The Free Electron Gas

Learning Objectives

Up to now we have built the stage and choreographed the symmetries. Chapter 1 gave us a real-space lattice; Chapter 2 gave us the algebra of its symmetry; Chapter 3 gave us reciprocal space, Brillouin zones, and Bloch's theorem — the language solids use to organise quantum states. With this section we finally invite the actors onto the stage: the electrons.

The free electron gas is the simplest quantum theory of metals you can write down. It throws away the lattice potential entirely, treats the conduction electrons as non-interacting plane waves, and yet correctly predicts conductivity, heat capacity, magnetic susceptibility, and the order of magnitude of the work function. It is the “hydrogen atom” of solid-state physics — the toy model every later result is benchmarked against.

By the end of this section you should be able to:

1. State exactly which approximations turn the full many-electron Schrödinger equation into the free electron gas, and explain which physical questions each approximation answers and which it forfeits.
2. Derive the allowed k-vectors of a box with periodic boundary conditions and explain why each occupies a volume $(2\pi/L)^3$ in k-space.
3. Sketch the dispersion $\varepsilon(k) = \hbar^2 k^2/(2m)$ and identify on it the Fermi level, the Fermi wavevector, and the occupied window.
4. Compute $k_F$, $\varepsilon_F$, and $v_F$ from the electron density $n$ in three independent ways and check that they agree.
5. Explain why the density of states in 3D scales as $\sqrt{\varepsilon}$ and how this single fact governs the heat capacity and conductivity of every simple metal.
6. Read a VASP OUTCAR, find the Fermi level, the plane-wave cutoff, and the smearing parameter, and connect every one of them to a quantity in this section.

The Question We Are Trying to Answer

A piece of sodium metal contains roughly $2.5 \times 10^{28}$ electrons per cubic metre that have detached from their parent atoms and are free to wander through the crystal. Why does it conduct electricity so well? Why does its heat capacity at room temperature obey the Dulong-Petit law of the ions but show no measurable contribution from the electrons? Why is its colour silver-grey and not transparent?

The classical answer — Drude (1900) treated the electrons as a classical ideal gas — gets the conductivity roughly right, but catastrophically overpredicts the heat capacity by a factor of about $T_F/T \approx 100$ at room temperature. The discrepancy lasted twenty-six years, until Sommerfeld (1927) replaced Maxwell-Boltzmann statistics with Fermi-Dirac statistics. That single change turned the disastrous classical electron gas into the wildly successful free electron gas.

The puzzle in one line. A classical electron gas at 300 K should contribute $\tfrac{3}{2} k_B$ per electron to the heat capacity of a metal. Experiment finds less than 1% of that. Where did the other 99% of the heat capacity go?

Sommerfeld's Leap: One Bold Simplification

The free electron gas is defined by four assumptions. List them explicitly so you know what we are buying and what we are throwing away.

The first three are approximations that we will systematically relax in later sections (Bloch → Section 4.2; many-body → Section 4.4; DFT → Section 4.5). The fourth is not negotiable: electrons are fermions, full stop. Sommerfeld's contribution was simply to honour that fact.

Setting It Up: Schrödinger in a Box

Drop a single non-interacting electron into a cubic box of side $L$ with no potential inside. The time-independent Schrödinger equation reduces to:

$-\frac{\hbar^2}{2m}\,\nabla^2 \psi(\mathbf{r}) = \varepsilon\,\psi(\mathbf{r})$

Inspect the equation. The left-hand side is curvature; the right-hand side is energy. Higher energy means more curvature, which in turn means smaller wavelength, which in turn means larger wavevector. The smallest possible amount of curvature gives the ground state, and additional curvature buys higher energies. This is the only physical content we are extracting, and it will guide every formula that follows.

