Chapter 7 Bose and Fermi statistics

§7-1 The statistical expressions of thermodynamic quantities 1 、 Bose systems: Define a macrocanonical partition function:

Now, let’s see the entropy and Lagrange Variable factor

Because:

So that,

Boltzmann relation

2 、 Fermi system (Bose)

Boltzmann relation

Addition: Define a new thermodynamic quantity –grand potential It is of great importance for the statistical treatment of thermodynamic problems.The total differential reads The remaining thermodynamic quantities can be calculated by differentiating the grand potential:

Because of Euler’s equation: The grand potential is identical with -pV

Because ：

§7-2 Bose and Fermi weak degeneracy ideal gas non-degeneracy condition: Now, we consider a condition, just as and is small,but can not be neglected. We called this weak degeneracy condition. Under this condition, we want to deal with the Bose ideal gas and Fermi ideal gas respectively.

Here g is the degeneracy caused by the particle’s spin. Photon g=2 Electron g=2

Bose Fermi This equation can be used to determine the Lagrange factor.

So that, we expand the into:

In the equation, the first term is the energy calculated according to the Boltzmann distribution, and the second term is the correlation energy caused by the quantum statistics.

0

§7-3 Bose-Einstein conglomeration From this section, we can see that when equals or is more than 2.612, the unique Bose-Einstein conglomeration phenomenon will appear. Because the can not be negative, So that

That is,here is the lowest energy level of Bose particles. If we assume the lowest energy of the Bose particle is 0, then, From the equation We can see that, the chemical potential is the function of temperature T and the density of particle number n.

If we substitute integration for the summation, there is —V

If the n is fixed, according to the equation above, we can see that the higher temperature, the smaller chemical potential. When the temperature reduces to a critical value T c,the chemical potential will reach its highest value 0. Because:

Then, if the temperature keep on reduces from Tc, what will happened? But, when

There is a contradiction in it, because : <

Here n 0 (T) is the density of particle number on the energy level, when temperature is T( ),.

When,Bose particles will accumulate on the energy level rapidly, and the density of particle number reach the same order of magnitude with the total particle number density n.This phenomenon is just the Bose-Einstein conglomeration.

T c is called conglomeration temperature, on the energy level of,we can see that the momentums of Bose particles are also 0, so, Bose-Einstein conglomeration is also called momentum conglomeration.

§7-4 Photon gas T: N(t) const. According to the idea of particles, we can regard the photon field in the cavity as a photon gas.

Bose statistics:

Here, Because the spin degeneracy of photon is 2 (1.-1),

Substitute for the equation

The equation above is called Planck’s formulation.

Integrating the equation above, we can obtain the total energy of the cavity.

We can also use another solution:

It is the same with what we have obtained before.

§7-5 The free-electron gas in the metal A further, very useful model system is that of a noninteracting non relativistic gas of Fermi particles. Nucleons in atoms, as well as electrons in metals, can be regard as an ideal Fermi gas to first approcimation. The case T= 0 has here a special importance. It stands for the mean particle number on the each quantum state.

V— However, since the particle possess 2s+1 different spin orientations which are energetically degenerate in the interaction free case, Equation above must be multiplied by an additional degeneracy factor g=2 We want to rewrite the sums in terms of integrals.

We can also see that the is the function of T and n. When T=0K Pauli principle requires an energetically higher state for each new particle, and the is the highest energy level of electrons.

Is called Fermi energy level. Is called Fermi momentum, which is the largest momentum.

For example: Cu (cuprum) 300K

T＞0T＞0

In the first term

Because the integration comes from,especially when the z is small. So, we can expand the numerator into power series.

We can also use another method:

If we integrate the term in the equation above by parts, it follows that:

Exercise: 8.3 Request the entropy and pressure of weak degeneracy Bose ideal gas and Fermi ideal gas.