Hello, welcome to the thirty ninth class in
this course the physics of materials. In our
in
the last several classes that we have seen
we have often encountered various particles
and try to understand what their behavior
is what it represents in terms of material
behavior.
So, throughout this course at several occasions
we ran into the photons. So, those are the
particles of lights so to speak. When light
indicates its behavior in the form of particles
we call them as photons. We also looked at
vibrations in the lattice and it is said that
you
know those are waves, but you could also look
at them as particles because the manner in
which they interact with respect to energy,
the way they pickup energy, they look the
way they lose energy that is also still quantized
and so, in a sense they behave like
particles, those particles of a lattice waves
are referred to as phonons and in the last
class
we looked at particular combination of particles
which basically we said that you know
pairs of electrons with opposite spin and
opposite wave vectors could combine together
and operate in a coordinated manner and in
that sense would behave like a combined
particles so to speak.
All though they are quite far off from each
other, but their behavior would appear like
that and those would be called the cooper
pairs. So, we saw these. So, this is what
we
saw last class and so these are all the or
some of the particles that we have seen through
the course. As it turns out and as I mentioned
in the last class, all of these have
characteristics that are common to them and
which makes them a certain type of particle
which is called a boson. So, this is a boson
each of them is a boson and 1 of the
requirements for particle to be a boson is
that it has to have integer spin. So, an electron
by itself has half integer spin. So, half
we say plus or minus half. Spin is plus or
minus
half we say.
So, it has half integer spin and therefore,
by itself it would actually qualify as a fermion.
So, Fermi Dirac statistics is what it would
follow, but cooper pairs for example, it to
the
extent that they behave as a combined entity,
they are actually having a net spin of 0 and
therefore, it now suddenly shows you an integer
spin and therefore, it satisfies the
requirement for it being a boson and so do
phonons and photons. So, they have integers
spin. So, they satisfy the criterion for being
boson. So, so this is what they are. We have
already seen these 3. I also mentioned that
for example, when I when I spoke about these
use of superconductors and the fact that they
are used often in this particle accelerator
and colliders. So, to speak particle colliders
then we have even there for example, at
CERN they are using effectively superconductors
which use phonons as well as cooper
pairs and therefore, or 2 bosons and they
are searching for another boson called the
Higgs bosons.
So, at the time of recording this lecture
that is what is being done. So, it is of interest
therefore, since we have discussed these in
detail to see what is a boson, what is that
behavior that characterizes boson, what is
that distribution because we always say you
know when you when you look at a set of particles
we would like to know in terms of
energy how are those particles distributed
because that gives you a very good feel for
what will that a system of such particles
do, when they are subject to certain condition.
So, that is why we need these kinds of information.
Bosons they get their name because
they are credited to Bose, Satyendranath Bose
who first proposed this and essentially
they follow statistical distribution referred
to as the Bose Einstein statistics.
So, they follow Bose Einstein statistics and
so, and hence the name bosons. So, today we
will look at Bose Einstein statistics. So,
that is that is our topic for today. Basically
for a
for 1 of the requirements for Bose Einstein
statistics is we are we are dealing with a
set
of particles which are showing us quantum
mechanical behavior, but in the specific case
of Bose Einstein statistics it applies to
a set of particles which are not subject to
the
Pauli’s exclusion principle. So, the particles
are not subject. So, these particles that
are
not subject to the Pauli’s exclusion principle.
They are quantum mechanical in the in
their behavior, but they, but they are not
subjected to the Pauli’s exclusion principle.
This
is very important because when we did the
Fermi Dirac statistical distribution this
was
the main principle that we incorporated.
We had the Pauli’s exclusion principle there.
So, it is in this fundamental sense that this
distribution differs from the from what we
are doing for the Fermi Dirac distribution.
So,
it exempts the particles from the Pauli’s
exclusion principle and when we say it exempts,
we are basically talking of particles which
do not care of I mean which are not affected
by the Pauli’s exclusion principle. So,
that is why this statistics which is based
on this
idea works for that those sets of particles.
So, they are not subject to the Pauli’s
exclusion principle meaning they are exempt
from the Pauling Pauli’s exclusion
principle. The direct result of this is that
see when we say Pauli’s exclusion principle
the
idea that it reduces to is that you cannot
have more than 1 particle in a quantum state.
Once, you define all the quantum numbers at
a finally, when you pick up a particular
state which if it includes if that definition
of the state includes the spin of the state
and so
on. If you put particles in their you cannot
put more than 1 particle per state, that is
the
definition of I mean that is what it reduces
to when you say Pauli’s exclusion principle
that is what it in terms of how you will enforce
it in the system that is the rule that you
will use to enforce it in the system.
