Equation of State of Ideal Quantum Gases

Heat any gas enough and it obeys $PV=Nk_BT$ . But cool it, or squeeze it, and identical quantum particles begin to feel each other — not through any force, but through the bookkeeping of their own indistinguishability. A gas of bosons pushes a little less hard than the ideal law predicts; a gas of fermions pushes a little more. Push the effect far enough and the two gases end up in completely different worlds: the fermions stiffen into the degeneracy pressure that holds up a white dwarf, while the bosons soften until, at a sharp threshold, they collapse into a Bose–Einstein condensate. The equation of state $P=P(n,T)$ is the single relation that tracks both journeys. The remarkable economy of an ideal quantum gas is that every thermodynamic function — pressure, density, energy, entropy — collapses onto one family of special functions, the polylogarithms $g_\nu(z)=\sum_{k\ge1}z^k/k^\nu$ (bosons) and their alternating cousins $f_\nu(z)$ (fermions). The single knob is the fugacity $z=e^{\mu/k_BT}$ , and the single dimensionless measure of "how quantum" the gas is, is the phase-space density $n\lambda^3$ — the number of particles packed into a thermal de Broglie volume. When $n\lambda^3\ll1$ the gas forgets its statistics and reverts to Maxwell–Boltzmann; as $n\lambda^3$ climbs toward unity, bosons and fermions part ways violently. We take from the quantum-statistics track the fugacity $z$ and the polylogarithm functions (recalled by name from Fugacity and Chemical Potential in Quantum Gases), and build the closed-form parametric equation of state. Along the way we prove the striking universal relation $PV=\tfrac23U$ that holds for any non-relativistic ideal gas — classical, Bose, or Fermi — and extract the leading quantum virial correction $P/nk_BT=1\pm n\lambda^3/2^{5/2}$ whose sign alone tells the two species apart.

Key takeaways

Every thermodynamic quantity of an ideal quantum gas with $\varepsilon=p^2/2m$ is a polylogarithm of the fugacity $z=e^{\mu/k_BT}$ : $n\lambda^3=g_{3/2}(z)$ and $P\lambda^3/k_BT=g_{5/2}(z)$ for bosons (alternating-sign $f_\nu$ for fermions).
Eliminating $z$ between those two gives the closed parametric equation of state $P=P(n,T)$ ; the degeneracy parameter $n\lambda^3$ controls how far the gas sits from the classical ideal law.
A universal relation $PV=\tfrac23U$ holds for any non-relativistic ideal gas — classical, Bose, or Fermi — because the exponent comes from the density of states $g(\varepsilon)\propto\varepsilon^{1/2}$ , not from the statistics.
The leading quantum virial correction is $\dfrac{PV}{Nk_BT}=1+\dfrac{n\lambda^3}{2^{5/2}}$ for fermions (statistical repulsion) and $1-\dfrac{n\lambda^3}{2^{5/2}}$ for bosons (statistical attraction).
The degenerate limits recover the $T=0$ Fermi degeneracy pressure $P\propto n^{5/3}$ and the Bose saturation at $z\to1$ that triggers condensation.
Key idea: one fugacity $z$ and the polylogarithm functions collapse the entire thermodynamics of an ideal quantum gas onto a single parametric curve $P(n,T)$ ; its universal $PV=\tfrac23U$ backbone and its $\pm n\lambda^3/2^{5/2}$ first correction already split fermions (repulsive) from bosons (attractive).

The core insight

For a non-relativistic ideal quantum gas, the pressure, density and energy are all polylogarithms of the fugacity $z=e^{\mu/k_BT}$ ; eliminating $z$ yields the parametric equation of state $P(n,T)$ , which obeys the universal $PV=\tfrac23U$ and whose first quantum correction $1\pm n\lambda^3/2^{5/2}$ distinguishes repulsive fermions from attractive bosons.

Where this lives today

The polylogarithmic equation of state is not a textbook relic — it is the working model for ultracold atomic gases. When experimentalists trap a cloud of $^{40}$ K or $^6$ Li atoms and cool it to nanokelvin, the gas is a nearly ideal Fermi or Bose gas at $n\lambda^3\sim1$ , exactly the crossover regime this article describes. The measured pressure, released by expanding the cloud and imaging it, is compared directly against $f_{5/2}(z)$ or $g_{5/2}(z)$ to extract the temperature and chemical potential — the equation of state is the thermometer. At the other extreme, the same relations underpin stellar structure codes: the electron equation of state in a white dwarf, or the neutron equation of state in a neutron star, is the degenerate limit of exactly these functions. And in condensed-matter simulations, the ideal-quantum-gas equation of state is the reference against which interacting corrections — the real virial series with genuine forces — are measured. The one parametric curve $P(n,T)$ built here is the common language of ultracold atoms, dense stars, and the electron liquid in every metal.

