Density of States for Free Particles

Every thermodynamic number you will ever compute for an ideal quantum gas — the particle number, the energy, the pressure, the condensate fraction, the electronic heat capacity — is one and the same integral, $\int_0^\infty g(\varepsilon)\,\bar n(\varepsilon)\,\mathrm{d}\varepsilon$ . The occupation $\bar n$ is supplied by Fermi or Bose statistics; the factor $g(\varepsilon)$ — the density of states — is what this article builds. It answers a purely kinematic question: how many single-particle quantum states have energy between $\varepsilon$ and $\varepsilon+\mathrm{d}\varepsilon$ ? For free particles in three dimensions the answer is the workhorse formula $g(\varepsilon)=\frac{g_s V}{4\pi^2}\big(\frac{2m}{\hbar^2}\big)^{3/2}\sqrt{\varepsilon}$ , and everything downstream in the Ideal Quantum Gases chapter is an integral against it. We take the microscopic input from the quantum-mechanics track. The allowed momenta and the box spectrum are owned by aq-qm-particle-in-3d-box, where the discrete levels "crowd together into a smooth density of states that scales as the volume — the very quantity that, filled with electrons up to a Fermi surface, becomes the free-electron model of a metal." The general machinery — that the mode count per $\mathbf k$ -cell is $V/(2\pi)^3$ and that this $g(E)$ is one object across contexts — is owned by aq-qm-density-of-final-states, whose punchline is that "the $\rho(E)$ that counts final states in a transition rate is identical to the $g(E)$ that appears in the partition function and the Fermi/Bose occupation sums — both count quantum states per unit energy." This article does not re-derive the box spectrum; it takes the quantization as given and owns the $g(\varepsilon)$ that results and feeds the SM5 sums. The single most useful fact is that the shape of $g(\varepsilon)$ is fixed by geometry alone. For a dispersion $\varepsilon\propto p^{s}$ in $d$ dimensions the density of states is a pure power law $g(\varepsilon)\propto\varepsilon^{d/s-1}$ . That exponent — falling in 1D, flat in 2D, rising in 3D — decides whether a Bose gas condenses, whether a Fermi gas has a finite density of states at its surface, and the temperature power of every heat capacity that follows.

Key takeaways

The density of states $g(\varepsilon)$ counts single-particle quantum states per unit energy; $g(\varepsilon)\,\mathrm{d}\varepsilon$ is the number of states in $[\varepsilon,\varepsilon+\mathrm{d}\varepsilon]$ .
It is obtained by counting momenta uniformly in $\mathbf k$ -space at density $V/(2\pi)^3$ (owned by aq-qm-density-of-final-states), then changing variables from $\mathbf p$ to $\varepsilon$ through the dispersion.
For free 3D particles, $g(\varepsilon)=\frac{g_s V}{4\pi^2}\big(\frac{2m}{\hbar^2}\big)^{3/2}\sqrt{\varepsilon}$ — the exact prefactor is identical to the quantum-mechanical $\rho(E)$ .
The shape is pure geometry: $g(\varepsilon)\propto\varepsilon^{d/s-1}$ for $\varepsilon\propto p^{s}$ in $d$ dimensions — 1D falls as $\varepsilon^{-1/2}$ , 2D is flat, 3D rises as $\sqrt{\varepsilon}$ .
Filling states to $\varepsilon_F$ gives $N=\int_0^{\varepsilon_F} g\,\mathrm{d}\varepsilon$ , which inverts to $\varepsilon_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ for a spin- $\tfrac12$ gas.
Key idea: count states uniformly in momentum space, then let the dispersion push them onto the energy axis — the Jacobian of that push, $g\,\frac{V}{(2\pi\hbar)^d}\Omega_d\,p^{d-1}(\mathrm{d}\varepsilon/\mathrm{d}p)^{-1}$ , IS the density of states, and its energy power $\varepsilon^{d/s-1}$ governs all of ideal-gas quantum statistics.

The core insight

Count states uniformly in momentum space at density $g_s V/(2\pi\hbar)^d$ , then change variables from momentum $\mathbf p$ to energy $\varepsilon$ using the dispersion relation. The Jacobian of that change of variables IS the density of states: an energy shell of thickness $\mathrm{d}\varepsilon$ contains a momentum-space volume $\Omega_d p^{d-1}(\mathrm{d}\varepsilon/\mathrm{d}p)^{-1}\mathrm{d}\varepsilon$ , and dividing by the cell volume gives $g(\varepsilon)$ .