Now, when you say that system is not subject
to Pauli’s exclusion principle, it means
immediately this restriction is no longer
true. It means in a given state you can put
as
many particles as you wish. So, the that is
the big difference whereas, previously you
could put only 1 particle per state now you
can put as many particles as you wish in the
state because they are not subject to the
Pauli’s exclusion principle. So, this is
the general
idea. On this basis we will do the statistical
distribution and see what kind of a result
we
get. So, again we have a system where we have
energy levels.
So, we let us say that the energy levels are
quantized. So, you have energy levels e 1,
e 2,
e 3, e 4 and so on. So, some up to some e
r and let us say it has a certain each of
them has
certain number of states. So, s 1, s 2, s
3, s 4 and so on s r. So, these are fixed
for the
system, given the system we it has a certain
set of energy levels and certain set of states.
So, again this is a constant volume system
and so on we are and at equilibrium that is
what we are looking at. Now, we would like
know we will say that up front we do not
know, but we will say that perhaps there are
n 1 particles at energy level e 1, given that
these are the states, n 2 at energy level
e 2, n 3 at energy level e 3, n 4 at energy
level e 4
and so on with a total number of particles
being n and the total number of total amount
of
energy being capital d. So, the total energy
is fixed at capital e these are the energy
states
that are available and the total number of
particles is fixed at capital n, these are
this is
the way in which that capital n particles
has been distributed across all these energy
levels.
We have n 1, n 2, n 3, n 4 at those respective
energy values. Now our problem of course,
as we have done before is to find out what
is the relationship that helps us figure out
what is the value of n 1, what is the value
n 2, what is the value of n 3 and so on. We
do
not know upfront what this values are, we
would like to find out what these values are
given that is the system subject to the rules
that we have just described. And again these
2 are fixed, this is the 1 that we have freedom
to change subject to this restriction the
total number is n. So, given this scenario
what we are interested in actually finding
out is
the effectively the information we are interested
in we are finding out is the probability
of occupancy of a state 
at energy level e i.
So, I said that we really want to know what
are these numbers? Mathematically it
reduces to this you want to find out what
is the probability that a particular state
at any
given energy level is occupied. If you have
a general expression for that probability
that
effectively will sort of give you all this
information that we are interested in. So,
then
you can go to any energy level you can get
an idea of what the n is at that level because
that will give you because we know this. So,
we have some probability of occupancy and
we can sort of figure it out and there. So,
we would like to get probability of occupancy
of a state at energy level e i. So, if you
take any energy level e i.
We basically have s 1 states and n 1 particles.
I am sorry s i states and n i particles at
energy level e i. So, actually the probability
of occupancy of a state is simply n i by s
i.
So, this is all we are, this is the probability
of you have s i state, you have n i particles.
So, what is the chances that a particular
state at this, 1 of this states is occupied
its simply
n i by s i. So, this is all we are interested
in. So, actually. So, the problem actually
reduces to finding out an expression for this.
We want an expression of for this based on
the conditions that are imposed on the
system. So, this is the expression we are
interested in once we get an expression for
this,
we sort of have the answer for the information
that we are looking for. So, that is that
is
all we are going to try and do in this class
now. So, let us get on to the mathematics
of
how we go about it. What we will do is we
are now faced with a slightly with a slightly
different situation than what we are we have
dealt with before and that is simply that
since Pauli’s exclusion principle does not
apply, any number of particles could be setting
at a given state.
So, therefore, what we actually have is we
have s 1 states and n 1, s i state and n i
particles and really you can mix them of any
which way you want at a given energy level
e i. So, there is since any number of them
could sit in 1 particular state, there is
we are
free to mix them up in any which way that
you wish. So, we will just put it put down
here we have let us say n i particles which
I will mark as x with small x is here and
then
we have some states, some arbitrary number
of states we have. These are the particles,
these are the states. We would like to put
this particles into the states in any which
way
that you want. Now, in general if you have.
So, you have what is this mathematically you
have s I, you have s i items of 1 kind and
n i items of another kind. What is the ways
in
which you can mix this 2 up that is in how
many ways can you arrange these given that
you have s i items of 1 kind and n i items
of another kind.
Normally, when you write such a problem when
you have you know A items of 1 kind
and B items of another kind and you would
like to know in how many ways you can
arrange these 2? The general answer is actually
A plus B factorial by A factorial times B
factorial. This is the general answer for
this kind of a problem. The only issue of
this
kind of an answer. So, in principle we can
apply this kind of an answer here, but there
is
1 small detail that we have to address before
we do this kind of an we apply it and that
is
when you do this kind of a problem you are
permitted to have a situation where you have
all the A items first. So, let us say A item.
