Download Chapter 11 statistical mechanics

Chapter 11. The microcanonical ensemble
The goal of equilibrium statistical mechanics is to calculate the diagonal elements of ρ̂eq
so we can evaluate average observables < A >= Tr{ Âρ̂eq } = A that give us fundamental
relations or equations of state.
Just as thermodynamics has its potentials U, A, H, G etc., so statistical mechanics has
its ensembles, which are useful depending on what macroscopic variables are specified.
We first consider the microcanonical ensemble because it is the one directly defined in
postulate II of statistical mechanics.
In the microcanonical ensemble U is fixed (Postulate I), and other constraints that are
fixed are the volume V and mole number n (for a simple system), or other extensive
parameters (for more complicated systems).
1. Definition of the partition function
The ‘partition function’ of an ensemble describes how probability is partitioned among
the available microstates compatible with the constraints imposed on the ensemble.
In the case of the microcanonical ensemble, the partitioning is equal in all microstates at
= 1 / W (U ) for each microstate
the same energy: according to postulate II, with pi = ρ (eq)
ii
“i” at energy U. Using just this, we can evaluate equations of state and fundamental
relations.
2. Calculation of thermodynamic quantities from W(U)
Example 1: Fundamental relation for lattice gas: entropy-volume part.
Consider again the model system of a box with M=V/V0 volume elements V0 and N
particles of volume V0, so each particle can fill one volume elements. The particles can
randomly hop among unoccupied volume elements to randomly sample the full volume
of the box. This is a simple model of an ideal gas. As shown in the last chapter,
M!
W=
(M − N )!N !
for identical particles, and we can approximate this, if M<<N by
N
1 ⎛V⎞
W≈
N ! ⎜⎝ V0 ⎟⎠
since M!/(M-N!)≈MN in that case. Assuming the hopping samples all microstates so the
system reaches equilibrium, we compute the equilibrium entropy, as proved in chapter 10
from postulate III, as
S = kB lnW ≈ S0 + NkB ln(V / V0 )
where S0 is independent of volume. This gives the volume dependence of the entropy of
an ideal gas. Note that by taking the derivative (∂S/∂V) = kBN/V = P/T we can
immediately derive the ideal gas law PV = NkBT = nRT.
Example 2: Fundamental relation for a lattice gas: entropy-energy part.
The above model does not give us the energy dependence, since we did not explicitly
consider the energy of the particles, other than to assume there was enough energy for
them to randomly hop around. We now remedy this by considering the energy levels of
 increases so
particles in a box. The result will also demonstrate once more that Ω
ferociously fast that it is equal to W with incredibly high accuracy for more than a
handful of particles.
Let the total energy U be randomly distributed among particles in a box of volume L3 =
V. The energy is given by
1 3N 2
U=
∑ pi ,
2m i =1
where i=1,2,3 are the x,y,z coordinates of particle #1, and so forth to i=3N-2,3N-1,3N are
the x,y,z coordinates of particle #N. In quantum mechanics, the momentum of a free
particle is given by p=h/λ, where h is Planck’s constant. Only certain waves Ψ(x) are
allowed in the box, such that Ψ(x) = 0 at the boundaries of the box, as shown in the figure
below.
Fig. 11.1 Particle in a box wavefunction can only have
wavelengths so that Ψ=0 at the boundaries. The state space (or
quantum number space) with 3N axes contains a hypersphere of
constant energy. In a large number of dimensions, the states
(black dots) in a layer at the surface of this sphere is essentially
equal to the total number of states within that surface.
The wavelengths λ=L/2, L, 3L/2 ... can be inserted in the equation for total energy,
yielding
2
3N
1 3N ⎛ hni ⎞
h 2 ni 2
U=
∑ ⎜ ⎟ = ∑ 8mL2 , ni = 1, 2, 3 ,
2m i =1 ⎝ 2L ⎠
i =1
the energy for a bunch of particles in a box. W(U) is the number of states at energy U.
Looking at the figure again, all the energy levels are “dots” in a 3N-dimensional cartesian
space, called the “state space”, or “action space” or sometimes “quantum number space.”
The surface of constant energy U is the surface of a hypersphere of dimension 3N-1 in
state space. The reason is that the above equation is of the form constant = x2+y2+...
where the variables are the quantum numbers. The number of states within a thin shell of
energy U at the surface of the sphere is
.
W (U ) where lim W (U ) = Ω
N →∞
 is the total number of states inside the sphere, which at a first glance would seem to be
Ω
much larger than W(U), the states in the shell. In fact, for a very high dimensional
 is
hypervolume, a thin shell at the surface contains all the volume, so in fact, Ω
essentially equal to W(U) and we can just calculate the former to a good approximation
when N is large.
