.
They are pointed to n x n y different integers
the k x k y k z values these are states on
which I can put electrons on each of these
states. I can put two electrons question is,
but I have only n electrons, so when I start
putting n electrons on these k states how
far do I go? I go in n x n y n z or k x k
y k z after which I run out of electrons and
therefore, it will tell me what is the maximum
energy those electrons can have and up to
how far they go. Those are the issues we want
solve now address though the states continue
to exist ad nauseam which is continue to exist.
Now, let us look at the filling part, so let
us see let us assume a example let me show
you example.
For example, let us assume n z is equal to
zero and if I draw this axis like this this
is k x this is k y axis k y axis then corresponding
to n x equal to 0 n y 0. I have a k x, I have
state right here, so here we have a n this
is corresponding to n x equal to 0, therefore
k x is equal to 0 corresponding n y is equal
to 0, therefore k y equal to 0, I have one
point. Now, corresponding to when n x is equal
to 1 n, n y continues to be 0, I will have
another state right here corresponding to
n x equal to 2 while k y is still 0.
I will have another state right here corresponding
to n x equal to 2 while k y is still zero
I will have another state here another state
here. Therefore, n x equal to minus 1 n x
equal to minus 2 n x equal to minus 2 where
k y equal to 0. Similarly, for n y equal to
plus 1 and n x equal to 0, I have a state
here and so on, I will have all the states
corresponding to different combination of
n x and n y. I will have all the states which
are I am denoting here and what is the spacing
between these state this between this and
this it is 2 pi by l, they are separated by
2 pi by L and what is this spacing 2 pi by
L of course.
You can see that even in three dimension in
3D also space between any two k states in
any of the three directions k directions is
equal to 2 pi by L space between any two neighboring
k states. In any of the three direction in
the k x direction k y direction k z direction,
the space is between 2 pi what do, What does
that mean, I can think like this that what
is the space what is the volume associated
with a k state this the this this in two dimension,
I have shown as a square each as length being
2 pi by l.
So, now what I am going to do is, I am going
to write this as follows, since the periodicity
is 2 pi by L all these points in all the in
the three dimensions. Then, volume and watch
out the words here carefully volume of k space
volume in k space not real volume in this
k space this is k space. I have drawn a k
x, k y space consisting of k x, k y and k
z, then in also any volume in that space is
the volume of the k space volume of k space
per k point. We take one k point, then that
obviously is 2 pi by L whole cube that is
the volume associated with each k point because
the periodicity in k space is 2 pi by l. That
implies that number of therefore, number of
k points per unit volume of k space is just
inverse of this this.
I am writing inverse of this that will be
l cube divided by this inverse of this relationship
which I have just written here inverse of
this is l cube 8 pi cube which is equal to
volume V real volume in the real space eight
pi cube. So, that is what this quantity is
now what have we done, therefore now have
density of k points in k space we have density.
Once I have a density of points. Then, if
I take large number of volume right in the
k space and sample over them, then I am all
I have to do is basically multiply with this
density as we go along we will see what we
are talking about.
Now, what we are saying is we are saying that
if I take notice that energy depends as goes
as h square by 2 m k square. So, all points
of constant k have same energy, so what surface
what if I think of this k space consisting
of k x axis. So, if I think of k x axis k
x k y and a k z axis, so I think of a sphere
in this, I think of a sphere in this about
this point and this has a constant k some
value k. According to this value on the surface
of the sphere all points have same energy,
but if think that they are Avogadro, since
I have to fill Avogadro number of electrons
n will be very large Avogadro number.
Then, if I am going to that means I will be
accessing all most 10 to 23 k states if i
accessing k that mean k state and I ask what
is the radius of the sphere in which I will
filling all the n since this number of sample
space. So, large we can forget that this is
a cuboid like arrangement; we can once you
calculated this density right here in the
k space I can take any volume in k space multiply
that by this density number of k points per
unit k space volume. So, I can take the volume
of the k space multiply with this density
and what I can get is therefore, number of
k states in that volume essentially that is
why we have calculated this density in here.
So, what I am going to do is now as follows,
so suppose this these n electrons fill up
to radius of this k sphere which I have just
drawn here k f. Thus, the radius of that sphere
then the volume of that sphere is of course,
4 by 3 pi k f cube, remember volume in the
k space is the volume in the k space in k
space multiplied by v by 8 pi cube density
of points in k points in k space. Then, that
gives me total number of k points, then that
gives me number of number of k points and
since each k point, so that is let us write
this quantity as equal to let us see 1 by
6 pi square k f cube multiplied by v, that
is what this quantity.