So, all the A items. So, let us say they are
squares, A squares. So, we can have all the
squares and let us say B is triangle.
So, we can have all the triangles after this.
This is a valid arrangement as far as this
kind
of a mathematical problem is there. You can
put all the squares first and then all the
triangles next similarly you can put all the
triangles first followed by all the squares.
So,
you this is allowed under this under the problem
that we have defined like this in the
mathematical sense and so this is. So, those
are 2 arrangements that are included in this
in this condition that you are looking at.
Now, in the condition that we are looking
at we
have particles and we have states. If you
have this kind of a situation here it is the
same
as saying you will have all the particles
first and then all states or you will have
all the
states first and then all the particles. That
creates a situation that that corresponds
to
what?
It corresponds to situation where all the
particles are outside all of the states. So,
that is
not a situation that is acceptable for us.
All the particles are free to be in any state
they
wish, but they have to been in some state
you cannot have. So, this is not a valid
condition to have all the particles outside
all of the states, like this kind of an
arrangement is not acceptable here. So, we
cannot simply write we cannot use this
equation in exactly this form you cannot simply
say A plus B factorial by A factorial
plus times B factorial. In this case you cannot
simply say s i plus n i factorial by s i
factorial times n i factorial, you cannot
do that because that would imply that you
are
allowing this system, this possibility that
all the particles are outside or several particles
are outside that is not allowed. All the particles
have to be somewhere in those states.
So, how do we do that? How do we handle that
situation? A simple way to do that is to
say that we will actually treat it as we will
we will eliminate 1 state in our count. So,
how
many ever states are there we will take 1
less state and then we will arrange all the
particles in the remaining states using the
same kind of a formula then whenever we say
that all the particles are before or after
it simply means they are in the last state
or in the
first state. So, or alternatively if you have
you know let us say for example, you have
here 1, 2, 3, 4, 5. 5 states you have here,
if you look at it the number of partitions
between these states is 1, 2, 3 and 4. So,
how many our states you have you have that
many minus 1 partitions. So, we can treat
the problem as though you are trying to
arrange s i minus 1 partitions and n i particles
in any which way that you can.
If you treat it as s i minus 1 partitions
you will never have a situation where all
the
particles are outside the states because if
all the partitions occur before the particles
then
you have all the particles sitting in this
last state because all the four partitions
are before
all the particles. The other example that
we have is all the four partitions are in
the end,
the particles are before that all the 4 partitions.
So, therefore, they are in the first state.
So, you can now not worry about this situation
that we just described, they particles will
never be outside these states, they will be
stuck within this state and regardless of
what
possibility you look at you can always come
back to this picture and you can see that
they are either in the first state or in the
last state or in some of the states in the
middle.
So, therefore, that issue is addressed. So,
instead of directly just using s i and n i
which
can create this kind of a controversy, we
will use s i minus one and n i. So, if you
use
that all the possibilities are addressed reasonably.
So, we will do that. So, therefore, we now
have, if you use now this formula we will
have n i plus s i minus 1 factorial by n i
factorial times s i minus 1 factorial. This
we will
designate as omega n i. So, this is the number
of ways in which you can attain you can
arrange n i particles in s i states subject
to all the conditions that the Bose Einstein
statistics requires. So, this is how we get
it. So, for the entire systems. So, this is
a for
entire system the for each energy level you
can do this, for every energy level this is
only
an i. So, you for energy 1 you can do it,
energy level 1, e 1, e 2, e 3, e 4 for every
1 of
them you can do this.
So, the total number of ways in which you
can accomplish a particular micro state. So,
so
for that micro state for example, we which
will designate with capital omega. So, on
particular distribution we have chosen where
this distribution represents n 1 at e 1, n
2 at
e 2, n 3 at e 3 and so on. The number of ways
you could do that is the product of all this
omega of n i n 1 times omega of n 2 times
omega of n 3 times omega of n 4 and so on.
For each number of particles at that particular
energy level whatever e i. So, n 1 is also
at
e 1. So, that is the other thing we are looking
at. So, if you took you write this formula
down for every 1 of those energy levels and
they are corresponding number of particles
and number of states each 1,each of these
terms will look like this and if you multiply
all
those terms that is a number of ways in which
you can attained that micro state. So, that
is that is what we are looking at.
So, that is the number of ways in which you
can attained the micro state. So, this is
simply as I using the same notation that we
have done before i equal to 1 to r omega n
i.
