We shall study Bose Einstein Condensation.
This is particularly relevant in the context
of 125th birth anniversary of Essen Bose the
Indian ah scientist, ah whose been associated
with this Bose Einstein condensation. We will
say a few things about him ah as we go along.
So, this is a special topic and it is distinguished
from ah the earlier topics where we have talked
about interacting systems mostly. However,
this happens in a this condensation happens
in a system of ideal bosons which means that
without any interaction.
So, this is the ah picture of S N Bose and
of course, Einstein ah here. So, this is Albert
Einstein 
and so we both of them are a responsible for
these ah you know discovery of this phenomena
and will let us learn about them.
So, ah what happened or rather how ah S N
Bose got interested ah in these things? So,
what happened is that their ah discussion
on the emission of electromagnetic radiation
as a function of temperature. You must have
seen that when ah a piece of iron or a metal
is heated it initially ah becomes red hot
and then it becomes white hot and so on. So,
the spectrum of the incident radiation changes
its frequency or wavelength ah from one to
another and experimentally it is found that
the ah spectrum the incident radiation intensity
of the radiation versus this wavelength. So,
this is the wavelength if you cannot see it
ah these are on small phones and this is the
intensity. So, that looks like this. However,
there is a non monotonicity at a given value
of the wavelength.
And this ah was discovered experimentally
and the classical existing classical theories
predicted that ah either it is like this or
it is like this which are ah according to
the Wien's displacement law or and the Rayleigh's
law. However this non monotonicity nobody
ah got theoretically. So, Bose understood
that ah there are a rather in fact, Planck
proposed ah with just conviction and knowing
ah sort of theoretical backing; however, Bose
said that there has to be a new statistics
for the photons or the incident ah or the
emitted radiation and this ah laid the ah
the Bose Einstein statistics to be proposed
which he did ah in consultation with Einstein
and with help from Einstein. So, ah it said
that the emitted radiation has an energy dependence
ah which goes as h nu or in quantum of h nu
and this was a birth of quantum mechanics
in some sense. However, the statistics governing
this photons are ah the statistics is given
by the Bose Einstein distribution
So, let us ah look at how the condensation
phenomena come into the picture. So, say that
ah there are indistinguishable particles.
So, bosons are indistinguishable particles.
So, bosons photons are bosons phonons are
Bosons. So, they are indistinguishable and
let us see that how a simple counting procedure
can give rise to a condensation like phenomena.
So, let us take 2 boxes and 2 balls marked
by ah as A and B here and let us consider
them as classical particle or Maxwell Boltzmann
particles, so the particles over Maxwell Boltzmann
statistics which means that they are distinguishable.
So, ah if they are Boltzmann particles then
ah we can have A to be in 1. So, the first
one refers to the particle ah nomenclature
or the name of the particle and one is the
corresponding to the box index. So, A could
be in 1, B could be in 1 as well because there
is no restriction on the number of particles
to be occupying ah any quantum state ah or
it could be that A could be in 2, B could
be in 2 or it could be that A could be in
1 and B could be in 2 or it could be that
A could be in 2 and B could be in 1.
So, there are 4 possibilities and if you look
at these 4 possibilities there are these 2
possibilities are that they are together.
So, they are bunched up in the same box. So,
there are 2 out of 4 is the possibility of
them being together for a classical particle
or a set of classical particles which are
ah represented by the Maxwell Boltzmann statistics
which means they are distinguishable.
Now, coming to the Bose particles which are
indistinguishable we could have ah there is
no now, no hm difference between A and B it
is just for our reference that we have written
as A and B, but they are just both of them
to be say A. So, both of them to be in box
1 is one possibility because the bosons do
not have that restriction of occupying the
same quantum state as the fermions. Fermions
obey exclusion principle which we have said
a number of times, during the course of this
particular advanced condensed matter physics
ah and then ah both of them could be hm in
the second box ah remember these nomenclatures
are just hypothetical in the sense for our
own convenience. They are both A's. So, both
of them are in 1, both of them are in 2 or
both 1 is in 1 the other is in 2 or the the
reverse happens.