For a plane-wave ansatz $\psi_{\mathbf{k}}(\mathbf{r}) = \frac{1}{\sqrt{V}}\,e^{i\mathbf{k}\cdot\mathbf{r}}$ the Laplacian gives $\nabla^2 \psi = -k^2 \psi$, and we read off the energy directly:

$\varepsilon(\mathbf{k}) = \frac{\hbar^2 |\mathbf{k}|^2}{2m} = \frac{\hbar^2}{2m}(k_x^2 + k_y^2 + k_z^2)$

The wavefunction is normalised on the box volume $V = L^3$, so the prefactor is $1/\sqrt{V}$. That is the entire dynamical content. What remains is the bookkeeping: how many $\mathbf{k}$ values are allowed?

The Allowed k-Vectors Form a Discrete Lattice

We impose periodic (Born-von Kármán) boundary conditions: the wavefunction repeats with period $L$ in each direction, $\psi(x + L, y, z) = \psi(x, y, z)$, and similarly for $y$ and $z$. This single trick removes the boundary without changing the bulk physics, and it forces:

$e^{i k_x L} = 1\ \Longrightarrow\ k_x = \frac{2\pi}{L} n_x,\quad n_x \in \mathbb{Z}$

and identical conditions for $k_y, k_z$. The allowed wavevectors form a cubic lattice in k-space with spacing $\Delta k = 2\pi/L$, and each point hosts exactly one orbital state — or, with spin, two electrons (one spin-up, one spin-down).

This is also our first encounter with the deep duality of the chapter: the box is finite in real space, but k-space is a discrete infinite lattice. As the box grows, the spacing shrinks and the k-lattice becomes a continuum. In the thermodynamic limit $L \to \infty$ we replace sums by integrals:

$\sum_{\mathbf{k}} \to \frac{V}{(2\pi)^3}\int d^3k$

The Dispersion Relation: ε(k) = ℏ²k²/2m

The function $\varepsilon(k) = \hbar^2 k^2 / 2m$ is a parabola in $k$. Three of its features carry the entire physics of the model.

1. It is isotropic. The energy depends only on $|\mathbf{k}|$, not on direction. Surfaces of constant energy in k-space are concentric spheres. Real metals have anisotropic Fermi surfaces precisely because the lattice potential breaks this isotropy.
2. It has zero curvature mass. Differentiating twice gives $d^2\varepsilon/dk^2 = \hbar^2/m$: the electron has its bare mass $m$. In a real band, that second derivative becomes $\hbar^2/m^*$ with an effective mass $m^*$ that captures all the lattice dressing.
3. It has no gap. Every energy $\varepsilon \geq 0$ is realised by some k. The free electron gas is automatically a metal — insulators only become possible once a periodic potential opens gaps at Brillouin-zone boundaries (Section 4.2).

Interactive: Dispersion and the Fermi Window

The visualisation below plots $\varepsilon(k) = \hbar^2 k^2 / 2m$ in dimensionless units ($\hbar = m = 1$, so the curve is simply $y = k^2$). Drag the slider to raise and lower the Fermi level $\varepsilon_F$. Notice three things:

• The Fermi wavevector $k_F = \sqrt{2 m \varepsilon_F}/\hbar$ is read directly off the parabola where the horizontal Fermi line cuts it.
• The occupied window $[-k_F, k_F]$ in 1D becomes a sphere of radius $k_F$ in 3D — the Fermi sphere we will explore next.
• Doubling $\varepsilon_F$ only multiplies $k_F$ by $\sqrt{2}$. The square root is the fingerprint of the parabolic dispersion.

Filling States: The Fermi Sphere

Now place $N$ electrons in the box. Pauli forbids two of them from sharing both a wavevector and a spin, so we fill states starting from the lowest energy and work upward. Because energy increases monotonically with $|\mathbf{k}|$, the occupied region is a ball in k-space. Its boundary is the Fermi surface; for free electrons this surface is a sphere of radius $k_F$, called the Fermi sphere.