So, this is simply pi over i of this, what
we have just written. This is omega n bar.
So, of
course, So, this is the expression that we
have. As we have always stated that for when
we use quantum mechanical steps, I mean when
we look at statistical mechanical
approach of dealing with such problems which
is what we are doing now. The idea is
that we would like to find the conditions
here which maximize this omega n bar.
So, that would represent the micro state with
the maximum number of ways that it is
possible to attain that micro state and that
would then represent the equilibrium state
of
the system because as we have discussed before
the most probable state is more probable
then all the other states combined. We discussed
that in detail before. So, will just accept
it now. So, that is what we are going to do.
So, we are going to try and maximize this
and
also in general if you look at omega n bar
the it is of the form that whether you maximize
this or the maximize the log of this, both
of them will give you the same result. The
2
functions are such that lone omega n bar and
omega n bar will behave in such a way that
when you maximize omega n bar lone omega n
bar maximizes. Similarly lone when lone
omega n bar reaches a maximum whatever is
that condition same condition omega n bar
would maximize and mathematically it is easier
to handle lone omega n bar.
So, that is what we will do. So, rewriting
it we have omega n bar is pi over I, n i plus
s i
minus 1 factorial by n i factorial times s
i minus 1 factorial. So, if you take the lone
of
this it is simply, now it becomes a somewhat
was product because you taken log natural
log, natural logarithm of this product you
get it becomes a sum, it is lone n i plus
s i
minus 1 factorial minus you can put this under
bracket because it is sum over the same i
is in for every time, it is the same i minus
lone n i factorial minus lone s i minus 1
factorial. So, this is what we have and in
general what is true is that given the values
of n
i and s i, n i plus s i is much larger than
1, in general n i plus s i will be much larger
than
1. So, n i plus s i minus 1 can be reasonably
approximated to n i plus s i because it is
larger than minus 1. The same will not be
true for this term, but for this term it will
be
true. So, therefore, here we will leave the
minus 1, here we will remove the minus 1
because n i plus n i is the very large number
generally. So, n i plus s i minus 1 is simply
n I, is simply n i plus s i.
So, we can do that here. So, this is simply.
So, we that is 1 thing, 1 change we will make
plus we will use Stirlings approximation which
simply says lone of x factorial is x lone
x
minus x. So, for every 1 of these term each
is a lone of some factorial, for every 1 of
those terms we will do this approximation
and x and we will change this n i plus s i
minus 1 to just simply n i plus s i. So, therefore,
this is simply equal to will write that
here lone of. So, this is n i plus s i. The
summation is still there. So, summation is
still
there we are only looking at the terms within
the bracket. Summation over i, n i plus s
i
lone n i plus s i minus n i plus s i 
minus you can put a bracket here curly bracket
n i lone
n i minus n i plus s i minus 1 lone s i minus
1 minus s i minus 1.
So, this is what we have. Now, if you look
at it there is a minus n i here and there
is a
minus of minus n i. So, that is a plus n i.
So, these n i and this n i will cancel. Again
there
is a minus s i here, there is a minus of minus
s i here. So, this s i will cancel. So, they
cancel out. So, we are just simply use Stirlings
approximation to get here and substituted
n i plus s i minus 1 as n i plus s i. So,
those are the changes we have made.
The summation is still there, this summation
that you see here that you see up here
continuous to remain there. So, that is what
we continue to have here. So, this is what
we
got. These terms go.
So, therefore, this is. So, we have the result
to be and then plus 1 will be there. Now,
you
will have minus of minus. So, its plus minus
1 here minus 1 there. So, they have 3 minus
here minus, minus and minus. So, you have
a minus 1 there. So, this is the result that
we
have got. Now, as we have always discussed
the you know if it if you are maximizing
this then effectively you know the differential
of this with respect to n I, the variable
you
have here is only n i.
So, if you maximizing this the differential
of this with respect to n i should be a 0
and it
we can also write it as simply as del lone
w n bar should be equal to 0 where you
effectively differentiating that with respect
to n i. So, if everything else is a constant
here
s i is a constant and of course, 1 is a constant.
So, now, when you differentiate it this is.
So, this is entirely a constant. 1 is a constant,
all of this term is a constant. So, this entire
things simply reduces to 0. So, we will only
have terms coming from the first term and
the second term here. This simply disappears.
So, if you write it down we get. So, this
implies that sigma over i, n i plus s i that
is
differentiate this term. So, that will be
one by n i plus s i times del n i because
that that n
i will be differentiated that is del n i and
you can also differentiate the n i here. So,
plus
del n i again s i is a constant. So, that
goes to 0 times lone n i plus s i. Now, you
take the
second term again you will have minus n i
by 1 by n i del n i and also minus del n i
lone
n i. So, that is the. So, this will give you
the 2 terms because you have 2 n i here and
this
also gives you 2 terms because you have 2
n i.
