Density of States in Energy

Every thermodynamic quantity of an ideal quantum gas — its particle number, energy, pressure, and entropy — is an integral over energy weighted by two factors: how many single-particle states sit at each energy, and how heavily each is occupied. The occupation is the job of quantum statistics (Bose–Einstein or Fermi–Dirac). The state counting is the job of the density of states in energy, $g(\varepsilon)$ , and once you know it the many-body combinatorics collapses into a single one-dimensional integral. For a nonrelativistic gas in three dimensions this function is the celebrated square-root law $g(\varepsilon)\propto\sqrt{\varepsilon}$ , and its shape decides which physics you get: whether a Bose gas condenses, whether a Fermi gas has a linear-in- $T$ heat capacity, and what spectrum a photon gas radiates. We take this machinery from quantum mechanics. In aq-qm-density-of-final-states the same object is built for transition rates: allowed wavevectors fill "a cell of volume $(2\pi/L)^d$ in $\mathbf k$ -space," giving $\mathrm dN=\frac{V}{(2\pi)^3}\,\mathrm d^3k$ and, after the change of variables, $\rho(E)=\frac{V}{(2\pi)^3}4\pi k^2\frac{\mathrm dk}{\mathrm dE}=\frac{V\,m\,k}{2\pi^2\hbar^2}$ . That article states the seam outright: "the density of 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." This article does not re-derive the box spectrum; it takes the uniform $\mathbf k$ -space density (owned by aq-sm-density-of-states-k-space) and pushes it forward through the dispersion relation to get $g(\varepsilon)$ , the object that feeds every quantum-gas sum downstream. The energy dependence is pure geometry. A constant-energy surface slices the uniform $\mathbf k$ -grid; where that surface is large or moves slowly with energy, states pile up. Dividing the shell area by the slope $\mathrm d\varepsilon/\mathrm dk$ gives the power law $g(\varepsilon)\propto\varepsilon^{d/2-1}$ for massive particles: rising as $\sqrt{\varepsilon}$ in 3D, flat in 2D, and diverging as $1/\sqrt{\varepsilon}$ in 1D. That single exponent is the switch that turns condensation, heat capacities, and spectra on or off.

Key takeaways

The density of states $g(\varepsilon)$ counts single-particle states per unit energy; it converts every mode sum into an energy integral, $\sum_{\mathbf k}\to\int g(\varepsilon)\,\mathrm d\varepsilon$ .
It is the uniform $\mathbf k$ -space density $V/(2\pi)^d$ (owned by aq-sm-density-of-states-k-space, mirroring QM's aq-qm-density-of-final-states) pushed through the dispersion $\varepsilon(\mathbf k)$ .
For nonrelativistic particles in 3D: $g(\varepsilon)=\dfrac{g_s V}{4\pi^2}\Big(\dfrac{2m}{\hbar^2}\Big)^{3/2}\sqrt{\varepsilon}$ — the exact prefactor matches QM's $\rho(E)=Vmk/2\pi^2\hbar^2$ .
Dimension and dispersion set the exponent: $g(\varepsilon)\propto\varepsilon^{d/2-1}$ (1D diverges, 2D flat, 3D $\sqrt{\varepsilon}$ ); photons give $\varepsilon^2$ .
Through $g(\varepsilon)$ every extensive quantity of an ideal Bose, Fermi, or Boltzmann gas becomes $\int g(\varepsilon)\,\bar n(\varepsilon)\,(\cdots)\,\mathrm d\varepsilon$ .
Key idea: the density of states is a change of variables — the density of quantum states is uniform in $\mathbf k$ -space; pushing it forward through $\varepsilon(\mathbf k)$ makes $g(\varepsilon)$ , whose energy dependence is purely the geometry of the constant-energy surface divided by the dispersion slope.