The physical picture

Start with the one number that decides everything: the phase-space density $n\lambda^3$ , where $\lambda=h/\sqrt{2\pi mk_BT}$ is the thermal de Broglie wavelength. It counts how many particles sit inside one "quantum blur" of volume $\lambda^3$ . When $n\lambda^3\ll1$ the blurs almost never overlap, the particles are effectively distinguishable, and the gas is a classical ideal gas: $PV=Nk_BT$ , statistics irrelevant. As you cool the gas, $\lambda$ grows like $T^{-1/2}$ and the blurs begin to touch. Now the symmetry of the many-body wavefunction — symmetric for bosons, antisymmetric for fermions — starts to matter, and it acts like a phantom interaction with no force behind it. The direction of that phantom interaction is fixed by the statistics. Identical bosons are more likely to be found close together (constructive symmetrization), which reduces the pressure a gas of them exerts — a statistical attraction. Identical fermions are forbidden from the same state (the exclusion hole), which raises the pressure — a statistical repulsion. The first figure makes this the headline: pressure divided by the classical value $Nk_BT$ , plotted against $n\lambda^3$ . The classical line sits flat at $1$ ; the Bose curve peels off below it, the Fermi curve above.

Figure

Equation of state PV/NkT versus the degeneracy parameter nλ³ — classical line at 1, Fermi curve rising above (statistical repulsion), Bose curve dipping below (statistical attraction)

The second picture is the machinery behind that split: the fugacity $z$ as a dial. Every occupation number, every thermodynamic function, is a power series in $z$ . Small $z$ (dilute, hot) means the linear term dominates and the gas is classical. As $z$ climbs toward $1$ , higher powers switch on — and here the two species diverge. For fermions $z$ can grow without bound (in the degenerate limit $z\sim e^{\varepsilon_F/k_BT}\to\infty$ ); for bosons $z$ is capped at exactly $1$ , because $g_{3/2}(z)$ hits its maximum finite value $\zeta(3/2)=2.612$ there. That ceiling is the origin of Bose–Einstein condensation.

Figure

Fugacity z as the master dial — for fermions z runs to infinity (degenerate limit), for bosons z is capped at z = 1 where g₃⁄₂ saturates at ζ(3⁄2) = 2.612 and condensation begins

Between these two limits lies the whole equation of state. The strategy is now clean: write $n\lambda^3$ and $P\lambda^3/k_BT$ each as a polylogarithm of $z$ , then eliminate $z$ to get $P(n,T)$ . That single elimination carries the gas continuously from the classical law, through the quantum virial series, into either degeneracy pressure (fermions) or condensation (bosons).

The core equations

Everything flows from the grand potential of a set of independent single-particle modes. In the grand canonical ensemble each mode of energy $\varepsilon$ contributes independently, and summing the logarithm of the single-mode grand partition function — then converting the mode sum to an energy integral with the free-particle density of states $g(\varepsilon)\propto\varepsilon^{1/2}$ — gives the pressure directly.

\frac{P\lambda^3}{k_BT} = \begin{cases} \; g_{5/2}(z), & \text{bosons}\\[4pt] \; f_{5/2}(z), & \text{fermions}\end{cases} \qquad n\lambda^3 = \begin{cases}\; g_{3/2}(z),&\text{bosons}\\[4pt]\; f_{3/2}(z),&\text{fermions}\end{cases}\qquad (1)

where: $z=e^{\mu/k_BT}$ is the fugacity; $\lambda=h/\sqrt{2\pi mk_BT}$ the thermal wavelength; and the polylogarithms are $g_\nu(z)=\sum_{k\ge1}z^k/k^\nu$ (bosons) and $f_\nu(z)=\sum_{k\ge1}(-1)^{k-1}z^k/k^\nu$ (fermions). Read as a pair of parametric equations: the second fixes $z$ from the density, the first then delivers the pressure. The whole equation of state is these two curves with $z$ eliminated. The functions $g_\nu$ and $f_\nu$ are the entire toolkit. They obey a differentiation ladder that makes the thermodynamics turn like clockwork: applying $z\,d/dz$ lowers the index by one.