So, these are the 4 terms that you get, of
this if you see again this reduces to simply
plus
del n i because this will cancel. So, that
is plus Del n i and this reduces to minus
del n i
this will cancel. So, minus del n i. So, again
this entire term will cancel with this entire
term. So, this is go, this will go and so,
you basically have del lone. So, this implies
that
you know of course, this the fact that this
is equal to 0 simply implies whatever we have
calculated here equals 0. So, that simply
implies that the summation over i the only
terms
that are left are this Del n I, of lone n
i plus s i minus lone n i equals 0 and since
it is lone
you can actually simply put that in denominator.
So, this implies sigma over i with marginal
and put this just rearranging the terms, we
will have or will write it write the beginning
del n i lone n i plus s i by n i equals 0.
So,
we have got this combination. We have got
simply taking the number of ways in which
we can arrange this system and trying to maximize
it we have found this 1 condition that
needs to be satisfied which we have got as
del n i lone n i plus s i by n i equals 0.
So, we will write that down here. So, that
is our first equation sigma over i, del n
i times
lone of s i plus n i by n i equals 0. So,
this is the condition that we have good got.
So, this
is the equation 1. As before the total energy
in the system is conserved and the total
number of particles in the system is conserved.
Conserved meaning you cannot increase
or decrease this, the number of particles
is already there, you can you have no choice
in
it. It is not going to change. So, therefore,
some of all the particles sigma over i n i.
So, n 1 plus n 2 plus n 3 plus n 4 plus n
5 etcetera equals the total number of particles
equals a constant. Therefore, if you just
differentiate this with respect to n i that
has to be
0. So, that simply means sigma over i del
n i equal 0, that is the same as saying you
know
you have a total number of particles, you
simply remove a few particles from 1 state
you
distribute it amongst the other states. So,
what is the change? You removed some
particles. So, you have reduced. So, some
minus term comes there. So, the change in
particles there is some minus 5 particles
let us say because you removed 5 particles,
the
same 5 particles you put in like 3 other states.
So, therefore, you have you know plus 2,
plus 2 and plus 1 let us say.
So, minus 5plus 2 plus 2 and plus 1 should
give you 0 that is basically all it says.
Whatever particles you remove from 1 state
you have to add some other state. Therefore,
this sum of all the changes has to be 0. Similarly,
the energy is a constant total energy of
the system is a constant. So, epsilon i n
i equals constant which we designated as capital
E. So, again here the each energy level in
the system is already fixed you do not have
a
freedom on that. So, this is already fixed
for you. So, therefore, only change that you
can
do is here. So, again the changes in energy
that you will do when you remove some
particles is the energy level times the number
of particles that you remove and when you
add increase energy when you add it to some
other state it will be the energy of that
state
times the number of particles that you have
adding to it.
So, therefore, changes in energy total change
in energy is simply sum over i epsilon i del
n i for all the states what is the number
of particles you are adding or removing times
the
energy of that state that gives you the total
change in energy with respect to that energy
level. Sum of all those changes has to be
0 because the sum of all the energies together
is
any way equal to a constant. So, this is equal
to 0. So, we now have 3 equations we
already identified equation 1 here, this is
equation 2 and this is equation 3.
So, we have three equations and what we will
do is effectively as we have done before
when we looked at Maxwell Boltzmann statistics
and we also looked at the Fermi Dirac
statistics. What we will do is we will use
the Lagrange method of undetermined
multipliers and then we will solve this system.
So, we are going to use the Lagrange
method of undetermined multipliers. So, we
are actually simply going to multiply this
by
this by a constant alpha, this by a constant
beta and add it to this equation up here.
So,
that is what is going to be there and there
all equal to 0. So, that is sum is also going
to
be 0. So, that each of them is 0. So, it does
not matter what you multiply it with it is
going to remain 0 and we are going to add
it to this.
Only thing is by convention if because the
result works out more convenient to interpret
instead of simply adding to this equation
we add it to we add these 2 to minus of this
equation and that is only a convention simply
to make the interpretation easier and in a
sense it makes again no difference because
its anyway its equal to 0. So, whether you
put
a minus sign front of it or plus sign front
of it makes no difference. So, we will add
a
minus sign here and then we will add these
terms together.