The core insight

$g(\varepsilon)$ is the uniform $\mathbf k$ -space density $g_s V/(2\pi)^d$ mapped through the dispersion $\varepsilon(\mathbf k)$ : the number of states in a shell is its area $S_d k^{d-1}$ divided by the slope $\mathrm d\varepsilon/\mathrm dk$ . The result $g(\varepsilon)\propto\varepsilon^{d/2-1}$ knows only geometry — dimension and dispersion — never statistics; bosons versus fermions enter solely through the occupation $\bar n(\varepsilon)$ that multiplies $g$ .

Where this lives today

The density of states is the workhorse that connects a material's microscopic spectrum to what an experiment measures. In metals, $g(E_F)$ fixes the Sommerfeld coefficient of the electronic heat capacity and the Pauli spin susceptibility; angle-resolved photoemission (ARPES) and scanning tunnelling spectroscopy map $g(\varepsilon)$ directly, exposing band gaps, van Hove singularities, and — in graphene — the linear "Dirac" DOS that vanishes at the neutrality point. In semiconductor devices the DOS at the band edges sets carrier statistics and threshold voltages; in cold-atom laboratories the trap geometry engineers a chosen $g(\varepsilon)$ to tune whether a Bose gas condenses. Wherever a mode sum is replaced by an energy integral — thermal transport, optical absorption, superconducting gaps — $g(\varepsilon)$ is the bridge.

The physical picture

Picture the allowed wavevectors as a uniform grid in $\mathbf k$ -space — one point per cell of volume $(2\pi/L)^d=(2\pi)^d/V$ , exactly the count aq-sm-density-of-states-k-space establishes and that aq-qm-density-of-final-states writes as $\mathrm dN=\frac{V}{(2\pi)^3}\mathrm d^3k$ . A surface of constant energy $\varepsilon(\mathbf k)=\varepsilon$ slices through this grid, and $g(\varepsilon)\,\mathrm d\varepsilon$ is simply the number of grid points caught between the shells at $\varepsilon$ and $\varepsilon+\mathrm d\varepsilon$ . Two things control how many that is: how big the surface is, and how fast it sweeps outward as you raise the energy. In 3D with $\varepsilon\propto k^2$ the constant-energy surface is a sphere of radius $k\propto\sqrt{\varepsilon}$ . Its area grows like $k^2\propto\varepsilon$ , but the shells also spread apart, since $\mathrm d\varepsilon/\mathrm dk\propto k\propto\sqrt{\varepsilon}$ . Area over spacing gives $\varepsilon/\sqrt{\varepsilon}=\sqrt{\varepsilon}$ — the square-root law. Lower the dimension and states shift toward low energy: in 2D the shell is a ring whose length $\propto k$ exactly cancels the $1/k$ from the widening spacing, leaving $g(\varepsilon)$ flat; in 1D the two endpoints $\pm k$ stay put while the spacing collapses, so states pile up and $g$ diverges as $\varepsilon\to0$ .

Figure

Density of states in 3D rises as √ε, is flat in 2D, and diverges as 1/√ε in 1D

The hero: one exponent $d/2-1$ sets the whole low-energy behaviour. The 3D $\sqrt{\varepsilon}$ curve rising from zero is the engine of the electron gas and of Bose–Einstein condensation.

Figure

The uniform k-grid sliced by a constant-energy shell becomes g(ε)

A shell in $\mathbf k$ -space (area $4\pi k^2$ , thickness $\mathrm dk$ ) maps to a slab under $g(\varepsilon)$ : divide the shell count by the dispersion slope $\mathrm d\varepsilon/\mathrm dk$ and the $\sqrt{\varepsilon}$ law falls out.

The core equations

The starting point is the sum-to-integral rule that aq-sm-density-of-states-k-space owns and that aq-qm-density-of-final-states writes as $\mathrm dN=\frac{V}{(2\pi)^3}\mathrm d^3k$ : because allowed wavevectors form a uniform grid with one state per cell $(2\pi)^d/V$ , a sum over modes becomes an integral over $\mathbf k$ with a constant weight.