So, now if you look at the bunching probability
then you will see that 2 out of 3 are bunched
in the same quantum state. And so this is
the crux of Bose Einstein condensation that
the statistics says that if they ah number
of particles that is a large number of particles
can actually occupy one given quantum state
then ah they will bunch up or rather they
will occupy crowd in that quantum state and
that quantum state at very low temperatures
would be the ground state of the system. So,
a macroscopic accumulation of particles in
the ground state is what is known as the Bose
Einstein condensation.
So, this is what is written that A 1 B 1 and
A 2 B 2 get two-third of the weight as opposed
to half. So, this is equal to half which is
50 percent and this is two-third is equal
to 67 percent. So, there is a larger possibility
and this could actually happen in a many particle
sector for us to understand Bose Einstein
condensation of course, we are going to go
to details into that, but this is a very simple
idea summarizing that how in-distinguishability
of particles ah can lead to a condensation
phenomena.
So, ah the more weightage to situations where
particles occupy the same energy state is
the ah central idea or the central focus basic
idea of a BEC. And so, Einstein upon receiving
a note from ah S N Bose ah he understood that
there is a lot of merit in Bose's derivation
of the Planck's formula, and he says that
Bose's derivation of the Planck's formula
appears to me an important step forward and
the method used here gives also the quantum
theory of an ideal gas as I shall show somewhere.
So, ah pictorially let us see what happens.
So, there are a, there is a ground state of
a system which corresponds to E equal 0 and
these are the spectrum of the quantum of the
excited states ah and we have just shown them
almost like a continuous spectrum because
in an infinite system or a thermodynamic system
ah they could be infinitesimally close to
each other. The reason that we have shown
the ground state to be separate from the excited
state is something that I am going to discuss
later because ah of the density of states
going as a particular fashion as a function
of E, which is E to the power half or square
root of E ah this is getting 0 weight which
it should not. So, a priori you cannot assign
a 0 way to any given quantum state, so we
are going to consider this ground state separately
ah as compared to the other excited state.
Now, see say there are N particles N bosons
in a given system where N 0 would occupy the
ground state ah just to let you know that
the ground state has in principle infinite
occupancy. We will show that at at lower temperature.
And N ex is the number of particles occupying
all the excited states put together.
So, now the number of bosons N can actually
be tuned using temperature. It is a function
of temperature which comes out of the Bose
distribution function. Now, N 0 is as I said
at very low temperatures is practically infinite.
Now, incidentally depending on certain conditions
and I will also speak about those conditions
N ex may be finite in certain circumstances
and in fact, to get any x to be ah filed rather
to get N naught to be infinity you need to
go to very low temperature and precisely lowering
that temperature lowering t to the desire
value to 71 years I will tell you why 71 years
because 1924 was the theoretical prediction
of BC by boson Einstein.
And it was 1995 as written here ah experimental
realization of BC came in ah sub system of
rubidium atoms and for the first time this
actually show a BEC. So, a very large amount
of development has taken place since then
and the field of what is called as the ultra
cold atoms has developed enormously. And,
so if the one of the example is that, so a
beam of atoms in the same quantum state is
called as an atom laser ah and this is possible
one has actually made atom laser and all these
are based on the condensation phenomena that
was put forward by Bose and Einstein.
So, let us see how to make BEC. So, you have
to heat the atoms say to vedyam atoms at about
600 Kelvin to make a gas of them it is somewhat
counterintuitive because we are talking about
cooling and then first you have to heat so
that we have a very low density of atoms.
So, that it forms something like a gas and
ah then confine them in a beam. So, this is
a laser beam, so when you, they say a a system
of atoms. So, if you confine them in such
you know and also from top and bottom.
So, then these atom loses its motion and when
it loses its motion it cools down and the
kinetic energy becomes less, and the temperature
corresponding temperature becomes less and
this is what is meant by confining them in
a beam. Use a magnetic trap will speak a little
about them as well ah you can ah also talk
about an optical trap this is something that
these are part of laser cooling we talked
a little about laser cooling. And of course,
there evaporative cooling which is ah opening
the you know or rather reducing the depth
of the trap such that more energetic atoms
would escape.
And then finally there were achievements that
1998 which we are not too keen on ah; however,
this is helium 4 was achieved ah the cooling
effects are not very important the cooling
effects that we are going to talk about because
it happens at 4 Kelvin ah or say a few Kelvin
not 4 Kelvin, but a few Kelvin about 3 3 Kelvin,
3.13 Kelvin. Then in 1995 as we said that
Eric Cornell and Curl Wieman the people here
they ah created the BEC in rubidium 87 atoms
and then Wolfgang Ketterle in ah he created
BEC in sodium 23 in 1995 just 2 months later
after this ah ah discovery of rubidium. So,
all 3 of them got Nobel Prize in 2001, and
these are Kapitsa Allen and Misener these
are the people who have liquefied helium in
2000, ah in ah 1938.