If this is hard to believe, consider an analogous example of a hypercube instead of a
hypersphere. Its volume is Lm, where m is the number of dimensions. The change in
volume with side length L is ∂V/∂L=mLm-1, so ΔV=mLm-1ΔL is the volume of a shell of
width ΔL at the surface of the cube. The ratio of that volume to the total volume is
ΔV/V=mΔL/L. Let’s take the example our intuition is built on, m=3, and assume
ΔL/L=0.001, just a 0.1% surface layer. Then ΔV/V=3.10-3<<V indeed. But now
consider m=1020, a typical number of particles in a statistical mechanical system. Now
ΔV/V=1020.10-3=1017. The little increment in volume is much greater than the original
volume of the cube, and contains essentially all the volume of the new “slightly larger”
cube. It may be “slightly” large in side length, but it is astronomically larger in volume.
Now back to our hypersphere in the figure. Its volume, which is essentially equal to
W(U), the number of states just at the surface of the sphere, is
3N /2
⎛U⎞
π 3N /2
.
Vhypersphere
R 3N = ⎜ ⎟
Γ(3N/ 2 + 1)
⎝ U0 ⎠
The (1/2)3N is there because all quantum numbers must be greater than zero, so only the
positive part of the sphere should be counted. The Gamma function Γ is related to the
factorial function, and R is the radius of the sphere, which is given by
⎛ 1⎞
W (U ) = ⎜ ⎟
⎝ 2⎠
3N
⎛ 1⎞
=⎜ ⎟
⎝ 2⎠
3N
8mL2U
,
h2
the largest quantum number, if all energy is in a single mode. They key is that R~√U, so
U is raised to the 3N/2 power, where N is the number of particles, 3 is because there are
three modes per particle, and the 1/2 is because of the energy of a free particle depends
on the square of the quantum number. Thus
3
3
S(U ) = kB lnW (U ) = S0 + NkB lnU = S0 + nR lnU ,
2
2
where the constant S0 is not the same as in the previous example. We used the volume
equation from the previous example to obtain an equation of state (PV=nRT), and we can
obtain another equation of state here:
1 3
1
3
⎛ ∂S ⎞
⎜⎝
⎟⎠ = = nR or U = nRT .
∂U V ,n T 2 U
2
This equation relates the energy of an ideal gas to its temperature. 3n is the number of
modes or degrees of freedom (3 velocities per particle time n moles of particles), whereas
the factor of 2 comes directly from the particle-in-a-box energy function – in case you
ever wondered where that comes from. So, for a harmonic oscillator, n~U
( E = ω (n + 1 / 2) as you may recall) instead of n~U1/2, and you might expect U=3nRT
for 3N particles held together by springs into a solid crystal lattice. And indeed, that is
true for an ideal lattice at high temperature (in analogy to an ideal gas at high
temperature). Unlike free particles, the energy of oscillators does not have the factor of
R = nmax =
1/2. The ‘deep’ reason is that an oscillator has two degrees of freedom to store energy in
each direction, not just one: there’s still the kinetic energy, but there’s also potential
energy.
Example 3: A system of N uncoupled spins sz=±1/2
The Hamiltonian for this system in a magnetic field is given by
N
NB
,
H = ∑ szj B +
2
j =1
where the extra term at the end is added so the energy equals zero when all the spins are
pointing down. At energy U=0, no spin is excited. For each excited spin, the energy
increases by B, so at energy U, U/B atoms are excited. These U/B excitations are
indistinguishable and can be distributed in N sites:
N!
Γ(N + 1)
.
W (U ) =
=
U⎞ ⎛U⎞
U⎞ ⎛U ⎞
⎛
⎛
⎜⎝ N − ⎟⎠ !⎜⎝ ⎟⎠ ! Γ ⎜⎝ N + 1 − ⎟⎠ Γ ⎜⎝ + 1⎟⎠
B
B
B
B
This is our usual formula for permutations; the right side is in terms of Gamma functions,
which are defined even when U/B is not an integer. Gamma functions basically
interpolate the factorial function for noninteger values.
This formula has a potential problem built-in: clearly, when U starts out at 0 and then
increases, W initially increases. But for U=NB (the maximum energy), W=1 again. In
fact, W reaches its maximum for U=NB/2. But if W(U) is not monotonic in U, then S
isn’t either, violating P3 of thermodynamics. Let’s see how this works out.
