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PROFESSOR: We looked at the macro state, where
the volume of the box, the temperature were given,
as well as the chemical potential.
So rather than looking at a fixed number of particles,
we were looking at the grand canonical prescription
where the chemical potential was specified.
We said that when we have a system
of non-interacting particles, we can construct
a many bodied state based on occupation of single particle
states such that the action of the Hamiltonian on this
will give me a sum over all single particle
states, the occupation number of that single particle
state times the energy of that single particle state, and then
the state back.
We said that in quantum mechanics,
we have to distinguish between bosons and fermions,
and in the occupation number prescription.
For the case of fermions, corresponding theta v minus 1
and k was 01.
For the case of bosons, corresponding theta
plus 1 and k equals [INAUDIBLE].
We said that within this planned canonical prescription,
these occupation numbers were independently given.
The probabilities were independently
distributed according to exponential rules
such that when we calculated the average, nk.
We commented on the statistics what
are the very simple forms, which was 1 over [INAUDIBLE]
into the beta epsilon k minus eta,
where z was constructed from the temperature
and the temporal potential as [INAUDIBLE] to the beta mu.
Now, for the case that we are interested,
for this gas that is in a box of volume v,
the one particle states were characterized
by [INAUDIBLE] waves and their energies
were f h bar squared k squared over 2m, basically just
the kinetic energy.
And then at various stages, we have
to perform sums over k's, and for a box of volume v,
the sum over k was replaced in the limit
of a large box with an integral over k times
the density of states, which was v divided by 2 pi cubed.
And we said that again, we recognized
that quantum particles have another characteristic, which
is their spin, and there is a degeneracy factor associated
with that that we can just stick over here
as if we had [INAUDIBLE] copies corresponding
to the different components of the spin.
Now, once we put this form of epsilon k over here,
then we can calculate what the mean number of particles
is in the grand canonical ensemble as sum
over k, the average nk, given the appropriate statistics.
Of course, this is a quantity that is extensive.
We then construct a density, which is intensive,
simply by dividing by v, which gets rid of that factor of v
over here, and we found that the answer could
be written as g over lambda cubed, again,
coming from this g, lambda cubed from the change of variables,
where we defined lambda and h over root 2 pi mkp.
And then a function that we indicated by s 3/2,
depending on the statistics of z.
And this was a member of our class of functions
we defined in general, f m h of z,
to be 1 over m minus 1 factorial, integral 0
to infinity, ex, x to the m minus 1, z inverse,
e to the x minus eta.
So for each m, we have a function of z.
In fact, we have two functions, depending
on whether we are dealing with the fermionic
or the bosonic variety.
We saw that these functions we could expand
for small z as z plus eta z squared over 2 to the m.
It's an alternating series for fermions.
All terms are positive for bosons, zq, 3 to the m,
eta z to the fourth, 4 to the m, and so forth.
And we also noted that these functions
have a nice property in that if you
were to take a derivative with respect to z and multiply by z,
you get the function with one lower index.
So this is, in fact, the same thing
as z divided by zz of f m plus 1 [INAUDIBLE].
This just made the connection between the number of particles
and the chemical potential that we
need to use when we go to the grand canonical prescription,
but we are interested in calculating various properties
of this gas, such as the pressure.
We found that the formula for the pressure
was simply g over lambda cubed, very similar to what
we had over here, except that, rather than having f 3/2,
we had f 5/2 of z to deal with, and if we
were interested in energy, we could simply use 3/2.
Indeed, this was correct for both fermions and bosons,
irrespective of the difficulties or the variations
in pressure that you would have, depending
on your bosons or fermions due to [INAUDIBLE] gases.
So what is the task here, if you are
interested in getting a formula for pressure
or the energy or the heat capacity?
What should we do if we are interested in a gas that
has a fixed number of particles or a fixed density?
Clearly, the first stage is to calculate
z as a function of density.
What we can do is to solve the first equation graphically.
We have to solve the equation f 3/2 8 of z
equals the combination m lambda cubed over g,
which we call the degeneracy factor.
So if I tell you what the temperature is,
you know lambda.
If I tell you what the density is,
you know the combination on the right hand side.
There is this function.
We have to plot this function and find
the value of the argument of the function.
That gives the value of m lambda cubed over g
as the value of the function.
So graphically, what we have to do
is to plot as a function of z these functions, f 3/2.
I will generically plot f m 8 of z,
but for the case that we are interested,
we really need to look at m equals 3/2.
As we will see in problem set, if you
were to solve the gas in d dimensions,
this m rather than the 3/2 would be g over 2.
So pictorially, the same thing would
work with slightly different forms
of these functions in different dimensions.
How do these functions look like?
Well, we see that initially, they start to be linear,
but then, the case of the bosons and fermions, the next order
term goes in different directions.