z\frac{\partial}{\partial z}\,g_\nu(z) = g_{\nu-1}(z), \qquad z\frac{\partial}{\partial z}\,f_\nu(z) = f_{\nu-1}(z) \qquad (2)

where: each derivative peels one power of $k$ off the denominator. This is why density ( $g_{3/2}$ ) is the $z\,d/dz$ derivative of pressure ( $g_{5/2}$ ) — pressure and density are one rung apart on the same ladder, and heat capacity is one rung further down at $g_{1/2}$ . The single most useful exact statement about the whole family is that pressure and energy are locked in a fixed ratio, no matter the temperature, density, or statistics — as long as the dispersion is $\varepsilon\propto p^2$ in three dimensions.

PV = \frac{2}{3}\,U \qquad (3)

where: $U$ is the internal energy. This is not a fermion fact or a boson fact; it is a density-of-states fact. The exponent $2/3$ comes entirely from $g(\varepsilon)\propto\varepsilon^{1/2}$ , so a classical gas, a degenerate Fermi gas, and a Bose gas below $T_c$ all obey it identically. Replace the dispersion with the ultrarelativistic $\varepsilon\propto p$ and the same argument gives $PV=\tfrac13U$ instead. Expanding equation (1) at small $z$ and eliminating it gives the equation of state as a virial series in the degeneracy parameter — the controlled bridge from the classical law into the quantum regime.

\frac{PV}{Nk_BT} = 1 \;\pm\; \frac{n\lambda^3}{2^{5/2}} \;+\; \Big(\tfrac18 - \tfrac{2}{3^{5/2}}\Big)(n\lambda^3)^2 \;+\;\cdots \qquad (4)

where: the upper sign ( $+$ ) is for fermions and the lower ( $-$ ) for bosons. The first correction is a pure exchange effect: fermions push harder than the classical law ( $+$ , statistical repulsion), bosons push softer ( $-$ , statistical attraction). The coefficient $2^{-5/2}=0.1768$ is exact, and the correction grows as the gas cools ( $\lambda^3\propto T^{-3/2}$ ), which is why quantum statistics is a low-temperature story.

Derivations

Derivation AThe parametric equation of state from the grand potential

The goal is to show that the pressure and density of an ideal quantum gas are the polylogarithms of equation (1), so that the equation of state $P(n,T)$ is fixed the moment we know the density. The route is the grand potential: write it as a sum over independent single-particle modes, convert the mode sum to an energy integral using the free-particle density of states $g(\varepsilon)=AV\varepsilon^{1/2}$ , expand the logarithm in powers of the fugacity, and integrate term by term. What to watch for is where the half-integer index $\nu=5/2$ comes from — it is the $\varepsilon^{1/2}$ density of states plus one power of $\varepsilon$ from the integration, and it is the same for both statistics, differing only by the alternating sign.

Step 1. Write the grand potential as a mode sum. For bosons $-\beta\Omega = -\sum_i\ln(1-ze^{-\beta\varepsilon_i})$ ; for fermions $-\beta\Omega=+\sum_i\ln(1+ze^{-\beta\varepsilon_i})$ , and $PV=-\Omega$ . Why: each single-particle mode is an independent grand-canonical subsystem, and its grand partition function is a geometric series (bosons) or a two-term sum (fermions).

Step 2. Convert the sum to an integral. With $g(\varepsilon)=AV\varepsilon^{1/2}$ , $A=\frac{2\pi(2m)^{3/2}}{h^3}$ , $PV=\mp k_BT\,AV\!\int_0^\infty\!\varepsilon^{1/2}\ln(1\mp ze^{-\beta\varepsilon})\,d\varepsilon$ . Why: the modes of a large box are dense, and the free-particle density of states in 3D scales as $\varepsilon^{1/2}$ ; the sign choice ( $\mp$ ) carries the boson/fermion distinction.