Where this lives today

The density of states is the single most-measured theoretical quantity in condensed-matter physics. Scanning tunneling spectroscopy maps $g(\varepsilon)$ point by point across a surface; angle-resolved photoemission (ARPES) images the occupied states $g(\varepsilon)\bar n(\varepsilon)$ directly in energy and momentum; and the low-temperature electronic heat capacity $C_e=\frac{\pi^2}{3}g(\varepsilon_F)k_B^2 T$ reads off $g(\varepsilon_F)$ from the slope of $C/T$ versus $T^2$ . The engineered version is the story of modern devices: quantum wells and heterostructures freeze one or two dimensions out, switching the DOS from the 3D $\sqrt\varepsilon$ to the 2D staircase of constants or the 1D $1/\sqrt\varepsilon$ spikes of a quantum wire, and each change reshapes the optics and transport. Graphene's linear-in-energy DOS (2D massless Dirac fermions, $d=2,s=1\Rightarrow g\propto\varepsilon$ ) that vanishes at the Dirac point is exactly the $d/s-1$ counting of this article, and it is what makes graphene a semimetal.

The physical picture

Picture momentum space tiled by a cubic lattice of allowed $\mathbf k$ -points, one per cell of volume $(2\pi/L)^3$ . The number of states inside any region is just its volume divided by that cell volume — this uniform $\mathbf k$ -space bookkeeping is the machinery quantum mechanics hands us. The density of states asks a different question: as the energy is raised by $\mathrm{d}\varepsilon$ , how much new momentum-space volume falls into the shell between the surfaces $\varepsilon$ and $\varepsilon+\mathrm{d}\varepsilon$ ? For free particles the constant-energy surfaces are spheres of radius $p=\sqrt{2m\varepsilon}$ , so the shell is a thin skin of area $4\pi p^2$ and thickness $\mathrm{d}p$ . Its volume, divided by the cell, is $g(\varepsilon)\,\mathrm{d}\varepsilon$ . The hero figure is the reason this article sits before Fermi energy, BEC, and the Sommerfeld expansion: the shape of $g(\varepsilon)$ is fixed by dimension and dispersion alone, and that shape is dramatically different in one, two, and three dimensions. The energy power $\varepsilon^{d/s-1}$ comes from two competing geometric effects — the shell area grows as $p^{d-1}$ while its thickness $\mathrm{d}p/\mathrm{d}\varepsilon$ shrinks as $p^{1-s}$ .

Figure

The density of states fan: 1D falls, 2D is flat, 3D rises as the square root of energy

The punchline: the same particle in different dimensions has a density of states that diverges, is constant, or vanishes at the band edge. That single power exponent controls whether a Bose gas condenses (needs 3D) and whether a Fermi surface has states to excite.

Where does the number itself come from? The second figure isolates the mechanism: a spherical shell in $\mathbf k$ -space of radius $k$ , area $4\pi k^2$ , thickness $\mathrm{d}k$ , tiled by cells of side $2\pi/L$ . Count the cells in the shell, convert $k$ to $\varepsilon$ , and the $\sqrt\varepsilon$ appears.

Figure

Counting states in a k-space shell of radius k and thickness dk

The area factor $4\pi k^2$ is what produces the $\sqrt{\varepsilon}$ : bigger energy means a bigger sphere with more surface, hence more states per unit energy.

The core equations

The first ingredient is the rule that converts a discrete sum over allowed wavevectors into an integral. It follows directly from the box quantization owned by aq-qm-particle-in-3d-box: periodic boundary conditions on a cube of side $L$ force $\mathbf k=(2\pi/L)\mathbf n$ , so each state occupies a $\mathbf k$ -cell of volume $(2\pi/L)^3=(2\pi)^3/V$ .

\sum_{\mathbf k}\;\longrightarrow\;\frac{V}{(2\pi)^{d}}\int \mathrm{d}^{d}k\;=\;\frac{V}{(2\pi\hbar)^{d}}\int \mathrm{d}^{d}p \qquad (1)

where: $V=L^d$ — volume of the confining box; $\mathbf k,\mathbf p=\hbar\mathbf k$ — wavevector and momentum; $d$ — spatial dimension. Physical reading: the prefactor $V/(2\pi)^d$ is the uniform number of states per unit $\mathbf k$ -volume; it is the same object quantum mechanics calls the mode density, and it becomes exact as $L\to\infty$ . The density of states repackages this uniform $\mathbf k$ -space count as a non-uniform count along the energy axis. Its definition is the volume of the thin energy shell, divided by the cell, per unit energy.