And in particular, for the case of fermions,
the quadratic term starts to bring you down,
whereas for the case of bosons, the quadratic term
is the opposite direction and will tend to make you large.
This is what these functions look.
For the case of fermions, I have additional terms in the series.
It's an alternating series.
And then the question is, what happens
to this at large values of z?
And what we found was that at large values,
this function satisfies the asymptotic form
that is provided by Sommerfeld formula, which
states that f m of z, for the case of fermions, so this is
eta equals minus 1, is log z to the power
of m divided by m factorial.
Of course, we are interested in m goes to 3/2.
1 plus pi squared over 6 m, m minus 1,
log z squared, and higher order inverse powers of log z, which
one can compute, but for our purposes,
these two terms are sufficient.
And if I'm interested in doing the inversion for fermions,
what I need to do is to plot a line that corresponds
to n lambda cubed over g and find
the intersection of that line, and that
would give me the value of z.
When I'm very close to the origin,
I start with the linear behavior,
and then I can systematically calculate the corrections
to the linear behavior, which would have been z equals n
lambda cubed over g in higher powers of n lambda cubed
over g.
And we saw that by doing that, I can gradually
construct a visual expansion that
is appropriate for these gases.
I can do the same thing, of course, for the case of bosons
also, but now I'm interested in the limiting form where
the density goes higher and higher or the temperature
goes lower and lower so that this horizontal line that I
have drawn will go to larger and larger values,
and I'm interested in the intersections that I have
when z is large and this asymptotic formula is
being satisfied.
So what do I need to do for the case of degenerate fermions?
I have this quantity, n lambda cubed over g, which I claim
is much larger than 1, so I have to look
for values of the function, f 3/2 minus of z that are large,
and those values are achieved when the argument is large,
and then it takes the form log z to the 3/2 divided
by 3/2 factorial, 1 plus pi squared over 6.
m in this case is 3/2, so I have 3/2 times m
minus 1, which is 1/2, divided by log z over 2.
And then I will have potentially higher order terms.
So I can invert this formula to find log
z as a function of density.
And to lowest order, what I have is
3/2 factorial n lambda cubed over g raised to the 2/3 power.
If I were to ignore everything here,
just arrange things this way, that's
the formula that I would get.
But then I have this correction, so I divide by this.
I have 1 plus-- this combination is equivalent to pi squared
over 8, 1 over log z squared.
I have taken it to the other side, so I put the minus sign
and raise again to the 2/3 power.
Now, the way that I have defined z up there, clearly log z
is proportional to beta nu.
We can see that lambda is proportional
to beta to the 1/2, so lambda cubed
is beta to the 3/2, which, when raised to the 2/3,
gives me a factor of beta.
So basically, the quantity that we have over here
is of the order of beta, and then we
saw that the remainder, which depends on properties
such as the mass of the gas, gh, et
cetera, which is independent of temperature, is a constant.
That is the Fermi energy that we can
compute by usual route of filling up a Fermi c.
And the value of epsilon f is h bar
squared over 2m times kf squared,
and the kf was 6 pi squared n over g to the 1/3 power,
so kf squared would be something like this.
So once we know the density of the fermion, its spin
and its mass, we can figure out what this quantity is.
The zeroth order solution for the chemical potential
is clearly this epsilon f.
What do I have to do if I want to calculate
the next correction?
Well, what I can do is I can put the zeroth order
solution for log z in this expression,
and then that will give me the first order correction.
I have to raise this whole thing to the minus 2/3 power,
so pi squared over 8 becomes 1 minus 2/3
of pi squared over h, which is minus pi squared over 12.
1 over log of z squared-- log z is again mu over kt,
so its inverse is kt over the zeroth order
value of mu, which is epsilon f raised to second power,
and we expect that there will be higher order terms.
So what have I stated?
If I really take this curve and try
to solve for mu as a function of temperature for a given
density, I find that at very low temperatures,
the value that I get is epsilon f,
and then the value starts to get reduced
as I go to higher temperatures and the initial fall
is quadratic.
Of course, the function will have higher and higher order
corrections.
It will not stay quadratic.
And I also know that at very large values,
it essentially converges to a form that is minus
kt log of n lambda cubed, because down here, I
know that this solution is z is simply n lambda cubed over g.
So at low densities and high temperatures, basically,
I will have something that is almost
linear with logarithmic corrections and negative,
and so presumably, the function looks something like this.
Of course, I can always convert.
We can say that what I see here is
the combination of kt over epsilon f.
I can define epsilon f over kv to be something
that I'll call tf, and this correction
I can write as t over tf squared.
So basically, I have defined epsilon f
to be some kv times a Fermi temperature.
Just on the basis of dimensional argument,
you would expect that the place where this zero is occurring
is of the order of this tf.
I don't claim that it is exactly tf.
It will have some multiplicative factor,
but it will be of the order of tf,
which can be related to the density, mass,
and other properties of the gas by this formula.