\sum_{\mathbf k}(\cdots)\;\longrightarrow\;\frac{g_s V}{(2\pi)^d}\int\mathrm d^d k\,(\cdots) \qquad (1)

where: $g_s=2s+1$ — spin/internal degeneracy (2 for electrons, 1 for a spinless boson); $V$ — volume (area/length in $d=2,1$ ); $d$ — spatial dimension; $\mathbf k$ — wavevector, quantized in units $2\pi/L$ . Physical reading: the density of states in $\mathbf k$ -space is a constant, fixed purely by the box; all the energy structure lives in the map to $\varepsilon$ . The density of states in energy collapses this integral onto a constant-energy shell. Writing it with a delta function makes the "count states at energy $\varepsilon$ " instruction explicit.

g(\varepsilon)=\frac{g_s V}{(2\pi)^d}\int\mathrm d^d k\;\delta\big(\varepsilon-\varepsilon(\mathbf k)\big) \qquad (2)

where: $\varepsilon(\mathbf k)$ — dispersion relation; $\delta$ — Dirac delta selecting the shell; $g(\varepsilon)$ — states per unit energy. Physical reading: $g(\varepsilon)\,\mathrm d\varepsilon$ is the number of grid points between the shells $\varepsilon$ and $\varepsilon+\mathrm d\varepsilon$ — the shell area divided by the dispersion slope. For a nonrelativistic gas in 3D, carrying out the shell integral gives the square-root law with its exact prefactor — the same object aq-qm-density-of-final-states records as $\rho(E)=Vmk/2\pi^2\hbar^2$ .

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

where: $m$ — particle mass; $\hbar$ — reduced Planck constant; $\varepsilon=\hbar^2k^2/2m\ge0$ — kinetic energy. Physical reading: the DOS vanishes at the band bottom ( $\sqrt{\varepsilon}\to0$ ) and rises without bound — most of the phase space lives at high energy, which is why a 3D gas can absorb particles into excited states. For an isotropic dispersion $\varepsilon=Ck^p$ in $d$ dimensions the exponent is fixed by geometry alone, independent of the prefactor.

g(\varepsilon)\propto\varepsilon^{\,d/p-1},\qquad \text{massive }(p=2):\;g(\varepsilon)\propto\varepsilon^{\,d/2-1} \qquad (4)

where: $p$ — dispersion exponent ( $p=2$ massive, $p=1$ photon/phonon). Physical reading: $d/2-1$ equals $+\tfrac12$ (3D), $0$ (2D), $-\tfrac12$ (1D) — the single number that decides whether low-energy states are scarce (3D) or overwhelming (1D). Finally, this is the object every quantum-gas calculation actually uses: every extensive quantity is $g(\varepsilon)$ weighted by the occupation and integrated over energy.

N=\int_0^\infty g(\varepsilon)\,\bar n(\varepsilon)\,\mathrm d\varepsilon,\qquad U=\int_0^\infty \varepsilon\,g(\varepsilon)\,\bar n(\varepsilon)\,\mathrm d\varepsilon \qquad (5)

where: $\bar n(\varepsilon)$ — mean occupation: Bose–Einstein $1/(e^{\beta(\varepsilon-\mu)}-1)$ , Fermi–Dirac $1/(e^{\beta(\varepsilon-\mu)}+1)$ , or Maxwell–Boltzmann $e^{-\beta(\varepsilon-\mu)}$ ; $N,U$ — particle number and energy. Physical reading: $g$ carries the geometry, $\bar n$ carries the statistics; multiply and integrate to get thermodynamics.

Derivations

Derivation ADensity of states from the k-space grid: the square-root law and its exact prefactor

The goal is to turn the uniform $\mathbf k$ -space count — one state per cell $(2\pi/L)^d$ , owned by aq-sm-density-of-states-k-space and written by aq-qm-density-of-final-states as $\mathrm dN=\frac{V}{(2\pi)^3}\mathrm d^3k$ — into the energy density of states $g(\varepsilon)$ , obtaining both the general power law $g(\varepsilon)\propto\varepsilon^{d/2-1}$ and the exact 3D prefactor. We do not re-derive the box spectrum (that is quantum-mechanics-owned); we change variables from $k$ to $\varepsilon$ through the dispersion and collect the geometric factors.

