Download Grand-canonical ensembles

Grand-canonical ensembles
As we know, we are at the point where we can deal with almost any classical problem (see below),
but for quantum systems we still cannot deal with problems where the translational degrees of
freedom are described quantum mechanically and particles can interchange their locations – in such
cases we can write the expression for the canonical partition function, but because of the restriction
on the occupation numbers we simply cannot calculate it! (see end of previous write-up). Even for
classical systems, we do not know how to deal with problems where the number of particles is not
fixed (open systems). For example, suppose we have a surface on which certain types of atoms can
be adsorbed (trapped). The surface is in contact with a gas containing these atoms, and depending
on conditions some will stick to the surface while other become free and go into the gas. Suppose
we are interested only in the properties of the surface (for instance, the average number of trapped
atoms as a function of temperature). Since the numbers of atoms on the surface varies, this is an
open system and we still do not know how to solve this problem.
So for these reasons we need to introduce grand-canonical ensembles. This will finally allow us to
study quantum ideal gases (our main goal for this course). As we expect, the results we’ll obtain at
high temperatures will agree with the classical predictions we already have, however, as we will see,
the low-temperature quantum behavior is really interesting and worth the effort!
Like we did for canonical ensembles, I’ll introduce the formalism for classical problems first, and
then we’ll generalize to quantum systems. So consider an open system in contact with a thermal and
particle reservoir.
system:
E,V,N,T,p, µ
reservoir: ER , VR , NR ,TR , pR , µ R
Figure 1: Model of an open thermodynamic system. The system is in contact with a thermal and particle reservoir
(i.e., both energy and particles get exchanged between the two, but E + ER = ET = const, N + NR = NT = const).
The reservoir is assumed to be much bigger than the system, E ≪ ER , N ≪ NR . In equilibrium T = TR , µ = µR .
As we know (see thermo review), the macrostate of such a system will be characterized by its
temperature T (equal to that of the reservoir), its volume V and its chemical potential µ (which will
also be equal to that of the reservoir) – again, I am using a classical gas as a typical example.
We call an ensemble of very many copies of our open system, all prepared in the same equilibrium
macrostate T, V, µ, a grandcanonical ensemble. As always, our goal is to find out the relationships that hold at equilibrium between the macroscopic variables, i.e. in this case to find out how
U, S, p, hN i, ... depend on T, V, µ. Note that because the number of particles is no longer fixed, we
can only speak about the ensemble average hN i (average number of particles in the container for
1
given values of T, V, µ) from now on.
1
Classical grand-canonical ensemble
As was the case for the canonical ensemble, our goal is to find the density of probability ρg.c. (N, q, p)
to find the system in a given microstate – once we know this, we can compute any ensemble average
and answer any question about the properties of the system.
Note that since the number of microsystems (atoms or whatever may be the case) that are inside
the system varies, we will specify N explicitly from now on: a microstate is characterized by how
many microsystems are in the system in that microstate, N , and for each for these microsystems we
need f generalized coordinates and f generalized momenta to describe its behavior, so we have a
total of 2N f microscopic variables q, p.
So we need to figure out ρg.c. (N, q, p). We will follow the reasoning we used for canonical ensembles
– have a quick look over that and you’ll see the parallels. In fact, I’m going to copy and paste text
from there, making only the appropriate changes here and there.
First, we reduce the problem to one that we already know how to solve, by noting that the total
system = system + reservoir is isolated, and so we can use microcanonical statistics for it. The
microstate of the total system is characterized by the generalized coordinates N, q, p; NR , qR , pR ,
where the latter describe the microsystems that are inside the reservoir. Clearly, N + NR = NT is the
total number of microsystems in the total system, which is a constant. So in fact the total system’s
microstate is characterized by: N, q, p; NT − N, qR , pR .
Its Hamiltonian is HT (N, q, p; NR , qR , pR ) = H(q, p) + HR (qR , pR ) – here I won’t write the