And if you look at a metal, such as copper,
where the electrons are approximately described
by this Fermi gas, this tf is of the order of, say,
10 to the 4 degrees Kelvin.
Yes?
AUDIENCE: So you were saying that at a high temperature,
the chemical potential should go as minus kvt
times ln of n lambda cubed, yeah?
PROFESSOR: Yes.
AUDIENCE: Now, the lambda itself has temperature dependence,
so how can you claim that it is behaving linearly?
PROFESSOR: I said almost linearly.
As you say, the correct behavior is something like minus 3/2 t
log t, so log t changes the exponent
to be slightly different from 1.
This is not entirely linear.
It has some curvature due to this logarithm.
In fact, something that people sometimes get confused at
is when you go to high temperatures, beta goes to 0.
So why doesn't z go to 1?
Of course, beta goes to 0 but mu goes
to infinity such that the product of 0 and infinity
is actually still pretty large, partly because
of this logarithm.
AUDIENCE: Could you just explain again why it's linear?
PROFESSOR: Down here it is linear,
so I know that z is n lambda cubed over g.
z is e to the mu over kt.
So mu is kt log of n lambda cubed over g.
Remember lambda depends on temperature,
so it is almost linear, except that really what is inside here
has a temperature dependence.
So it's really more like t log t.
AUDIENCE: When z is small?
PROFESSOR: When z is small and only when z is small,
and z small corresponds to being down here.
Again, that's what I was saying.
Down here, beta goes to zero but mu
goes to minus infinity such that the product of 0 and minus
infinity is still something that is large and negative because
of this.
AUDIENCE: If the product is large, then isn't z large?
PROFESSOR: If the product is large and negative,
then z is exponentially small.
In fact, if I were to plot this function just
without the higher order corrections,
if I plot the function, kt log of n lambda cubed over g,
you can see that it goes to 0 when the combination n
lambda cubed over g is 1 because log of 1 is 0.
So that function by itself, asymptotically this
is what I have, it will come and do something like this,
but it really is not something that we are dealing with.
It's in some sense the classical version.
So classically, your mu is given by this formula,
except that classically, you don't
know what the volume factor is that you have to put there
because you don't know what h is.
So that's the chemical potential,
but the chemical potential is really,
for our intents and purposes, a device
that I need to put here in order to calculate
the pressure, to calculate the energy, et cetera.
So let's go to the next step.
We are again in this limit.
Beta p is g over lambda cubed f 5/2 plus z.
That's going to give me g over lambda cubed in the limit where
z is large-- we established that z is large-- log z to the 5/2
divided by 5/2 factorial using the Sommerfeld
expansion, 1 plus pi over 6.
Now my m is 5/2.
m minus 1 would be 3/2 divided by log z squared
in higher order terms.
AUDIENCE: Question.
PROFESSOR: Yes?
AUDIENCE: Why is that plus?
PROFESSOR: Because I made a mistake.
Just to make my life and manipulations easier,
I'll do the following.
I write under this the formula for n
being g over lambda cubed, f 3/2 minus z, which
is g over lambda cubed log z to the 3/2
divided by 3/2 factorial, 1 plus-- I calculated
the correction-- pi squared over 6 divided
by various combination, which is pi squared over 8,
1 over log z squared.
Now what I will do is to divide these two equations.
If I were to divide the numerator by the denominator,
what do I get?
The left hand side becomes beta p over n.
The right hand side, I can get rid of the ratio of log
z 5/2 to log z.
3/2 is just one factor of log z.
What do I have when I divide 5/2 factorial by 3/2 factorial?
I have 5/2.
This division allows me to get rid of some unwanted terms.
And then what do I have here?
This combination here is 5 pi squared over 8 minus 1 pi
squared over 8.
Because of the division, I will put a minus sign here,
so that becomes 4 pi squared over 8,
which is pi squared over 2.
So I will get 1 plus pi squared over 2,
and at the order that we are dealing,
we can replace 1 over log z squared
with kt over epsilon f squared.
Now, for log z over here, I can write beta mu,
and for mu, I can write the formula that I have up here.
So you can see that once I do that, the betas cancel
from the two sides of the equation, and what I get
is that p is 2/5, the inverse of 5/2.
The n I will take to the other side of the equation.
To the lowest order mu is epsilon
f, so I will put epsilon f, but mu is not exactly epsilon f.
It is 1 minus pi squared over 12, kt over epsilon f squared.
And actually, there was this additional factor of 1
plus pi squared over 2, kt over epsilon f squared.
So what do we find?
We find that to the zeroth order,
pressure is related to the density
but multiplied by epsilon f, so that even at zero temperature,
there is a finite value of pressure left.
So we are used to ideal gases that are classical,
and when we go to zero temperature,
they start to stop moving around.