Step 1. Count states in a thin shell. In $d$ dimensions the number of grid points between $k$ and $k+\mathrm dk$ is the shell volume $S_d k^{d-1}\mathrm dk$ divided by the cell $(2\pi)^d/V$ and multiplied by $g_s$ : $\mathrm dN=\frac{g_s V}{(2\pi)^d}S_d k^{d-1}\mathrm dk$ , with $S_d=2\pi^{d/2}/\Gamma(d/2)$ so $S_1=2$ , $S_2=2\pi$ , $S_3=4\pi$ . Why: isotropy lets us do the angular integral once and for all, leaving a radial count.

Step 2. Change the variable to energy. With $\varepsilon=Ck^p$ , invert to $k=(\varepsilon/C)^{1/p}$ and differentiate, $\mathrm dk/\mathrm d\varepsilon=\frac{1}{pC}(\varepsilon/C)^{1/p-1}$ . Then $g(\varepsilon)=\mathrm dN/\mathrm d\varepsilon=\frac{g_s V S_d}{(2\pi)^d}\,\frac{1}{pC^{d/p}}\,\varepsilon^{d/p-1}$ . Why: the Jacobian $\mathrm dk/\mathrm d\varepsilon$ carries all the dispersion physics; the leftover $\varepsilon^{d/p-1}$ is pure geometry.

Step 3. Specialize to 3D massive particles. Set $d=3$ , $p=2$ , $C=\hbar^2/2m$ , $S_3=4\pi$ . The powers give $\varepsilon^{3/2-1}=\sqrt{\varepsilon}$ and the constants collapse to $g(\varepsilon)=\frac{g_s V}{4\pi^2}\big(\frac{2m}{\hbar^2}\big)^{3/2}\sqrt{\varepsilon}$ . Why: this is the standard square-root DOS quoted in every quantum-gas chapter.

Step 4. Confirm the seam: match the quantum-mechanics owner. Setting $g_s=1$ and writing $k=\sqrt{2m\varepsilon}/\hbar$ , our $g(\varepsilon)$ equals $Vmk/2\pi^2\hbar^2$ — the very expression aq-qm-density-of-final-states records as $\rho(E)=\frac{V}{(2\pi)^3}4\pi k^2\frac{\mathrm dk}{\mathrm dE}=\frac{Vmk}{2\pi^2\hbar^2}$ . The two are the same object, as that article states: the density of final states in a transition rate is identical to the $g(E)$ in the partition function and the Fermi/Bose sums. Why: both are just the geometry of counting quantum states; the physical context (a golden-rule rate versus a thermal integral) does not change the count. Verified: the invert $\varepsilon(k(\varepsilon))=\varepsilon$ , the exact 3D prefactor, the match to $Vmk/2\pi^2\hbar^2$ , and the exponents $\{+\tfrac12,0,-\tfrac12,2\}$ for $(d,p)=(3,2),(2,2),(1,2),(3,1)$ are checked symbolically in verify_density-of-states-energy.py.

Go deeper · Derivation B · 1 minFrom g(ε) to the Fermi energy and the classical limit

Derivation BFrom g(ε) to the Fermi energy and the classical limit

The goal is to show that $g(\varepsilon)$ is not a bookkeeping curiosity but the object that closes the two flagship calculations of the chapter: the zero-temperature Fermi energy of an electron gas (via the Fermi–Dirac occupation, aq-sm-fermi-dirac-distribution) and the emergence of the thermal de Broglie wavelength in the dilute limit (aq-sm-thermal-de-broglie-wavelength-sm4). Both follow by plugging the $\sqrt{\varepsilon}$ DOS into the master integral $N=\int g\,\bar n\,\mathrm d\varepsilon$ and doing one integral.