dependence of the Hamiltonian on N explicitly, but we know it’s there: for example, in the kinetic
energy we sum over the energies of the microsystems inside the system, and the number of terms
in the sum is N , i.e. how many microsystems are contributing to the energy in that particular
microstate.
Based on the microcanonical ensemble results, we know that the density of probability to find the
total system in the microstate N, q, p; NT − N, qR , pR is:
R
R
ρmc (N, q, p; NT −N, q , p ) =
(
1
ΩT (ET ,δET ,V,VR ,NT )
0
if ET ≤ HT (N, q, p; NT − N, qR , pR ) ≤ ET + δET
otherwise
where ΩT (ET , δET , V, VR , NT ) is the multiplicity for the total system. This is the same as saying
that the total probability to find the system in a microstate with N microsystems that are located
between q, q + dq and p, p + dp AND the reservoir to have the remaining NT − N microsystems
located between qR , qR + dqR and pR , pR + dpR is:
dqdpdqR dpR
ρmc (N, q, p; NT − N, qR , pR ).
GN GNT −N hN f +(NT −N )fR
Note that even though the microsystems are the same, they may have different number of degrees
of freedom in the system and the reservoir – think about the previous example with atoms adsorbed
on a surface. It is very likely that we need different degrees of freedom to describe the state of an
atom when on the surface (inside the system) than when it’s in the gas (the reservoir). However, the
total number of atoms on the surface plus inside the gas is fixed, and that’s all that matters.
We do not care/need to know what the reservoir is doing, all we want to know is the probability
that the system is in a microstate which has N microsystems between q, q + dq and p, p + dp. To
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find that, we must simply sum up over all the possible reservoir microstates, while keeping the system
in the desired microstate. Therefore
Z
dqdpdqR dpR
dqdp
=
ρmc (N, q, p; NT − N, qR , pR ),
ρg.c. (N, q, p)
GN hN f
Res GN GNT −N hN f +(NT −N )fR
where the integral is over all the reservoir’s degrees of freedom. After simplifying, we have:
ρg.c. (N, q, p) =
=
Z
Res
dqR dpR
ρmc (N, q, p; NT − N, qR , pR )
GNT −N h(NT −N )fR
dqR dpR
1 Z
ΩT ET ≤H(q,p)+HR (qR ,pR )≤ET +δET GNT −N h(NT −N )fR
since ρmc = Ω1T when this condition is satisfied, and zero otherwise. However, we can rewrite the
condition as:
ρg.c. (N, q, p) =
1 Z
dqR dpR
→
ΩT ET −H(q,p)≤HR (qR ,pR )≤ET −H(q,p)+δET GNT −N h(NT −N )fR
ρg.c. (N, q, p) =
ΩR (ET − H(q, p), δET , VR , NT − N )
ΩT (ET , δET , V, VR , NT )
since the integral is by definition just the multiplicity for the reservoir to be in a macrostate of energy
ET − H(q, p), with VR and NT − N microsystems. If you have a quick look at the canonical ensemble
derivation, you’ll see that so far everything went very similarly, except that we kept track carefully
of how many microsystems are where.
Using the link between the entropy of the reservoir and its multiplicity SR = kB ln ΩR (because
the reservoir is so big and insensitive to what the system does, all microcanonical formulae valid for
an isolated system apply to the reservoir), we then have:
ΩR (ET − H(q, p), δET , VR , NT − N ) = e
SR (ET −H(q,p),δET ,VR ,NT −N )
kB
SR (ET ,δET ,VR ,NT )−H(q,p)
≈e
∂SR
∂SR
−N
∂ER
∂NR
kB
where I used the fact that the energy of the system H(q, p) ≪ ET and N ≪ NT and performed
a Taylor expansion. So the appearance of the second derivative is where the difference between
canonical and grandcanonical ensembles shows up.
But we know that at equilibrium
∂SR
1
1
=
= ;
∂ER
TR
T
∂SR
µR
µ
=−
=−
∂NR
TR
T
so we find the major result:
1 −β[H(q,p)−µN ]
e
(1)
Z
where Z is a constant (what we obtain when we collect all terms that do not depend on (N, q, p)),
called the grand-canonical partition function. We can find its value from the normalization
condition:
ρg.c. (N, q, p) =
1=
∞ Z
X
N =0
∞ Z
X
dqdp
dqdp −β[H(q,p)−µN ]
ρg.c. (N, q, p) → Z(T, V, µ) =
e
N
f
GN h
GN hN f
N =0
3
(2)
Note that here “sum over all microstates” means to sum over all possible numbers N of microsystems
in the system, and for each N to “sum” over all possible locations/momenta of the microsystems. Of
course, for classical systems we know that the “sum” over locations/momenta is really an integral,
because these are continuous variables. You might argue that we should stop the sum over N at
NT , but since NT is by definition much much bigger than the average number of microsystems in the
system, it will turn out that we can use the upper limit to be infinity – it makes calculations easier
and the “error” can be made arbitrarily small by making the reservoir bigger, i.e. NT → ∞.
Note also that Z is a function of T (through the β in exponent), of V (the integrals over positions
are restricted to the volume V ), and µ (again through the exponent).