Step 3. Expand the logarithm. Using $\mp\ln(1\mp x)=\sum_{k\ge1}(\pm1)^{k-1}x^k/k$ with $x=ze^{-\beta\varepsilon}$ and integrating $\int_0^\infty\varepsilon^{1/2}e^{-k\beta\varepsilon}d\varepsilon=\tfrac{\sqrt\pi}{2}(k\beta)^{-3/2}$ gives $\frac{P\lambda^3}{k_BT}=\sum_{k\ge1}\frac{(\pm1)^{k-1}z^k}{k^{5/2}}$ . Why: each power of $z$ contributes an independent Gaussian integral, and the factor $k^{-3/2}$ from the integral combines with the $1/k$ from the log to give $k^{-5/2}$ — the polylogarithm $g_{5/2}$ (bosons, all $+$ ) or $f_{5/2}$ (fermions, alternating).

Step 4. Get the density by the ladder. Since $n=\frac1V\,z\,\frac{\partial}{\partial z}(\beta PV)$ , applying $z\,d/dz$ lowers the index by one: $n\lambda^3=g_{3/2}(z)$ (bosons) or $f_{3/2}(z)$ (fermions). Why: differentiating $\ln\Xi$ with respect to $\mu$ gives $\langle N\rangle$ , and $z\,d/dz$ turns $k^{-5/2}$ into $k^{-3/2}$ — pressure and density are one rung apart, equation (2).

Step 5. Eliminate $z$ . Equations (1) are a parametric pair $(n\lambda^3, P\lambda^3/k_BT)$ indexed by $z$ ; solving the density equation for $z(n\lambda^3)$ and substituting into the pressure equation gives $P=P(n,T)$ in closed (implicit) form. Why: $z$ is not a thermodynamic variable, only a parameter along the equation-of-state curve; eliminating it leaves the physical relation between $P$ , $n$ , and $T$ . *Verified: the $k^{-5/2}$ and $k^{-3/2}$ series, the ladder relation $z\,d/dz\,g_\nu=g_{\nu-1}$ , and the classical limit $z\to0\Rightarrow PV=Nk_BT$ are checked symbolically in verify_equation-of-state-quantum-gases.py against Pathria & Beale §7.1, §8.1.*

Go deeper · Derivation B · 1 minThe universal relation PV equals two-thirds U

Derivation BThe universal relation PV equals two-thirds U

The goal is to prove that $PV=\tfrac23U$ holds for any non-relativistic ideal gas in three dimensions — classical, Bose, or Fermi, at any temperature — and to trace the factor $2/3$ to its true source. The plan is to write $PV$ as the density-of-states integral of the previous card and integrate it by parts once. The payoff is conceptual and load-bearing for the whole subject: the $2/3$ is not a statistical accident but a geometric consequence of $g(\varepsilon)\propto\varepsilon^{1/2}$ , which is why the identical relation governs the classical ideal gas, a metal's degenerate electrons, and a Bose gas below its condensation point.

Step 1. Start from $PV=\mp k_BT\,AV\int_0^\infty\varepsilon^{1/2}\ln(1\mp ze^{-\beta\varepsilon})\,d\varepsilon$ . Integrate by parts with $u=\ln(1\mp ze^{-\beta\varepsilon})$ and $dv=\varepsilon^{1/2}d\varepsilon$ , so $v=\tfrac23\varepsilon^{3/2}$ . Why: the antiderivative of $\varepsilon^{1/2}$ is $\tfrac23\varepsilon^{3/2}$ — the pure number $2/3$ enters here and nowhere else.

Step 2. The boundary term vanishes at both ends ( $\varepsilon^{3/2}\to0$ at $0$ , log $\to0$ at $\infty$ ), leaving $PV=\tfrac23\,AV\int_0^\infty\varepsilon^{3/2}\,\frac{1}{z^{-1}e^{\beta\varepsilon}\mp1}\,d\varepsilon$ . Why: differentiating the log reproduces the Bose/Fermi occupation $\langle n(\varepsilon)\rangle=1/(z^{-1}e^{\beta\varepsilon}\mp1)$ , converting the pressure integral into an energy-weighted occupation integral.

Step 3. Recognize the internal energy. Since $U=\int_0^\infty\varepsilon\,g(\varepsilon)\langle n(\varepsilon)\rangle\,d\varepsilon=AV\int_0^\infty\varepsilon^{3/2}\langle n\rangle\,d\varepsilon$ , the integral in Step 2 is exactly $U$ . Hence $PV=\tfrac23U$ . Why: the extra factor of $\varepsilon$ that integration by parts produced is precisely the energy weight; the pressure integral and the energy integral are the same object up to the factor $2/3$ .

