Density of States: First Encounter

The density of states counts how many quantum states are available per energy interval. The simplest useful derivation comes from a free particle in a large box: replace the continuum by discrete box-normalized momentum states, count points in momentum space, and then take the large-box limit.

This page is a first encounter. It derives the one-particle, spinless, three-dimensional free-particle density of states. Many-body density of states, band density of states, phonon density of states, and thermodynamic applications belong in later application volumes.

What Is Being Counted

For a finite system with discrete energy levels, the exact density of states can be written formally as

$$D_{\mathrm{exact}}(E) = \sum_\alpha \delta(E-E_\alpha),$$

where $\alpha$ labels independent energy eigenstates. This expression is a distribution: it is a set of spikes at the exact energy levels.

In a large box, the levels become very dense. Instead of tracking each spike, one counts the number of states in a small energy window:

$$D(E)\,\Delta E \approx \text{number of states with energies between }E\text{ and }E+\Delta E.$$

Equivalently,

$$D(E)=\frac{dN}{dE},$$

where $N(E)$ is the number of states with energy less than or equal to $E$. The notation $g(E)$ is also common, but this page uses $D(E)$ to avoid confusing density of states with finite-level degeneracy.

Periodic Box Regulator

Put a spinless free particle in a cubic periodic box of side length $L$ and volume

$$V_{\mathrm{box}}=L^3.$$

The normalized plane waves are

$$\psi_{\mathbf n}(\mathbf r) = \frac{1}{\sqrt{V_{\mathrm{box}}}} e^{i\mathbf k_{\mathbf n}\cdot\mathbf r},$$

with allowed wavevectors

$$\mathbf k_{\mathbf n} = \frac{2\pi}{L} (n_x,n_y,n_z), \qquad n_x,n_y,n_z\in\mathbb Z.$$

Each allowed state occupies one cell of volume

$$\Delta k_x\,\Delta k_y\,\Delta k_z = \left(\frac{2\pi}{L}\right)^3 = \frac{(2\pi)^3}{V_{\mathrm{box}}}$$

in $\mathbf k$-space. Therefore a large-box sum becomes

$$\sum_{\mathbf n} \longrightarrow \frac{V_{\mathrm{box}}}{(2\pi)^3} \int d^3k.$$

The box is only a regulator. The physical continuum result is obtained after converting sums to integrals and keeping finite densities or finite rates.

Counting Points Inside a Sphere

For a free particle,

$$E = \frac{\hbar^2k^2}{2m}, \qquad k=\lVert\mathbf k\rVert.$$

States with energy less than or equal to $E$ correspond to $\mathbf k$ points inside a sphere of radius

$$k_E = \frac{\sqrt{2mE}}{\hbar}.$$

In three dimensions, fixed energy means a sphere in momentum space. Density-of-states counting comes from the number of allowed $\mathbf k$ points in a thin spherical shell.

The number of states inside the sphere is approximately

$$N(k) = \frac{V_{\mathrm{box}}}{(2\pi)^3} \frac{4\pi k^3}{3}.$$

Thus

$$N(k) = \frac{V_{\mathrm{box}}}{6\pi^2}k^3.$$

Equivalently, the number of states in a thin shell from $k$ to $k+dk$ is

$$dN = \frac{V_{\mathrm{box}}}{(2\pi)^3} 4\pi k^2\,dk.$$

The factor $4\pi k^2$ is the surface area of the sphere in $\mathbf k$-space. It is the geometric reason that the three-dimensional free-particle density of states grows with energy.

Converting From k to Energy

Using

$$k = \frac{\sqrt{2mE}}{\hbar},$$

the cumulative state count becomes

$$N(E) = \frac{V_{\mathrm{box}}}{6\pi^2} \left( \frac{2mE}{\hbar^2} \right)^{3/2}.$$

Differentiate with respect to energy:

$$D(E) = \frac{dN}{dE} = \frac{V_{\mathrm{box}}}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt E.$$

This is the spinless three-dimensional free-particle density of states for a large volume. Per unit volume,

$$\frac{D(E)}{V_{\mathrm{box}}} = \frac{1}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt E.$$

The units are states per energy. The volume factor is physical for an extensive system: doubling the box volume doubles the number of available one-particle states in the same energy interval.

Spin and Internal Degeneracy

The result above counts one spatial state for each allowed $\mathbf k$ value. If there is an internal degeneracy that does not change the energy, multiply by that factor.

For a spin-$s$ particle with no spin-dependent splitting,

$$g_{\mathrm{int}}=2s+1,$$

and

$$D_{\mathrm{total}}(E) = g_{\mathrm{int}}D(E).$$

For an electron in a model where spin is included but magnetic and spin-orbit splittings are ignored, $g_{\mathrm{int}}=2$. If a magnetic field or spin-dependent interaction splits the spin states, this simple multiplier is no longer the right description.