Counting argument

The number of orbital k-states inside a sphere of radius $k_F$ is its volume divided by the volume per state:

$N_{\text{orbitals}} = \frac{(4/3)\pi k_F^3}{(2\pi)^3 / V} = \frac{V k_F^3}{6 \pi^2}$

Each orbital holds two electrons (spin up and down), so the total electron count is $N = V k_F^3 / (3\pi^2)$. Dividing by $V$ gives the density-to-Fermi-wavevector relation:

$\boxed{\,n = \frac{k_F^3}{3\pi^2}\,}\ \Longleftrightarrow\ \boxed{\,k_F = (3\pi^2 n)^{1/3}\,}$

From here, every other Fermi-level quantity follows by substitution. The Fermi energy:

$\varepsilon_F = \frac{\hbar^2 k_F^2}{2m} = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$

The Fermi velocity (group velocity at the Fermi surface):

$v_F = \frac{\hbar k_F}{m} = \frac{\hbar}{m}(3\pi^2 n)^{1/3}$

And the Fermi temperature, comparing εF to thermal energy:

$T_F = \varepsilon_F / k_B$

Interactive: Building the Fermi Sphere

Drag, rotate, and zoom the visualisation below. Each dot is one allowed orbital k-state of a finite cubic box. The translucent gold sphere is the Fermi surface; states inside it are occupied (bright gold) and states outside are empty (dim slate). Use the sliders to:

• Increase the box size $N$: the k-mesh becomes denser. In the limit $L \to \infty$, the discrete count $\sum_{\mathbf{k}} \theta(k_F - k)$ collapses into the continuous integral $\frac{4}{3}\pi k_F^3$ — the readout below the panel shows both numbers, and they match better as $N$ grows.
• Increase $k_F$: more dots turn gold. This is what physically happens when you add electrons to the metal (e.g. by chemical doping or a gate voltage).

Numerical Walk-Through: Sodium Metal

Let's convert the abstract formulas into a real number. Sodium has mass density $\rho = 0.971\,\text{g/cm}^3$ and atomic weight $M = 22.99\,\text{g/mol}$. Each Na atom contributes one valence electron (the 3s electron) to the free electron gas.

Step 1: Electron density

Start from atoms per unit volume, times one electron per atom:

$n = \frac{\rho N_A}{M} = \frac{0.971 \times 6.022 \times 10^{23}}{22.99} \,\text{cm}^{-3} = 2.54 \times 10^{22}\,\text{cm}^{-3} = 2.54 \times 10^{28}\,\text{m}^{-3}$

Step 2: Fermi wavevector

Plug into $k_F = (3\pi^2 n)^{1/3}$:

$k_F = (3 \pi^2 \cdot 2.54 \times 10^{28})^{1/3} = (7.52 \times 10^{29})^{1/3} = 9.10 \times 10^9\,\text{m}^{-1} = 0.91\,\text{Å}^{-1}$

For comparison, the size of the first Brillouin zone of sodium (BCC, lattice constant $a = 4.29$ Å) is of order $2\pi/a \approx 1.46\,\text{Å}^{-1}$. The Fermi sphere of Na fits comfortably inside the first Brillouin zone — which is precisely why the free electron model describes Na so well. (In Cu, by contrast, the Fermi sphere bulges into the next zone along $\langle 111 \rangle$ directions, creating the famous “necks”.)

Step 3: Fermi energy

$\varepsilon_F = \frac{\hbar^2 k_F^2}{2 m_e} = \frac{(1.055 \times 10^{-34})^2 (9.10 \times 10^9)^2}{2 \cdot 9.109 \times 10^{-31}}\,\text{J} = 5.06 \times 10^{-19}\,\text{J} = 3.16\,\text{eV}$

Step 4: Fermi velocity and temperature

$v_F = \frac{\hbar k_F}{m_e} = 1.05 \times 10^6\,\text{m/s} \approx 0.0035\,c$

$T_F = \varepsilon_F/k_B = 3.66 \times 10^4\,\text{K}$

Pause and absorb. A conduction electron in sodium moves at one million metres per second — Mach 3000 — even at absolute zero. That is purely quantum kinetic energy forced on it by Pauli exclusion. The Fermi temperature $T_F \approx 37{,}000\,\text{K}$ is two orders of magnitude above room temperature, which means at 300 K the gas is about 1% warm. That single ratio explains the heat-capacity puzzle from the start of the section.