g(\varepsilon)\,\mathrm{d}\varepsilon = g_s\,\frac{V}{(2\pi\hbar)^{d}}\!\!\int_{\varepsilon\le\varepsilon(\mathbf p)\le\varepsilon+\mathrm{d}\varepsilon}\!\!\mathrm{d}^{d}p \qquad (2)

where: $g_s$ — internal degeneracy (e.g. $2s+1$ spin states, or $2$ photon polarizations); $\varepsilon(\mathbf p)$ — dispersion relation. Physical reading: $g(\varepsilon)$ is literally a Jacobian — the momentum-space volume that maps into a unit slice of energy. For an isotropic dispersion the shell is spherical, so the shell integral is the surface area $\Omega_d p^{d-1}$ times the radial thickness $\mathrm{d}p=(\mathrm{d}\varepsilon/\mathrm{d}p)^{-1}\mathrm{d}\varepsilon$ . This is the master formula from which every specific case drops out.

g(\varepsilon)=g_s\,\frac{V}{(2\pi\hbar)^{d}}\,\Omega_d\,p^{d-1}\left(\frac{\mathrm{d}\varepsilon}{\mathrm{d}p}\right)^{-1} \qquad (3)

where: $\Omega_d$ — surface area of the unit sphere: $\Omega_1=2,\ \Omega_2=2\pi,\ \Omega_3=4\pi$ ; $p$ evaluated at $\varepsilon$ via $\varepsilon(p)$ ; $(\mathrm{d}\varepsilon/\mathrm{d}p)^{-1}$ — the shell-thickness factor. Physical reading: the area factor $p^{d-1}$ steepens the DOS with dimension, the inverse group-slope flattens it with faster dispersion — together they give the exponent $d/s-1$ . Specializing to $d=3$ with the free-particle dispersion $\varepsilon=p^2/2m$ gives the single most-used result in the whole chapter. Here $p=\sqrt{2m\varepsilon}$ and $\mathrm{d}\varepsilon/\mathrm{d}p=p/m$ .

g(\varepsilon)=\frac{g_s V}{4\pi^{2}}\left(\frac{2m}{\hbar^{2}}\right)^{3/2}\sqrt{\varepsilon}\,,\qquad \varepsilon\ge 0 \qquad (4)

where: $m$ — particle mass; $\varepsilon=p^2/2m$ — kinetic energy. Physical reading: the $\sqrt\varepsilon$ growth is the spherical-shell area $4\pi p^2$ shining through; the prefactor is fixed exactly, with no free constants, and equals the quantum-mechanical $\rho(E)$ from aq-qm-density-of-final-states. The one-line reason all of ideal-gas quantum statistics reduces to energy integrals is that the DOS converts every momentum sum into an integral against the occupation function.

N=\int_0^\infty g(\varepsilon)\,\frac{\mathrm{d}\varepsilon}{e^{(\varepsilon-\mu)/k_BT}\pm 1},\qquad U=\int_0^\infty \varepsilon\,g(\varepsilon)\,\frac{\mathrm{d}\varepsilon}{e^{(\varepsilon-\mu)/k_BT}\pm 1} \qquad (5)

where: $\mu$ — chemical potential; $+$ gives Fermi-Dirac, $-$ gives Bose-Einstein. Physical reading: swap the occupation for $\bar n$ owned by aq-sm-fermi-dirac-distribution or aq-sm-bose-einstein-distribution and every thermodynamic quantity is one integral against $g(\varepsilon)$ . Integrating the $\sqrt\varepsilon$ law gives the cumulative count of states below an energy, the object that fixes the Fermi energy.

\mathcal N(\varepsilon)\equiv\int_0^{\varepsilon} g(\varepsilon')\,\mathrm{d}\varepsilon'=\frac{g_s V}{6\pi^{2}}\left(\frac{2m\varepsilon}{\hbar^{2}}\right)^{3/2} \qquad (6)

where: $\mathcal N(\varepsilon)$ — total number of states with energy below $\varepsilon$ ; it equals the $\mathbf p$ -sphere volume of radius $p(\varepsilon)$ divided by the cell volume. Physical reading: setting $\mathcal N(\varepsilon_F)=N$ and inverting is exactly how the Fermi energy $\varepsilon_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ is obtained (owned in full by aq-sm-fermi-energy-momentum-temperature).