Boundary Conditions and the Large-Box Limit

Periodic boundary conditions are convenient because every allowed $\mathbf k$ point has the same cell volume. Hard-wall boxes use standing waves and positive mode numbers instead. For a large cubic hard-wall box, the allowed spacing is $\pi/L$ in each positive direction, so one counts one octant of $\mathbf k$-space.

The leading volume term is the same:

$$N(k) \sim \frac{V_{\mathrm{box}}}{6\pi^2}k^3.$$

The finite-size corrections differ. Boundary conditions affect surface terms, edge terms, and low-lying levels, but the leading bulk density of states is independent of the regulator when the large-box limit is taken correctly.

Density of States Versus Degeneracy

Finite-level degeneracy and density of states are related but not identical.

Degeneracy asks how many independent states have exactly the same energy. In a finite cubic box, for example, different integer triples may give the same value of $n_x^2+n_y^2+n_z^2$.

Density of states asks how many states lie in an energy interval. In the large-box limit, neighboring energies become so close that the smooth shell count is more useful than exact degeneracy at a single energy.

The connection is the energy shell: exact degeneracies are discrete repeated points on a shell, while the density of states counts how many allowed points lie in a thin shell.

Dimensional Comparison

The energy dependence changes with spatial dimension because the shell geometry changes.

For a free particle in $d$ spatial dimensions, the number of states below wavenumber $k$ scales as

$$N(k)\propto k^d.$$

Since $E\propto k^2$,

$$D(E)=\frac{dN}{dE} \propto E^{d/2-1}.$$

Thus a one-dimensional free-particle density of states scales like $E^{-1/2}$, a two-dimensional one is constant, and the three-dimensional one scales like $\sqrt E$. The proportionality constants depend on the normalization volume, boundary convention, and internal degeneracies.

Common Mistakes

• Forgetting the $4\pi k^2$ shell factor in three dimensions.
• Counting only positive $\mathbf k$ components when using periodic boundary conditions.
• Double-counting spin by both multiplying by $2$ and separately listing spin labels.
• Treating the smooth $D(E)$ as accurate for a small box with widely spaced levels.
• Dropping the volume factor before converting to density per unit volume.
• Confusing exact finite-level degeneracy with the number of states in an energy interval.
• Mixing $k$-space and $p$-space counting without the Jacobian $\mathbf p=\hbar\mathbf k$.

Where This Is Used

• Free Particle in Three Dimensions introduces momentum vectors, energy shells, and the periodic-box regulator.
• Plane Waves and Delta Normalization explains why box-normalized sums become continuum integrals.
• Three-Dimensional Box provides the hard-wall comparison.
• Degeneracy in Separable Systems distinguishes exact degeneracy from large-box state density.
• Normalization Conventions records the box and delta normalization conventions used in the derivation.
• Landau Levels gives a magnetic example where state counting is tied to flux rather than ordinary free-particle shells.
• Density of States in Transition Rates develops weighted final-state measures, normalization cancellation, threshold laws, and the continuum time window.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Academic Press, 2011.

Exercises

1. Derive the cumulative count $N(k)$ for a spinless free particle in a periodic cubic box.

Solution

Allowed $\mathbf k$ points are separated by $2\pi/L$ in each direction, so one state occupies $\mathbf k$-space volume

$$\left(\frac{2\pi}{L}\right)^3 = \frac{(2\pi)^3}{V_{\mathrm{box}}}.$$

The volume of a sphere of radius $k$ is $4\pi k^3/3$. Therefore

$$N(k) = \frac{4\pi k^3/3}{(2\pi)^3/V_{\mathrm{box}}} = \frac{V_{\mathrm{box}}}{6\pi^2}k^3.$$

2. Starting from $N(E)$, derive the three-dimensional free-particle density of states.

Solution

Use

$$N(E) = \frac{V_{\mathrm{box}}}{6\pi^2} \left( \frac{2mE}{\hbar^2} \right)^{3/2}.$$

Differentiating gives

$$D(E) = \frac{dN}{dE} = \frac{V_{\mathrm{box}}}{6\pi^2} \frac{3}{2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt E.$$

Thus

$$D(E) = \frac{V_{\mathrm{box}}}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt E.$$

3. How does including unsplit spin-$1/2$ states change the result?

Solution

For spin $1/2$ with no spin-dependent splitting, each spatial momentum state has two spin states. Therefore

$$g_{\mathrm{int}}=2,$$

and

$$D_{\mathrm{total}}(E)=2D(E).$$

This multiplier is valid only when both spin states have the same energy.