Physics Asked by CVJM on January 15, 2021
I want to understand the thermodynamic properties of the free fermi gas and for that I have the following two basic problems:
The free fermi gas is treated in the quantum mechanics setting. In this setting and in the derivation of the thermodynamic properties the important quantity is number of "states". I am very confused about that, since each linear combination of two given states is in quantum mechanics also a valid state, such that the number of possible states are even for a discrete system infinite.
However, it seems that one is just interested in the number of eigenstates. Is that right?
And if so, why the density of these states is the interesting one in the thermodynamic limit.
Thank you!
The density of states (in energy space) counts the number of linearly independent states in a given energy interval. Therefore, even you may have an infinite number of states by combining two of them, the counting of the independent states stops at 2.
Of course, since the counted states have a well definite energy, they must correspond to a well-defined energy (eigenstates, if the spectrum is discrete).
We are interested in the density of states because it corresponds to a well-defined concept even at the thermodynamic limit and also because it is a convenient a simple way to encode the information needed to evaluate thermal and transport properties. Moreover, it is a measurable quantity.
Correct answer by GiorgioP on January 15, 2021
Let's take as an example of a free fermi gas, that of electrons in a metal. The first attempt at understanding the properties of metals is to assume a free electron fermi gas, so this is a good system to study.
So by using the word "free", we are ignoring the periodic potential of the ions and any interactions between electrons themselves. So the energy of the electrons are just kinetic and the states are defined by $$E_k= frac{hbar k^2}{2m}$$ where if we are working in 3 dimensions, k is the sum of the squares of the three components of the k vector.
With no interaction between the electrons with the ion potential, or themselves (or other things like phonons), at T=0 one just fills up these energy levels with the electrons. The Fermi-Dirac distribution has a value of 1 up to the Fermi level and then it is 0 after that. As the temperature changes, the FD distribution traces the changes in the occupation of the various energy states. Many text books show how the FD distribution looks at various temperatures.
So yes, for "free" fermions, like electrons, we are just interested in the occupation of the single-particle states. As the temperature changes, only those electron states that are within about kT of the Fermi level can participate in things such as heat capacity or electrical conductivty, because the Pauli principle forbids the rest of the electrons from participating as there are no empty free states available. That is why we are only interested in the density of states at the Fermi level.
As an example, the heat capacity of a free electron gas in a metal is: $$C_{el}=0.3pi^2D(epsilon_f)k_b^2T$$ Here $D(epsilon_f)$ is the density of states at the Fermi level.
A next step in forming a more complex model might be to form the fully antisymmetric state using a Slater determinant. This would be a sum of states similar to what you mention above. Usually this is cumbersome and leads directly into the introduction of the number representation.
Answered by CGS on January 15, 2021
Maybe it is easier to understand lattice vibrations. The Einstein model is independent oscillators. Debye included interactions, but that does not change the number of modes. At high temperature, both models result in the Dulong & Petit value of $c_v = 3 R.$
Answered by Pieter on January 15, 2021
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