Derivations

Derivation AThe 3D free-particle density of states and its exact prefactor

The goal is to derive the workhorse result $g(\varepsilon)=\frac{g_s V}{4\pi^2}(2m/\hbar^2)^{3/2}\sqrt{\varepsilon}$ from the uniform $\mathbf k$ -space count, taking as given only the box quantization owned by the quantum-mechanics track (aq-qm-particle-in-3d-box: the discrete box levels "crowd together into a smooth density of states that scales as the volume"). The route is to count states in a spherical $\mathbf k$ -shell, then push the radial variable from $k$ to $\varepsilon$ through the free dispersion. The one thing to watch is the exact numerical prefactor — every $2\pi$ , every power of two, must survive, because this constant sets the absolute scale of every metal's Fermi energy and heat capacity. This card owns the resulting $g(\varepsilon)$ ; it does not re-derive the box spectrum.

Step 1. Count states in a $\mathbf k$ -shell. From Eq. (1) the number of states with wavevector magnitude in $[k,k+\mathrm{d}k]$ is the shell volume times the cell density: $\mathrm{d}N=g_s\,\frac{V}{(2\pi)^3}\,4\pi k^2\,\mathrm{d}k$ . Why: the allowed $\mathbf k$ -points sit on a lattice of cell volume $(2\pi/L)^3=(2\pi)^3/V$ , so the count is the isotropic shell volume $4\pi k^2\,\mathrm{d}k$ divided by that cell, with $g_s$ internal states per point.

Step 2. Change variables from $k$ to $\varepsilon$ . The free dispersion $\varepsilon=\hbar^2 k^2/2m$ gives $k=\sqrt{2m\varepsilon}/\hbar$ and $\mathrm{d}k/\mathrm{d}\varepsilon=\frac{1}{\hbar}\sqrt{m/2\varepsilon}$ . Why: the density of states is the count per unit energy, so we must trade $\mathrm{d}k$ for $\mathrm{d}\varepsilon$ using the dispersion — this is the Jacobian that turns uniform-in- $k$ into non-uniform-in- $\varepsilon$ .

Step 3. Assemble $g(\varepsilon)=\mathrm{d}N/\mathrm{d}\varepsilon$ . Substituting $k^2=2m\varepsilon/\hbar^2$ and $\mathrm{d}k=\frac{1}{\hbar}\sqrt{m/2\varepsilon}\,\mathrm{d}\varepsilon$ into Step 1, $g(\varepsilon)=g_s\frac{V}{2\pi^2}k^2\frac{\mathrm{d}k}{\mathrm{d}\varepsilon}=g_s\frac{V}{2\pi^2}\frac{2m\varepsilon}{\hbar^2}\cdot\frac{1}{\hbar}\sqrt{\frac{m}{2\varepsilon}}$ . Why: the $4\pi/(2\pi)^3=1/2\pi^2$ collects the geometric constants; the leftover $\varepsilon\cdot\varepsilon^{-1/2}=\sqrt\varepsilon$ is the signature.

Step 4. Collect the prefactor. The energy powers combine to $\sqrt\varepsilon$ and the constants to $\frac{g_s V}{2\pi^2}\cdot\frac{2m}{\hbar^2}\cdot\frac{\sqrt m}{\sqrt 2\,\hbar}=\frac{g_s V}{4\pi^2}\big(\frac{2m}{\hbar^2}\big)^{3/2}$ , giving Eq. (4). Why: rewriting $\frac{2m}{\hbar^2}\cdot\frac{\sqrt{m}}{\sqrt2\,\hbar}=\frac{1}{2}\big(\frac{2m}{\hbar^2}\big)^{3/2}$ folds all mass and $\hbar$ dependence into the single group $(2m/\hbar^2)^{3/2}$ , and the factor $\tfrac12\cdot\tfrac{1}{2\pi^2}=\tfrac{1}{4\pi^2}$ is the exact coefficient. Verified: the symbolic reduction of $g_s\frac{V}{2\pi^2}k^2(\mathrm{d}k/\mathrm{d}\varepsilon)$ to $\frac{g_s V}{4\pi^2}(2m/\hbar^2)^{3/2}\sqrt\varepsilon$ , and its identity with the quantum-mechanical $\rho(E)=g_s Vmk/2\pi^2\hbar^2$ when re-expressed in $k$ , are checked in verify_density-of-states-free-particles.py.