Source: Ashcroft & Mermin, Solid State Physics, Table 2.1. Aluminium contributes 3 valence electrons per atom; copper, silver, gold each contribute 1 (the s electron).

The Density of States g(ε)

The dispersion tells us the energy of one k. The density of states tells us how many states sit at a given energy — the most useful quantity for thermodynamics, because every thermal observable is an energy integral weighted by $g(\varepsilon)$.

Definition: $g(\varepsilon)\,d\varepsilon$ is the number of orbital states (per unit volume) whose energies lie between $\varepsilon$ and $\varepsilon + d\varepsilon$. We get it by counting k-points in a thin spherical shell.

Derivation in three lines

1. The number of orbital states with $|\mathbf{k}| \leq k$ per unit volume is $N(k)/V = k^3/(6\pi^2)$.
2. Substitute $k = \sqrt{2m\varepsilon}/\hbar$ to get the integrated DOS $G(\varepsilon) = \dfrac{1}{6\pi^2}\!\left(\dfrac{2m\varepsilon}{\hbar^2}\right)^{3/2}$.
3. Differentiate. Including the spin factor of 2, we obtain:

$\boxed{\,g(\varepsilon) = \frac{1}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{\varepsilon}\,}$

Three takeaways pop out of the formula:

• The square root. In 3D, $g(\varepsilon) \propto \sqrt{\varepsilon}$. In 2D it would be a step function (constant), and in 1D it would diverge as $1/\sqrt{\varepsilon}$. The dimensionality of the system is encoded in the power of $\varepsilon$.
• The mass dependence. $g \propto m^{3/2}$: heavier carriers (large effective mass) give a denser DOS. This is exactly why flat bands and heavy fermions have such a strong response.
• Self-consistency. Integrating up to $\varepsilon_F$ must reproduce the electron density: $n = \int_0^{\varepsilon_F} g(\varepsilon)\,d\varepsilon = \dfrac{1}{3\pi^2}\left(\dfrac{2m\varepsilon_F}{\hbar^2}\right)^{3/2} = \dfrac{k_F^3}{3\pi^2}$. Reassuringly identical to our earlier counting argument.

Interactive: g(ε) and Counting Electrons

The plot below shows $g(\varepsilon) = C\sqrt{\varepsilon}$ (we set $C = 1$). The amber-shaded region from $0$ to $\varepsilon_F$ equals the total electron density. As you slide $\varepsilon_F$, watch the readout update: the integral grows like $\varepsilon_F^{3/2}$, not $\varepsilon_F$.

Finite Temperature: The Fermi-Dirac Distribution

At $T = 0$ all states with $\varepsilon < \varepsilon_F$ are filled and all states above are empty — a step function. At finite temperature, electrons can be thermally excited from just below the Fermi level to just above it. The probability that a state of energy $\varepsilon$ is occupied becomes the celebrated Fermi-Dirac distribution:

$f(\varepsilon) = \frac{1}{e^{(\varepsilon - \mu)/k_B T} + 1}$

Read it as follows. Far below $\mu$, the exponent goes to $-\infty$, so $f \to 1$ (state is occupied). Far above $\mu$, the exponent goes to $+\infty$, so $f \to 0$ (state is empty). Right at $\mu$ the exponent is zero, so $f = 1/2$: half-filled. The transition zone has width on the order of $k_B T$ — a thermal smear of width about $4 k_B T$ centered on $\mu$.

The chemical potential $\mu$ is fixed by particle-number conservation: $n = \int g(\varepsilon) f(\varepsilon)\,d\varepsilon$. For temperatures far below $T_F$, the Sommerfeld expansion gives $\mu(T) \approx \varepsilon_F \left[\,1 - \dfrac{\pi^2}{12}(T/T_F)^2 \right]$ — for sodium at 300 K this correction is one part in $10^4$. We will routinely use $\mu \approx \varepsilon_F$ below.