So, this is that old picture that which is
connecting ah the 1925 story when the theory
was proposed and 1995 there are experimental
evidences. Will try to say a little on this,
but just a priori this is still not a condensate
here and it starts becoming a condensate and
this is a condensate here. So, this is not
a condenser 
and this starts becoming a condensate and
so on.
So, what happens is that. So, they have cooled
the atoms and then they release the trap in
which the atoms were held and when the trap
is released the atoms fly away and then a
high speed camera device images them and they
are the image is converted into Fourier transformed
into k space and there is a case phase picture.
So, a peak here corresponds to a macroscopic
accumulation of particles corresponding to
a given k value which is momentum value of
the particles and this case the momentum is
equal to k equal to 0. So, there is a ah macroscopic
accumulation of particles in a given quantum
state which is what BEC is all about.
So, just to summarize ah and the developments
that had you know ah come across in ah various
ah fields um or rather in this trapped ultra
cold atoms. So, 87 rubidium was cooled in
1995 and it was in Jila ah in ah by Eric Conell.
Then it was lithium in July 1995 in Rice University
by Randy Hulet, then sodium in September 1995
by Ketterle, hydrogen in ah ah MIT ah and
ah then of course, helium again in 2001 by
the group of Allen aspect.
So, now let us start ah discussing something
more ah you know concrete. So, that you understand
the ah concepts of a BEC. So, let us just
talk about the basic features of BEC. So,
that things are put in perspective.
We have said a few things, but will still
repeat it. So, the condensation occurs in
the momentum space. So, it is a K equal to
0 is ah macroscopically occupied. So, this
is an important key concept rather. So, it
is not a real space accumulation of particles.
So, it is ah K equal to 0 is occupied, and
it is very important that the particles are
non-interacting and we are talking about an
ideal Bose gas. In fact, whether the ah the
liquefaction of ah helium is a BEC that is
the question that one had to understand, because
helium is still an interacting ah fluid and
possibly. So, there are ah confusion whether
the lambda transition in helium actually falls
into the class of a BEC and you will not get
into that. Rather let us talk about a few
ah things such as the cooling techniques.
And to begin with let us talk about two, one
is called as a magnetic evaporative cooling
and the second is let us call it as a laser
cooling ok.
So, we will briefly describe both of them.
So, this is the magnetic evaporative cooling.
So, this you must be knowing that a system
of two coils where the currents are actually
moving in the same ah direction ah produces.
So, this is there are two coils where there
is the same current is moving in ah same direction
and the distance between the radii of these
two coils is say a or rather ah this is say
2 a. So, this is 0 to, so this is minus a
and this is at a. And then what happens is
that then the ah magnetic field variation
because there are current flowing in the coil
there will be generation of magnetic field.
Now, because there are 2 coils, the magnetic
field or the magnetic induction would be a
superposition of the effect from both the
coils. And this would be ah like 
symmetrically it will, so these fall off as
one over r whereas, it is fairly constant
here and this is a shear minus a to plus a.
So, this is the situation for a Helmholtz
coil or Helmholtz double coil as it is said.
Now, what ah person in MIT called David Pritchard
did he produced a similar coil with again
to such coils; however now, the currents are
flowing in different directions in both the
coils. So, one with respect to the other and
now you will have a magnetic field variation
to be having a minima, as opposed to a maxima
and which you can do a simple calculation
to see that this is of course, symmetric about
0 probably did not draw it as symmetric what
it is.
So, this if you can load the atoms here then
this acts like a trap for it it is a magnetic
trap for it. So, any atom which has more energy
will actually escape. So, atoms will escape
and leaving only say something like 10 to
the power 9 number of atoms in the system
ah which is still you know it is ah far away
from BEC, because BEC requires something around
10 to the power 6 number of atoms. But at
least this does one step ah and it makes a
lot of atoms which have which are energetically
ah more I mean they have higher energy they
energetically more favorable atoms would escape
and leaving behind the slower ones. So, we
have a gas of cool atoms and then so this
is a one technique, the other one has we say
that the laser cooling.