Now that we know the grandcanonical density of probability, we can calculate the internal energy
U = hH(q, p)i =
∞ Z
X
N =0
∞ Z
1 X
dqdp
dqdp
ρg.c. (N, q, p)H(q, p) =
H(q, p)e−β[H(q,p)−µN ]
N
f
GN h
Z N =0 GN hN f
Here we have to be a bit careful. We can’t simply use the trick with the derivative with respect to
β, since this will bring down both H (which we want), but also µN (which we don’t want):
−
∂ −β[H(q,p)−µN ]
e
= [H(q, p) − µN ] e−β[H(q,p)−µN ] 6= H(q, p)e−β[H(q,p)−µN ]
∂β
So here’s what we do. Let me define:
α = βµ
and use this instead of µ as a variable, so that Z = Z(β, V, α) and
ρg.c. (N, q, p) =
1 −βH(q,p)+αN
e
Z
Now, we have to pretend that α and β are independent variables, i.e. we “forget” for a bit what is
definition of α, we pretend that it’s just some quantity totally unrelated to β. If this was true, then
we could use:
∂
− e−βH(q,p)+αN = H(q, p)e−βH(q,p)+αN
∂β
and we could then write:
∞ Z
∂ X
1
dqdp −βH(q,p)+αN
−
U=
e
→
Z
∂β N =0 GN hN f
"
#
1
∂
∂
Z(β, V, α) = −
ln Z(β, V, α)
Z(β, V, α) ∂β
∂β
So the point is to treat α as a variable independent of β while we take the derivative, and only after
we’re done with taking the derivative to remember that α = βµ.
This “forgetfulness” is very useful since it allows us to calculate another ensemble average very
simply, namely:
U =−
hN i =
∞ Z
X
N =0
∞ Z
X
dqdp
1
dqdp
ρg.c. (N, q, p)N =
N e−βH(q,p)+αN
N
f
GN h
Z(β, V, α) N =0 GN hN f
Again, treating α and β as independent variables while we take derivatives, we have:
∂ −βH(q,p)+αN
e
= N e−βH(q,p)+αN
∂α
4
so that we avoid doing the integrals and we find:
hN i =
1
∂
∂
Z(β, V, α) =
ln Z(β, V, α).
Z(β, V, α) ∂α
∂α
So we can easily also calculate the average number of microsystems in the system. We will look
at some examples soon and you’ll see that doing this is simple in practice, it just requires a bit of
attention when taking the derivatives.
This approach can be extended easily to find (check!) that:
hH2 i =
1
∂2
Z(β, V, α)
Z(β, V, α) ∂β 2
and
∂2
1
Z(β, V, α).
Z(β, V, α) ∂α2
Of course, we would need these quantities to calculate standard deviations.
Now, this trick I described above, with using α and β, is what people usually do, and what is
given in textbooks etc. All is needed is that while you take the derivatives, you treat α and β as
independent variable. For reasons which escape me, some students think that this is too fishy and
refuse to use this trick. So here is an alternate trick, which is almost as good (takes just a bit more
work) and gives the precise same answers at the end of the day. Let’s look again at:
hN 2 i =
U = hH(q, p)i =
∞ Z
X
N =0
∞ Z
dqdp
dqdp
1 X
ρ
(N,
q,
p)H(q,
p)
=
H(q, p)e−β[H(q,p)−µN ]
g.c.
N
f
GN h
Z N =0 GN hN f
Clearly we’d like to not have to do the integrals explicitly, so we have to get rid of the H somehow.
If you do not like the trick with introducing α, then we can do this. First, introduce an x in front
of the Hamiltonian, in the exponent, and ask that x be set to 1 at the end of the calculation, since
clearly:
∞ Z
dqdp
1 X
−β[xH(q,p)−µN ] H(q,
p)e
U=
Z N =0 GN hN f
x=1
Now we can take the derivative with respect to x, so that we have:
"
1 ∂
1
Z(x)
−
U=
Z
β ∂x
where
Z(x) =
∞ Z
X
N =0
#
x=1
dqdp −β[xH(q,p)−µN ]
e
GN hN f
can be quickly calculated (this is where the extra work comes in), just like you calculated Z = Z(x =
1) (it’s just a matter of tracking where the extra x goes in some gaussians).
So, putting these together, and since we set x = 1 at the end, we have:
and similarly
1 ∂
U =−
ln Z(x)
β ∂x
x=1
1 ∂2
Z(x)
hH2 i = 2
2
β ∂x
x=1
5
and here the only “trick” is to first take all the derivatives, and then set x = 1 as the very last step.
To calculate ensemble averages of N we do not need to introduce any new variable, since we
already have µ there, so we can also write:
∞ Z
X
1
dqdp
1 1 ∂
1 ∂
hN i =
N e−βH(q,p)+βµN =
Z =
ln Z
N
f
Z(β, V, α) N =0 GN h
Z β ∂µ
β ∂µ
#
"
and similarly
hN 2 i =
∞ Z
X
1 1 ∂2
dqdp 2 −βH(q,p)+βµN
1
N
e
=
Z
Z(β, V, α) N =0 GN hN f
Z β 2 ∂ 2µ
If you think about it, taking these derivatives is just the same as taking derivatives with respect to
α, in the previous notation.
How about calculating other ensemble averages, for example the entropy? Using the expression
of ρg.c. in the general Boltzmann formula, we find that:
S = −kB hln ρi = −kB
∞ Z
X
N =0
= −kB
∞ Z
X
N =0
dqdp
ρg.c. (N, q, p) ln ρg.c. (N, q, p)
GN hN f
dqdp 1 −β[H(q,p)−µN ]
e
[− ln Z − βH(q, p) + βµN ] →
GN hN f Z
S = kB ln Z + kB βU − kB βhN i
since the first integral is related to the normalization condition, while the second and third are just
the ensemble averages of H, N . From this we find that the grand-canonical potential is:
φ(T, V, µ) = U − T S − µhN i = −kB T ln Z(T, V, µ)
