Chemical Potential of a Fermi Gas vs Temperature

Ask a metallurgist for the "energy of the electrons" in a copper wire and you will get a number near 7 eV — the Fermi energy. Ask for the chemical potential at room temperature and you will get, to five decimal places, the same 7 eV. The two are not the same quantity, yet in a metal they agree so closely that physicists use the words interchangeably. Understanding why is the whole story of this article, and the answer turns out to be a statement about the shape of the density of states. The chemical potential $\mu$ is the energy cost of adding one more fermion at fixed temperature and volume, and it is the parameter that sits inside the Fermi–Dirac distribution. For a gas held at fixed density $n=N/V$ , $\mu$ is not something you dial in — it is fixed implicitly by the demand that the occupied states always add up to exactly $N$ particles. Tracking how $\mu$ must slide as the temperature changes is the master step from which every thermodynamic function of the Fermi gas — its heat capacity, its pressure, its response to a field — then follows. At absolute zero the distribution is a cliff: every state below the Fermi energy is full, every state above is empty, and $\mu(0)=\varepsilon_F$ . Warming the gas smears that cliff into a slope of width $k_BT$ . Because the density of states rises with energy in three dimensions, the smearing is lopsided — more states open up above $\mu$ than are vacated below — so to keep the particle count fixed, $\mu$ must edge downward. Push the temperature past the Fermi temperature and $\mu$ crosses zero, going large and negative, smoothly rejoining the classical Maxwell–Boltzmann gas. This behaviour is the mirror image of the ideal Bose gas, where $\mu$ climbs to a ceiling and condensation begins.

Key takeaways

The chemical potential $\mu(T)$ of an ideal Fermi gas is fixed by the number constraint $N=\int_0^\infty g(\varepsilon)f(\varepsilon)\,d\varepsilon$ at constant density — it is an output, not an input.
At $T=0$ it equals the Fermi energy, $\mu(0)=\varepsilon_F$ ; the sharp Fermi surface splits filled from empty states.
For $T\ll T_F$ the Sommerfeld expansion gives $\mu(T)\simeq\varepsilon_F[1-\frac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$ — a quadratic decrease.
The decrease happens because $g(\varepsilon)\propto\sqrt\varepsilon$ rises with energy, so symmetric thermal broadening would over-count particles; $\mu$ falls to compensate.
Near $T\approx 0.99\,T_F$ the chemical potential passes through zero; above it $\mu\to k_BT\ln(n\lambda_T^3/g_s)<0$ , the classical result.
In metals $T_F\sim 10^4$ – $10^5$ K, so at room temperature $\mu$ sits within $\sim 10^{-5}$ of $\varepsilon_F$ — which is why the two terms are used interchangeably.
Key idea: hold the density fixed and ask where the Fermi–Dirac step must sit. Because $g(\varepsilon)$ grows with energy, symmetric broadening would add particles, so $\mu$ must fall below $\varepsilon_F$ — quadratically at first, then crossing zero into the classical regime.

The core insight

Hold the density fixed and ask where the Fermi–Dirac step must sit. Because the density of states grows with energy, symmetric thermal broadening would add particles, so the chemical potential must fall below the Fermi energy — quadratically at first, then crossing zero into the classical regime at high temperature.

Where this lives today

Tracking $\mu(T)$ is the daily bread of semiconductor engineering, where the chemical potential is called the Fermi level. Its position relative to the band edges — and precisely how it drifts with temperature toward midgap — sets the carrier concentrations that make a transistor switch or a diode rectify; every device simulator solves a version of the number constraint (2) at each temperature and bias. In compact stars, the enormous Fermi temperature ( $T\ll T_F$ by orders of magnitude) means $\mu\approx\varepsilon_F$ is effectively frozen, and that near-constancy is what makes the degeneracy pressure supporting a white dwarf temperature-independent. The cleanest modern laboratory for the full curve, though, is ultracold atoms. Trapped $^6$ Li and $^{40}$ K gases are cooled well below their Fermi temperatures, and using the local-density approximation, experimentalists read $\mu$ directly off the shape of the atomic cloud. Watching the quadratic departure of $\mu$ from $\varepsilon_F$ , and then its plunge through zero, turns the hero figure of this article into measured data — a direct equation-of-state and thermometry probe of Fermi degeneracy that no solid-state system can offer, because no metal can be warmed to a meaningful fraction of its Fermi temperature.