Step 1. Zero-temperature Fermi gas. At $T=0$ the Fermi–Dirac occupation is a step: every state below $\mu=E_F$ is filled, every one above empty. So $N=\int_0^{E_F}g(\varepsilon)\,\mathrm d\varepsilon=\frac{g_s V}{6\pi^2}\big(\frac{2mE_F}{\hbar^2}\big)^{3/2}$ . Why: the step function turns the occupation-weighted integral into a plain integral of the DOS up to $E_F$ .

Step 2. Solve for $E_F$ . Inverting with $n=N/V$ gives $E_F=\frac{\hbar^2}{2m}\big(\frac{6\pi^2 n}{g_s}\big)^{2/3}$ ; for spin- $\tfrac12$ electrons ( $g_s=2$ ) this is the textbook $E_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ . The DOS at the surface, $g(E_F)=\frac{3N}{2E_F}$ , sets the linear-in- $T$ electronic heat capacity of aq-sm-electronic-heat-capacity-metals. Why: only the states near $E_F$ can be thermally excited, so $g(E_F)$ is the number that matters at low $T$ .

Step 3. Dilute (classical) limit. When occupations are tiny the mean occupation reduces to Maxwell–Boltzmann, $\bar n\to e^{(\mu-\varepsilon)/k_BT}$ (aq-sm-maxwell-boltzmann-classical-limit owns this reduction). Then $N=\frac{g_s V}{4\pi^2}\big(\frac{2m}{\hbar^2}\big)^{3/2}e^{\mu/k_BT}\int_0^\infty\sqrt{\varepsilon}\,e^{-\varepsilon/k_BT}\mathrm d\varepsilon$ . Why: the fugacity $z=e^{\mu/k_BT}$ factors out, leaving a pure Gaussian-type moment.

Step 4. Do the integral and read off $\lambda$ . Using $\int_0^\infty\sqrt{\varepsilon}\,e^{-\varepsilon/k_BT}\mathrm d\varepsilon=\Gamma(\tfrac32)(k_BT)^{3/2}=\frac{\sqrt\pi}{2}(k_BT)^{3/2}$ , the assembly gives $N=g_s V\big(\frac{mk_BT}{2\pi\hbar^2}\big)^{3/2}e^{\mu/k_BT}=\frac{g_s V}{\lambda^3}e^{\mu/k_BT}$ with $\lambda=h/\sqrt{2\pi mk_BT}$ — the thermal de Broglie wavelength of aq-sm-thermal-de-broglie-wavelength-sm4 emerging directly from the DOS integral, and the degeneracy parameter $n\lambda^3$ of aq-sm-quantum-concentration-classicality-criterion following at once. Why: the $\sqrt{\varepsilon}$ DOS is exactly the weight that turns the fugacity into $g_sV/\lambda^3$ , tying the state count to the quantum length scale. Verified: the Fermi-energy closure $E_F=(\hbar^2/2m)(3\pi^2n)^{2/3}$ (with copper $\approx7.0$ eV), $g(E_F)=3N/2E_F$ , the moment $\int_0^\infty\sqrt\varepsilon e^{-\varepsilon/k_BT}\mathrm d\varepsilon=\frac{\sqrt\pi}{2}(k_BT)^{3/2}$ , and the classical closure $N=g_sV/\lambda^3\cdot e^{\mu/k_BT}$ are checked in verify_density-of-states-energy.py.

Limiting cases

For an isotropic massive dispersion $\varepsilon=Ck^2$ in $d$ dimensions, $g(\varepsilon)\propto\varepsilon^{d/2-1}$ ; for linear (photon/phonon) dispersion the exponent shifts. The low-energy behaviour decides condensation, heat capacities, and spectra.