Step 4. Check the classical and relativistic corners. In the classical limit $\langle n\rangle\to ze^{-\beta\varepsilon}$ and $U=\tfrac32Nk_BT$ , so $PV=\tfrac23\cdot\tfrac32Nk_BT=Nk_BT$ — the ideal gas law. Replacing $\varepsilon^{1/2}\to\varepsilon^2$ (linear dispersion) makes $v=\tfrac13\varepsilon^3$ and gives $PV=\tfrac13U$ . Why: the same one-line integration by parts with a different power of $\varepsilon$ produces $\tfrac13$ instead of $\tfrac23$ — the exponent is set entirely by the density of states, not the statistics. *Verified: the integration-by-parts factor $\int\varepsilon^{1/2}d\varepsilon=\tfrac23\varepsilon^{3/2}$ (and $\int\varepsilon^2d\varepsilon=\tfrac13\varepsilon^3$ for the relativistic contrast) is checked symbolically in verify_equation-of-state-quantum-gases.py; matches Kardar SPP §7.1 and Pathria §7.1.*

Go deeper · Derivation C · 1 minThe leading quantum virial correction and the sign that splits the species

Derivation CThe leading quantum virial correction and the sign that splits the species

The goal is to extract the first quantum correction to the ideal-gas law, $PV/Nk_BT=1\pm n\lambda^3/2^{5/2}$ , and to establish that the sign — $+$ for fermions, $-$ for bosons — is exact and statistics-driven. The strategy is a controlled small- $z$ inversion: keep the first two terms of $n\lambda^3=z\pm z^2/2^{3/2}$ , invert to get $z(n\lambda^3)$ , and substitute into $P\lambda^3/k_BT=z\pm z^2/2^{5/2}$ . The subtlety to watch is that the density series and the pressure series have the same leading sign but different $2^{-3/2}$ versus $2^{-5/2}$ second coefficients, and it is the interplay of the two that fixes the final sign of the correction to the pressure.

Step 1. Write the two series to second order (upper sign fermions, lower bosons): $n\lambda^3=z\mp\frac{z^2}{2^{3/2}}+\cdots$ and $\frac{P\lambda^3}{k_BT}=z\mp\frac{z^2}{2^{5/2}}+\cdots$ . Why: these are the first two terms of $f_{3/2},f_{5/2}$ (fermions, alternating $-$ ) and $g_{3/2},g_{5/2}$ (bosons, all $+$ ); I write both with $\mp$ so the fermion case is the upper sign throughout.

Step 2. Invert the density series. Write $z=n\lambda^3+a(n\lambda^3)^2+\cdots$ and demand consistency: $a=\pm2^{-3/2}$ , so $z=n\lambda^3\pm\frac{(n\lambda^3)^2}{2^{3/2}}$ . Why: substituting the ansatz into Step 1 and matching the $(n\lambda^3)^2$ coefficient fixes $a$ ; the fermion sign flips to $+$ on inversion because moving the term to the other side of the equation reverses it.

Step 3. Substitute into the pressure. $\frac{P\lambda^3}{k_BT}=z\mp\frac{z^2}{2^{5/2}}=n\lambda^3\pm\frac{(n\lambda^3)^2}{2^{3/2}}\mp\frac{(n\lambda^3)^2}{2^{5/2}}=n\lambda^3\pm(n\lambda^3)^2\big(\frac{1}{2^{3/2}}-\frac{1}{2^{5/2}}\big)$ . The bracket is $\frac{1}{2^{3/2}}-\frac{1}{2^{5/2}}=\frac{2-1}{2^{5/2}}=\frac{1}{2^{5/2}}$ . Why: the inversion correction ( $2^{-3/2}$ ) and the direct pressure correction ( $2^{-5/2}$ ) combine; the arithmetic $2/2^{5/2}-1/2^{5/2}=1/2^{5/2}$ leaves exactly $2^{-5/2}$ .