(3)
As you see, things again fit very nicely together. If you remember, when we reviewed thermodynamics,
we decided that for an open system we need to be able to compute the grand-canonical potential
φ(T, V, µ), since then we can use
dφ = −SdT − pdV − hN idµ
(4)
to find S, p and hN i as its partial derivatives, which tells us how they depend on T, V, µ.
This seems to give us an alternate way to find hN i, but in reality we find the same formula with
the derivative from ln Z (not surprisingly – we should get the same result no matter how we go about
it). There is also an alternative way to find U , from using U = φ + T S + µhN i, where all the terms
on the right hand side are known once Z is calculated. Of course, this gives the same result as the
tricks with derivatives, but it involves a bit more work.
So let us summarize how we solve a classical grandcanonical ensemble problem, once T, V, µ, ...
(the macroscopic variables) are given to us. As always, we begin by identifying the number of degrees
of freedom f per microsystem, and all needed generalized coordinates q, p that fully characterize a
microstate that has N microsystems in the system, and then the Hamiltonian of the system H(q, p).
Then we calculate the partition function from Eq. (2). Once we have it, we know φ = −kB T ln Z.
Once we have φ, we can find
∂φ
S=−
∂T
!
∂φ
;p = −
∂V
V,µ
6
!
∂φ
; hN i = −
∂µ
T,µ
!
T,V
We can also calculate the internal energy and hN i from
U =−
∂
ln Z(β, V, α);
∂β
hN i =
∂
ln Z(β, V, α);
∂α
where α, β are treated as independent variables while we take the derivatives, after which we can set
α = βµ. Similarly, we can calculate averages of hH2 i, hN 2 i, hHN i, ..., but using the proper number
of derivatives with respect to α and β. If you do not like this, use the trick with introducing the x,
and then setting it to be 1 after all derivatives were taken.
Any other ensemble averages are calculated starting from the definition of an ensemble average
and the known density of probability for the grandcanonical ensemble, and by doing all integrals
over positions/momenta and sum over N .
One last thing. Remember that for the canonical partition function and non-interacting systems,
we could use the factorization theorem to simplify the calculation. It turns out we can do a similar
thing here. First, start with the general definition of the grand-canonical partition function:
Z(T, V, µ) =
∞ Z
X
N =0
Z
∞
X
dqdp −βH(q,p)
dqdp −β[H(q,p)−µN ]
βµN
e
e
=
e
GN hN f
GN hN f
N =0
Now we recognize that the integrals are simply the canonical partition function for a system with N
particles, so:
Z(T, V, µ) =
∞
X
eβµN Z(T, V, N )
N =0
So in fact, if we know how to calculate Z (which we do) there isn’t much left of the calculation.
Let’s simplify even further. For non-interacting systems where particles can move and exchange
positions (such as gases), we know from the factorization theorem that:
Z(T, V, N ) =
1
[z(T, V )]N
N!
Using this in the above sum, we find:
Z(T, V, µ) =
∞
X
1 h
N =0
N!
z(T, V )eβµ
iN
= exp eβµ z(T, V )
since the sum is just the expansion of the exponential function. Those of you with good memory will
be delighted to learn that eβµ has its own special name, namely fugacity.
For problems where the microsystems are distinguishable because they are located at different
spatial location (crystal-type problems), Gibbs’ factor is 1 and:
Z(T, N ) = [z(T )]N
and therefore:
Z(T, V, µ) =
∞ h
X
eβµ z(T )
N =0
iN
=
1
1−
eβµ z(T )
if |eβµ z(T )| < 1 (otherwise the geometric series is not convergent). Of course, we’ll find that this
condition is generally satisfied.
So the conclusion is that once we calculate z (just as we did it for canonical ensembles), we
immediately have Z, i.e. dealing with a grandcanonical classical system really does not involve any
more work/math that dealing with a canonical system – at least so far as classical problems are
concerned.
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1.1
Classical ideal gas
Let’s check how this works for a classical ideal gas – our favorite classical model. Assume a system
of volume V in contact with a thermal and simple atom reservoir with temperature and chemical
potential T, µ. Let’s calculate the average number of particles in the container, hN i, their internal
energy U and their pressure p.
Because these are simple, non-interacting atoms, they have f = 3 degrees of freedom, and we can
use factorization theorem, and GN = N !.
Of course, we know that:
Z
p
~2
1 Z
z(T, V ) = 3 d~r d~pe−β 2m = V
h
2πmkB T
h2
!3
zN
N!
2
→ Z(T, V, N ) =
(three identical gaussians plus one integral over the volume). Therefore (see formula above):
Z=
∞
X
eβµN Z(T, V, N ) =
∞
X
1 h
N =0
N =0
N!
eβµ z
and so:
iN