Since they are not moving around,
there is no pressure that is exerted, but for the Fermi gas,
you cannot say that all of the particles are not moving around
because that violates this exclusion of n being zero
or one.
You have to give particles more and more momentum
so that they don't violate the condition that they should have
different values of the occupation numbers, the Pauli
exclusion.
And therefore, even at zero temperature,
you will have particles that in the ground state
are zipping around, and because they're moving around,
they can hit the wall and exert pressure on it,
and this is the value of the pressure.
As you go to higher temperature, that pressure gets modified.
We can see that balancing these two terms,
there is an increase in pressure, which
is 1 plus 5 pi squared over 6, kt over epsilon f squared.
So you expect the pressure, if I plot it
as a function of temperature, at zero temperature,
it will have this constant value, pf.
As I put a higher temperature, there
will be even more energy and more kinetic energy,
and the pressure will rise.
Eventually, at very high temperatures,
I will regain the classical, rather
pressure is proportional to temperature.
That's the pressure of this ideal Fermi gas.
The energy.
Well, the energy is simply what we
said over here, always 3/2 pv.
So it is 3/2, the pressure over there multiplied
by v. When I multiply the density by v,
I will get the number of particles.
3/2 times 5/2 will give me 3/5, so I will 3/5 and epsilon
f at the lowest order.
Again, you've probably seen this already
where you draw diagrams for filling up a Fermi c.
Up to some particular kf, you would
say that all states are occupied.
When epsilon f is the energy of this state that
is sitting right at the edge of the Fermi c,
and clearly, the energies of the particles
vary all the way from zero up to epsilon f,
so the average energy is going to be less than epsilon f.
Dimensionally, it works out to be 3/5 of that.
That's what that says.
But the more important part is how
it changes as a function of temperature.
What you find is that it goes to 1 plus 5 pi squared over 6,
kt over epsilon f squared in high order terms.
AUDIENCE: Question.
Should the 6 be 12?
PROFESSOR: 6 should be 12?
Let's see.
This is 6/12 minus 1/12 is indeed 5/12.
Thank you.
So if I ask what's the heat capacity at constant volume,
this is dE by dT, so just taking the derivative.
The zeroth order term, of course, is irrelevant.
It doesn't vary the temperature.
It's this t squared that will give you the variations.
So what does it give me?
I will have n epsilon f.
I have 3/5 times 5 pi squared over 12.
The derivative of this combination
will give me 2kb squared t divided by epsilon f squared.
There will be, of course, higher order,
but this combination you can see I can write as follows.
It is extensive.
It is proportional to n.
There is one kb that I can take out, which is nice,
because natural units of heat capacity,
as we have emphasized, are kb.
This combination of numbers, I believe the 5 cancels.
3 times 2 divided by 12 gives me 1/2,
so I have pi squared over 2.
And then I have the combination, kt over epsilon
f, which I can also write as t over tf.
So the heat capacity of the Fermi gas
goes to 0 as I go to zero temperature,
in accord with the third law of thermodynamics, which
we said is a consequence of quantum mechanics.
So we have this result, that the proportionality
is given by the inverse of tf.
If I were to plot the heat capacity in units of nkb
as a function of t, at very high temperatures,
I will get the classical result, which is 3/2,
and then I will start to get corrections to that.
Those corrections will gradually reduce this,
so eventually, the function that I will get starts linearly
and then gets matched to that.
So the linear behavior is of the order of t over tf,
and presumably at some temperature
that is of the order of tf, you will
switch to some classical behavior.
And as I said, if you look at metals, this tf is very large.
So when you look at the heat capacity of something
like copper or some other metal, you
find that there is a contribution from the electrons
to the heat capacity that is linear.
Now, the reason for this linear behavior
is also good to understand.
This calculation that I did is necessary
in order to establish what this precise factor of pi squared
over 2 or appropriate coefficient
is, but the physical reason for the linearity you
should be able to know.
Basically, we saw that the occupation numbers
for the case of fermions have this form that
is 1 over z inverse, e to the epsilon plus 1.
If I plot the occupation numbers as a function of energy,
for a case that is at zero temperature,
you can see that the occupation number is 1 or 0,
so you have a picture that is like this.
You switch from 1 to 0 at epsilon f.
Oops, there was a beta here that I forgot, beta epsilon.
When I go to finite temperature, this becomes fuzzy
because this expectation value for the number,
rather than being a one zero step function,
becomes smoothed out, becomes a function
such as this, and the width over which this smoothing takes
place is of the order of kt.
So what happens is that rather than having this short Fermi c,
you will start having less particles here
and more occupied particles over here.
To do that, you have created energies
going from here to here that you have stored
that is of the order of this kt.
What fraction of the total number
have you given this energy?
Well, the fraction is given from here.
It is the ratio of this to the entirety,
and that ratio is t, or kt, divided
by epsilon f is the fraction that
has been given energy this.
