Density Of States
Density of Levels / Density of States
Motivation
How many electron states are available near a given energy?
One often needs to calculate quantities that are weighted sums over electron levels. These quantities have the form
\[ Q = 2 \sum_{n,\mathbf{k}} Q_n(\mathbf{k}) \]
where:
\(n\): band index
\(\mathbf{k}\): crystal wave vector
\(Q_n(\mathbf{k})\): property of the state \((n,\mathbf{k})\)
The factor \(2\) accounts for spin degeneracy.
For example, for the total energy,
\[ Q_n(\mathbf{k}) = \varepsilon_n(\mathbf{k}) f(\varepsilon_n(\mathbf{k})) \]
where \(f\) is the Fermi function.
For the total number of electrons,
\[ Q_n(\mathbf{k}) = f(\varepsilon_n(\mathbf{k})) \]
Replacing Sums by Integrals
For a macroscopic crystal, allowed values of \(\mathbf{k}\) are very closely spaced. Therefore, sums over \(\mathbf{k}\) can be replaced by integrals.
In 1D,
\[ \Delta k = \frac{2\pi}{L} \]
and
\[ \sum_k f(k) = \sum_{n=-\infty}^{\infty} f\left(\frac{2\pi}{L}n\right) \]
Insert
\[ 1 = \frac{\Delta k}{\Delta k} \]
so
\[ \sum_k f(k) \simeq \int_{-\infty}^{\infty} \frac{dk}{\Delta k} f(k) \]
Using
\[ \Delta k = \frac{2\pi}{L} \]
we obtain
\[ \sum_k f(k) \simeq \frac{L}{2\pi} \int_{-\infty}^{\infty} f(k)\,dk \]
In 3D
In three dimensions,
\[ \sum_{\mathbf{k}} f(\mathbf{k}) = \sum_{n_x,n_y,n_z} f\left( \frac{2\pi}{L}(n_x,n_y,n_z) \right) \]
Insert
\[ 1 = \frac{(\Delta k)^3}{(\Delta k)^3} \]
and recognize the 3D \(k\)-space volume element:
\[ \sum_{\mathbf{k}} f(\mathbf{k}) \simeq \frac{1}{(\Delta k)^3} \int f(\mathbf{k})\,d^3k \]
Since
\[ \Delta k = \frac{2\pi}{L} \]
we get
\[ \sum_{\mathbf{k}} f(\mathbf{k}) \simeq \frac{L^3}{(2\pi)^3} \int f(\mathbf{k})\,d^3k \]
Because \(V = L^3\),
\[ \sum_{\mathbf{k}} f(\mathbf{k}) \simeq \frac{V}{(2\pi)^3} \int f(\mathbf{k})\,d^3k \]
Therefore, per unit volume,
\[ q = \lim_{V\to\infty} \frac{Q}{V} \]
Using
\[ Q = 2\sum_{n,\mathbf{k}} Q_n(\mathbf{k}) \]
we obtain
\[ q = 2\sum_n \int_{\text{primitive cell}} \frac{d^3k}{(2\pi)^3} Q_n(\mathbf{k}) \]
The integral is over one primitive cell in reciprocal space, usually the first Brillouin zone.
Often,
\[ Q_n(\mathbf{k}) = Q(\varepsilon_n(\mathbf{k})) \]
Then the sum over \(k\)-states can be rewritten as an integral over energy:
\[ q = \int d\varepsilon \, g(\varepsilon) Q(\varepsilon) \]
where \(g(\varepsilon)\) is the density of states per unit volume.
Thus,
\[ g(\varepsilon)d\varepsilon \]
is the number of electron states per unit volume with energies between
\[ \varepsilon \]
and
\[ \varepsilon + d\varepsilon \]
Density of States
The total density of states is the sum over bands:
\[ g(\varepsilon) = \sum_n g_n(\varepsilon) \]
For band \(n\),
\[ g_n(\varepsilon) = \int \frac{d^3k}{4\pi^3} \delta\left( \varepsilon - \varepsilon_n(\mathbf{k}) \right) \]
Here,
\[ \frac{1}{4\pi^3} \]
comes from
\[ 2 \times \frac{1}{(2\pi)^3} \]
where the factor \(2\) is due to spin degeneracy.
Geometrical Interpretation
Another way to understand \(g_n(\varepsilon)\) is to work in \(k\)-space.
The equation
\[ \varepsilon_n(\mathbf{k}) = \varepsilon \]
defines a constant-energy surface,
\[ S_n(\varepsilon) \]
Similarly,
\[ \varepsilon_n(\mathbf{k}) = \varepsilon + d\varepsilon \]
defines a nearby surface,
\[ S_n(\varepsilon + d\varepsilon) \]
The states with energies between \(\varepsilon\) and \(\varepsilon+d\varepsilon\) lie in the thin shell between these two surfaces.
Therefore,
\[ g_n(\varepsilon)d\varepsilon = \text{volume of thin shell in \(k\)-space} \times \frac{1}{4\pi^3} \]
The shell volume is the thickness \(ds\) times the surface area element:
\[ g_n(\varepsilon)d\varepsilon = \int_{S_n(\varepsilon)} \frac{dS}{4\pi^3} ds \]
Since
\[ \nabla_{\mathbf{k}}\varepsilon_n(\mathbf{k}) \]
points perpendicular to the constant-energy surface, its magnitude tells how fast the energy changes.
For a small displacement \(ds\) normal to the surface,
\[ d\varepsilon = \left| \nabla_{\mathbf{k}}\varepsilon_n(\mathbf{k}) \right| ds \]
Therefore,
\[ ds = \frac{d\varepsilon} { \left| \nabla_{\mathbf{k}}\varepsilon_n(\mathbf{k}) \right| } \]
Substituting this gives
\[ g_n(\varepsilon) = \int_{S_n(\varepsilon)} \frac{dS}{4\pi^3} \frac{1} { \left| \nabla_{\mathbf{k}}\varepsilon_n(\mathbf{k}) \right| } \]
So flat bands give a larger density of states because
\[ \left| \nabla_{\mathbf{k}}\varepsilon_n(\mathbf{k}) \right| \]
is small.