Step 4. Divide by $n\lambda^3$ to get the pressure per particle: $\frac{PV}{Nk_BT}=\frac{P\lambda^3/k_BT}{n\lambda^3}=1\pm\frac{n\lambda^3}{2^{5/2}}+\cdots$ . Fermions ( $+$ ) push harder — a statistical repulsion from Pauli exclusion; bosons ( $-$ ) push softer — a statistical attraction from constructive symmetrization. Why: the second virial coefficient is $B_2=\pm\lambda^3/2^{5/2}$ , positive for fermions and negative for bosons, and it enters the pressure through $PV/Nk_BT=1+B_2 n$ . *Verified: the full symbolic inversion, the exact coefficient $2^{-5/2}=0.17678$ , the signs ( $+$ fermion, $-$ boson), and a numerical spot-check at $z=0.2$ (Bose $0.9617$ , Fermi $1.0330$ ) are in verify_equation-of-state-quantum-gases.py; matches Pathria §7.1, §8.1 and Huang §8.*

Limiting cases

The equation of state interpolates smoothly between four regimes, each governed by the degeneracy parameter $n\lambda^3$ or, equivalently, the fugacity $z$ .

Case	Result	Meaning
Classical, $n\lambda^3\ll1$ ( $z\to0$ )	$PV=Nk_BT$	Blurs don't overlap; statistics irrelevant, Maxwell–Boltzmann law recovered.
Leading quantum correction	$PV/Nk_BT=1\pm n\lambda^3/2^{5/2}$	First exchange effect: $+$ fermion repulsion, $-$ boson attraction.
Degenerate Fermi, $T\to0$ ( $z\to\infty$ )	$P\to\tfrac25 n\varepsilon_F\propto n^{5/3}$	$f_{5/2}(z)\to\tfrac{8}{15\sqrt\pi}(\ln z)^{5/2}$ ; the $T=0$ degeneracy pressure.
Bose ceiling, $z\to1$	$g_{3/2}(1)=\zeta(3/2)=2.612$	Density saturates at fixed $T$ ; surplus condenses (Bose–Einstein condensation).

Common trapThe PV = ⅔U relation and the ideal-gas law PV = NkT are the same statement

They are not. $PV=\tfrac23U$ is exact for every non-relativistic ideal quantum gas at all temperatures — including a fully degenerate Fermi gas, where $PV=\tfrac23U$ but $U=\tfrac35N\varepsilon_F$ has nothing to do with $Nk_BT$ . The classical law $PV=Nk_BT$ is a special case that emerges only when $U=\tfrac32Nk_BT$ , i.e. when equipartition holds. Assuming $PV=Nk_BT$ inside a white dwarf, where the electrons are degenerate, underestimates the pressure by the enormous factor $\varepsilon_F/k_BT$ . The robust relation is $PV=\tfrac23U$ ; the fragile one is $PV=Nk_BT$ .

Worked examples

1 · The exchange correction to helium gas at room temperature (confidence builder)

We compute how far a real gas of helium-4 at room temperature and atmospheric pressure sits from the ideal-gas law due to quantum statistics alone (ignoring the real van der Waals interaction).

Step 1. The thermal wavelength of $^4$ He at $T=300$ K, with $m=6.6465\times10^{-27}$ kg: $\lambda=h/\sqrt{2\pi mk_BT}=6.626\times10^{-34}/\sqrt{2\pi(6.6465\times10^{-27})(1.381\times10^{-23})(300)}$ . The denominator is $\sqrt{1.731\times10^{-46}}=1.316\times10^{-23}$ , giving $\lambda=\mathbf{5.03\times10^{-11}\ m}$ (about half an ångström).

Step 2. At atmospheric pressure the number density is $n=P/k_BT=1.013\times10^{5}/(1.381\times10^{-23}\times300)=2.45\times10^{25}\ \mathrm{m^{-3}}$ .

Step 3. The degeneracy parameter is $n\lambda^3=2.45\times10^{25}\times(5.03\times10^{-11})^3=2.45\times10^{25}\times1.27\times10^{-31}=\mathbf{3.1\times10^{-6}}$ .

Step 4. Helium-4 is a boson, so the correction is $PV/Nk_BT=1-n\lambda^3/2^{5/2}=1-3.1\times10^{-6}/5.66=1-\mathbf{5.5\times10^{-7}}$ . The statistical attraction lowers the pressure by less than one part in a million — utterly negligible at room temperature, which is exactly why helium behaves classically until you cool it to a few kelvin. The quantum correction is $-5.5\times10^{-7}$ , confirming the gas is deep in the classical regime.

2 · Where does the Fermi correction reach 1%? (exam level)

We find the density at which an electron gas at $T=300$ K departs from the ideal-gas law by 1% due to statistical repulsion.

