Filling Fermions: The Non-Interacting Ground State

The Fermi Sea and the Ground State

Since our Hamiltonian describes non-interacting fermions, finding the many-body ground state is wonderfully straightforward. We simply fill the single-particle energy levels, $E(k)$, starting from the lowest energy available, until we have accommodated all $N_e$ electrons. This is much like the Aufbau principle for filling atomic orbitals.

The Pauli exclusion principle dictates the rules of this filling procedure: each distinct momentum-spin state $(k,\sigma )$ can be occupied by at most one electron. Since there are two spin states ($\sigma =\uparrow ,\downarrow$), each momentum level $k$ can hold up to two electrons. At zero temperature ($T=0$), the electrons settle into the lowest possible energy configuration. The collection of occupied states is called the Fermi sea. The energy of the highest occupied level is the Fermi energy, denoted $E_F$, and its corresponding momentum is the Fermi momentum, $k_F$.

The following reading formalizes this procedure.

As the text explains, the many-body ground state, $|{\Psi }_0⟩$, can be written by applying the creation operators for all the filled states to the vacuum state $|\text{vac}⟩$, which contains no electrons:

The total energy of this ground state is simply the sum of the energies of all the occupied single-particle states:

The boundary in momentum space that separates occupied states from unoccupied states is called the Fermi surface. In one dimension, this "surface" is just two points, $\pm k_F$. In higher dimensions, it becomes a line (2D) or a surface (3D), whose shape is determined by the band structure $E(\mathbf{k})$ and the electron filling.

To see a more general case, let's briefly consider a 2D square lattice.

The Role of Temperature: The Fermi-Dirac Distribution

So far, we have been working at zero temperature, where the boundary between filled and empty states is perfectly sharp. In this scenario, the occupation number of a state with energy $E$ is exactly 1 if $E\le E_F$ and exactly 0 if $E>E_F$.

At any finite temperature ($T>0$), thermal energy allows electrons near the Fermi surface to be excited into states above $E_F$, leaving empty states (or "holes") behind. This "smears out" the sharp boundary. The probability that a state with energy $E$ is occupied at a given temperature is described by the Fermi-Dirac distribution:

Here, $\mu$ is the chemical potential, which sets the energy scale for filling. At $T=0$, the chemical potential is equal to the Fermi energy ($\mu =E_F$), and the Fermi-Dirac distribution becomes a perfect step function, confirming our $T=0$ picture.

The following video provides an excellent derivation of the Fermi-Dirac distribution from the principles of statistical mechanics, a topic with which you are already familiar. It clearly illustrates the behavior at both zero and finite temperatures.

For a complementary text-based derivation, you can also consult the following resource, which uses a series of exercises to guide you to the same result.

Example: Ground State of the 1D Chain

Let's make this concrete by applying our findings to the 1D tight-binding chain we analyzed in the last lesson. The band structure is $E(k)=-2t\cos (ka)$, and the Brillouin zone is $k\in [-\pi /a,\pi /a]$.

Suppose we have $N_e$ electrons on a chain of $N$ sites. The electron density is $n=N_e/N$, which is also called the filling. At $T=0$, we fill all states with $|k|\le k_F$. To find the Fermi momentum $k_F$, we need to relate it to the total number of electrons.

In the thermodynamic limit ($N\to \infty$), the discrete $k$ points become a continuum. The density of $k$-states per unit length is $1/(2\pi )$, so for a chain of length $L=Na$, the number of states in a small interval $dk$ is $\frac{L}{2\pi }dk$. The total number of electrons is found by integrating this density over all occupied states, times 2 for spin:

Dividing by $N$, we find the relationship between the filling $n$ and the Fermi momentum $k_F$:

The Fermi momentum is directly proportional to the electron filling. The Fermi energy is then simply $E_F=E(k_F)=-2t\cos (k_{F}a)=-2t\cos (n\pi /2)$.

Let's consider two important cases:

1. Half-filling ($n=1$): This corresponds to one electron per site.

   • $k_F=\frac{1\cdot \pi }{2a}=\frac{\pi }{2a}$. This is exactly halfway to the Brillouin zone edge.
   • $E_F=-2t\cos (\pi /2)=0$. The band is filled from its bottom at $E=-2t$ up to the middle at $E=0$. The system is a metal, because there are empty states infinitesimally close in energy to the highest filled states, allowing for easy electronic excitations.

2. Quarter-filling ($n=0.5$):

   • $k_F=\frac{0.5\cdot \pi }{2a}=\frac{\pi }{4a}$.
   • $E_F=-2t\cos (\pi /4)=-\sqrt{2}t$.

This procedure gives us a complete description of the ground state for any filling.

Conclusion

In this lesson, we have moved from the single-particle picture to the many-body reality. By applying the Pauli exclusion principle to the energy band derived in the previous lesson, we have constructed the non-interacting ground state.

The key takeaways are:

• The Fermi Sea: The ground state of non-interacting fermions is formed by filling all single-particle energy levels from the bottom up to the Fermi energy, $E_F$.
• The Ground State Wavefunction: This state, $|{\Psi }_0⟩$, is formally constructed by acting on the vacuum with creation operators for all states within the Fermi sea.
• The Fermi-Dirac Distribution: At finite temperature, the sharp boundary at $E_F$ is smeared out. The occupation of any state is given by the Fermi-Dirac distribution function, which reduces to a step function at $T=0$.
• Filling and the Fermi Level: The Fermi energy $E_F$ and Fermi momentum $k_F$ are determined by the electron filling $n$. For a 1D chain, $k_F=n\pi /(2a)$.

We now have a well-defined ground state for our non-interacting system. In the next lesson, we will learn how to use this state to calculate important physical observables, such as the electron density at each site and the total kinetic energy of the system.