So, this is a very nice technique to slow
down atoms and hence a strip their kinetic
energies and thereby reducing the temperature.
So, what happens is that when an electromagnetic
radiation of frequency omega 
falls on an atom, atom whose energy levels
are given by omega naught or h cross omega
naught then this I will do without proof,
then the probability amplitude for absorption
is proportional to 1 divided by omega minus
omega not plus some i gamma by 2 where gamma
is the line width line width of the radiation
this is a natural line width.
So, this happens when the atom is at rest,
but when the atom is moving then the Doppler
effect comes into play and if ah we write
this. So, ah what happens when a Doppler effect
comes into place, that when the atom moves
towards light source, towards light source
which means electromagnetic radiation, so
the omega increases the effective omega increases.
And this is called as blue shift and when
the reverse happens that is atoms moves away
from light source omega decreases and this
is called as red shift.
So, ah this shift in frequency 
is say equal to delta omega which is of has
a form which is v dot k with or a minus v
dot k where v is the velocity of the atom
and k is the momentum. So, if you need to
slow down atoms then it is beneficial for
the atoms to be moving towards the light source
or the incident radiation. So, if you can
make the atoms ah move towards the light source
then they will slow down and slowing down
means the kinetic energy becomes less and
when the kinetic energy becomes less the equivalent
temperature because the energy is always ah
expressible in terms of temperature with the
Boltzmann relation they equal to k t. So,
the t decreases.
So, what happens is that the ah that the denominator
here. So, this denominator it takes a form
of omega minus omega naught minus v dot k
ah plus i gamma by 2 and so the quantity so
delta is equal to minus omega plus omega 0
rather omega naught minus omega 0 is called
as the detuning parameter.
So, ah the Doppler effect yields 
delta prime equal to delta plus v k, v our
v dot k and so basically now, an atom moving,
moving along to counter propagating laser
beams this is what I had shown in one of the
earlier slides that the atoms are actually
held between counter propagating laser beams.
Counter propagating means a propagating in
opposite directions with detuning. Then for
both of them 
for both ah delta 1 prime that is this one
is equal to delta plus ah v k. So, this is
a 0 angle between them. So, it is a v k cos
theta, but cos theta is equal to 0 because
they are moving towards each other and the
atom is moving along the line. So, there is
no angle that it is making and delta 2 prime
for the other beam it is equal to delta minus
v k.
So, it is clear that ah the some more energy
is actually absorbed from one of the beams
as corresponding to the other. So, we are
talking about the 2 laser beams. So, one of
them is absorbing more energy from the atom
as compared to the other, but this further
means that the atomic momentum is reduced
by twice of this k v divided by C, ok. So,
this forces ah exerted by the beams on the
atoms are not balanced. So, there is a resulting
in a net force ah opposite to v and thereby
reducing the kinetic energy and so, reduces
kinetic energy 
and causing cooling.
Now, this ends the story on cooling from our
side, but you understand that all these engineering
of having cold atoms possible in 70 years
and they are the best minds that were working
in the subject it is still took a very long
time, for ah getting them down to a temperature
which is extremely small and BEC really happens
in ah sort of micro Kelvin in the micro Kelvin
range. However, well do a calculation and
well show that ah actually for ah the ah atoms
such as helium the the condensation or rather
the condensation like temperature is just
about a few Kelvin maybe 2 to 3 Kelvin. However,
understand that actually the rubidium atoms
required a temperature to be ah of the order
of a micro Kelvin and so let me write that
the temperature is a of the order of finally,
and there is only a part of the cooling techniques
that I have talked about there are the sisyphus
cooling, and ah ah and sympathetic cooling
and so on, and so this temperature comes to
about 10 to the power minus 6 Kelvin and the
number of atoms or number of particles that
should be left is about 10 to the power 6.
So, this is the condition for BEC to be 
formed in lab and. However, as I said that
will not, ah will not discuss the exact magnitude
of the temperature, but well do an order of
magnitude calculation.
Now, since we are done with the cooling part
let us understand that why ah is that ah we
are talking about bosonic atoms when the atoms
are actually new neutral or rather the charge
neutral and if we are talking about the electronic
ah the spin then that electronic spin is always
half and these alkali atoms rubidium being
an alkali atom. So, this that that should
correspond to electrons having spin half and
that cannot be put in the same class as bosonic
atoms. So, let us understand why they are
called as bosonic atoms.