So the excitation energy that you have
is this number, the total times this fraction
times the energy that you have, which is kt.
And if I take the derivative of this,
I will get precisely this formula over
here up to this factor of pi squared over 2,
which I was very cavalier, but the scaling everything
comes from this.
And to all intents and purposes, this
is the characteristic of all Fermi systems,
that as you go to low temperatures,
there is a small fraction of electrons, if you like,
that can be excited.
That fraction goes to 0 at low temperatures
as kt over epsilon f, and the typical energies that they have
is of the order of kt.
So essentially, many of the results for the degenerate
Fermi gas you can obtain by taking classical results
and substituting for the number of particles
nkt over epsilon f.
You will see in the problem set that this kind of argument
tells you something also about the magnetic response
of a system of electrons, what is the susceptibility
and why does the susceptibility saturate at low temperatures.
It's just substituting for the number of particles
this fraction, nkt over epsilon f,
will give you lots of things.
This picture is also valid in all dimensions,
whereas next, we'll be discussing the case of bosons,
where the dependence is on temperature.
There is an exponent here that determines
which dimension you are or depends on dimensions.
For the case of fermions, this linearity exists independently.
So now let's think about bosons and what's
going to happen for bosons.
For bosons, we saw that the series, rather than
being alternating, all of the terms
are adding up together so that the parabolic correction,
rather than reducing the function
for the case of bosons, will increase it.
So for a particular value of density,
we find that the z of fermions is larger than the z of bosons.
So for a given density, what we will
find is that the chemical potential, whereas for the case
of fermions was deviating from the classical form by going up,
for the case of bosons, we'll start
to deviate from the classical form and go down.
Now, there is something, however,
that the chemical potential cannot do for the case
of bosons which it did it for the case of fermions,
and that was to change sign.
Why is that?
Well, we have that over there, the occupation number
for a boson is 1 over z inverse, which
is e to the beta epsilon k minus mu.
z inverse is e to the minus meta mu minus 1.
Now, this result was obtained from summing
a geometric series, and the condition certainly
is that this object, that is, the inverse of what
was multiplying different terms in the series,
has to be larger than 1.
Of course, it has to be larger than 1
so that I get a positive occupation number.
If it was less than 1, the geometric series
was never convergent.
Clearly, the negativity of the occupation number
has no meaning.
So this being positive strictly immediately implies
that mu has to be less than epsilon k for all k.
So it has to be certainly less than the minimum of epsilon k
with respect to all k.
And for the case that I'm looking at where my epsilon
k's are h bar squared k squared over 2m,
the lowest one corresponds to k equals 0,
so mu has to be less than 0.
It can never go to the other side of this.
Or alternatively, z has to be less than
or approaches maybe 1.
So there is certainly a barrier here
that we are going to encounter for the case of bosons at z
equals 1.
We should not go beyond that point.
So let's see what's happening for z equals 1
to this function.
You see, my task remains the same.
In order to find z, I have to solve graphically
for the intersection of the curve that corresponds
to f plus 3/2 and the curve that corresponds to the density n
lambda cubed over g.
So what happens as I go to higher and higher values
of n lambda cubed over g?
You can see one possibility is that this curve just
diverges as z approaches 1.
Then for every value of n lambda cubed over g,
you will find some value of z that will gradually
become closer and closer to 1.
But is that the scenario?
For that, we need to know what the value of this f function
is at z equals 1, so the limit f m plus of z as z goes to 1.
How do you obtain that?
Well, that is 1 over m minus 1 factorial, integral 0
to infinity, dx, x to the m minus 1, z inverse, which is 1,
e to the x minus 1.
I have to integrate this function.
The integrand, how does it look like?
x to the m minus 1, e to the x minus 1.
Well, at large x, there is no problem.
It goes to 0 exponentially.
At small x, it goes to 0 as x to the m minus 2.
So basically, it's a curve such as this
that I have to integrate.
And if the curve is like I have drawn it, there is no problem.
I can find the integral underneath it
and there actually is a finite value that has a name.
It's called a zeta function, so this is some tabulated function
that you can look at.
But you can see that if m, let's say, is 0,
then it is dx over x squared.
So the other possibility is that this
is a function that diverges at the origin, which
then may or may not be integrable.
And we can see that this is finite and exists only
for m that is larger then 1.
So in particular, we are interested in the case
of m equals 3/2.
So then at z equals 1, I have a finite value.
It is zeta of 3/2.
So basically, the function will come up to a finite value at z
equals 1, which is this zeta of 3/2,
which you can look up in tables.
It's 2.612.
Now, I tried to draw this curve as if it comes and hugs
the vertical line tangentially, that is, with infinite slope,
and that is the case.
Why do I know that?
Because the derivative of the function
will be related to the function at one lower index.