Step 1. Set the fermion correction to $0.01$ : $n\lambda^3/2^{5/2}=0.01$ , so $n\lambda^3=0.01\times5.657=0.0566$ .

Step 2. For electrons at $300$ K, $m_e=9.109\times10^{-31}$ kg, $\lambda=h/\sqrt{2\pi m_ek_BT}=6.626\times10^{-34}/\sqrt{2\pi(9.109\times10^{-31})(1.381\times10^{-23})(300)}$ . The denominator is $\sqrt{2.372\times10^{-50}}=1.540\times10^{-25}$ , so $\lambda=4.30\times10^{-9}$ m ( $43$ Å — electrons are light, so their thermal wavelength is large).

Step 3. The density is $n=0.0566/\lambda^3=0.0566/(4.30\times10^{-9})^3=0.0566/7.96\times10^{-26}=\mathbf{7.1\times10^{23}\ m^{-3}}$ .

Step 4. Compare with a metal, where $n\sim10^{29}\ \mathrm{m^{-3}}$ — about $10^{5}$ times denser. A metal is $\sim10^5$ times past the 1% mark, so the virial expansion is hopeless there: metallic electrons are fully degenerate ( $n\lambda^3\sim10^4$ ), and one must use the $z\to\infty$ Sommerfeld limit, not this small- $z$ series. The virial expansion is the tool for dilute gases, not for degenerate matter.

Origins · 1 minquantum pressure and how this was discovered

quantum pressure

concept: R. H. Fowler 1926, Pathria & Beale ch. 7–8 · Quantum pressure is the pressure an ideal gas carries purely because its particles are indistinguishable quantum objects, above and beyond the classical

Nk_BT/V

. The polylogarithmic equation of state

PV/Nk_BT=g_{5/2}(z)/g_{3/2}(z)

(bosons) or

f_{5/2}(z)/f_{3/2}(z)

(fermions) is exactly the machine that tracks it: at low

n\lambda^3

the ratio

\to1

and the quantum pressure vanishes, while as

z\to\infty

the fermion branch stiffens into the

T=0

pressure

P\propto n^{5/3}

that holds up white dwarfs and metals. R. H. Fowler recognized in 1926 that dense stellar matter is a degenerate Fermi gas, converting Fermi–Dirac statistics — published only months earlier — into a mechanism against gravitational collapse; this equation of state is the finite-temperature engine whose zero-temperature limit is that degeneracy pressure.

History

1924

— Bose derives the Planck law from photon statistics; Einstein extends it to a massive ideal gas.

1926

— Fermi and, independently, Dirac formulate Fermi–Dirac statistics for the electron gas.

1926

— Fowler applies degenerate Fermi statistics to white dwarfs, launching the quantum equation of state.

1927

— Pauli and Sommerfeld build the free-electron theory of metals on the same degenerate gas.

1938

— London interprets the ideal Bose gas condensation as the mechanism behind superfluid helium.

1995

— Cornell, Wieman and Ketterle realize Bose–Einstein condensation in dilute atomic gases.

Going further

Standard treatments: Pathria & Beale, Statistical Mechanics, 3rd ed., §7.1 (ideal Bose systems) and §8.1 (ideal Fermi systems); Kardar, Statistical Physics of Particles, §7.1–7.3; Huang, Statistical Mechanics, §8; Landau & Lifshitz, Statistical Physics Part 1, §56. The polylogarithm identities are collected in Robinson's classic tables. Provenance: adapted and upgraded from the AllQFT base article "Equation of State of Ideal Quantum Gases" (ref STAT.IQGAS.EOSQG.1239; Pathria & Beale, 3rd ed., ch. 7–8), with every distribution formula, the $\pm2^{-5/2}$ exchange sign, and the $PV=\tfrac23U$ relation independently python-verified (verify_equation-of-state-quantum-gases.py, 14 checks). Prerequisites: Quantum Statistics of Ideal Gases: Occupation Numbers · Density of States for Free Particles · Grand Canonical Ensemble. Leads to: Degenerate Fermi Gas at Zero Temperature · Degeneracy Pressure of Fermions · Bose-Einstein Condensation in the Ideal Gas · Sommerfeld Expansion.

Why does the same PV = ⅔U hold for a classical gas and a fully degenerate Fermi gas?

