Next, let us find the occupancy (mean occupation number) fk,ms:
Now, we note that for |x|<1,
This tells us that the sum in the numerator is thus , and so
Compare this to the occupancy for a fermion fluid ; the difference is a change of sign in the denominator. Note that unlike the fermion case, for a boson fluid, the occupancy can be greater than one.
Archive for September, 2009
Next, let us find the occupancy (mean occupation number) fk,ms:
Let us consider a metal. When metal atoms are brought together, the few electrons in the outermost shell break away from the atoms, and can move (mostly) freely through the solid. On average, the electrical repulsive forces between these electrons tend to be neutralized by the electrical interactions with the background positive charges of the fixed ions. Thus, these “conduction band” electrons can be approximated by an ideal spin-½ fermi fluid. For an alkali metal, such as sodium or potassium (where the approximation works best), we have a concentration of (approximately) one conduction electron per ion. With an interionic distance of ~5 Å, we have an electron density on the order of 1028 to 1029 electrons per cubic meter.
Now, let us find the Fermi energy μ0, which you may remember from here is the value of the Fermi level at T=0. Here, the occupation probability is 1 for and zero for . Using our integral formula for the number of particles in an ideal Fermi fluid (see here), we see:
For our metal, this is
The temperature TF for which μ0=kTF (the Fermi temperature), is then on the order of 104 to 105 K. For room temperature, we are at T much lower than this, placing us clearly in the quantum regime
(For 300 K, is approximately 43 Å). A Fermi fluid in such a state is referred to as a “degenerate” Fermi gas.
Now, the energy of the Fermi fluid at absolute zero is
so the energy per particle is . For our electron gas in our metal, this is on the order of 104 K in equivalent temperature units. For nucleons in the nucleus of an atom of a heavy element, the fermi energy even higher, on the order of 30 MeV, so the average kinetic energy is on the order of 18 MeV, which is still in the non-relativistic domain.
Note that the density of states increases as energy increases, so, using our result from here that the Fermi level is “repelled” by energies with higher densities of states, we should expect the Fermi level to decrease as the temperature rises. Similarly, an increasing number of electrons at energies just below the Fermi energy will be promoted to states with energies above the Fermi energy.
For an arbitrary function , we consider the integral
Using the fact that resembles a step function at low temperatures, one can, with some math, expand this integral in a power series in temperature, getting:
where the upper limit in the right-hand side integral and the points of evaluation in φ‘ and φ”’ is the temperature dependent Fermi level μ, not the Fermi energy μ0.
To find the dependence of the Fermi level on temperature, we first use . This latter integral is the above integral with , meaning
To find μ to second order in T, let us replace μ by μ0 in the second-order term of the above expansion, and using , we solve for μ to get:
confirming the Fermi level decreases with energy.
Similarly, we can develop a series expansion for internal energy as a function of temperature:
This tells us that the heat capacity at constant volume is
, where the term is the classical value; the factor in parentheses is thus a quantum correction factor. For our metal at room temperature, this factor is on the order of 10-1. This drastic reduction in the heat capacity of the conduction band electrons from what classical theory predicts was one of the motivations in the development of the Fermi-Dirac distribuion discribing the occupancy of states in a Fermi gas (see here).
Now, recall that for our ideal Fermi fluid, we have equation of state . For small T, the internal energy will be roughly the value at absolute zero, ; thus,
, where n is the particle density. Note that this is substantially larger than the P→0 as T→0 of the classical ideal gas; for our metal, this pressure is on the order of 1010 Pa! (In our metal, this enormous pressure is countered by the Coulomb attraction between the electrons and the positively charged fixed ions).
This additional, non-classical pressure, which ultimately derives from the Pauli exclusion principle, is known as degeneracy pressure. For massive astronomical objects without internal fusion, most notably white dwarfs, the object is supported against gravitational collapse by the degeneracy pressure of a degenerate electron gas. Similarly, neutron stars are supported by the degeneracy pressure of a degenerate neutron gas.
Combining and , we see ; this gives the dependency of the degeneracy pressure on density (for non-relativistic particle velocities). Further, for , the degeneracy pressure is effectively independent of the temperature.
It took almost two weeks for the replacement part to get here, but my computer is fixed, and back in my possession.
Continuing from last week’s discussion of the thermodynamics of the ideal Fermi fluid, we now explore the classical limit.