Now, each of this each of this this is number
of k points, each k point can take up to two
electrons, therefore two times number of k
points k f cube divided by 6 pi square multiplied
v should therefore be equal to total number
of electrons that I have. Total number of
electrons that I have what that implies is
that we can write n which is equal to n divided
by v as be equal to k f cube by 3 pi square.
So, my relationship important relationship
is 
k f is radius of what we call as Fermi sphere
the sphere up to which the electrons fill.
So, if I have n electrons per unit volume
in a system small n number of electrons per
unit volume then they fill up k states up
to that radius k f and that is called the
Fermi sphere. Now, since you already we already
know that energy depends on since energy,
now if you ask the question up to the why
where the Fermi sphere is what is the energy.
We know that this energy Fermi energy we will
call as Fermi energy be equal to h square
by 2 m k f square. Thus, by same definition
e f is Fermi energy is the highest energy
which electrons have because why the k f is
the radius which is the largest radius since
k f is the largest radius, so the electron
energy is highest there all other electrons
imagine like this.
If I have a k x k y k z space in that, there
is a sphere in this sphere if I put electrons,
I start putting electrons at the lowest energy,
I put one on each k state, I start putting
two and keep doing this keep doing this keep
doing this. We reach up to k f a sphere were
we put our last electron and they are done
all k states after that radius are, therefore
unoccupied all electron and all states below
this k f up to k f rather are all occupied
each having two electrons on them. So, what
is the energy of the electron energy of electron
is this e f and that energy is called the
Fermi energy and which, now we can write as
e f.
We can write also as substitute value of k
f in terms of n is from this equation here,
take this equation and substitute in there.
So, what you get is this comes out as h square
by 2 m e 3 pi square n to power 2 by 3, now
you know that if you take, remember if we
take a metal, we know it is density you have
seen in first lecture. If we know it is density,
if we know it is molecular weight which we
do, if we assume how many electrons per atom
it conducts, then we can calculate what the
value of n is. If we can calculate the value
of n, then we can calculate the Fermi energy
up to which these electrons completely fail.
That is one thing we can calculate, second
thing what we can calculate of that electron
that highest energy electron momentum of highest
energy electron at k equal to k f.
What is that quantity equal to, of course
we know that momentum is h k f in this case
and that therefore will be the h times we
substitute k f in there, we know that what
k f is, so 3 pi square n to the power 1 by
3 in terms of n. So, if I know number of electrons,
though I can predict what the momentum will
be once I know the momentum, then I can divide
it by mass of I can say what the velocity
of this highest velocity of this electrons
is which is h k f divided by mass of electron.
Then, it can be calculated as h bar by m e
three pi square n to power 1 by 3, if you
now remember what we did in Drude’s theory
we calculated estimate of velocity v 0 by
classical equi partition of energy in context
in context of kinetic theory of gases.
We wrote that as 3 by 2 Boltzmann constant
multiplied by temperature and that we equated
to half m v square and we calculated velocity.
Now, we count that velocity to 10 to power
7 per centimeter per second, now as we calculate
this velocity, you will find that this v f
is on order of 10 to power 8 centimeters per
second. Then, at that time I had mentioned
that we have underestimated velocity by a
order of magnitude that is what this is that
the quantum mechanics is of actual velocity
is only 0.01.
The Fermi velocity is almost one percent within
1 percent of the velocity of light, so that
is why I will stop at this lecture here on
free electron theory. We have built the bases
what we have done is that we have built the
bases defined what free electron is, then
we have built the bases and we have calculated
that we have k states quantized k states,
but very narrow narrowly packed. Therefore,
you can almost think of them as quasi continuous
on these states we start putting two electrons
and we see how far we can fill this up.
We can fill this up to the energy what is
called as Fermi sphere k f and corresponding
to that k f is Fermi energy. That is the highest
energy electron can have corresponding to
that what is the momentum of the electron
what is the velocity of an electron. We have
derived all those expressions and explained
to you next time, I will start with density
of electron states and then from that on,
I will give it quick view of how we can think
of conduction in thick stuffs.
Thank you.