Because the factor $2/3$ comes from the density of states $g(\varepsilon)\propto\varepsilon^{1/2}$ , not from the occupation statistics. Integrating the pressure by parts always turns $\varepsilon^{1/2}$ into $\tfrac23\varepsilon^{3/2}$ , and the leftover integral is the energy — regardless of whether the occupation is a Maxwell–Boltzmann exponential, a Fermi step, or a Bose distribution. The statistics change what $U$ is (thermal $\tfrac32Nk_BT$ classically, $\tfrac35N\varepsilon_F$ for a degenerate Fermi gas), but the ratio $PV/U$ stays pinned at $2/3$ for any 3D non-relativistic gas.

What exactly is the fugacity z, and why not just use the chemical potential μ?

The fugacity is $z=e^{\mu/k_BT}$ , a repackaging of $\mu$ that makes the grand partition function a clean power series. Each single-mode factor is $1/(1-ze^{-\beta\varepsilon})$ (bosons) or $1+ze^{-\beta\varepsilon}$ (fermions), and expanding in powers of $z$ gives the polylogarithms $g_\nu,f_\nu$ directly — the $k$ -th power of $z$ counts $k$ -particle exchange effects. Writing the same expansion in $\mu$ is a mess of nested exponentials. Physically, in the classical limit $z\approx n\lambda^3$ , so $z$ is literally the "how quantum is this gas" number: small $z$ is dilute and classical, $z$ of order $1$ is where statistics takes over.

The Bose pressure is lower than classical, the Fermi pressure higher — but there are no forces. What is pushing?

Nothing is pushing; it is pure statistics. The requirement that the many-body wavefunction be symmetric (bosons) raises the amplitude for two particles to sit close together, so they spend more time near each other than distinguishable particles would — an effective attraction that lowers the pressure. Antisymmetry (fermions) creates an exclusion hole around each particle, keeping them apart — an effective repulsion that raises the pressure. Both effects come from the free-particle wavefunctions alone, with zero interaction potential. The pair correlation $g^{(2)}(0)$ is $2$ for bosons (bunching) and $0$ for fermions (the exclusion hole), and the exchange virial $B_2=\pm\lambda^3/2^{5/2}$ is the thermodynamic face of that difference.

Why can the fugacity of a Bose gas never exceed 1, while a Fermi gas has no such limit?

For bosons the density is $n\lambda^3=g_{3/2}(z)=\sum z^k/k^{3/2}$ , and this series converges only for $z\le1$ ; at $z=1$ it reaches its maximum finite value $\zeta(3/2)=2.612$ . If you try to add more particles at fixed temperature, $z$ cannot rise past $1$ — the excited states saturate, and the surplus particles are forced into the single ground state, which is Bose–Einstein condensation. For fermions the series $f_{3/2}(z)=\sum(-1)^{k-1}z^k/k^{3/2}$ is alternating and converges for all $z$ , growing without bound as $z\to\infty$ (the degenerate limit $z\sim e^{\varepsilon_F/k_BT}$ ). No saturation, no condensation — the exclusion principle already spreads the fermions across many states.

The virial expansion works beautifully for a dilute gas but fails for a metal — why?

The virial expansion (4) is a power series in $n\lambda^3$ , so it converges only when $n\lambda^3\ll1$ . For a room-temperature gas that is easily satisfied ( $n\lambda^3\sim10^{-6}$ ). But metallic conduction electrons are light (large $\lambda$ ) and dense, giving $n\lambda^3\sim10^4$ — the series diverges catastrophically. There the correct tool is the opposite limit, $z\to\infty$ , handled by the Sommerfeld expansion, which is an expansion in $k_BT/\varepsilon_F$ rather than in $n\lambda^3$ . Virial for dilute gases, Sommerfeld for degenerate matter: they are the two ends of the same equation of state, and neither works in the other's regime.

Does PV = ⅔U mean the pressure of a Bose gas below Tc doesn't depend on volume?

Yes, and it is a genuine signature of the condensation. Below $T_c$ the fugacity is stuck at $z=1$ , so the pressure $P=k_BT\,g_{5/2}(1)/\lambda^3$ depends only on temperature, not on volume or total density. Compressing the gas at fixed $T$ pushes more particles into the condensate without changing the pressure — the isothermal compressibility diverges, exactly as at a first-order phase coexistence. The relation $PV=\tfrac23U$ still holds, but now both $P$ and $U/V$ are functions of $T$ alone. This volume-independence of the pressure is one of the sharpest thermodynamic fingerprints of the Bose–Einstein transition.