We are just talking about alkali atoms because
more it started with alkali atoms as I have
shown you rubidium, sodium etcetera potassium
lithium again. So, what happens is that ah
even if the electrons have ah the spin half
it is the nuclear spin that comes into picture,
but because the nuclear magnetic moment, ah
nuclear magnetic moment ah mu N is far lower
than mu e which is the nuclear magnetic moment
for electrons by at least of 4 orders of magnitude.
And thus to have nuclear spins to play an
important role we really need very low temperature.
So, let us say that ah F is the atomic spin
I is the nuclear spin 
and J is the electronic spin. So, f is equal
to I plus J and so it has a range ah I plus
J to I minus J and for J equal to ah which
is a electronics spin equal to half. So, we
have F and for rubidium ah ok, let us for
all the alkali atoms 
such as rubidium, sodium, lithium etcetera
we have I which is the nuclear spin is 3 by
2. So, the atomic spin F is equal to ah 1
or 2, so they are bosonic atoms with integral
spins. And as I said for the nuclear spin
to play a role ah the temperature has to be
low.
Now, let us start doing some calculations
in order to understand the condensation phenomena
in an ideal Bose gas. So, let us start with
the ideal Bose gas.
And the discussion is ah you can follow a
statistical mechanics book very good statistical
mechanics book by Pathria ah which gives a
very good ah description of this phenomenon
ah before we proceed let us write down the
grand canonical partition function. So, the
word ah grand canonicals grand canonical partition
function is that where ah we also allow in
a in addition to exchange of energy among
the particles ah we also allow for exchange
of particles between the system and the bath.
So, it is the canonical is about exchanging
energies between the system and the bath and
this is also in addition to that allowing
the number of particles. So, the number of
particles is not constant. So, a grand canonical
partition function as you know that the partition
function is written with ah Z and this Z comes
from a a German word probably it is called
as; there could be a problem with the spelling,
but it means zustandssumme means that ah it
is a sum over states. And this G, ah subscript
G ah that corresponds to the ah grand canonical.
So, this is equal to ah N equal to 0 to infinity
as we just said that will allow the particles
to be exchanged between the system and the
bath, so that is that term and exponential
beta mu N and this is equal to an exponential
minus beta epsilon i n i and sum over all
states let us ah you know you can write n
i and so on. ah
So, this is the canonical partition function
and. So, it is basically ah N equal to 0 to
infinity exponential ah beta mu N Z C is pretty
much the formula for the the grand canonical
partition function. And you can follow path
via to see that the grand canonical partition
function the log of that which is related
to the free energy a minus K T of that is
equal to F which is equal to a PV. So, this
is equal to ah PV over K T and which is equal
to a minus sum over i log of 1 minus Z f l
right for Z f is and beta epsilon i and so
on.
So, let us make this notations all clear.
So, P is equal to pressure ah v 2 be volume
ah t is of course, temperature, ah Z f is
called as the fugacities which is equal to
exponential beta mu beta is 1 over K T and
mu is the. So, let us write it here mu is
chemical potential 
and ah beta equal to 1 over K T. And, so this
is by and large the expression for this. So,
this can actually be the equation of state
for an ideal boost gas because we are writing
PV by K T which you know for a classical ideal
gas PV by K T is equal to some r or ah it
is n r or something like that.
So, it is a constant whereas, in an ideal
boost gas at quantum gas which follows an
ah in distinguishability constraint is given
by this. And similarly the number of particle
which is equal to sum over i n i this is equal
to sum over i and a 1 divided by ah Z f inverse
or you can simply write it as exponential
minus beta mu and exponential minus beta epsilon
i ah plus ah this is minus 1 sorry. So, for
boson is minus 1 and for fermions it is equal
to plus 1 epsilon i's are the single particle
states. So, what I mean by signal particle
states is that if there are free particles.
So, there they will go as the in k space they
will go as h cross square k square over 2
m.
Now, before we proceed what we have to do
is that we have to convert these or rather
these are the working equations for any system
whether you are talking about condensation
or you are talking about studying other ah
thermodynamic or statistical mechanics mechanical
properties you have to start with these equations
for both fermions and boson.