Case	Result	Meaning
3D nonrelativistic ( $d=3,p=2$ )	$g(\varepsilon)\propto\sqrt{\varepsilon}$	electrons and atoms; supports BEC and the Sommerfeld $\sqrt{\varepsilon_F}$ structure
2D nonrelativistic ( $d=2,p=2$ )	$g(\varepsilon)=\dfrac{g_s mA}{2\pi\hbar^2}=$ const	energy-independent DOS; ideal 2D Bose gas does not condense at finite $T$
1D nonrelativistic ( $d=1,p=2$ )	$g(\varepsilon)\propto\varepsilon^{-1/2}$	van Hove divergence at the band bottom; strong infrared weight
Photons ( $d=3,p=1$ )	$g(\varepsilon)\propto\varepsilon^{2}$	linear dispersion $\varepsilon=\hbar ck$ ; yields the Planck spectrum and Stefan–Boltzmann $T^4$
At $E_F$	$g(E_F)=\dfrac{3N}{2E_F}$	controls the linear-in- $T$ electronic heat capacity

Common trapThe density of states knows nothing about statistics

It is tempting to think fermions and bosons have "different densities of states." They do not — $g(\varepsilon)$ counts single-particle states and is a purely kinematic object; a spin- $\tfrac12$ Fermi gas and a spinless Bose gas of the same mass differ only by the degeneracy factor $g_s$ , not by the $\sqrt{\varepsilon}$ shape. All the statistical difference (Pauli exclusion versus Bose enhancement) enters through the occupation $\bar n(\varepsilon)$ that multiplies $g$ . Two further traps: do not forget $g_s=2s+1$ (electrons carry the factor 2), and do not confuse $g(\varepsilon)$ (per unit energy) with the cumulative count $G(\varepsilon)=\int_0^\varepsilon g\,\mathrm d\varepsilon'$ (states below $\varepsilon$ ).

Worked examples

Example 1 — The Fermi energy of copper from the DOS.

Step 1. Setup. Copper contributes one conduction electron per atom, giving $n=8.49\times10^{28}\ \mathrm{m^{-3}}$ ; electrons are spin- $\tfrac12$ so $g_s=2$ , and $m=m_e=9.11\times10^{-31}$ kg.

Step 2. Assemble $E_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ . Compute $3\pi^2 n=3\times9.8696\times8.49\times10^{28}=2.514\times10^{30}\ \mathrm{m^{-3}}$ , so $(3\pi^2 n)^{2/3}=1.847\times10^{20}\ \mathrm{m^{-2}}$ .

Step 3. Multiply by $\hbar^2/2m=\dfrac{(1.0546\times10^{-34})^2}{2\times9.11\times10^{-31}}=6.104\times10^{-39}\ \mathrm{J\,m^2}$ : $E_F=6.104\times10^{-39}\times1.847\times10^{20}=1.128\times10^{-18}\ \mathrm{J}$ .

Step 4. Convert: $E_F=1.128\times10^{-18}/1.602\times10^{-19}=\mathbf{7.04\ eV}$ , matching the measured Fermi energy of copper ( $\approx7.0$ eV). The corresponding Fermi temperature $T_F=E_F/k_B\approx\mathbf{8.2\times10^4\ K}$ is why room-temperature electrons are highly degenerate: $T\ll T_F$ .

Example 2 — The most-probable energy versus the mean energy in a classical gas.

Step 1. Setup. In the dilute limit the energy distribution is $g(\varepsilon)\bar n(\varepsilon)\propto\sqrt{\varepsilon}\,e^{-\varepsilon/k_BT}$ . Find where it peaks and its mean.

Step 2. Peak. Maximize $\ln(\sqrt\varepsilon\,e^{-\varepsilon/k_BT})=\tfrac12\ln\varepsilon-\varepsilon/k_BT$ ; set the derivative $\frac{1}{2\varepsilon}-\frac{1}{k_BT}=0$ , giving the most-probable energy $\varepsilon_{\rm mp}=\mathbf{\tfrac12 k_BT}$ .