So the derivative of f 3/2 is really an f 1/2.
f 1/2 does not exist.
It's a function that goes to infinity.
So essentially, this curve comes with an infinite slope.
Now, it will turn out that this is the scenario
that we have in three dimensions.
If you are in two dimensions, then what you need to do
is look at f that corresponds to m equals 2 over 2 or 1,
and in that case, the function diverges.
So then, you have no problem in finding some intersection point
for any combination of n lambda cubed over g.
But currently in three dimensions,
we have a problem because for n lambda cubed
over g that falls higher than zeta of 3/2,
it doesn't hit the curve at any point,
and we have to interpret what that means.
So for d equals 3, encounter singularity
when n lambda cubed over g is greater than
or equal to zeta of 3/2.
This corresponds at a fixed density to temperatures
that are less than some critical temperature that depends on n,
and that I can read off as being 1 over kb.
This is going to give me a combination.
Lambda cubed is proportional inversely to 3/2,
so this will give me nz 3/2 over g to the 2/3,
and then I have h squared 2 pi m.
I put the kb over here.
So the more dense your system is,
the lower-- I guess that's why I have it the opposite way.
It should be the more dense it is, the higher the temperature.
That's fine.
But what does that mean?
The point is that if we go and look at the structure
that we have developed, for any temperature that
is high enough, or any density that is low enough, so that I
hit the curve on its continuous part,
it means that I can find the value of z that is strictly
less than 1, and for that value of z,
I can occupy states according to that probability,
and when I calculate the net mean occupation,
I will get the actual density that I'm interested in.
As I go to temperatures that are lower,
this combination goes up and up.
The value of z that I have to get gets pushed more towards 1.
As it gets pushed more towards 1, I see that mu going to 0,
the state that corresponds to the lowest energy, k equals 0,
gets more and more occupied.
But still, nothing special about that occupation except what
happens when I am at higher values.
The most natural thing is that when I am at higher values,
I should pick z equals 1 minus a little bit.
And if I choose z to be 1 minus a very small quantity, then
when I do the integration over here,
I didn't get a value for the density, which
is g over lambda cubed, the limiting value
of this f function, which is zeta of 3/2.
I will call this n star because this is strictly
less than the total density that I have.
That was the problem.
If I could make up the total density with z equals 1,
I would be satisfied.
I would be making it here, but I'm not
making it up with the spectrum that I have written over here.
But on the other hand, if epsilon is incredibly small,
what I find is that the occupation number of the k
equals to 0 state-- let's write it n of k equals 0.
What is that?
It is 1 over z inverse, which is the inverse of this quantity,
which is 1 plus epsilon, which is approximately
e to the epsilon minus 1.
Actually, maybe what I should do is to write it in the form e
to the beta mu minus 1 and realize that this beta mu is
a quantity that is very small.
It is the same thing that I was calling epsilon before.
And if it is very small, I can make it to be 1 over beta mu.
So by making mu arbitrarily close to the origin or epsilon
mu arbitrarily close to 1, I could in principle
pump a lot of particles in the k equals 0 state,
and that does not violate anything.
Bosons, you can put as many particles as you like in the k
equals 0 state.
So essentially, what I can do is I can make this, but you say,
well, isn't this already covered by this curve
that you have over here?
Are you doing something different?
Well, let's follow this line of thought a little bit more.
How much do I have to put here?
This n is the total number of particles divided by volume,
and so this is going to be this quantity, n star, which
is gv over lambda cubed divided by volume.
Let's write it in this fashion, z over lambda cubed zeta of 3/2
and then whatever is left over, and the leftover
I can write as n0.
So what we know is if we take z equals 1,
this function will tell me how many things
I have put in everything except k equals 0
that I will calculate separately.
That amount is here.
And I will put some more in k equals 0,
and you can see that the amount I
have to put there is going to be n minus g over lambda
cubed zeta of 3/2.
Actually, I will have to put a volume here.
Why is that?
The reason I have to put the volume is because the volume
that I had here, that ultimately I
divided the number of particles by volume to get the density,
came from replacing the sum with an integration.
So what I have envisioned now is that there are all of these
points that correspond to different k's.
I'll replace the sum with a integration,
but then there was one point, k equals 0,
that I am treating separately.
When I'm treating that separately, it
means that is really is the same as the total number
expectation value without having to divide by the volume, which
comes from this density of state,
and I have something like this.
Now, the problem is that you can see that suddenly, you
have to pick a value of mu that is inversely
proportional to the volume of the system.
This is a problem in which the thermodynamic limit
is taken in this strange sense.
You have all of these potential values of the energy, epsilon
of k, that correspond to h bar squared, k squared over 2m.
There is one that is at 0, and then choosing k equals 2 to pi
over l, you will have one other state, another state,
all of these states.
All of these states are actually very finely spaced.