Go deeper · Derivation B · 1 minThe universal power law and the Fermi energy it fixes

Derivation BThe universal power law and the Fermi energy it fixes

The goal is to prove two structural facts that make the density of states a chapter-opening tool: that for any isotropic power-law dispersion $\varepsilon=\alpha p^{s}$ in $d$ dimensions the DOS is a pure power $g(\varepsilon)\propto\varepsilon^{d/s-1}$ (so its shape needs no computation, only counting of exponents), and that filling the 3D $\sqrt\varepsilon$ states up to $\varepsilon_F$ inverts the cumulative count into $\varepsilon_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ for a spin- $\tfrac12$ gas. These are the owned structural results; the second hands the Fermi energy to its dedicated home (aq-sm-fermi-energy-momentum-temperature) with the exponent already established here.

Step 1. Insert a power-law dispersion into the master formula. With $\varepsilon=\alpha p^s$ , $p=(\varepsilon/\alpha)^{1/s}$ and $\mathrm{d}\varepsilon/\mathrm{d}p=s\alpha p^{s-1}$ , so Eq. (3) gives $g(\varepsilon)=g_s\frac{V\Omega_d}{(2\pi\hbar)^d}\frac{p^{d-1}}{s\alpha p^{s-1}}=g_s\frac{V\Omega_d}{(2\pi\hbar)^d}\frac{p^{d-s}}{s\alpha}$ . Why: the shell area $p^{d-1}$ and inverse group-slope $p^{1-s}$ combine to $p^{d-s}$ — the whole dimensional and dispersive content sits in this one exponent.

Step 2. Convert $p^{d-s}$ to energy. Since $p^{d-s}=(\varepsilon/\alpha)^{(d-s)/s}$ and $(d-s)/s=d/s-1$ , we get $g(\varepsilon)=g_s\frac{V\Omega_d}{(2\pi\hbar)^d}\frac{1}{s\,\alpha^{d/s}}\,\varepsilon^{\,d/s-1}$ . Why: this exposes the universal exponent $d/s-1$ : it is $\tfrac12$ for 3D nonrelativistic ( $d=3,s=2$ ), $0$ for 2D ( $d=2,s=2$ , flat), $-\tfrac12$ for 1D ( $d=1,s=2$ , divergent), and $2$ for 3D ultrarelativistic ( $d=3,s=1$ , photons).

Step 3. Count occupied states at $T=0$ in 3D. A Fermi gas fills every state up to $\varepsilon_F$ , so $N=\mathcal N(\varepsilon_F)=\frac{g_s V}{6\pi^2}(2m\varepsilon_F/\hbar^2)^{3/2}$ from Eq. (6). Why: the cumulative count $\mathcal N$ is just the $\mathbf p$ -sphere volume of radius $p_F=\sqrt{2m\varepsilon_F}$ divided by the cell — the integrated form of the $\sqrt\varepsilon$ law.

Step 4. Invert for the Fermi energy. Solving $(2m\varepsilon_F/\hbar^2)^{3/2}=6\pi^2 n/g_s$ gives $\varepsilon_F=\frac{\hbar^2}{2m}(6\pi^2 n/g_s)^{2/3}=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ for $g_s=2$ . A useful corollary follows immediately: since $\mathcal N\propto\varepsilon^{3/2}$ , $g(\varepsilon_F)=\mathrm{d}\mathcal N/\mathrm{d}\varepsilon|_{\varepsilon_F}=\tfrac32 N/\varepsilon_F$ . Why: the $\tfrac32$ ratio between the DOS at the surface and $N/\varepsilon_F$ is exactly what makes the electronic heat capacity $C_e=\frac{\pi^2}{3}g(\varepsilon_F)k_B^2 T$ come out as $\frac{\pi^2}{2}Nk_B(T/T_F)$ downstream. Verified: the exponent $d/s-1$ for $(d,s)=(3,2),(2,2),(1,2),(3,1)$ , the inversion $\varepsilon_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ with numeric copper $\varepsilon_F=7.0$ eV, and the corollary $g(\varepsilon_F)=\tfrac32 N/\varepsilon_F$ are checked in verify_density-of-states-free-particles.py.