So, or when we do this would when we do what
I just said when we are multiplying by 2
constants and adding them this is what we
will get, summation over i. We will write
the
first term down this same way minus will put
a bracket here minus lone n i plus s i by
n i
is bracket covers everything including the
n i plus alpha epsilon i I am sorry plus alpha
plus beta epsilon i del n i equals 0. So,
we had 3 terms we had sum of epsilon i del
n i
equals 0 we have multiplied that by beta and
brought it here.
We had as some of del n i equals 0 we have
multiplied that by alpha and brought it here
and I kept the original equation. So, it is
equation 1 plus alpha time’s equation 2
plus
beta time’s equation 3 which is what we
have done. So, that is equal to 0 and as we
discussed before the point is we have a bunch
of energy levels, we have a bunch of states
of those energy levels and when you maximize
it regardless of what changes you are
making in those states adding a little, subtracting
a little the sum should always be 0. In
other words this sum has be to 0 regardless
of the values of these n i because you can
keep making minor changes here and there still
it should be a 0. So, independent in other
words we have to make it independent of these
del n i and the only way you can do that .
So, therefore, the only way you can guarantee
that the sum is 0 regardless of where you
are adding, what you are adding, what minor
changes you have making how the only
way you can guarantee that this is 0 is to
say that every individual term here within
this
bracket has to become I mean this whatever
is within the bracket should always be 0.
Only when you do that you will you will be
able to guarantee that this sum is equal to
0.
So, therefore, we say lone of this is the
term within the bracket as long as you guarantee
that this term is 0 regardless of which energy
level you are looking at i, e i you are
looking at you can guarantee that this sum
remains 0 otherwise there is it is difficult
to
guarantee because you have all sorts of various
values of del n i and then it will become
difficult to get. This is only way you can
guarantee it 0. So, now, we have this expression
we will rearrange it marginally. So, basically
it says that lone of this is n i plus s i
by n i.
So, we this is we can simply write it as 1
plus s i by n i and we will move that other
2
terms on the other side. So, therefore, that
is why this minus sign is now suddenly
disappeared, is alpha plus beta epsilon i
and marginally and therefore, once you
simplified it further 1 plus s i by n i equals
e power alpha plus beta epsilon i and this
implies s i by n i equals e power alpha plus
beta epsilon i minus 1.
We started off by saying we want to know the
probability of occupancy of a state at
energy level e i that is simply the inverse
of this. We want n i by s i, we have now got
s i
by n i. So, that is what we want. So, we wants
actually n i by s i. So, n i by s i is simply
the inverse of this it is 1 by e power alpha
plus beta epsilon i minus 1. So, this equation
that you get here n i by s i is 1 by e power
alpha plus beta epsilon i minus 1 this equation
that we have got here.
This is the Bose Einstein distribution and
we designated also by f which subscripts b
e,
this tells you for every given energy level
what is a probability of occupancy of a state
at
that energy level. So, this is what it is.
Again this can be ah this alpha and beta can
be
interpreted as in terms of the chemical potentials
of the system and so on. So, that is not
something we are immediately interested in,
but this is the form of the equation that
we
get. This is the Bose Einstein statistical
distribution. So, when you look at phonons,
you
look at photons, you look at cooper pairs
they as long as they are to the extent that
they
are you know they are bosons, they are classified
as bosons they are following the Bose
Einstein statistical distribution which is
here.
So, through this course through this course
we have actually looked at 3 statistical
distributions. One was the Maxwell Boltzmann
distribution and that basically was that
we got we could write it as P equals, P of
E equals some P naught E to the power minus
E i by K p T if you just write it in the denominator
it will be E power E i by K p T. So,
this is the Maxwell Boltzmann distribution.
This applies to classical particles, we also
got the Fermi Dirac distribution. This works
for fermions. So, electrons would classify
as
fermions. There we got f of E is or f of E
is 1 by 1 plus E power E minus E f by K p
T.
So, this is where Fermi energy and so on we
have there.
So, this is what we got and now we have got
the Bose Einstein distribution. So, f of.
So,
will call this f f d Fermi Dirac and this
is Bose Einstein and that we have simply called
as
1 by 1 sorry E power alpha plus beta epsilon
i 
minus 1. So, this is a essentially a plus
1
and this is a minus 1.
So, so these are the three distributions that
we have got and I mean if you did not know,
only when you plot this up you realize that
they are very different and or when you
realize the kinds of assumption that have
gone it to them they look very different.
Some
wave similarity is there in terms of how they
layout, but they are big difference in how
they behave in actual behavior and In fact,
in a sense this is the Maxwell Boltzmann
distribution is considered the extreme approximation
of these two. So, under the right
kind of conditions these kinds of particles
would actually reduce to looking like classical
particles under the right kind of conditions,
but in general these are separate sets of
particles. So, they are very distinctly different
in how they behave, how they are
distributed across energy and therefore, what
you can expect from a system consisting of
particles of any one kind when they are subject
to certain experimental conditions.
