The average occupation numbers

• In the above case, it means the average number of students in room 1 or room 2.

• Let j = 1 or 2 , where N1 is the number of students in room 1 and N2 is the number of students in room 2; Let Njk be the number of students in room j for the kth case

kk

kkjk

j W

WNN

The value of thermodynamic probability WN1 will become extremely large as the values of N and N1 are increased.

To facilitate the calculation, Stirling’s approximation becomes useful:

ln (n!) = n ln(n) – n

When n is more than 50, the error in using Striling’s approximation becomes very small.

Extend the above example into the situation where there are multiple rooms (n rooms)

Now, the macrostate will be defined by the # of students in each room, say: N1, N2 …Nn

Note that

N1 + N2 + N3 + … + Nn-1 + Nn = N

The number of microstates for the above macrostate can be calculated from

W= x x ... x

=

!!!!

!

321 nNNNN

N

12.4 Thermodynamic Probability and Entropy

Boltzman made the connection between the classical concept of entropy and the thermodynamic probability

S = f (w)f (w) is a single-valued, monotonically increasing function (because S increases monotonically)

For a system which consists of two subsystems A and B

Stotal = SA + SB (S is extensive)Or…

f (Wtotal) = f (WA) + f (WB)

The configuration of the total system can be calculated as Wtotal = WA x WB

thus: f (WA x WB) = f (WA) + f (WB)

The only function for which the above relationship is true is the logarithm. Therefore:

S = k · lnW

where k is the Boltzman constant with the units of the entropy.

12.3 Assembly of distinguishable particles

• An isolated system consists of N distinguishable particles.

• The macrostate of the system is defined by (N, V, U).

• Particles interact sufficiently, despite very weakly, so that the system is in thermal equilibrium.

• Two restrictive conditions apply here (conservation of particles)

(conservation of energy)

where Nj is the number of particles on the energy level j with the energy Ej.

Example: Three distinguishable particles labeled A, B, and C, are distributed among four energy levels, 0, E, 2E, and 3E. The total energy is 3E. Calculate the possible microstates and macrostates.

NNn

jj

1

UENn

jjj

1

Solution: The number of particles and their total energy must satisfy

(here the index j starts from 0)

# particles on Level 0

# Particles on Level 1 E

# particles on Level 2E

# particles on Level 3E

Case 1 2 0 0 1

Case 2 1 1 1 0

Case 3 0 3 0 0

33

0

j

jN

EENj

jj 33

0

So far, there are only THREE macrostates satisfying the conditions provided.

Configurations for case 1

Thermodynamic probability for case 1 is 3

Level 0 Level 1E Level 2E Level 3E

A, B C

A, C B

B, C A

Configurations for case 2

Configuration for case 3

Therefore, W1 = 3, W2 = 6, and W3 = 1

.

Level 0 Level 1E Level 2E Level 3E

A B C

A C B

B A C

B C A

C A B

C B A

Level 0 Level 1E Level 2E Level 3E

A, B and C

• The most “disordered” macrostate is the state with the highest probability.

• The macrostate with the highest thermodynamic probability will be the observed equilibrium state of the system.

• The statistical model suggests that systems tend to change spontaneously from states with low thermodynamic probability to states with high thermodynamic probability.

• The second law of thermodynamics is a consequence of the theory of probability: the world changes the way it does because it seeks a state of probability.

12.5 Quantum States and Energy Levels

To each energy level, there is one or more quantum states described by a wave function Ф.

When there are several quantum states that have the same energy, the states are said to be degenerate.

The quantum state associated with the lowest energy level is called the ground state of the system. Those that correspond to higher energies are called excited states.

The energy levels can be thought of as a set of shelves of different heights, while the quantum states correspond to a set of boxes on each shelf

• E3 g3 = 5

• E2 g2 = 3

• E1 g1 = 1

• For each energy level the number of quantum states is given by the degeneracy gj

a particle in a one-dimensional box with infinitely high walls

• A particle of mass m• The time-independent part of the wave function

Ф (x) is a measure of the probability of finding the particle at a position x in the 1-D box.

• Ф (x) = A*sin(kx) 0 ≤ x ≤ 1with k = n (п/L), n = 1, 2, 3

n=1

n=2

The momentum of the particle is P = · k (de Broglie

relationship)

where k is the wave number and h is the Planck constant.

The kinetic energy is:

2

2222

42

1

2

1

2

1

kh

mm

PmvE

• Plug-in

Therefore, the energy E is proportional to the square of the quantum number, n.

Lnk

2

22

2

22

2

2

8

8

mL

nh

L

n

m

hE

In a three-dimensional box

E = h2 ( + + )

8m

where any particular quantum state is designated by three quantum numbers nx, ny and nz .

If Lx= Ly = Lz = L

then, we say nj2 = nx

2 + ny2 + nz

2

Where nj is the total quantum number for states whose energy level is EJ.

An important result is that energy levels depends only on the values of nj

2 and not on the individual values of integers (nx, ny, nz). nj is the total quantum number for energy level Ej.

Since

L3 = V L2 = V⅔

therefore,

For the ground state, nx = 1, ny = 1, nz = 1There is only one set of quantum number leads to this energy,

therefore the ground state is non-degenerate.

)(1

8222

2

2

zyxj nnnLm

hE

)(1

8222

3/2

2

zyxj nnnVm

hE

For the excited state, such as nx2 + ny

2 + nz2 = 6

it could be nx = 1 ny = 1 nz = 2

or nx = 1 ny = 2 nz = 1

or nx = 2 ny = 1 nz = 1

They all lead to

Note that increasing the volume would reduce the energy difference between adjacent energy levels!

For a one-liter volume of helium gas

nJ ≈ 2 x 109

61

8 3/2

2

Vm

hE j

12.6. Density of Quantum States

When the quantum numbers are large and the energy levels are very close together, we can treat n’s and E’s as forming a continuous function.

For

nx2 + ny

2 + nz2 = · V⅔ ≡ R2

The possible values of the nx, ny, nz correspond to points in a cubic lattice in (nx, ny, nz) space

)(1

8222

3/2

2

zyxj nnnVm

hE

Let g(E) · dE = n (E + dE) – n(E) ≈

Within the octant of the sphere of Radius R;

Therefore,

dEdE

Edn )(

dEEmh

VdE

dE

EdndEEg 2/12/3

3

24)()(

Note that the above discussion takes into account translational motion only.

Where γs is the spin factor, equals 1 for zero spin bosons and equals 2 for spin one-half fermions

dEEmh

VdEEg s

2/12/33

24)(

• Example (12.8) Two distinguishable particles are to be distributed among nondegenerate energy levels, 0, ε, 2ε , such that the total energy U = 2 ε.

(a)What is the entropy of the assembly?

(b)If a distinguishable particle with ZERO energy is added to the system, show that the entropy of the assembly is increased by a factor of 1.63.