Step 3. Mean. $\langle\varepsilon\rangle=\dfrac{\int_0^\infty\varepsilon^{3/2}e^{-\varepsilon/k_BT}\mathrm d\varepsilon}{\int_0^\infty\varepsilon^{1/2}e^{-\varepsilon/k_BT}\mathrm d\varepsilon}=\dfrac{\Gamma(5/2)}{\Gamma(3/2)}k_BT=\dfrac{3/2}{1}\cdot\tfrac{1}{1}k_BT=\mathbf{\tfrac32 k_BT}$ .

Step 4. Meaning. The mean sits at three times the mode ( $\tfrac32 k_BT$ versus $\tfrac12 k_BT$ ): the long high-energy tail of $\sqrt{\varepsilon}\,e^{-\varepsilon/k_BT}$ drags the average far above the peak. This is why reaction rates — governed by the tail beyond an activation energy — are so temperature-sensitive even though the most-probable molecule carries only $\tfrac12 k_BT$ ; the mean $\tfrac32 k_BT$ is the equipartition value for three translational directions.

Origins · 1 minfermi energy and how this was discovered

fermi energy

g(ε) integrated to the Fermi sea · The energy density of states

g(\varepsilon)

finds its sharpest application in the Fermi energy: integrating

g(\varepsilon)

up to

\varepsilon_F

counts exactly the

N

electrons of a metal at zero temperature, giving

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

. The counting function itself began in Debye's 1912 lattice-vibration model — as aq-qm-density-of-final-states notes, it is "older than the golden rule that made it famous in transition theory" — and Sommerfeld's 1927 free-electron theory used exactly this

g(\varepsilon)\propto\sqrt\varepsilon

to explain why only electrons within

\sim k_BT

\varepsilon_F

contribute to the heat capacity. The same

g(\varepsilon)

counts phonon modes in a solid and photon modes in a cavity.

History

1900

— Planck counts electromagnetic modes in a cavity to derive the blackbody spectrum

1907

— Einstein models a solid as identical oscillators, quantizing the vibrational energy

1912

— Debye counts acoustic modes with a linear dispersion, giving the T³ heat capacity

1927

— Sommerfeld applies the Fermi–Dirac occupation to the free-electron gas in a metal

1928

— Bloch extends the mode count to electrons in a periodic potential (band structure)

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §1.4, §6.1, §8.1; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.9–9.10; Sethna, Statistical Mechanics, Ch. 7; Landau & Lifshitz, Statistical Physics Part 1, §55–58.
Quantum-mechanics owner: the $\mathbf k$ -space cell $(2\pi/L)^d$ , $\mathrm dN=\frac{V}{(2\pi)^3}\mathrm d^3k$ , and $\rho(E)=Vmk/2\pi^2\hbar^2$ — with the seam statement that this is the same $g(E)$ as in the partition function and Fermi/Bose sums — are owned by aq-qm-density-of-final-states; the $\mathbf k$ -space count is owned by aq-sm-density-of-states-k-space.
Leads to: the Bose–Einstein and Fermi–Dirac occupations that $g(\varepsilon)$ multiplies (aq-sm-bose-einstein-distribution, aq-sm-fermi-dirac-distribution), the classical limit (aq-sm-maxwell-boltzmann-classical-limit), the quantum-concentration criterion (aq-sm-quantum-concentration-classicality-criterion), and downstream the electronic heat capacity, BEC, and the Planck spectrum (SM5).
Provenance: AllQFT `sm` export (Pathria & Beale), aq-sm-density-of-states-energy, ref STAT.QSTAT.DOSE.1212. All DOS prefactors, exponents, the Fermi-energy closure, and the classical-limit integral are python-verified (19 checks, all PASS). Standard constants used: $\Gamma(3/2)=\sqrt\pi/2$ , $\varepsilon_F=\hbar^2(3\pi^2 n)^{2/3}/2m$ , copper $E_F\approx7.0$ eV.

Why does the density of states vanish at zero energy in 3D but diverge in 1D?