2πmkB T
h2
2πmkB T
h2
!3
= exp eβµ V
βµ
φ(T, V, µ) = −kB T ln Z = −kB T e V
!3 
2

2
This is indeed an extensive quantity. It is maybe not so obvious that it has the right (energy) units,
but you should be able to convince yourselves that that is true (remember that z is a number, and
µ is an energy).
To calculate hN i we have two alternatives, either as a partial derivative of φ with respect to µ
(I’ll let you do this) or using the trick with α and β. Let’s do it by the second method. First, we
replace βµ → α everywhere where they appear together. We then find:

α
Z(β, V, α) = exp e V
2πm
βh2
!3 
2
 → ln Z(β, V, α) = eα V
2πm
βh2
!3
2
Now, assuming α and β to be independent variables, we have:
∂
hN i =
ln Z(β, V, α) = eα V
∂α
2πm
βh2
!3
2
This looks a bit strange, but let’s not loose heart – this tells us how hN i depends upon T, V, µ
(the variables which characterize the macrostate for the open system) and it is not something we’ve
looked at before. Notice that you can extract how µ depends on T, V, hN i from this – you should
do that and see that the result agrees with what we obtained for canonical ensembles, if we replace
N → hN i.
The internal energy is:
∂
3 α
U =−
ln Z(β, V, α) =
e V
∂β
2β
So yeeeii! It works!
8
2πm
βh2
!3
2
3
→ U = hN ikB T
2
Finally, we find the pressure from the partial derivative of φ with respect to V :
∂φ
p=−
∂V
!
= kB T eβµ
T,µ
2πmkB T
h2
!3
2
→ pV = hN ikB T
So, yeeeiii again! As you can see, we do obtain the same results we got using micro- and canonical
ensemble formalisms for this problem, except here we must use hN i instead of N .
Why is this, why do we get the same predictions even though now the number of particles if not
fixed? Exactly like in the case of the equivalence between microcanonical and canonical formulations’
results, this is due to the fact that the systems are very large, hN i ∼ 1023 . Unlike in a canonical
ensemble, in a grandcanonical ensemble the number of particles is not fixed, and it could be anything,
in principle. However, because the system is so big, one can show (just as we did for the energy of the
canonical ensembles) that the probability of finding the system to have anything but hN i particles
is an extremely small number (basically zero), i.e. it is overwhelmingly likely that the system always
has precisely hN i atoms inside. One can show that the relative standard deviation
σN
=
hN i
q
hN 2 i − hN i2
hN i
∼q
1
hN i
→0
(I’ll probably give this to you as an assignment exercise).
So, as I’ve said all along, for (large) thermodynamic systems we can use whichever formalism is
most convenient to solve a problem. We will see more examples of classical problems to be solved by
grandcanonical formalism (the kinds of problems where the number of particles can really change) in
the next assignment. We will also consider problems where we have more than one species of atoms
in the container – in that case, it is possible that one species can pass through the wall (system is
open, from its point of view) and another species cannot go through the wall (system is closed from
its point of view). In this case, we need to use a mixed description, using canonical formalism for the
atoms whose number is conserved, and grandcanonical formalism for those whose number can vary.
This probably sounds more complicated than it is in practice, as you’ll see.
But now it is finally time to concentrate on quantum gas -type of problems, which we could not
solve by any other formalism. We are finally able to study them.
2
Quantum grandcanonical ensembles
First, we need to figure out how to properly characterize microstates of the quantum system, since
the number of microsystems is not fixed. If you think about it, we can’t just say that the microstates
are eigenstates of the Hamiltonian (the way we did for quantum canonical systems), because after all
even the Hamiltonian is not unique – microstates with different numbers of particles have different
Hamiltonians! (number of terms, e.g. kinetic energy contributions, in the Hamiltonian changes in
proportion to how many microsystems are in the microstate). So we have to be a bit careful how we
go about characterizing all the microstates.
What will simplify things tremendously is the fact that we will only deal with non-interacting
systems. For interacting systems, one really needs to use more formal objects like density matrices,
but we won’t go there.
Let eα be the energies of all (discrete) energy levels, if there was a single microsystem in the
system. We’ll always assume the energy to be zero if there is no microsystem in the system (empty
9
system). α are all the needed quantum numbers to describe these levels – degeneracies are very
important! These energies (and the single-particle wavefunctions associated with them) are usually
called single-particle energies or single-particle orbitals.
To make things more clear, let’s look at some examples as we go along. First, a simple “crystal”like example. Assume we have a surface with a total number NT of sites where simple atoms could
be trapped. If a trap is empty, its energy is zero. If it catches an atom, its energy is lowered to
−ǫ0 < 0. This surface is in contact with a gas of atoms (the reservoir), with known T and µ. The
question could be, for instance, what is the average number hN i of atoms trapped on the surface.
What are the single-particle orbitals, in this case? Well, if we have a single atom in the system=
surface with traps, it must be trapped in some site or other, and the energy will be −ǫ0 . We could
use as “quantum number” an integer 1 ≤ n ≤ NT which tells us at which site is the atom trapped –
so here we have a NT degenerate spectrum of single-particle states, all with the same energy −ǫ0 .
As a second example, let’s consider a quantum gas problem. Assume simple atoms with quantum
dynamics in a cubic box of volume V = L3 . Of course, we’ll generally want to know what are the
properties of the quantum gas when the system is in contact with a thermal and particle reservoir
with known T, µ. However, right now all we want to know, is what are the single particle levels.
For this, we must find the spectrum (the eigenstates) when there is just one atom in the system.
This we know how to do. In this case the Hamiltonian is:
!
h̄2
d2
d2
d2
ĥ = −
+
+
2m dx2 dy 2 dz 2
and the eigenstates are:
h2
(n2 + n2y + n2z )
8mL2 x
where nx , ny , nz = 1, 2, ... are strictly positive integers. So here three quantum number α = nx , ny , nz
characterize the single-particle orbitals (if you don’t remember where this formula comes from, it is
a simple generalization of the 1d case we solved when we looked at multiplicities, when we discussed