Limiting cases

With dispersion $\varepsilon\propto p^{s}$ in $d$ dimensions the DOS exponent is $d/s-1$ . The four diagnostic cases:

Case	Result	Meaning
3D nonrelativistic ( $d=3,s=2$ )	$g(\varepsilon)\propto\sqrt{\varepsilon}$	Vanishes at the band edge; the standard Fermi/Bose gas — condensation possible, finite $g(\varepsilon_F)$
2D nonrelativistic ( $d=2,s=2$ )	$g(\varepsilon)=\dfrac{g_s mA}{2\pi\hbar^2}=\text{const}$	Energy-independent; the ideal 2D Bose gas has no finite- $T$ BEC; clean closed-form thermodynamics
1D nonrelativistic ( $d=1,s=2$ )	$g(\varepsilon)\propto\varepsilon^{-1/2}$	Diverges at the band edge — a 1D van Hove singularity
3D ultrarelativistic ( $d=3,s=1$ )	$g(\varepsilon)\propto\varepsilon^{2}$	Massless quanta (photons, $g_s=2$ ); seeds the Planck law and $u\propto T^4$

Common trapThe continuum density of states gives zero weight to the ground state

In 3D the free-particle DOS vanishes at $\varepsilon=0$ because the momentum-space shell around $p=0$ has vanishing area, $g(\varepsilon)\propto\sqrt\varepsilon\to0$ . For fermions this is harmless. For bosons it is catastrophic: below the condensation temperature a macroscopic fraction of particles occupies the single $\mathbf k=0$ state, which the continuum integral weights as zero. The fix is to restore that ground-state term by hand — $N=N_0+\int_0^\infty g(\varepsilon)\bar n\,\mathrm{d}\varepsilon$ — never trust $\int g\,\bar n\,\mathrm{d}\varepsilon$ alone below $T_c$ . A second trap: the internal degeneracy $g_s$ is a multiplicative factor that is easy to drop; for spin- $\tfrac12$ electrons $g_s=2$ , and forgetting it shifts $\varepsilon_F$ by $2^{2/3}\approx1.59$ .

Worked examples

Example 1 — Count the states in a small energy window and read off the density of states.

Step 1. Setup. Take a cube of side $L=1\ \mathrm{nm}$ ( $V=10^{-27}\,\mathrm{m^3}$ ) holding spin- $\tfrac12$ electrons ( $g_s=2$ , $m=9.109\times10^{-31}$ kg). Ask: how many single-particle states lie below $\varepsilon=1\ \mathrm{eV}=1.602\times10^{-19}$ J?

Step 2. Apply the cumulative count. $\mathcal N(\varepsilon)=\frac{g_s V}{6\pi^2}(2m\varepsilon/\hbar^2)^{3/2}$ . Compute $2m\varepsilon/\hbar^2=\frac{2(9.109\times10^{-31})(1.602\times10^{-19})}{(1.0546\times10^{-34})^2}=2.625\times10^{19}\,\mathrm{m^{-2}}$ , so $(2m\varepsilon/\hbar^2)^{3/2}=1.345\times10^{29}\,\mathrm{m^{-3}}$ .

Step 3. Evaluate. $\mathcal N=\frac{2\times10^{-27}}{6\pi^2}\times1.345\times10^{29}=\frac{2\times10^{-27}}{59.22}\times1.345\times10^{29}=\mathbf{4.54}$ states below 1 eV. A cubic nanometer of free electrons has only a handful of states in the first electron-volt — states are sparse in a small box.

Step 4. Read off the DOS at 1 eV. $g(\varepsilon)=\tfrac32\mathcal N/\varepsilon=1.5\times4.54/(1\ \mathrm{eV})=\mathbf{6.8\ eV^{-1}}$ for this box. The relation $g=\tfrac32\mathcal N/\varepsilon$ is exact for the $\sqrt\varepsilon$ law and is the fastest way to get $g(\varepsilon_F)$ once you know $N$ and $\varepsilon_F$ .

Example 2 — Density of states at the Fermi surface of copper.

Step 1. Setup. Copper has one conduction electron per atom, $n=8.49\times10^{28}\,\mathrm{m^{-3}}$ , $g_s=2$ . First get the Fermi energy from Eq. (6) inverted: $\varepsilon_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ .

Step 2. Fermi energy. $3\pi^2 n=3\pi^2(8.49\times10^{28})=2.514\times10^{30}$ , so $(3\pi^2 n)^{2/3}=(2.514\times10^{30})^{2/3}=1.848\times10^{20}\,\mathrm{m^{-2}}$ ; then $\varepsilon_F=\frac{(1.0546\times10^{-34})^2}{2(9.109\times10^{-31})}\times1.848\times10^{20}=1.128\times10^{-18}\,\mathrm{J}=\mathbf{7.04\ eV}$ , matching the measured copper value.