Now, we have spoken about the Bose Einstein
distribution because we have already
encountered 3 bosons in our discussion and
there are some peculiarities of it about of
this
bosons which is what we will familiarizes
ourselves with the. As I mentioned the Bose
Einstein distribution since it consist of
a it is addressing a set of particles that
are
displaying quantum mechanical behavior, but
are exempt from the Pauli exclusion
principle there is no limit to the number
of particles you can place in a given state.
So, it
turns out that given this basic idea if you
actually go to very low temperatures you can
expect that all the particles will keep going
to lower and lower levels of energy and in
fact, can condense to a single state.
So, can condense to a single state. This kind
of situation where all the particles in the
system actually settle down to a single state
at very low temperatures is called Bose
Einstein condensate. So, this is called a
Bose Einstein condensate. So, as you go to
very
low temperatures the particles will settle
down to a single state and that is called
a Bose
Einstein condensate. Now, the beauty of the
Bose there are several things that have very
nice and interesting about the Bose Einstein
condensate, the first and I would say the
most startling thing about it is this was
predicted in 1925. So, when the Bose Einstein
statistics statistical distribution was put
I mean written down and introduced to the
world
at that point this the possibility that such
a thing could occur because of the fact that
this
it is allowed in that system that was propounded
by Bose Einstein, Bose and Einstein.
So, this is called the Bose Einstein condensate.
It was predicted in 1925.
For a very long time after that experimental
facilities where not available to enable
people to investigate the those bosons at
those very low temperatures. People were
unable to investigate these bosons at extremely
low temperatures to see whether or not
such a condensate forms. So, this was predicted
in 1925, only 70 years later in 1995, 70
years after the prediction was made that is
such a such a state of matter could exist,
it is
considered as a new state of matter 70 years
after it was predicted that such a state of
matter could exist. In 1995 in a collection
of rubidium atoms it was shown that when you
go to very low temperatures, when you talk
of very low temperatures please remember
this is not I mean already talking of 2, 3
Kelvin is very low temperatures, but where
they
are in this kind of an experiment they are
talking in terms of 10 power minus 7 Kelvin,
extremely close to absolute 0.
It is a huge experimental achievement to even
get a experimental set up where you can
reach this temperature. So, this kind it is
not a simple thing to reach this kind of a
temperature. So, phenomenal piece of work
to reach this kind of a temperature and in
1995 a group, I mean scientist where able
to actually accomplish this temperature with
and subject rubidium atoms to this kind of
a temperature. When they did that they were
able to get create this condensate. So, they
created this Bose Einstein condensate for
a set
of rubidium atoms. For doing this in 1995
the 3 people involved who were Cornell,
Wieman and Ketterle. Cornell, Wieman and Ketterle
they got their Nobel prize in 2001
they got a Nobel prize for accomplishing this
in 2001 for something that they did they
managed do in 1925.
So, till 1995 70 years after this state of
matter was predicted nobody was able to
accomplish this, with a lot of difficulty
they managed to reach this kind of a temperature
this group I mean this these scientist and
they were able to show in 1995 that this
condensate actually exists, that it can actually
be formed and for that they got the Nobel
prize. The point is that. In fact, if you
look at it for example, that there I would
like to
point out couple of things here 1 is just
a matter of historical interest and sort of
reflection of how sometimes things work in
science. Bose Einstein predicted this in
1925, 70 years later somebody managed to actually
show it, for doing this they got a
Nobel prize, but strangely enough Bose who
was involved in all this and who was very
instrumental in this, it was this original
idea strangely enough he never got a Nobel
prize
for the for his work. So, Satyendranath Bose
somehow missed getting a Nobel prize, but
people who later proved his prediction to
be correct actually got a Nobel prize. So,
that is
1 of these strange things about science and
just wants to point that out.
The other thing is actually when you look
at this kind of a temperature, I think it
is
important to understand what this represents.
So, if you see we spoke about black body
radiation, right at the beginning of a course
and through the course we have touched
upon it on a few occasions. Black body radiation,
if you look at the universe around you
people have done experiments and I mentioned
this earlier where you can looked at the
background radiation in the universe. Based
on the background radiation in the universe
you can say what is the current temperature
of the universe. The current temperature of
the universe is of the order of 2.7 Kelvin.
So, 2.7 Kelvin is the background temperature
of the universe. The current understanding
of the universe is that it is started off
at a very
high temperature and then it has been cooling
steadily.
