Fermi gas

A Fermi gas is a model for a system of many non-interacting fermions, the simplest example of which is a fermionic ideal gas. The key property is that fermions are subject to the Pauli exclusion principle and can't simultaneously occupy the same state, which leads to a "pseudo-potential" that pushes fermions away from each other.

Zero temperature and Fermi energy#

In the low-temperature limit $T\to 0$ and due to the exclusion principle, the Fermi-Dirac distribution approximately becomes

What this means is that all states with energy lower than $\mu$ will be occupied, and all states with higher energy will be vacant. We can use the Heaviside step function $\Theta$ to represent this symbolically:

The specific energy threshold is known as the Fermi energy $E_{F}$. This energy leads to a momentum called the Fermi momentum $p_{F}=\sqrt{ 2mE_{F} }$. In phase space, this momentum occupies a spherical region, which contains all the occupied states discussed before. The surface of this sphere is known as the Fermi surface. This odd state is known as quantum degeneracy (not to be confused with state degeneracy, although the two are closely related). Despite being under no potential, fermions in this state are subject to an effective pressure, known as degenerate pressure, which prevents them for compacting further. This pressure is a considerable factor for fermionic systems in extreme environments, such as white dwarfs, which rely on electron degeneracy pressure, and neutron stars, which rely on neutron degeneracy pressure.

Fermi energy can be calculated purely from knowledge of the exclusion principle, that fermions are non-interacting and by choosing some appropriate boundary conditions.

1D#

Consider a one dimensional system composed in a length $L$ of space.

Open boundary conditions#

Starting from the time-independent Schrödinger equation, we have

The solution depends on the boundary. If we set open boundary conditions (also known as a hard wall, that is, the edges of the system are impassable, so $\psi(x=0)=\psi(x=L)=0$), our system turns into a infinite square well, so the solution is $\psi_{n}\propto \sin kx$. We also have $kL=n\pi$, which means

If our $N$ fermions have two possible Spin eigenvalues, $N/2$ will be spin up and $N/2$ will be spin down due to the exclusion principle. (For the rest of the article, we'll only consider fermions with two possible spins, which are by far the most common, but it is possible to generalize to $\ell$ spins by multiplying all formulas directly dependent on spin count by $\ell$.) Since each state has precisely one fermion in it, the highest occupied state must be $N/2$. Thus the Fermi energy is found at $n=N/2$:

Periodic boundary conditions#

With periodic boundary conditions, our system becomes a looping ring defined by $\psi(x=0)=\psi(x=L)$. Since fermions are free, their eigenfunctions are plane waves $\propto e^{ikx}$, so this is like saying $e^{ikx}=e^{ik(x+L)}$ and therefore $e^{ikL}=1$. This is true when $2\pi n=kL$, where $n=0,\pm 1, \pm 2,\ldots$. Since the energy is given by the wave vector, we get

The Fermi energy is

since $n_\text{last occupied}=N/4$, unlike in the open boundary case. This is because, beyond spin degeneracy, here we have to consider both signs of $n$, whereas in the square well we drop the negative $n$'s. This is balanced out by the additional $2$ in the periodic condition (before it was $\pi n=kL$, now it is $2\pi n=kL$).

Density of states#

We can also use a density of state (DOS) function through the Thomas-Fermi approximation. The state counting function for momentum is

This counts the number of states up to an energy $E$. In one dimension, this becomes

since the measure in 1D is the length of the interval, which in our case is the diameter of the sphere, i.e. $2p$. The state count at Fermi energy for each spin is given as before by the number of particles, divided by $2$:

The single spin Fermi energy returns the previous result:

The DOS for both spins is

where the $2$ is due to considering both spins. Due to the one state = one particle clause, we can find the Fermi energy exclusively from the DOS, if we somehow have access to it directly:

Reversing the formula gives the familiar result

Zero temperature internal energy#

The total internal energy of a Fermi gas at zero temperature, denoted $U_{0}$, can be found from the Fermi energy. The internal energy is just the sum of the energies of each occupied state. For doubly-degenerate spins in open boundary conditions, that is

We know the sum

The square is

where the approximation holds at large $s$ (proven by taking an integral instead of a sum, which are essentially the same for large enough numbers). Back to the energy, we have

The $1/3$ factor depends on the dimension, but $N$ and $E_{F}$ are always the same.

The same result can be found from the DOS by energy:

3D#

We now move on to 3D, in a volume $V=L^{3}$.

Open boundary conditions#

Using open boundary conditions in 3D is more challenging. The Hamiltonian is

and we're still in an infinite square well, just a 3D one (an infinite square box?). Our position eigenfunctions are

where $n_{x,y,z}=1,2,3,\ldots$. Our energy eigenvalues are

Reasonable enough so far. The issue is now finding what the last occupied value is. In 1D it was easy: the number of particles divided by $2$. But here it's complicated, as there are three separate degrees of freedom. For instance, a singly-excited state can be any of $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. At higher excitations, there are more and more combinations. In fact, this is a combinatorics problem that we'd need to solve. It's easier to just pick another set of boundary conditions in which this does not happen.

Periodic boundary conditions#

Using periodic boundary conditions and with wavefunctions $\psi_{\mathbf{k}}(\mathbf{r})=\frac{1}{\sqrt{ V }}e^{i\mathbf{k}\cdot \mathbf{r}}$ for a direction $\mathbf{k}=\frac{2\pi}{L}\hat{\mathbf{n}}$, the Schrödinger equation

is solved by

Again, now we need to figure out what the last occupied $\mathbf{k}$ is, which is the same problem as before since $k\propto n$. Our best bet is to just go with density of states directly.