Now, we need in order to calculate them we
need to, ah calculate them analytically we
need to convert them into integrals. And you
know when a sum is converted to an integral
you need the density of states you need the
number of states ah in the energy range E
and E plus B E and that is an important quantity
in condensed matter physics as I might have
already told a number of times because ah
this is what creates a difference between
a 2 dimensional the properties of a 2 dimensional
system with that of a 3 dimension because
the density of states go down. In fact, the
nano the whole branch of nanoscience and nanotechnology
when we talk about quantum wires which means
that we are ah confining electrons or the
charge carrier in one dimension ah while in
quantum dots we do it in 0 dimension and in
a 2 dimensional electron gas this is in 2
dimension and so on.
So, ah we need the density of states in order
to calculate or rather convert the sums into
integrals. Now, this is I will not go into
that, but it is a very simple calculation
first course of solid state physics would
do it that the density of states in short
called as ah DOS.
So, that goes as epsilon to the power half
for a epsilon equal to h cross square k square
over 2 m this is in 3D it goes as constant
also let us write it as a epsilon to the power
0 which is also which is a constant in 2D
and it goes as epsilon to the power minus
half in 1D ok. So, for this dispersion a free
particle dispersion.
So, if you see this that the density of say,
so we are talking about a 3 dimensional system
with k square dispersion. So, we are talking
about a d-dimensional system, where d equal
to 3. So, d equal to 3 for us and we have
a dispersion which is s going as, dispersion
going as k to the power s, so s is equal to
also, s is equal to 2 for us. In fact, whether
a BEC will occur or not will crucially depend
upon this value d and s and a you can actually
convert it into a single parameter. So, either
you call it as by d or d by s. So, depending
on certain ah values of s by d or d by s it
is Bose Einstein condensation is possible.
So, we convert using this density of states,
we convert the ah the first equation let us
call them as first equation and second equation
and let us write 1 gives. So, 1 means equation
1 and so this is equal to ah minus 2 pi over
h cube ah 2 m whole to the power 3 by 2 m
is the mass of the particle mass of the bosons
here ah and this comes from the density of
states we have not written here a a coefficient
and that coefficient would involve all these
ms and so on m h etcetera etcetera. And as
I said its epsilon to the power half ah d
epsilon and I ah convert the Z f inverse and
exponential beta epsilon minus 1.
Now, I should have actually done it ah for
ah all epsilons, but however, you see if we
have epsilon to the power half and for the
ground state which is epsilon equal to 0,
you would attach 0 weight. This is what I
was talking about earlier that you cannot
have a 0 weight assigned to a given state
then that state is completely unimportant
for our ah computation. So, we will have to
take that state out and write it for ah for
that state will write this thing as ah rather
it is a minus sign ah it is a minus 1 by V
ah log of 1 minus Z f. So, this is that contribution
for the epsilon equal to 0 and this is for
epsilon not equal to 0.
Now, ideally I in this integral we should
ah separate out the epsilon equal to 0, but
since this is an integral just having one
point less does not make any difference in
this integral. So, we can still put it as
ah 0 to infinity. So, so the integral becomes
0 to infinity and this integrand and then
this is the epsilon equal to 0 component of
that equation equation number ah 1.
So, now let us call this as equation number
3, and 2 gives in the same manner ah N by
V is equal to 2 pi over h cube ah 2 m by 3
by 2, and a 0 to infinity just in the same
way and d epsilon and you have a Z f inverse
exponential beta epsilon minus 1 and the plus
1 over V, Z f by 1 minus Z f let us call this
as equation 4.
Now, if you look at these 2 equations we have
clearly as separated them into two terms,
one corresponding to epsilon equal to 0 the
contribution to the pressure corresponding
to epsilon equal to 0 and the contribution
to the density of particles or the number
of particles ah to be ah f for epsilon equal
to 0 and for epsilon not equal to 0. This
is in one of my slides that I have shown the
ground state energy to be separate as compared
to the excited state energies.
So, now, let us see the second terms of both
of both 3 and 4.
So, see what the second terms are. Let us
write that second term as this its equal to
Z f ah 1 minus Z f and that is equal to N
naught by V. You understand N naught, N naught
is the occupancy or the number of particles
that the ground state can hold and this is
equal to exponential beta mu 1 minus exponential
beta mu which is nothing, but equal to exponential
minus beta mu minus 1. So, this is the N 0
by V that is the second term in equation 4.