microcanonical ensembles). Strictly speaking, atoms also have some spin S, so we should actually
include a 4th quantum number α = nx , ny , nz , m where m = −S, −S + 1, ...S − 1, S is the spin
projection. For any other problem we can figure out the single-particle orbitals (or states) similarly.
Now let’s go back to our general description, where eα are the energies of all possible single-particle
orbitals. What happens if there are more microsystems in the system? Well, let’s start with two.
Because these are non-interacting microsystems, they occupy the same set of single-particle orbitals.
So to characterize the state, we now need two pairs of quantum numbers, say α and α′ , to tell us
which two states are occupied. The energy is simply eα + eα′ . For example, if there is a second
atom on the surface, it must also be trapped in some trap or other, just like the first one, so I could
specify the state by saying atom 1 is trapped at site n while atom 2 is trapped at site n′ . Of course,
the energy is −2ǫ0 . Similarly, a second atom in the box will be in some eigenstates en′x ,n′y ,n′z and the
energy will be the sum of the two. The wavefunction for the total system is now:
enx ,ny ,nz =
Ψ(~r1 , ~r2 ) = φα (~r1 )φβ (~r2 )
where φα (~r) is the single-particle wavefunction associated with the single-particle state eα .
Right? WRONG!
What quantum mechanics tells us is that if the particles are identical, their wavefunction must be
either symmetric (for so-called bosonic particles, i.e. whose spin S is an integer) or antisymmetric
(for so-called fermionic particles, i.e. whose spin S is half-integer ) to exchanges of the two, i.e.
Ψ(~r1 , ~r2 ) = ±Ψ(~r2 , ~r1 ) → |Ψ(~r1 , ~r2 )|2 = |Ψ(~r2 , ~r1 )|2
10
This is what indistinguishability really means. If the particles are truly identical, then there is no
measurement whatsoever that we can perform to tell us which of the two particles is at ~r1 and which
at ~r2 , so if we exchanged their positions we should see no difference (same probability to find them
at those locations). So going back, it follows that for two fermions, the two-particle wavefunction
must be:
ΨF (~r1 , ~r2 ) = φα (~r1 )φβ (~r2 ) − φα (~r2 )φβ (~r1 )
while for bosons, we must have:
ΨB (~r1 , ~r2 ) = φα (~r1 )φβ (~r2 ) + φα (~r2 )φβ (~r1 )
(there is actually an overall normalization factor, but that is just a constant that has no relevance
for our discussion).
This immediately tells us that we cannot have two fermions occupy the same single-electron orbitals, i.e. α 6= β always. If α = β we find ΨF (~r1 , ~r2 ) = 0, which is not allowed (wavefunctions must
normalize to 1). This is known as Pauli’s exclusion principle. There is no such restriction for
bosons, there we can have any number of bosons occupying the same single-particle state.
If we now look at these two-particle wavefunctions, we see that it’s just as likely that particle 1 is
in state α and 2 in β, as it is to have 1 in state β and 2 in state α. Therefore, it makes much more
sense to characterize this state by saying that there is a particle in state α and a particle in state β,
and not attempt anymore to say which is which – they’re identical and either one could be in either
state with equal probability.
We can rephrase this by saying that of all single-particle orbitals, only α and β are occupied, while
all other ones are empty. In fact, we can define an occupation number which is an integer nα
associated to each single-particle level. For empty levels nα = 0, while for occupied levels nα counts
how many particles are in that particular orbital.
For bosons, nα = 0, 1, 2, 3, ... – could be any number between 0 and infinity. For fermions, nα = 0
or 1! Because of the exclusion principle, we cannot have more than 1 fermion occupying a state. This
is the difference between fermions and bosons. It might not look like much, but as we will see, it
will lead to extraordinarily different behavior of fermions vs. bosons at low temperatures, i.e. where
quantum behavior comes into play.
We can now generalize. For any number of microsystems (particles) present in the system, we can
specify a possible microstate by giving the occupation numbers for all the single-particle orbitals (if
there is an infinite number of orbitals, such as for a particle in a box, lots and lots of occupations
numbers will be zero; but we still have to list all – infinite number – of them). So the microstate
is now specified through the values {nα } of all occupation numbers in that microstate. Allowing all
numbers {nα } to take all their possible values will generate all possible microstates for all possible
numbers of microsystems.
Of course, the total number of microsystems (particles) in the microstate must be
N{nα } =
X
nα
α
i.e. we go through all levels and sum how many particles are occupying each of them – clearly this
sum gives the total number of particles in that microstate. The energy of this microstate is:
E{nα } =
X
nα e α
α
again we go through all levels, for each one which is occupied (nα 6= 0) we add how many particles
are in that level, nα , times the energy of each one of them, eα . Again, I think this should be quite
11
an obvious equation for non-interacting systems. For interacting systems, things are much more
difficult, because the energy is no longer the sum of single-particle energies.
This was the hard part. Now that we know how we describe the microstates, and how many
particles are in a microstate and what is their energy, we’re practically done. Following the same
reasoning as for classical ensembles, we find that the probability to be in a microstate is
1 −β(Eµstate −µNµstate )
e
Z
As in the canonical case, the only difference is that classical degrees of freedom are continuous, so we
can only talk about density of probability to be in a microstate. For quantum systems, microstates
are discrete so we can talk about probability to be in a microstate. The reason the formula above
holds is because nowhere in its derivation did we have to assume that the energy is continuous, so
things proceed similarly if the energy is discrete.
Of course, the grand-canonical partition function Z must be such that the normalization condition
holds:
X
X
pµstate = 1 → Z =
e−β(Eµstate −µNµstate )
pµstate =
µstates
µstates
In terms of occupation numbers which characterize the microstates, these formulae become:
P
P
P
1
1
1 −β(E{nα } −µN{nα } )
e
= e−β( α nα eα −µ α nα ) = e−β α nα (eα −µ)
Z
Z
Z
To find Z we must sum over all microstates. Since the allowed values for the occupation numbers
are different, let’s do this separately for the two cases,
For a fermionic system, each nα = 0 or 1. As a result:
p{nα } =