U
For large N, and temperature neither so low that
~ O(1) , nor so high that
B
U
~ O(N ) , we can use the Stirling expansion lnN! ≈ NlnN-N, yielding
B
S
U⎞ ⎛
U⎞
U U U U
⎛
= ln Ω ≈ N ln N − N − ⎜ N − ⎟ ln ⎜ N − ⎟ + N − ln +
⎝
kB
B⎠ ⎝
B⎠
B B B B
U⎞ ⎛
U⎞ U U U
U
⎛
≈ N ln N − ⎜ N − ⎟ ln ⎜ N − ⎟ − ln + ln N − ln N
⎝
B⎠ ⎝
B⎠ B B B
B
U ⎞ U ⎛
U ⎞ U ⎛ U ⎞
⎛
≈ −N ln ⎜ 1 −
+ ln ⎜ 1 −
⎟
⎟ − ln ⎜
⎟
⎝
NB ⎠ B ⎝
NB ⎠ B ⎝ NB ⎠
U ⎞ U ⎛ U ⎞
⎛U
⎞ ⎛
≈ ⎜ − N ⎟ ln ⎜ 1 −
⎟ − ln ⎜
⎟
⎝B
⎠ ⎝
NB ⎠ B ⎝ NB ⎠
after canceling terms as much as possible. We can now calculate the temperature and
obtain the equation of state U(T):
1 ⎛ ∂S ⎞
k ⎛ NB ⎞
NB
=⎜
≈ ln ⎜
− 1⎟ ⇒ U ≈
⎟
⎠
T ⎝ ∂U ⎠ N B ⎝ U
1 + e B / kT
In this equations, at T ~ 0, U → 0; and as T → ∞,U → NB / 2 . So even at infinite
temperature the energy can only go up to half the maximum value, where W(U) is
monotonic. The population cannot be ‘inverted’ to have more spins point up than down.
At most half the spins can be made to point up by heating. This should come as no
surprise: if the number of microstates is maximized by having only half the spins point up
when energy is added, then that’s the state you will get (this is true even in the exact
solution). Note that this does not mean that it is impossible to get all spins to point up. It
is just not an equilibrium state at any temperature between 0 and ∞. Such nonequilibrium states with more spins up (or atoms excited) than down are called “inverted”
populations. In lasers, such states are created by putting the system (like a laser crystal)
far out of equilibrium. Such a state will then relax back to an equilibrium state, releasing
a pulse of energy as the spins (or atoms) drop from the excited to the ground state.
The heat capacity of the above example system is
NB 2
e B / kT
⎛ ∂U ⎞
cv = ⎜
≈
peaked at 4kBT/B,
⎝ ∂T ⎟⎠ N kT 2 1 + e B / kT 2
(
)
so we can calculate thermodynamic quantities as input for thermodynamic manipulations.
As we shall see in detail later (actually, we saw it in the previous example!), in any real
system the heat capacity must eventually approach cv = NkB / 2 , where N is the number
of degrees of freedom. However, a broad peak at 4k BT / B is a sign of two low-lying
energy levels spaced by B. Levels at higher energy will eventually contribute to cv ,
making sure it does not drop.
Example 4: Let us check that T derived from
1
⎛ ∂S ⎞
⎛ ∂kB lnW ⎞
⎜⎝
⎟⎠ = ⎜⎝
⎟⎠ =
∂U N
∂U
T
N
indeed agrees with the intuitive concept of temperature. Consider two baths within a
closed system, so U = U1 + U 2 = const. ⇒ dU = 0 ⇒ dU1 = −dU 2 . If we know
∂Wi
Wi (U i ) ⇒ dWi =
dU i
∂U i
for each bath, then
Wtot = W1W2
dWtot = (W1 + dW1 ) ⋅ (W2 + dW2 ) − W1W2 + O(dW 2 )
= W1dW2 + W2 dW1
⎛
∂W2
∂W1 ⎞
= ⎜ −W1
+ W2
dU1 = 0
∂U 2
∂U1 ⎟⎠
⎝
at equilibrium because the maximum number of states is already occupied. For this to be
true for any infinitesimal energy flow dU1,
1 ⎛ ∂W2 ⎞
1 ⎛ ∂W1 ⎞
⇒
=
W2 ⎜⎝ ∂U 2 ⎟⎠ V , N W1 ⎜⎝ ∂U 2 ⎟⎠ V , N
⎛ ∂ lnW2 ⎞
⎛ ∂ lnW1 ⎞
⎛ ∂S2 ⎞
1 ⎛ ∂S1 ⎞
1
⇒⎜
=
or
=
=⎜
=
⎟
⎟
⎜
⎟
⎜
⎟
⎝ ∂U 2 ⎠ V , N ⎝ ∂U 2 ⎠ V , N
⎝ ∂U 2 ⎠ V , N T2 ⎝ ∂U ⎠ V , N T1
At equilibrium, the temperatures are equal, fitting out thermodynamic definition that
“temperature is equalized when heat flow is allowed.”