So, this is 4.
Now, when ah your Z f is much smaller than
1, now Z f is actually between 0 and 1 ok.
So, it can take a maximum value 1 and can
take a minimum value 0. So, if Z f is close
to 0 that is a smaller much smaller than 1
then we have, this corresponds to very large
temperature. And if it ah talks about very
large temperature then ah you have ah ah the
number of particles ah is very low because
this f is equal to f is a very small and so
N 0 is very small and of course, that corresponds
to a no Bose Einstein condensation because
you have a classical physics that is taking
over. So, thermal effects should you know
drive all the particles away.
Now, you try to understand that as you reduce
the temperature N 0 increases. So, at very
low temperature N 0 is very high and so as
ah Z f goes to 0. So, that is, so mu is actually
a chemical potential which is negative. So,
if Z f goes to 0 ah then ah N 0 by V is significantly
large. In fact, it is so large that your N
equal to N ex plus N 0 somehow if you can
show that your N ex is equal to a finite number
then since N 0 is infinitely large. So, if
your N is infinite or rather very large then
ah all the particles will go to the ground
state because it is has infinite occupancy
whereas, a very smaller number of particles
would actually go to the ah the excited states
because it has limited occupancy.
Now, if you look at 3 then you will see that
this is equal to the the 0 epsilon equal to
0 contribution is equal to this and this tells
that for a Z f is much smaller then this is
equal to negligible. So, this minus 0 over
V log of this thing is ah very small that
of course, we know that the pressure due to
the all the particles that would go to the
ground state is very small because there is
no almost no particles at large temperatures
this is the ah the limit for large temperature.
However if you ah go to the other limit that
is Z f going to 1, I am sorry the Z f should
go to one here not not 0. So, Z f is 1. So,
if Z f goes to 1, so this is for Z f going
to 1. So, Z f going to 1; however, this thing
would take a form which looks like 1 over
V ah log of N naught. Now, N naught could
be large, but log of that would be you know
still small. So, the second term in equation
3 can still be neglected even if at low temperature
that is Z f going to 1, of the fugacities
going to 1; however, that cannot be neglected
in equation 4. This is the main central message
of this discussion; that even the pressure
contribution from all the particles in the
ground state of the system could be infinitesimally
small; however, the number of the number density
is significant.
So, now ah well write this. So, basically
our 3 becomes ah its P over K T its equal
to minus 2 pi over h cube or 2 m K T whole
to the power 3 by 2 and 0 to infinity x to
the power half. So, ah its log of um 1 minus
Z f exponential minus x, dx, where x equal
to exponential minus sorry x equal to beta
E, that is the thing.
And so beta has come out and the 4 a gives
let us call this as maybe 5 and 4 gives N
minus N 0 by V, which is N ex by v E x, N
ex is the occupancy or the number of particles
in the excited states and this is equal to
2 pi 2 m K T by h cube ah and this is equal
to, so this is 3 by 2 and this is equal to
x to the power half d x ah Z f inverse exponential
x minus 1.
Now, these are called as the oh science stein
integral and a g n of z. So, this these are
Bose Einstein integral lifts in short called
as BE integral and this BE integral takes
a form its equal to 1 minus gamma N ah x to
the power N minus 1 you can write it as Z
f and this is equal to dx divided by 0 to
infinity and its equal to Z f inverse exponential
x minus 1. So, these integrals look very similar
to that excepting for its N equal to say ah
3 by 2 and so on. So, and this is equal to
ah it can be shown ah this I leave it to you
it is a matter of you know doing a partial
integral integration, sorry integration by
parts of this integral and a express it in
the form of this and finally, what one gets
is, so let us call this a 6.
So, 5 becomes equal to P over K T its equal
to 1 over lambda cube and g 5 by 2, Z f. We
are less interested in this formula though
it is important nevertheless; however, we
are interested in this formula the one that
we are going to write later. So, let us call
this as 7 and this is equal to N ex by V which
is equal to 1 by lambda cube g 3 by 2 Z f.
That is the second equation that is equation
6. So, 6 yield this.
So, our the number of particles which is what
we wanted in the excited states is given by
some quantity which is this where lambda is
equal to ah root over h over h divided by
root over 2 pi m ah 2 m K T sorry not 2 pi
m, 2 m K T it is called as a thermal de Broglie
wavelength. So, this is equation a thermal
de Broglie wavelength.