Step 3. DOS per unit volume at $\varepsilon_F$ . Using $g(\varepsilon_F)/V=\tfrac{3}{2}n/\varepsilon_F$ (the exact corollary), $g(\varepsilon_F)/V=1.5\times8.49\times10^{28}/(1.128\times10^{-18}\,\mathrm{J})=1.129\times10^{47}\,\mathrm{J^{-1}m^{-3}}$ .

Step 4. Convert to friendly units. Multiplying by $1.602\times10^{-19}\,\mathrm{J/eV}$ gives $g(\varepsilon_F)/V=\mathbf{1.81\times10^{28}\ eV^{-1}m^{-3}}=1.81\times10^{22}\ \mathrm{eV^{-1}cm^{-3}}$ . This number, plugged into $C_e=\frac{\pi^2}{3}g(\varepsilon_F)k_B^2T$ , predicts copper's Sommerfeld coefficient $\gamma\approx0.7\ \mathrm{mJ\,mol^{-1}K^{-2}}$ — within a factor of the measured $0.69$ , the small residual being the band-structure effective mass.

Origins · 1 minfelix bloch and how this was discovered

felix bloch

1905–1983 · Felix Bloch showed in 1928 that electrons in a periodic potential occupy Bloch waves, replacing the free-particle spectrum with energy bands — turning the smooth free-electron density of states derived here into the structured density of states of real solids, complete with van Hove singularities and band gaps. His theorem is why the free-particle DOS is the backbone on which solid-state physics builds. He later founded nuclear magnetic resonance, sharing the 1952 Nobel Prize in Physics.

History

1900

— Drude models metals as a classical electron gas, but the electronic heat capacity comes out far too large

1926

— Fermi and, independently, Dirac formulate the quantum statistics of exclusion-obeying particles

1927

— Sommerfeld replaces the classical gas with a degenerate free-electron gas, resolving the heat-capacity puzzle using the density of states at the Fermi surface

1928

— Bloch proves that periodic potentials give energy bands, generalizing the free-electron density of states to real crystals

1953

— Van Hove classifies the singularities that band structure imprints on the density of states

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §§6.1, 7.1, 8.1; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.9; Ashcroft & Mermin, Solid State Physics, Ch. 2; Landau & Lifshitz, Statistical Physics Part 1, §§56–57.
Microscopic owner: the box quantization and the $\rho(E)$ machinery are owned by aq-qm-particle-in-3d-box (the box levels "crowd together into a smooth density of states that scales as the volume") and aq-qm-density-of-final-states ("the $\rho(E)$ that counts final states … is identical to the $g(E)$ that appears in the partition function and the Fermi/Bose occupation sums"). This article owns the resulting $g(\varepsilon)$ and its power law.
Builds on: Quantum Statistics of Ideal Gases: Occupation Numbers · Particle in a box quantization · the SM4 sibling Density of States in Energy (which first built $g(\varepsilon)$ in the quantum-statistics chapter) · Density of States in k-Space.
Leads to: Fermi Energy, Momentum and Fermi Temperature · Sommerfeld Expansion · Electronic Heat Capacity of Metals · Bose-Einstein Condensation in the Ideal Gas · Photon Gas and Planck Radiation Law.
Provenance: AllQFT `sm` export (Pathria & Beale, §§6.1/7.1/8.1), aq-sm-density-of-states-free-particles, ref STAT.IQGAS.DOSFP.1221. All prefactors, exponents, and the copper numbers are python-verified (17 checks, all PASS). The copper $\varepsilon_F=7.04$ eV and $g(\varepsilon_F)/V=1.81\times10^{28}\ \mathrm{eV^{-1}m^{-3}}$ reproduce the standard Ashcroft-Mermin values.

What is the difference between the density of states here and the one in the quantum-mechanics track?

There is none — it is the same object viewed in two contexts. The quantum-mechanics article aq-qm-density-of-final-states states it directly: "the $\rho(E)$ that counts final states in a transition rate is identical to the $g(E)$ that appears in the partition function and the Fermi/Bose occupation sums — both count quantum states per unit energy." Quantum mechanics owns the machinery (the box spectrum, the $V/(2\pi)^3$ cell density, how $\rho(E)$ enters a golden-rule rate). This statistical-mechanics article takes that machinery as given and owns the specific free-particle $g(\varepsilon)=\frac{g_s V}{4\pi^2}(2m/\hbar^2)^{3/2}\sqrt\varepsilon$ that gets plugged into the occupation sums.

Why does the square root of energy appear, physically?