ZF = 
1
Y X

 e−β
α nα =0
P
α
nα (eα −µ)
Throughout the remainder of the course, I’ll use this shorthand notation:


1
1
X
Y X
≡

α nα =0
1
X
···
nα1 =0 nα2 =0
where there is a sum for each single particle orbital (if there’s an infinite number of them, there’s
an infinite number of sums). The short-hand notation simply says that for each possible α, there’s
a sum sign in the product.
This is actually very simple to calculate, because the exponential factorizes in terms each of which
depend on a single occupation number nα . So we can group each exponential with its sum, and find:
ZF =
Y
α
But now each sum is trivial,
P1
nα =0


1
X
nα =0

e−βnα (eα −µ) 
e−βnα (eα −µ) = 1 + e−β(eα −µ) , and so:
ZF =
Y
1 + e−β(eα −µ)
α
Just to make sure you followed this, let us do this for the case of the atoms trapped on the surface
(we’ll study the quantum gases in detail starting next lecture). In this case, we decided that we have
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a finite number of single-particle levels indexed by the integer α → n, 1 ≤ n ≤ NT which tells us at
which site is the single atom trapped.
Let’s assume that the atoms are fermionic, i.e. there can be at most 1 atom in any trap (you
might want to ask me some questions here ....). The microstate is now described by the occupation
numbers n1 , ..., nNT where ni is zero if trap i is empty and 1 if trap 1 is occupied by an atom.
P T
P NT
The number of atoms in the microstate
is N = N
i=1 ni , and the energy is E =
i=1 (−ǫ0 )ni =
PN T
−ǫ0 N , and E − µN = (−ǫ0 − µ) i=1 ni . In this case:
ZF =
1
1
X
X
···
n1 =0 n2 =0

=
1
X
n1 =0
e

−β(−ǫ0 −µ)n1  
1
X
e
−β(E−µN )
n2 =0
e
···

−β(−ǫ0 −µ)nNT
=
n1 =0 n2 =0
nNT =0
1
X
1
1
X
X

−β(−ǫ0 −µ)n2 
···
1
X
e
1
X
e−β(−ǫ0 −µ)(n1 +···+nNT )
nT =0
nNT =0

h
iN
 = 1 + e−β(−ǫ0 −µ) T
In this case, there is a finite number of single particle orbitals, and each contributes the same since
they all have the same energy −ǫ0 . In the general case, each orbital contributes 1 + e−β(eα −µ) , and
we must multiply over all the orbitals. That’s precisely what the general formula for ZF means.
All the other formulae we have derived for classical grand-canonical system hold unchanged, in
particular:
i
X h
φF = −kB T ln Z = −kB T
ln 1 + e−β(eα −µ)
α
For different fermionic systems we’ll have different energies ǫα and number of levels α, but this
formula will always hold.
Let us do the same for a bosonic system. In that case:
ZB


∞
P
Y X
 e−β α nα (eα −µ)
=
α nα =0
since occupation numbers can now be anything. Again we can factorize the product:
ZB =
Y
α