Now, we are almost done, we have obtained
an equation for the number of particles in
the excited states. If this quantity is finite
is not infinitely large which would depend
upon certain criteria then we are done we
would again leave this thing and show that
your g n, Z f is equal to sum over l equal
to 1 to infinity, it is a Z f to the power
l, l to the power n which is equal to Z f
plus Z f square by 2 to the power n plus Z
f cubed by 3 to the power n and so on.
Now, for small z that is that if much smaller
than one the classical limit you can be satisfied
with Z f, ok. So, if you put that equal to
a Z f then of course, that becomes ah will
depend on the ah the the excited state occupancy
ah is ah is you know is goes as exponential
ah beta mu.
However, at low temperature that is when Z
f goes to in the limit goes to 1, you have
to take all of these terms into consideration
and cannot stop at a finite terms, but fortunately
this is equal to for g 3 by 2 said f g 3 by
2 and Z f equal to 1 which is of interest
to us this is interesting to us and this its
equal to a Riemann zeta function which is
3 by 2 which has a value 2.612.
So, this is a finite value this is what we
were hinting at time and again that the excited
state occupancies which are coming out from
this equation number 8 has an the excited
state occupancy has a finite value. So, if
there are more number of particles they will
all go to the ground state. If there are a
macroscopically large number of particles
they would all go to the ground state and
this is what the condensation. So, N greater
than you know a V T to the power 3 by 2 and
2 pi m k divided by 3 by 2 h cube etcetera
it becomes equal to a 3 by 2 and the psi is
ah this is not the way to write it is like
this, it is called as the Riemann zeta function.
So, the Bose Einstein integral is ah related
to the Riemann zeta function when you take
this entire sum. The entire sum has a closed
form which is called as Riemann zeta function
and for an argument equal to 3 by 2 it has
a value equal to 2.612. So, if N is ah greater
than, this N is the total number of particles
then BEC occurs and for the BEC to occur ah
the critical condition is that your N by V
has to be equal to some T C 3 by 2 it will
happen at that ah value. So, this is equal
to a k ah 3 by 2 and by h cube and this value
which is equal to 2.612.
So, if you put everything there then what
we get is that ah a T C has an expression
which is equal to ah h square by 2 pi m k
and ah its equal to N by V and its equal to
3 by 2 and its equal to a 2 by 3. So, this
is the expression for T C which means the
temperature if its lowered below this then
all the particles will go to the ground state
and the excited state because the excited
states have a a finite occupancy they will
avoid the excited states and go to the ground
state because the temperature is also very
slow.
Now, this we will just do it for helium 4
which as we said that is not a a prototype
case for Bose Einstein condensation, but still
if you take this values 10 to the power minus
24 gram. And so the density or the inverse
density which is called as a specific volume
V by N equal to 27.6 as centimeter cube by
per mole ah then if you put all these things
T C comes out to be equal to 3.13 Kelvin and
13 Kelvin and this is close to the observed
lambda point transition 
of helium liquid helium, liquid helium ah
which is equal to 2.17 Kelvin.
So, there was a initially misconception that
the ah liquification of helium is actually
or rather they heal the lambda point transition
in helium is actually a BEC transition. We
will not elaborate on that, but what we have
got is a condition for the Bose Einstein condensation
also we have explained various steps that
are associated with the cooling process and
finally, when the cooling happens the whole
ah atom the the 10 to the power 6 atoms number
of atoms.
They ah form a structure ah which has very
low ah temperature which is like 10 to the
power minus 6 to minus 7 Kelvin which is like
a less than a nano, I mean about ah micro
Kelvin temperature which is the probably the
coldest temperature in the universe and they
are imaged as I said that releasing the traps
and letting them fly apart. So, when they
fly they are imaged and they are finally,
Fourier transformed into k space to see there
is a macroscopic accumulation of particles
in the k space.
So, this is a k space phenomena. It is a real
momentum space accumulation of particles B
ah the BEC is a a a example of that and so
this the credit goes to Bose and Einstein
who have proposed this in 1924, nearly a 100
years ah from now, 100 years earlier. However,
as I said the realization had take place a
very large number of years.
