We had begun looking at the Sommerfeld's picture
of how an electron moves through the solid
and the important thing with Sommerfeld's
realized was that the electron does not really
follow the kinetic theory of gases. Its distribution
is not like a Maxwell Boltzmann distribution,
but the distribution follows quantum statistics.
Inside the solid you have states energy states
which are available and the electron fills
up these energy states and the way they these
states are filled up is governed by the Fermi
Dirac distribution.
So, there is a chemical potential mu up to
which the states are occupied with our occupancy
or with a probability of 1, and above it becomes
0. The 0 temperature chemical potential is
an energy level which is close to the maximum
energy which the particles have at 0 temperature
and that is what we call as the Fermi energy.
And the general way to determine your chemical
potential is that if you count all the number
of states which are occupied, then these are
all the states which are occupied up to the
chemical potential multiplied by two particles
per state then the total number of particles
you will get the total number of particles.
So, if you solve this equation you will get
your chemical potential.
So, we consider that the solid is a cube of
sides of length L we had seen this already
and you write down the Schrodinger's equation
for the free electron, whose energy is given
by this E k is equal to h cross square k square
by 2 m and psi k is the wave function of the
electron.
Being free electron we consider it like a
plane wave. So, this is your plane wave solution
for the free electron, which is moving through
the solid. And you consider the solid has
a finite volume, you normalize it and you
get the constant the amplitude which is 1
by square root of V for this wave function
and then you have to put some boundary conditions
on this wave function.
Now, is it that all momentum states because
you want to generate the energy states which
are available inside the solid, using the
plane wave you want to generate what are the
energy states available are there a continuum
of states or are there some finite discrete
set of states which are available. And for
that some of will used the periodic boundary
conditions.
So, in your problem of just a single electron
inside a box you consider that the wave function
becomes 0 at the two edges, but this has a
problem because it gives rise to standing
waves inside the solid which you want a propagating
mode, you want electrons to actually propagate
through the solid you do not want them to
be static and you do not want them to be standing
wave type solutions.
So, what Sommerfeld considered was this very
important periodic boundary condition. That
the wave function at x is equal to the wave
function at x plus L and this leads to solutions
which are of the travelling wave form where
this is not a solution where you have nodes,
but instead this is a solution, so you do
not have nodes at the edges of the sample.
And if you use this sort of a periodic boundary
condition in one dimension then you will get
a condition on your wave function which is
e raised to i k into L is equal to 1 which
gives you a condition on your momentum, that
the momenta is going to be integral multiples
of 2 pi by L.
The net wave function psi k x, y, z is k psi
k y y, psi k z. So, these are the mementos
in the x, y and z direction, ok. So, we write
the net wave function as the product of these
wave functions and then you can show that
you can write down the periodic boundary condition
as psi x plus L, y plus L and z plus L is
equal to psi x, y and z. Namely, there is
a periodic boundary condition along the x
direction along the y direction as well as
along the z direction. And all of this leads
to the condition on k x, k y and k z. So,
let us write down the conditions on k.
So, you will have e raise to i k x into L
we consider that the cube is symmetric it
has a side of length L, length, width and
height are all of length L, and so if you
use the periodic boundary conditions you will
have this e raise to i k y into L will be
1 and e raise to i, k z into L will be 1.
And this will put conditions on k x, these
will give you your momentum states or the
energy states, k y is equal to 2 pi n y over
L and k z is equal to 2 pi and z over L. These
are your different momentum states, where
n x is 0, 1, 2 and so on; n y is 0, 1, 2;
n z is equal to 0, 1, 2 and so on. So, these
are your different momentum states.
So, you generate your momentum or energy states
inside the solid which is a cube with sides
of length L. And now the states that you generate
you will have to start filling up. So, can
we use this expression, N is equal to summation
of all the momentum states this is the Fermi
Dirac distribution e k minus mu, which at
0 temperature at t equal to 0 Kelvin this
expression N is equal to twice of summation
of all momentum states f D, E k minus E F
as mu is equal to E F at 0 Kelvin. So, can
we use this expression to evaluate what is
the fermi energy of the system for this solid?
If I look at the momentum of my electron in
the solid it has momentous k x, k y and k
z. And these will be typically for n x n y
you know you will have 0, 0, 0; you will have
2 pi by L 0 0; 0 2 pi by L 0; 0 0 2 pi by
L, 2 pi by L 2 pi by L 0. And like this you
will have a discrete set of states you will
have a discrete set of momentum states inside
the solid, each of them have their own kinetic
energy which is given by E k is equal to h
cross square by 2 m, k x, k x square plus
k y square plus k z square. So, from these
discrete momenta states you will get discrete
energy states which will then be filled up
based on Fermi Dirac distribution.
So, now, let us look at this momentum space
k x, k y and k z, this is the 0 momentum state
which is the 0, 0, 0 and every subsequent
state either in the k y direction or in the
k z direction or in the k x direction all
the momentum states are spaced by a distance
of 2 pi by L. These are the discrete momentum
states or the energy states in the momentums.
These are the discrete momentum states that
we have we are drawing here which we have
obtained just now and the spacing between
any two points in either direction is either
2 pi by L.
And you can draw 
what is the smallest. So, there is a point
in the k x, k y plane which will have 2 pi
by L, this is this point which is in the k
x, k y plane which will be this similarly
there will be another point here, and you
will have another point here, and you will
have another point here.
So, you will have one such cube like this
there are no momentum states which are available.
So, the smallest volume in the momentum space
is equal to 2 pi by L the whole cube. This
is the smallest volume because this distance
has 2 pi by L this is 2 pi by L and similarly
this length is 2 pi by L, ok.
You can show for yourself that this is the
smallest volume. Below this volume there is
no states which are available.