Both follow from $g(\varepsilon)\propto\varepsilon^{d/2-1}$ with $p=2$ . In 3D the exponent is $+\tfrac12$ , so $g\to0$ : the constant-energy sphere shrinks to a point faster than the shells narrow. In 1D the exponent is $-\tfrac12$ , a van Hove divergence: the two turning points $\pm k$ stay at finite count while the slope $\mathrm d\varepsilon/\mathrm dk=\hbar^2k/m\to0$ , so states pile up at the band bottom. The 2D case sits exactly between, with the ring's length cancelling the slope to give a constant.

Is the density of states the same thing as the partition function?

No, but it is the ingredient the partition function is built from. $g(\varepsilon)$ counts single-particle states per unit energy; the (grand) partition function sums over how those states are occupied. For an ideal quantum gas the connection is direct: $\ln\Xi=\pm\int g(\varepsilon)\ln(1\pm e^{-\beta(\varepsilon-\mu)})\,\mathrm d\varepsilon$ , so once you know $g(\varepsilon)$ every thermodynamic quantity is a one-dimensional integral. The DOS is the geometry; the partition function is the geometry weighted by statistics.

Do I need to include the spin degeneracy factor?

Yes, whenever the particle carries spin. The factor $g_s=2s+1$ multiplies the whole DOS: $g_s=2$ for electrons (spin- $\tfrac12$ ), $g_s=1$ for a spinless boson, $g_s=2$ for photons (two transverse polarizations, though photons are usually handled by their own $\varepsilon^2$ DOS). Forgetting it is the single most common numerical error in quantum-gas problems — it changes $E_F$ , $T_c$ , and the heat capacity by fixed powers of 2.

How does this connect to the density of final states in the golden rule?

It is literally the same object. As aq-qm-density-of-final-states states, "the density of 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." A golden-rule rate weights $g(E)$ by a matrix element; a thermal integral weights it by an occupation function. The counting — one state per $\mathbf k$ -cell, pushed through the dispersion — is identical, which is why the same $\rho(E)=Vmk/2\pi^2\hbar^2$ appears in both contexts.

Why does the photon gas have an ε² DOS instead of √ε?

Because photons have a linear dispersion $\varepsilon=\hbar ck$ rather than the quadratic $\varepsilon=\hbar^2k^2/2m$ of massive particles. The general rule $g(\varepsilon)\propto\varepsilon^{d/p-1}$ with $d=3$ , $p=1$ gives $\varepsilon^{2}$ . That extra factor of $\varepsilon$ (relative to the massive $\sqrt\varepsilon$ ) is exactly what produces the Stefan–Boltzmann $T^4$ law: the energy integrand $\varepsilon\cdot g(\varepsilon)\cdot\bar n\propto\varepsilon^3/(e^{\beta\varepsilon}-1)$ scales as $T^4$ after rescaling. The DOS exponent is the fingerprint of the dispersion.

Why does a 2D ideal Bose gas fail to condense while a 3D one does?

The condensation criterion is whether the excited-state integral $N_{\rm exc}=\int_0^\infty g(\varepsilon)/(e^{\beta(\varepsilon-\mu)}-1)\,\mathrm d\varepsilon$ stays finite as $\mu\to0$ . In 3D, $g\propto\sqrt{\varepsilon}$ suppresses the $1/\varepsilon$ infrared singularity of the Bose factor enough that the integral converges to a finite maximum — extra particles must go into the ground state, giving BEC. In 2D the constant DOS is not enough suppression, so the integral diverges logarithmically and can absorb any number of particles; no macroscopic ground-state occupation is forced at finite $T$ . The dimensionality is read straight off the DOS exponent.

Does the density of states depend on the shape of the box?

Not in the thermodynamic limit. The bulk DOS depends only on the volume $V$ , the dispersion, and the dimension — not on whether the box is a cube, a sphere, or has periodic versus hard walls (that boundary-independence is the subject of aq-sm-boundary-conditions-mode-counting). Shape enters only through surface corrections that scale as area/volume $\sim1/L$ and vanish as the system grows. For a macroscopic sample $g(\varepsilon)$ is universal; for a nanostructure the actual discrete spectrum matters.