Interactive: Fermi Smearing

Sweep the temperature slider below and watch the step soften into a sigmoid. The two values you should anchor in your mind are:

• $k_B T = 0$ (slider at 0): perfect step, T = 0 limit.
• $k_B T \sim \mu$ (slider near 1): the step is fully washed out and electrons populate all energies — this is the classical (Boltzmann) limit, never reached in real metals because $T_F$ is so high.

Real metals at room temperature live in the regime $k_B T / \varepsilon_F \sim 0.01$: an almost-perfect step with a tiny fuzz at the edge. That fuzz is responsible for all finite-temperature effects we attribute to electrons.

Why It Works: Heat Capacity Demystified

We can finally answer the question we opened with. In a classical ideal gas every electron contributes $\tfrac{3}{2} k_B$ to the heat capacity, regardless of temperature. Why doesn't that happen in a metal?

Pauli exclusion. At any given temperature, only the electrons within roughly $k_B T$ of the Fermi level can absorb heat — everyone deeper down is locked in by the electrons sitting just above them, who already occupy the only states they could be promoted to. The fraction of “active” electrons is therefore $\sim T/T_F$, and each of them contributes the classical $\sim k_B$. So:

$C_v^{\text{electronic}} \sim n \, k_B \cdot (T/T_F) = \frac{n k_B^2 T}{\varepsilon_F}$

The full Sommerfeld result puts a precise prefactor on this back-of-the-envelope argument:

$C_v = \frac{\pi^2}{3}\, g(\varepsilon_F)\, k_B^2 T = \gamma T$

Two predictions you can take to the lab:

1. Linear in T. The electronic contribution to $C_v$ is proportional to $T$, not constant. Confirmed in every metal ever measured.
2. The coefficient γ measures g(εF). Plot $C/T$ vs $T^2$ and read off the intercept — you have just measured the density of states at the Fermi level. A factor-of-10 anomaly in $\gamma$ is what gave heavy-fermion materials their name.

Where It Fails (and Why That Is Fine)

Three things the free electron gas cannot tell you, and each one sets up a section of this chapter:

Yet for the bulk thermodynamics of simple metals (the alkalis, Cu, Ag, Au, Al), the free electron gas predicts heat capacities, magnetic susceptibilities, electrical conductivities, and bulk moduli to within a factor of two. That is a remarkable triumph for a model that throws away the entire crystal structure.

VASP Connection: Free-Electron Echoes in Real Calculations

Even though VASP solves the full Kohn-Sham problem with a crystalline potential, every line of its INCAR and every column of its OUTCAR still carries the fingerprints of the free electron gas. Here is the dictionary.

1. ENCUT and the plane-wave kinetic energy cutoff

VASP expands every Kohn-Sham orbital in plane waves up to a maximum kinetic energy:

$\frac{\hbar^2 |\mathbf{G}_{\max}|^2}{2 m_e} \leq E_{\text{cut}}$

That is precisely the free-electron dispersion turned inside out. Setting ENCUT = 400 eV in INCAR means VASP keeps every plane wave whose free-electron energy is below 400 eV. The number of plane waves per atom typically runs into the thousands, dwarfing any localised-basis approach.

2. ISMEAR and SIGMA — choosing your Fermi smearing

Computing observables like total energy or DOS requires evaluating the integral $\int g(\varepsilon) f(\varepsilon)\,d\varepsilon$ numerically, on a finite k-mesh. The step function at $T = 0$ gives miserably slow convergence, so VASP smears it. Pick your weapon with ISMEAR:

1# INCAR — choose smearing for the Fermi step
2ISMEAR =  0    # Gaussian smearing of width SIGMA
3ISMEAR = -1    # Fermi-Dirac smearing (this section's f(eps))
4ISMEAR =  1    # Methfessel-Paxton order 1 (default for metals)
5ISMEAR =  2    # Methfessel-Paxton order 2
6ISMEAR = -5    # Tetrahedron method with Bloechl corrections (best for DOS)
7SIGMA  = 0.10  # smearing width in eV  (keep <~ kT for good accuracy)

For metals VASP recommends ISMEAR = 1 with SIGMA chosen so the entropy term per atom is below about 1 meV. For semiconductors and insulators, the gap eliminates the smearing problem and you can use ISMEAR = 0 with a small SIGMA, or ISMEAR = -5 for accurate DOS.

