next up previous
Next: 5.1 Ideal quantum gas: Up: sp Previous: 4.4 Problems for Chapter


5. Statistical Quantum Mechanics

Counting the states in phase space takes some consideration if we want to apply quantum mechanics. We have already mentioned the fact that quantum particles are not distinguishable, and we took this into account in a heuristic manner, by introducing the permutation factor $1/N!$ in the partition function.

Another feature of the quantum picture is that state space is not continuous but consists of finite ``raster elements'': just think of the discrete states of an ideal quantum gas. Finally, it depends on the ``symmetry class'' of the particles how many of them may inhabit the same discrete microstate. Fermions, with a wave function of odd symmetry, can take on a particular state only exclusively; the population number of a raster element in pahese space can be just $0$ or $1$. In contrast, even-symmetry bosons may in any number share the same microstate.

What are the consequences of these additional counting rules for Statistical Thermodynamics? To seek an answer we may either proceed in the manner of Boltzmann (see Section 2.2) or à la Gibbs (see Chapter 4). For a better understanding we will here sketch both approaches.


Subsections
next up previous
Next: 5.1 Ideal quantum gas: Up: sp Previous: 4.4 Problems for Chapter
Franz Vesely
2005-01-25