Density of states#

The state function is given by the usual Thomas-Fermi approximation method. For each spin, we have:

At Fermi energy, the state count per spin is still

This is the saving grace. The Pauli exclusion principle doesn't care about dimensionality and will give the same number of states regardless. The only thing that matters is the number of states for each energy level, i.e. the degeneracy, i.e. the number of spin states. If we compare the two equations, we find

where we defined the Fermi wavevector as

The DOS function for both spins can be found to be

We can find the number of particles from the DOS:

Zero temperature internal energy#

The zero temperature internal energy is

By using $E_{F}^{5/2}=E_{F}^{3/2}+E_{F}$ and substituting the previous formula for $E_{F}$ in $E_{F}^{3/2}$ we get

As we can see, only the numerical factor changed from the 1D version.

Nonzero temperature#

Virial expansion of the equation of state#

The equation of state of a fermion ideal gas can be found through a quantum ensemble, with the quantum grand canonical ensemble being the most convenient. For a derivation, see Ideal gas > Equation of state. The equation is

where $k_{B}$ is the Boltzmann constant, $T$ is the temperature, $\lambda$ is the de Broglie thermal wavelength and $f_{5/2}(z)$ is a Fermi function. We can approximate it by using its series representation:

($v=V/N$ is the volume per particle). We know that

If we substitute $z$ with its first order approximation $\lambda ^{3}_{T}/v$ we get¹

and if we put this back in the equation of state we get

where we "calculated" the square of the expansion but only kept the first term and hid all the higher order ones behind the dots, since we don't care about them. Note the consequence here. The classical ideal gas is

Meanwhile, a fermion ideal gas is

This expansion is a form of the virial expansion and the terms are called virial coefficients. Specifically, the term after $1$ is the second virial coefficient and is usually the most relevant one. This series isn't a new thing in statistical mechanics: the virial expansion normally occurs when you generalize an ideal gas to an interacting gas, which also includes internal interactions. But there are no interactions here, it's just inert fermions. The pseudo-interaction is the manifestation of the Pauli exclusion principle, which can be interpreted as a sort of repulsive force between fermions that prevents them from occupying the same state. It's not really a potential, but it certainly behaves like one.

Fermi energy#

All virial coefficients except the first are only relevant at temperatures that are low compared to the Fermi temperature. Far above and the Fermi-Dirac distribution converges to the Maxwell-Boltzmann statistic, which gives the classical results. But since we are at low temperature anyway, we might as well use the Sommerfeld expansion. This lets us explore the behavior of the Fermi energy at low-but-not-zero temperatures. We should start off by saying two things: one, the Fermi energy is not well defined in $T>0$. Formally, it is only defined at $T=0$, so we can only hope to get an approximation; and two, recall that the Fermi energy matches the chemical potential at $T=0$, so if we are to seek $E_{F}$, it is sensible to begin by looking for $\mu$, or equivalently $\ln z$. The latter is found in Fermi functions, which as we've already said can be approximated with the Sommerfeld expansion.

Starting from the density equation and using only the first term of the expansion

The logarithm is

But we also have

As such, the Fermi energy is

The average occupation number is

If $\varepsilon_{\mathbf{p}}<E_{F}$ we get $\langle n_{\mathbf{p}} \rangle\simeq1$, otherwise $\langle n_{\mathbf{p}} \rangle\simeq0$.

To higher orders#

We again employ the Sommerfeld expansion, but this time to the second order

We can invert the terms to find

This is a much harder problem to solve than the first order approximation and only the end result is given. We can now substitute $\ln z=\beta \mu$, call $\mu \simeq E_{F}$ and solve for $\mu=E_{F}[1+c(k_{B}T/E_{F})^{2}+\ldots]$ using perturbation theory:

where $T_{F}=E_{F}/k_{B}$ is the Fermi temperature. The impact of these corrections is dependent on the ratio between temperature and Fermi temperature, which means that "high" or "low" temperature here is entirely define with respect to the Fermi temperature. The latter is found on a material-by-material basis. For instance, in metals it is $T_{F}\sim 10^{4}\text{ K}$, which means that for room temperature metals ($T\sim 300\text{ K}$), all corrections are tiny expect at most the first. In neutron stars, $T_{F}> 10^{7}\text{ K}$.

Dealing with spin#

Up until now, we've ignored the presence of Spin. If we call the spin $s$, the spin degeneracy is going to be

and the following condition must hold:

A $g$-degenerate gas of $N$ fermions behaves like $g$ independent Fermi gases stacked on top of each other.

Zero temperature#

At $T=0$, the Fermi energy is found, for instance, with the density of states.

Higher degeneracy reduces the Fermi energy due to there being more states to fill.

Internal energy#

The internal energy $U$ of a Fermi gas could be derived by direct integration:

That said, this integral is a mess. It's much easier to use something like a quantum grand canonical ensemble and use its partition function:

and use the relation

The $f_{5/2}(z)$ function is independent from $\beta$, so we only need

We can finally write

This is essentially an extended equipartition theorem. Unfortunately, actually calculating the two $f$ function in closed form is not possible. We can however use a Sommerfeld expansion for both of them at $z\gg 1$. The $f_{3/2}$ expansion is above. The $f_{5/2}$ expansion is

Using the first term of each we get

For $T=0$ we can see $k_{B}T\ln z=\mu=E_{F}$ and we have

which proves that this is a self-consistent description. If we add more terms we get

and so

From this, we can find other thermodynamic quantities.