Because in 3D the constant-energy surfaces are spheres, and a sphere's surface area grows as $p^2$ . When you raise the energy by $\mathrm{d}\varepsilon$ , the shell that enters has area $4\pi p^2$ and thickness $\mathrm{d}p$ ; the number of states in it is proportional to that area. Since $p=\sqrt{2m\varepsilon}$ , the area $\propto p^2\propto\varepsilon$ , but the shell thickness $\mathrm{d}p/\mathrm{d}\varepsilon\propto1/p\propto\varepsilon^{-1/2}$ trims it down, leaving $g\propto\varepsilon\cdot\varepsilon^{-1/2}=\sqrt\varepsilon$ . The $\sqrt\varepsilon$ is the net of a growing sphere and a thinning shell.

Where does the internal degeneracy factor g_s come from and when is it 2?

$g_s$ counts the number of internal states each momentum mode carries. For electrons (and any spin- $\tfrac12$ fermion) there are two spin projections, $g_s=2s+1=2$ . For photons it is the two transverse polarizations, $g_s=2$ . For spin-1 phonons in a solid it is the three acoustic branches (one longitudinal, two transverse). It multiplies $g(\varepsilon)$ linearly, so it also enters $\varepsilon_F$ : since $\varepsilon_F\propto(n/g_s)^{2/3}$ , dropping the factor of 2 for electrons would raise the predicted Fermi energy by $2^{2/3}\approx1.59$ — a large and common error.

Does the density of states depend on the boundary conditions of the box?

Only through terms that vanish in the thermodynamic limit. Periodic boundary conditions give $\mathbf k=(2\pi/L)\mathbf n$ over all of $\mathbf k$ -space; hard walls give $\mathbf k=(\pi/L)\mathbf n$ restricted to the positive octant. The finer spacing of the hard-wall lattice (factor $2^d$ more points) is exactly cancelled by the restriction to one octant (factor $2^d$ fewer), so the bulk $g(\varepsilon)$ is identical. The differences are surface corrections of order (area/volume), which scale as $1/L$ and are negligible for macroscopic samples — the reason we can use the simpler periodic form freely.

Why is the density of states introduced before the Fermi energy or BEC?

Because it is the common ingredient in all of them. The Fermi energy is $\int_0^{\varepsilon_F}g\,\mathrm{d}\varepsilon=N$ inverted; Bose-Einstein condensation is decided by whether $\int_0^\infty g\,\bar n_{\mathrm{BE}}\,\mathrm{d}\varepsilon$ stays finite as $\mu\to0$ ; the electronic heat capacity is $\frac{\pi^2}{3}g(\varepsilon_F)k_B^2T$ ; the Planck law is the photon $g(\varepsilon)\propto\varepsilon^2$ against the Bose occupation. Every one of these is $\int g(\varepsilon)\,f(\varepsilon)\,\mathrm{d}\varepsilon$ for some $f$ . Build $g(\varepsilon)$ once and the rest of the chapter is one-dimensional energy integrals.

How does the exponent d/s−1 decide whether a Bose gas condenses?

For a Bose gas the number of particles in excited states is $N_{\mathrm{exc}}=\int_0^\infty g(\varepsilon)/(e^{\varepsilon/k_BT}-1)\,\mathrm{d}\varepsilon$ at $\mu=0$ . Near $\varepsilon=0$ the occupation blows up as $k_BT/\varepsilon$ , so the integrand behaves as $g(\varepsilon)/\varepsilon\propto\varepsilon^{d/s-2}$ . This is integrable at the lower limit only if $d/s-2>-1$ , i.e. $d/s>1$ . The 3D nonrelativistic gas has $d/s=3/2>1$ , so $N_{\mathrm{exc}}$ is bounded and a condensate forms once $N$ exceeds it; the 2D gas has $d/s=1$ exactly, the integral diverges logarithmically, and there is no finite-temperature BEC. The whole condensation question is settled by one exponent.

Is the ultrarelativistic photon result really the same formula?

Yes — it is the same master formula with a different dispersion. Photons have $\varepsilon=cp$ (so $s=1$ ) and two polarizations ( $g_s=2$ ), which the general result $g\propto\varepsilon^{d/s-1}$ turns into $g(\varepsilon)=\frac{V\varepsilon^2}{\pi^2(\hbar c)^3}$ in 3D. That $\varepsilon^2$ growth, fed into the Bose occupation at $\mu=0$ , is precisely what produces the Planck spectral shape and the Stefan-Boltzmann $u\propto T^4$ . The free-particle $\sqrt\varepsilon$ and the photon $\varepsilon^2$ are two evaluations of the identical geometric counting.