
Appendix B
Quantum statistics and Boltzmann kinetics
B.1 Grand partition function
If the number of particles is large, the life history of a single particle is not
so important. What really determines the physical properties of a macroscopic
system is the average distribution of particles in real and momentum space.
A statistical approach combined with quantum mechanics allow for a full
description of the macroscopic system. Within this approach, we first assume
that we know the exact eigenstates |Q and energy levels U
Q
of the many-
particle system. The particles interact with the walls of their container (i.e. with
a thermostat) so as to establish a thermal equilibrium but not to such an extent
as to affect the whole set of quantum numbers Q and energy levels. Due to the
interaction with the thermostat, the energy and the total number of particles (N)
fluctuate. The system takes some time staying in every allowed quantum state Q.
The statistical probability P of finding the system in a particular quantum state
Q with the total number of particles N is proportional to this time. It depends on
U
Q
and N: P = P(U
Q
, N).
It is not hard to find the probability P(U
Q
, N) in thermal equilibrium. Let
us consider a system containing two independent parts 1 and 2 with the energies
U
1,2
and the number of particles N
1,2
, respectively. If the two parts do not interact
with each other, the probability P
1
(U
1
, N
1
) that system 1 should be in the state
U
1
, N
1
is independent of the probability P
2
(U
2
, N
2
). The probability for two
independent events to occur is equal to the product of their separate probabilities.
This means that the probability P(U, N) of finding the whole system in the state
U, N is the product of P
1
(U
1
, N
1
) and P
2
(U
2
, N
2
),
P(U, N) = P
1
(U
1
, N
1
)P
2
(U
2
, N
2
). (B.1)
The thermal equilibrium is described by a universal probability, P
1
(U, N) =
P
2
(U, N) = P(U, N), which depends only on U and N but not on the particular
system. Then taking the logarithm of both parts of equation (B.1), we obtain
ln P(U, N) = ln P(U
1
, N
1
) + ln P(U
2
, N
2
). (B.2)
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