The big bang theory that we talk of effectively
implies this kind of a situation with
respect to temperature, extremely high temperatures
were there in the first nano seconds
after the universe was formed and ever since
it has been cooling and what you have this
background temperature that we are talking
of is the current background temperature of
the universe. So, that that is in another
words this is how cool the universe has become,
anything else in the universe is only hotter
than this. Nothing in the universe in the
sense
in the sense of naturally occurring parts
of the universe whatever is occurred naturally
in
the universe over time has been cooling steadily
and this is the temperature that we have
in the background.
So, whenever you are doing some experiment
in the lab, if you manage to reach a
temperature lower than this in your experiment,
if you go to a temperature lower than
this which you will be doing artificially
you are doing using some compressors, you
are
using some pumps, you are using something
else, you are using much more sophisticated
equipment were used here, they used all sorts
of different means to slow down atoms
because the energy of the atom is effectively
its temperature. So, they used very
interesting methods to slow down the atoms
and get them and effectively lower their
temperature. So, whenever you do this complicated
thing you are doing it artificially,
you are imposing it on the system, you have
sort of forcing the system to undergo all
those things. So, whenever you draw below
this temperature in a sense you are dealing
with a situation that does not naturally exist
anywhere else in the universe.
That is a remarkable statement to make to
say that in my lab I have an experimental
situation which is not naturally occurring
anywhere else in the universe that too we
can
say with confidence is whenever you are doing
this kind of thing. If you are going to
higher temperatures and higher pressures invariably
there is you will find something in
the universe which is having that kind of
a temperature or pressure, but to get it to
below
this point you are already talking of something
that does not naturally exist anywhere. I
mean what it to the extent that if it exist
anywhere else chances are, that means, there
is
intelligence life elsewhere which is managing
to do this the same kind of experiment.
The fact that you are actually able to accomplishing
this in the lab means you have done
something that nature has not accomplished
of its own accord in the universe and this
is
just 2.7 Kelvin.
We are talking in here of 10 power minus 7
Kelvin which is remarkably lower than this
and as I mentioned you know in material science
and metallurgy and so on. We talk of
temperatures say from say minus 100 degree
c to few 1000 degree c that is a range of
temperatures that we routinely access, we
do not think too big about it, we just routinely
access that scale of temperatures, but if
you draw from about minus 100 degree c to
minus 273 degrees c which is what this 0 Kelvin
is. So, that is that looks that is only
about 173 degrees centigrade or of the order
of say you can even go to a narrower
window of the order of say 100 Kelvin, liquid
nitrogen will get it about 77 Kelvin. So,
from there if you want to drop down it starts
getting more and more difficult.
So, liquid helium will get you liquid hydrogen,
liquid helium will all get you into that 10
Kelvin or there about. So, from there to drop
down to 0 Kelvin the last 10 degrees starts
getting to be very difficult, it is not easy
to get you down the last 10 degrees. Suddenly,
as you start dropping in to the last few degrees
it is it is remarkable piece of work to get
to that kind of temperature. That is the reason
it took 70 years to go from whatever was
predicted to actually demonstrate it in the
lab. Simply, not being able to cool a system
to
that temperature that was the experimental
difficulty. So, you talk of you know highly
sophisticated experiments and what not, this
is highly sophisticated in experiment, but
what were they doing?
They were simply lowering the temperature,
that is the basic that is the fundamental
thing that they were doing. They were of course,
lot of other things that they were doing
but that is the fundamental thing that they
were doing, lowering the temperature. So,
lowering the temperature it itself a very
phenomena likes experimental accomplishment.
So, in this process they were able to show
the Bose Einstein condensate. So, to sum up
today what we have seen is a we have looked
at a bosons in much greater detail, we have
looked at the conditions that are relevant
to the particles that are classified as bosons.
We
have looked and we have used those conditions
and looked at the distribution across
energy levels, that the bosons will demonstrate
and so, we have therefore, worked our
way through the Bose Einstein statistical
distribution and we see we have also briefly
seen how it looks with respect to Fermi Dirac
statistics and Maxwell Boltzmann statistics
and we have also looked at this phenomenal
and very interesting scientific feed of some
prediction that was made in 1925 and then
experimentally proven 70 years later in 1995.
So, with that we will halt today. This sort
of covers our discussion on the bosons and
its
sort of relates to all the other things that
we have discussed in the last few classes.
So, we
will some we will halt here today at this
stage and we will pick it up in the next class
as
we look at a just a few more interesting points
before we wind up this course. Thank you.
.
.