The difference between the ground state
and the first excited state is h bar squared over 2m,
2 pi over l squared.
So this distance over here is of the order of 1 over l squared.
But I see that in order to occupy
this with the appropriate number,
I have to choose my chemical potential to be as close as 1
over the volume.
So in the limit where I take the size of the system
go to infinity, this approaches this.
It never touches it, but the distance that I have here
is much, much less than the distance
that I have in the spacings that I made this replacement.
So I can indeed treat these separately.
I can give a particular weight to this state
where the mu has come this close to it as 1 over v,
treat all of the other ones as part
of this replacement of the summations with the integral.
You can also see that this trick is going to be problematic
if I were to perform it in two dimensions, because rather
than 1 over l to the third power,
that is, the volume in three dimensions,
I would have had 1 over l squared.
And indeed, that's another reason
why two dimensions is special.
And as I said, in two dimensions,
the curve actually does go all the way to infinity.
You won't have this problem.
So this is essentially what happens
with this Bose-Einstein condensation,
that the occupation number of the excited state for bosons
is such that you encounter a singularity in three
dimensions beyond a particular density or temperatures
lower than a certain amount or highly degenerate case,
you have to separately treat all the huge number of particles
that is proportional to the volume that have now piled up
in the single state, k equals 0, the ground state in this case,
but whatever the ground state may
be for your appropriate system, and all
the other particles go in the corresponding excited states.
Now, as you go towards zero temperature,
you can see that this combination
is proportional to temperature goes to zero.
So at 0 temperature, essentially all of the particles
will need to be placed in this k equals 0 state.
So if you like, a rough picture is that the net density.
This is the density, n, at high temperature
is entirely made up as being part of the excited state.
Let's draw it this way.
When you hit Tc, you find that the excited state can no longer
accommodate the entire density.
This is a coordination that, as we said,
goes to zero at zero temperature as T to the 3/2.
So there's a curve that goes like T to the 3/2,
hits 1 at exactly Tc, which is a function of the density
that you choose.
This is the fraction that corresponds to the excited
states.
And on top of that, there is a macroscopic fraction
that is occupying the ground state, which
is the compliment to this curve.
Basically, it behaves something like this.
One thing that I should emphasize
is that what I see here makes it look like at high temperatures,
there is no occupation of k equals 0 state.
That is certainly not correct because if you
look at this function at any finite z
as a function of epsilon, you can
see that the largest value of this function
is still at epsilon equals 0.
It is much more likely that there
is occupation of the ground state than any other state,
except that the fraction that is occupying here,
when I divide by the total number, goes to zero.
It becomes macroscopic when I'm below the BEC transition.
Now, the properties of what is happening in this system
below Tc is actually very simple because below Tc,
the chemical potential in the macroscopic sense
is stuck at 1.
Essentially, what I'm saying is that in this system,
the chemical potential comes down.
It hits Tc of n, and then it is 0.
Now of course, 0, as we have discussed here,
if I put a magnifier here and multiply by a factor of n
or a factor of volume, then I see
that there is a distinction between that and 0.
So it doesn't quite go to 0, but the distance
is of the order of 1 over volume or 1
over the number of particles.
Effectively, from the thermodynamic perspective,
it is 0.
But again, for the purposes of thermodynamics, pressure,
everything comes from the particles
that are moving around.
The particles in the ground state are frozen out.
They don't do anything.
So the pressure is very simple.
Beta p is g over lambda cubed, f 5/2, eta of z
equals 1, which is some number.
It is g over lambda cubed, the zeta
function at 5/2, which again has some value that you can read
of in tables, 1.34, something like that.
So you can see that this lambda cubed is inversely
proportional to t to the 3/2, so pressure
is proportional to t to the 5/2.
If I were to plot the pressure as a function of temperature,
what I find is that at low temperatures,
the pressure is simply given as t to the 5/2.
And you can see that this pressure knows nothing
about the overall density of the system
because this is a pure number.
It is only a function of temperature, mass, et cetera.
There is no factor of density here.
So what is happening?
Because we certainly know that when
I am at very high temperatures, ideal gas behavior
dictates that the pressure should
be proportional to temperature times density.
So if I have two different gases where the density here
is greater than the density here,
these will be the form of the isotherms
that I have at high temperatures.
So what happens is that you will start with the ideal gas
behavior.
We saw that for the case of bosons,
there is some kind of an effective attraction that
reduces the pressure, so the pressure starts to go down.
Eventually, it will join this universal curve
at the value that corresponds to Tc of n2.
Whereas if I see what's happening with the lower
density, at high temperatures, again, I
will have the linear behavior of the ideal gas with a lower
slope because I have a lower density.
As I go to lower temperatures, quantum corrections
will reduce the pressure.
Eventually, I find that this curve
will join my universal curve at the point that would correspond
to Tc of n1, which in this case would be lower.