∞
X
nα =0
e

−βnα (eα −µ) 
Each sum is an infinite geometric series. Note that we must have:
e−β(ǫα −µ) ≤ 1
in order for each of these series to be convergent. Since β > 0, it follows that for a bosonic system,
we must always have µ ≤ eα for all single particle levels and therefore
for bosons, we must have: µ ≤ eGS
where eGS is the energy of the single-particle ground-state. This restriction will turn out to have
important consequences. For fermions we have no restrictions for P
the chemical potential.
n
If the restriction µ ≤ eGS is satisfied, then each geometric series ∞
n=0 x = 1/(1−x) is convergent,
and we find:
Y
1
ZB =
1 − e−β(eα −µ)
α
13
and
φB = −kB T ln ZB = +kB T
X
h
ln 1 − e−β(eα −µ)
α
i
In fact, because of the similarities of the formulae, we can group together the results for both
fermions and bosons and write:
±1
Y
Z=
1 ± e−β(eα −µ)
α
and
φ = −kB T ln Z = ∓kB T
X
ln 1 ± e−β(eα −µ)
α
where the upper sign is for fermions, and the lower sign is for boson systems. From partial derivatives
of φ we can calculate S, p, hN i, as usual, since dφ = −SdT − pdV − hN idµ. We can also use the
tricks with α and β to find U and hN i. Let’s remember them, and check that they still hold.
First, we replace βµ → α everywhere this product appear. In terms of α and β, we have:
pµstate =
1 −βEµstate +αNµstate
e
Z
where from normalization,
Z(α, β, ...) =
e−βEµstate +αNµstate
X
µstates
By definition,
U=
X
pµstate Eµstate =
µstates
1 X
∂
1 ∂
Z(α, β, ...) = −
ln Z(α, β, ...)
Eµstate e−βEµstate +αNµstate = −
Z µstates
Z ∂β
∂β
if, while taking the derivative, we pretend that α and β are independent variables. Similarly,
hN i =
X
pµstate Nµstate =
µstates
1 ∂
∂
1 X
Nµstate e−βEµstate +αNµstate =
Z(α, β, ...) =
ln Z(α, β, ...)
Z µstates
Z ∂α
∂α
So indeed, wePhave precisely the same formulae as before. This is because the only difference is what
is meant by µstates . For classical systems that implies a sum over N and many integrals over all
classical degrees of freedom; for quantum systems, this is a sum over all possible occupation numbers.
But nothing in the derivation depended on such details.
For our quantum system, after replacing βµ → α, we have:
ln Z(α, β, ...) = ±
X
γ
ln 1 ± e−βeγ +α
(again, upper sign for fermions, lower sign for bosons). I prefer to call the quantum numbers γ this
time, since α = βµ is now taken. If we take the derivatives, we find that:
U =−
X ∓eγ e−βeγ +α
X
1
∂
ln Z(α, β, ...) = −(±)
=
eγ β(eγ −µ)
−βe
+α
γ
∂β
e
±1
γ 1±e
γ
and
hN i =
X e−βeγ +α
X
∂
1
ln Z(α, β, ...) = (±)
=
−βeγ +α
β(eγ −µ) ± 1
∂α
γ 1±e
γ e
(we can go back to α → βµ after we took the derivatives. Results are generally in terms of µ.
14
But, on the other hand, using the expressions for number of particles and energy of a microstate
in terms of occupation numbers, we have:
hN i = h
X
nγ i =
γ
X
hnγ i
γ
and
U =h
X
e γ nγ i =
γ
X
eγ hnγ i
γ
Comparing these with the equations above, we see that we must have the average occupation
number of level γ to be:
1
hnγ i = β(eγ −µ)
e
±1
These are extremely important results, which will come up time and time again. So let’s discuss
them separately, in some detail.
For fermions, the average occupation number of a level with energy ǫγ is:
hnγ i =
1
eβ(eγ −µ) + 1
This is called the Fermi-Dirac distribution. Let’s analyze it a bit. First, since eβ(eγ −µ) ≥ 0 no
matter what values β, µ, eγ have, it is clear that always 0 ≤ hnγ i ≤ 1. This makes perfect sense. In
any microstate, nγ can only be 0 or 1, so its average must be a number between 0 and 1!
At T = 0 → β → ∞, we see that if eγ < µ → eβ(eγ −µ) → 0 and so hnγ i → 1. In other words,
levels whose energy is below the chemical potential µ are certainly occupied at T = 0.
If, however, eγ > µ → eβ(eγ −µ) → ∞ and so hnγ i → 0. Therefore, levels whose energy is above
the chemical potential µ are certainly empty at T = 0. If the temperature is low, but not
zero, the occupation number will be somewhat changed for levels whose energy is within about kB T
of µ (see figure below). But levels with eγ − µ ≪ −kB T are still certainly occupied, and levels with
eγ − µ ≫ kB T are still certainly empty. It’s just within kB T of µ the transition from occupied to
empty is no longer abrupt, instead the average occupation numbers continuously go from 1 to 0.
<n(e)>
Fermi−Dirac
1
T=0
kT
T>0
µ
eGS
e
Figure 2: Fermi-Dirac distribution, showing the average occupation number of a level of energy e, as a function of
e. There are no levels below eGS . At T = 0, all levels between eGS and µ are fully occupied, hni = 1, while all
levels above µ are completely empty, hni = 0. At finite-T, the average occupation numbers for energies roughly in the
interval [µ − kB T, µ + kB T ] are changed, and the decrease is now gradual.
15
For bosons, the average occupation number of a level with energy ǫγ is:
hnγ i =
1
eβ(eγ −µ)
−1
This is called the Bose-Einstein distribution. Let’s analyze it a bit. First, remember that for
bosons we must always have µ ≤ eGS → eγ − µ ≥ 0. Now we see that this is very necessary, since
with this restriction eβ(eγ −µ) ≥ 1 and the average occupation numbers are positive! They must be
positive – the average of any number whose only allowed values are 0, 1, 2, ... cannot be negative.
Unlike for fermions, however, we see that an average occupation number could be anything between
0 and infinity. In fact, let us consider T = 0 behavior. Here we have two cases: (1) µ < eGS . In this
case eβ(eγ −µ) →
∞ as β → ∞, T → 0, so all average occupation numbers become vanishingly small.
P
Since hN i = γ hnγ i, if all hnγ i → 0 then hN i → 0. This can happen if conditions are such that
particles would rather not stay in the system (they prefer to be in the bath at low-temperatures).
But this is a rather boring case. The more interesting case is when (ii) µ = eGS . In this case, the
average occupation numbers are still zero for all higher energy levels, but we see that the ground-state
itself has an infinite occupation number! Clearly, that can’t be quite right – indeed, we’ll have to
do this analysis more carefully when we study the so-called Bose-Einstein condensation. What
this result tells us, though, is that for bosons, at T = 0 all particles that are in the system (however
many they may be) occupy the ground-state orbital. This does make sense! We know that at T = 0
we expect the system to go into its ground-state. Of course, the lowest total energy is obtained when
we place the particles in the orbitals with the lowest possible energies. For bosons, we are allowed
to place all of them in the single-particle ground-state orbital, and that is indeed the lowest total
energy possible. For fermions, we cannot put more that 1 particle in a state, therefore to get the
lowest total energy possible, we occupy all the lowest energy states available with one particle in
each – which is what the Fermi-Dirac distribution predicts at T = 0.
So you see how the change of sign in the denominator leads to extremely different results!
You might now wonder how is it possible that a gas of bosonic atoms and a gas of fermionic atoms
will behave the same at high-temperatures, given how differently they behave at low-T. After all,
we know that at high-T we should have agreement with the classical predictions, for e.g. find that
pV = N kB T , etc. Since there is only one set of relationships for classical gases, it follows that both
bosons and fermions should behave the same at high temperatures. For this to happen, clearly these
average occupation numbers should also be equal at high-temperatures, otherwise we should be able
to tell the difference somehow.
Interestingly enough, this indeed happens. What we will show a bit later on (you’ll have to just
believe me for now) is that at high-temperature we have to set µ to be an extremely large negative
number, µ ≪ −kB T , if we want to have large numbers of particles in the box, hN i ∼ 1023 . All
one-particle levels eγ = enx ,ny ,nz in a box have positive energies, and therefore now β(eγ − µ) ≫ 1 →
eβ(eγ −µ) ≫ 1 → eβ(eγ −µ) ± 1 ≈ eβ(eγ −µ) . So the sign doesn’t make any difference anymore, and at
high temperatures we find both for fermions and bosons that:
hnγ i ≈ e−β(eγ −µ) ≪ 1
As we will show soon, in this limit we indeed find agreement with the expected classical results.
16