First, I introduce the physical property known as fugacity. Fugacity is defined as , and can be seen as an alternative parameter to μ. Note that constant ξ is equivalent to holding the product βμ constant. In this post on the grand canonical ensemble, I showed that , where ; thus, using the chain rule to rewrite in terms of fugacity,
The key difference between the ideal Fermi fluid and the classical ideal gas is that a given fermion particle is not free to occupy an arbitrarily chosen state, as the Pauli exclusion principle forbids it from states that are already occupied. At high temperature or low density, though, the probability of occupation for each orbital state becomes small, thus reducing the effect of the Pauli exclusion principle; so we see the gas becomes classical when the density is sufficiently low or temperature sufficiently high, so that the occupancy for each state is small; this happens when . For this to happen for all energies, we need , or in other terms, ; the classical regime occurs when the fugacity is small. (Since β>0, we see that in the classical regime, the Fermi level must be at very negative values). In this range, we see that since ,
Note that the energy dependence in this classical approximation is the Boltzmann factor, confirming that in the classical limit, we approach the Maxwell-Boltzmann distribution of the classical ideal gas.
Now, we find the needed physical conditions (density and temperature) so that the fugacity is small. We found last week that the number of particles is
In our classical limit of small fugacity, we see that . Thus, we see that
where is a quantity with units of length, and g=2S+1 is the number of different spin orientations, here equal to 2.
which is the classical result for an ideal gas (with only translational modes for the gas molecules), and which combines with our to give the ideal gas law .
Solving our particle number formula for the fugacity, we see: , where is the particle density. Thus, the classical condition means that , and the classical quantum boundary occurs when .
Now, let us examine the physical interpretation of the quantity . This is called the “thermal de Broglie wavelength,” and is equivalent to the de Broglie wavelength of a particle of mass m and velocity , roughly that of the average particle of an ideal gas at temperature T. The thermal de Broglie wavelength decreases with increasing temperature.
Note that for a gas of particle density n, the average interparticle distance will be approximately , and so the classical condition is equivalent to saying that the thermal de Broglie wavelength must be significantly smaller than the average interparticle distance, again confirming the need for high temperature or low density.
Next week, we will explore what happens when an ideal Fermi fluid is firmly in the quantum regime.
I’m likely going to be without a computer for the next week or so (depending on time for repair and how long it takes to ship parts here to Alaska). I have scheduled the Physics Friday and Monday Math posts for the next week or so, so those should be here at their appropriae times, but don’t expect any other posts for a bit here.
Previously, I used the grand canonical formalism to derive the fundamental thermodynamic relation for a system of spin-½ fermions in a quantum system with a small number of orbital states. Now, let us extend this analysis to the ideal Fermi fluid, the quantum analog to the classical ideal gas for fermions; a system of fermion particles with no (or negligible) interaction forces, such as a collection of neutrons in the interior of a neutron star, or a low-temperature gas of 3He atoms.
As with our previous work, we will use the grand canonical formalism; the fundamental relations will be independent of these particular boundary conditions. We consider spin-½ fermions, so that for each orbital state we have two states, given by ms=±½. The orbital states available can be specified by the wavenumber vector k of the wavefunction (see here). As before, our grand partition sum factors: . As we are considering fermions, each orbital state is either empty or singly occupied. The energy of an empty state is zero, and the energy of an occupied state (k,ms) is (independent of the spin direction). This means the partition sum for a state (k,ms) is
In turn, we can consider the product of the two states of orbital k and opposite spin directions as the “partition sum of mode k” , with terms for the empty mode, singly occupied mode (with two possible orientations), and doubly occupied mode.
Each state is independent, so we have mean occupation number
(in analogy to here).
Now, we can find the grand potential:
Now, consider our particle confined to a cubical box of volume V=L3; I demonstrated the quantum particle in a cubical box here; we have the allowed orbital states given by , , where nx, ny, and nz are positive integers.
Thus, the number of states of energy less than or equal to E is the number of triplets of positive integers (nx, ny, nz) for which , where . Considering the three dimensional abstract space with coordinates (nx, ny, nz), we see our points are those contained in the eighth of the sphere of radius n in the first octant. As each point corresponds to a cube of unit volume in this space (visualize the lattice of integer points), for large n, we can approximate the number of orbital states of energy less than or equal to E by the volume of this eighth of the sphere:
Differentiating this with respect to energy, we can obtain a “density of (orbital) states”:
If we multiply this by two, since we have two spin orientations per each orbital state, we can approximate our sum in the grand potential with an integral:
There is no closed-form expression for the integral (though it can be expressed in terms of polylogarithms), so the physical properties derived from Ψ must also be expressed in terms of integrals (or polylogarithms), which can be computed numerically or via approximation methods.
We can, though, immediately determine the number of particles Ñ and internal energy U:
(which we could also have obtained via and our integral expression for Ψ),
[These integrals may also be expressed in terms of polylogarithms:
. See here for more of the math involved.]
Now, if we perform integration by parts on the integral for Ψ, with and , we obtain
We also recall from the end of this post, that for simple systems, the equation of state is . Thus, we see that
. Note that this is exactly the result for a classical ideal gas whose molecules have only translational degrees of freedom available (see here and here).
Next week, I intend to show that this reduces to the classical ideal gas for high temperature, and extract the criterion separating the quantum and classical regimes.