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. .
(conservation of particles)
(conservation of energy)
where Njis 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.
Solution: The number of particles and their total energy must satisfy (here the index j starts from 0)
So far, there are only THREE macrostates satisfying the conditions provided.
Detailed configurations (i.e. microstate) for case 1
Thus, thermodynamic probability for case 1 is 3
Microstates for case 3
Therefore, W1 = 3, W2 = 6, W3 = 1, and Ω = 10
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)
f (Wtotal) = f (WA) + f (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 entropy.
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
with k = n (π/L), n = 1, 2, 3
P = · k (de Broglie relationship)
where k is the wave number and h is the Planck constant.
The kinetic energy is:
Therefore, the energy E is proportional to the square of the quantum number, n.
E = h2 ( + + )
where any particular quantum state is designated by three quantum numbers x, y, and z.
then, we say nj2 = nx2 + ny2 + nz2
Where is the total quantum number for states whose energy level is EJ.
only on the values of nj2 and not on the individual
values of integers (nx, ny, nz).
(nj is the total quantum number for energy level Ej)
L3 = V L2 = V⅔
For the ground state, nx = 1, ny = 1, nz = 1
There is only one set of quantum number leads to this energy, therefore the ground state is non-degenerate.
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
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.
nx2 + ny2 + nz2 = · V⅔ ≡ R2
The possible values of the nx, ny, nz correspond to points in a cubic lattice in (nx, ny, nz) space
Within the octant of the sphere of Radius R;
n(E) = ⅛ · πR³ = V( ) · E
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