3. NELECT and the Fermi level

NELECT sets the total electron count $N$. VASP then adjusts the Fermi level $\varepsilon_F$ at every SCF step so the integrated DOS up to $\varepsilon_F$ equals NELECT. You can read the converged Fermi level from OUTCAR:

1grep "E-fermi" OUTCAR
2# E-fermi :   3.1742     XC(G=0):  -9.4031     alpha+bet : -16.2117
3
4# For sodium with the right computational settings, this should be
5# within ~10% of our analytical 3.16 eV — the residual difference is
6# the band-structure effect we'll add in section 4.2.

4. A jellium sanity check (optional but illuminating)

VASP can run a literal free electron gas: give it a cubic cell with no atoms and use NELECT to specify how many electrons to put in. The resulting DOS, Fermi level, and total energy must reproduce the analytical formulas of this section. This is one of the cleanest ways to check that your VASP build is sane before committing to a Mn:CdSe simulation.

1# INCAR for a jellium sanity check
2SYSTEM = Free electron gas, n = 2.5e22 cm^-3 (Na density)
3PREC   = Accurate
4ENCUT  = 400
5ISMEAR = -1     # Use exact Fermi-Dirac for direct comparison
6SIGMA  = 0.025  # = kT at 300 K
7NELECT = 32     # 32 electrons in a cubic cell of side a
8                # gives n = 32/a^3; choose a to match Na density

5. Reading band structures back to free-electron behaviour

When you plot a real band structure (Section 5.1) along a high-symmetry path, the lowest band of a simple metal often looks almost exactly like the parabola $\hbar^2 k^2 / 2m$, gently dressed by the lattice. The deviation is small near the zone centre and grows toward the Brillouin-zone boundary, where the free-electron model finally breaks down and band gaps open. We will pick up the story there in the next section.

Summary

1. The free electron gas treats conduction electrons as non-interacting plane waves in a box. Four assumptions: no lattice potential, no electron-electron interactions, finite box with periodic boundary conditions, Fermi-Dirac statistics.
2. Allowed wavevectors form a cubic lattice in k-space with spacing $\Delta k = 2\pi/L$; each orbital state occupies a k-volume $(2\pi)^3/V$.
3. The dispersion is parabolic and isotropic: $\varepsilon(\mathbf{k}) = \hbar^2 k^2 / 2m$.
4. Filling states up to $\varepsilon_F$ gives a Fermi sphere of radius $k_F = (3\pi^2 n)^{1/3}$. Everything else ($\varepsilon_F, v_F, T_F$) follows by substitution.
5. The 3D density of states is $g(\varepsilon) \propto \sqrt{\varepsilon}$; its value at the Fermi level controls every linear-response coefficient of the metal.
6. At finite temperature, occupation follows the Fermi-Dirac distribution, with thermal smearing of width $\sim k_B T \ll \varepsilon_F$ at ordinary temperatures. Only this thin shell of electrons participates in thermal physics — resolving the heat-capacity puzzle.
7. In VASP, ENCUT, ISMEAR, SIGMA, and NELECT are the free-electron dictionary. The Fermi level reported in OUTCAR is the same $\varepsilon_F$ we computed by hand.

Next section we re-introduce the lattice potential as a small perturbation. The Fermi sphere will deform, gaps will open at Brillouin-zone boundaries, and the simple parabola will fracture into the bands that distinguish metals, semiconductors, and insulators.