| Gibbs Paradox | Quantum microcanonical ensemble | Quantum ideal gas |
-> Gibbs Paradox: Fact: If we have two containers of the SAME gas at the SAME temperature and density separated by a partition. Removing the partition does not change the system and hence there can be no entropy change. However, if we use the entropy function for a classical gas (Eq. 6.59), we get an entropy increase, presumablly the final volume is larger. This is incorrect, and is called the Gibbs paradox. |
->Analysis of paradox: Fundamentally, all particles are either Fermions or Bosons. If the gas particles are Fermions, then the particles cannot occupy the same momentum and position (Pauli exclusion principle). But, in our classical treatment, we assume that the phase point (p1=p2=...;q1=q2=...) is an allowable state. If the particles are Bosons, then all the particles should condensed into one state. The error of treating the particles as distinguishable persists even at high temperature where normally classical statistical mechanics is valid. When we calculate the allowable phase space, we have V**N because we allow each particle to occupy the entire volume. This CANNOT be so for indistinguishible particles (e.g. Pauli exclusion). -> Resolving the paradox: Gibbs fixed the problem empirically. He proposed that there was an over-counting of the number of allowable states, one should divide the number by N factorial. One could hand-wave an agrument for the N factorial, but it is useless. The N! correction is only valid at high temperature. At low temperature, the correct treatment must come from quantum mechanics (to be discussed later in Ch. 8, 11, & 12). -> Entropy of mixing If the gas in two containers are distinguishable (say, helium and oxygen), then there is an entropy of mixing. (H.W#1, Problem 5) Question: In a real experimental setup, what do we mean by the two gas are distinguishable? |
In most undergraduate text, the postulate for quantum microcnaonical ensemble is state as followed: "In a closed system with energy E, all the quantum (energy-eigen) states with that energy E are equally probable" Why do use the energy-eigenstate as the preferred basis in the case of thermal equilibrium? In quantum mechanics, the quantum state of a system is represented by a many-body wavefunction which need not be an energy-eigenstate. Huang implicitly stated the reason of using the energy-eigenstates (sections 8.2-8.2) but it is not very clear. In fact, he seems to give the impression that any basis set would do. Here is my opinion: At thermal equilibrium, all macroscopic measureable quantities are time-indepedent. In quantum mechanics, all measurables can be expressed in terms of the density matrix operator. In order for the measurables to be time-indepedent, the density matrix operator must commute with the Hamiltonian. => the density matrix is a diagonal matrix in the basis of energy-eigenstates. Can we prove using quantum dynamics that starting from any arbirary initial condition, the system will evolve in such a way that the density matrix operator eventually commutes with the Hamiltonian? No one has such a proof so far. We have to postulate it. -> At thermal equilibrium, the wavefunction can be regarded as an "incoherent" superposition of energy-eigenstates. This is called the postulate of random phase. See Section 8.1. Eq. (8.6) This postulate will imply that the density matrix is diagonal in the energy-eigenstate basis. Note: Most undergraduate text neglect the mention of the postulate of random phase. Postulate of equal probability: For a closed system with energy E, we impose the postulate of equal probability just as in the classical treatment. "Every energy-eigenstate with energy E (actually within the range of E and E + delta) is equally probable". Eq. (8.5). If we let Gamma(E) = the number of energy-eigenstates within energy E and E + delta, then the probability of each eignestate is 1/Gamma. Entropy function: Same formula as before: S = k log Gamma(E) The difference between classical statistical mechanics and quantum statistical mechanics is the way Gamma(E) is calculated. Density Matrix Operator: Note: Huang expressed the partition function in terms of density matrix operator, it is good to be exposed to this notation (although it is not necessary to use the density matrix because the density matrix is diagonal and the diagonal elements are simply probabilities ). |
Using microcanonical ensemble to treat the quantum ideal gas is actually rather complicated (it is easier to use the grand canonical ensemble, to be discussed later). We are doing this as a comparison with the treatment of classical ideal gas. The difference between quantum ideal gas and classical ideal gas is that the particles are indistinguishable (both have continuum energy spectrum). Fermion - anti-sysmmetric wavefunctions Boson - symmetric wavefunctions When we calculate Gamma(E), we should only included anti-symmetric (symmetric) energy-eigenstates for Fermion (Boson) system. The best way to keep track of anti-symmetric or symmetric wavefunction is to use occupation number for each single particle orbital (second quantization method). For Fermions: n sub p= 0, 1 For Bosons: n sub p=0,1,2,3,... Since the orbitals are very close in energy (as V -> infinity, the energy spectrum approaches a continuum), we should consider a group of orbitals within an energy range and ask what is the average occupation number for the group. Results are the famous Fermi-Dirac distribution and Bose-Eistein distribution (Eq. 8.45). At "high temperature", both distributions approaches the so-called Boltzmann distribution. Q. How high is high temperature? Since the energy spacing is zero, isn't kT always greater than energy spacing and hence any non-zero temperature is considered high temperature? Or there another energy scale to compare with? |