But beyond that, it will forget what the density was,
the pressure would be the same for all of these curves.
And I have kind of indicated, or tried to indicate,
that the curves join here in a manner
that there is no discontinuity in slope.
So an interesting exercise to do is
to calculate derivatives of this pressure
as a function of temperature and see
whether they match coming from the two sides.
And you will find that if you do the algebra correctly,
there is a matching of the two.
Now, if I have the pressure as a function of temperature,
then I also have the energy as a function of temperature, right?
Because we know that the energy is 3/2 PV.
So if I know P, I can immediately
know what the energy is.
I know, therefore, that in the condensate,
the energy is proportional to T to the 5/2.
Take a derivative, I know that the heat capacity
will be proportional to T to the 3/2.
That T to the 3/2 is a signature of this condensate
that we have.
Again, heat capacity will go to 0 as T goes to 0, as we expect.
And as opposed to the case of the fermions,
where the vanishing of the heat capacity was always linear,
the vanishing of the heat capacity for bosons
will depend on dimensions, and this vanishing as T to the 3/2
is in general T to the d over 2, as can
be shown very easily by following this algebra.
I wanted to do a little bit of work
on calculating the heat capacity because its shape is
interesting.
Let's write the formula a little bit more accurately.
I have 3/2 v. Pressure is kTg over lambda cubed, f
5/2 plus of z.
This formula is valid, both at high temperatures
and at low temperatures.
At high temperatures, z will be varying
as a function of temperature.
Let me also write the formula for the number of particles.
The number of particles is V times g over lambda
cubed f 3/2 plus of z.
Once more, I can get rid of a number of things
by dividing these two.
So the energy per particle is 3/2 kT, f 5/2
of z divided by f 3/2 of z.
Let's make sure I didn't make any mistake.
Now, I want to calculate the heat capacity, let's say,
per particle, which is d by dT of the energy per particle.
Now, I have one factor out here, which is easy to evaluate.
There's an explicit temperature dependence here,
so I will get 3/2 kb.
I will have f 5/2 plus of z divided by f 3/2 plus of z.
Something here that I don't like.
Everything that I have written here is clearly valid only
for T greater than Tc because for T less than Tc,
I cannot write n in this fashion.
So what I'm writing for you is correct for T greater than Tc.
Point is that there is also an implicit dependence
on temperature because z is a function of temperature.
So what I can do is I can do 3/2 kbT,
and this is a function of z, and z is a function of temperature.
So what I have it is dz by dT done at constant volume
or number of particles times the derivative of this function
with respect to z.
Now, when I take the derivative with respect to z,
I know that I introduce a function
with one lower derivative up to a factor of z.
So I divide by the 1 over z, and then
the derivative of the numerator is f 3/2 plus divided by f 3/2
plus in the denominator minus the derivative
of the denominator, which is f 1/2 plus of z
times the numerator divided by f 3/2 plus of z squared.
So what you would need to evaluate in order
to get an expression that is meaningful
and we can eliminate z's as much as possible is what dz by dT
is.
Well, if this is our formula for the number of particles
and the number of particles or density is fixed,
then we take a derivative with respect to temperature,
and I will get that dN by dT, which is 0, is v.
And then we have to take the temperature
derivative of this combination, g over lambda
cubed, f 3/2 plus of z.
Now, lambda cubed scales like 1 over t to the 3/2,
so when I take a derivative of this,
I will get 3/2 T to the 1/2, which I can again
combine and write in this fashion, 1 over T,
f 3/2 plus of z.
Or then I take the implicit derivative that I have here,
so just like there, I will get dz by dT at constant density
times 1 over z.
The derivative of f 3/2 will give me f 1/2 plus of z.
Setting it to zero, we can immediately
see that this combination, T over z dz by dT
at constant density over the number of particles
is minus 3/2, f 3/2 plus divided by f 1/2 plus of z.
So what I need to do is to substitute this over here.
And then when I want to calculate the various limiting
behaviors as I start with the high temperature and approach
Tc, all I need to know is that z will go to 1.
f 1/2 of 1 is divergent, so this factor will go to 0.
This factor will disappear, and so then these things
will take nice, simple forms.
Once you do that-- we will do this more correctly next time
around-- we'll find that the heat
capacity of the bosons in units of kb
as a function of temperature has a discontinuity at Tc of n.
It approaches the classical result,
which is 3/2 at high temperatures.
At low temperatures, we said it is simply proportional
to T to the 3/2, so I have this kind of behavior
just a simple T to the 3/2 curve.
And one can show through arguments,
such as the one that I showed you above,
that the heat capacity is continuous
but its derivative is discontinuous at Tc.
So the behavior overall of the heat capacity of this Bose gas
is something like this.
So next time, we'll elucidate this a little bit better
and go on and talk about experimental realizations
of Bose-Einstein condensation.