The physical picture

Picture the occupied states as water poured into a tank whose cross-section widens with height — that widening is the density of states $g(\varepsilon)\propto\sqrt\varepsilon$ . At $T=0$ the water is flat-topped at the Fermi energy $\varepsilon_F$ : a sharp, mirror-still surface. Warming the gas softens that surface into a transition layer of thickness $k_BT$ . Here is the crucial asymmetry: because the tank is wider above the surface than below it, softening the level symmetrically would spill more water in than out. To keep the total amount of water — the particle number — fixed, the average level has to sink a little. That small sinking is exactly $\mu(T)<\varepsilon_F$ . One landmark never moves. The energy $\varepsilon=\mu$ is always exactly half-occupied, $f(\mu)=\tfrac12$ , at every temperature — this is a symmetry of the Fermi function, not an accident. So $\mu$ marks the centre of the thermal smear, not the top of the filled region. Watching where the half-filling energy sits is the same thing as watching $\mu(T)$ . The first figure is the whole article in one curve: the chemical potential, in units of $\varepsilon_F$ , plotted against temperature in units of $T_F$ . On the far left it clings to $1$ , dipping as a gentle downward parabola — the Sommerfeld regime of every metal. Then it accelerates, plunges through zero near $T\approx 0.99\,T_F$ , and heads off into the negative — the classical branch, where the gas has forgotten it was ever degenerate.

Figure

The chemical potential slides down and crosses zero — degenerate parabola on the left, classical branch on the right

In the opposite corner — the dilute, hot limit — the fermions hardly ever land in the same state, exclusion becomes irrelevant, and the gas behaves classically. There the occupation must be tiny, $f\ll 1$ , which forces $e^{-\mu/k_BT}\gg 1$ : $\mu$ has to be large and negative. A negative chemical potential is the hallmark of a non-degenerate gas, and it is where every classical ideal-gas formula you already know lives.

Figure

Warming the Fermi–Dirac step: the cliff at ε_F softens over a width k_B T about μ, and μ sits just below ε_F

The core equations

The Fermi–Dirac distribution is where $\mu$ does its work. It gives the mean occupation of a single-particle state of energy $\varepsilon$ , and its shape — a step softened over a width $k_BT$ — is set entirely by the two energies $\varepsilon-\mu$ and $k_BT$ .

f(\varepsilon)=\frac{1}{e^{(\varepsilon-\mu)/k_BT}+1} \qquad (1)

where: $f(\varepsilon)$ — mean occupation ( $0\le f\le 1$ , the exclusion ceiling); $\mu$ — chemical potential; $k_B$ — Boltzmann constant; $T$ — temperature. Read it at three points: far below $\mu$ , $f\to 1$ (states full); at $\varepsilon=\mu$ , $f=\tfrac12$ exactly; far above, $f\to 0$ . The chemical potential is fixed by insisting the occupied states sum to the actual particle number. This number constraint is the equation you invert, at each temperature, to find $\mu$ .

N=\int_0^\infty g(\varepsilon)\,f(\varepsilon)\,d\varepsilon \qquad (2)

where: $N$ — total particle number (held fixed); $g(\varepsilon)$ — density of single-particle states. The physics of " $\mu$ slides to conserve $N$ " lives entirely in this integral: fix the left side, and the right side dictates where $\mu$ must go. Everything hinges on the shape of $g(\varepsilon)$ . For free particles in three dimensions it rises as the square root of energy — that rise is the reason $\mu$ falls. (This density of states is derived and owned by Density of States for Free Particles; we take it as given here.)

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

where: $g_s$ — spin degeneracy ( $2$ for electrons); $V$ — volume; $m$ — particle mass; $C$ — the shorthand prefactor. The single fact we use downstream is $g\propto\sqrt\varepsilon$ : more room at higher energy. At $T=0$ the constraint (2) is trivial to invert — the step fills states up to $\varepsilon_F$ — giving the Fermi energy and the Fermi temperature that set the scale for everything. (These are owned by Fermi Energy, Momentum and Fermi Temperature; we recall them by name.)

\varepsilon_F=\frac{\hbar^2}{2m}\left(\frac{6\pi^2 n}{g_s}\right)^{2/3},\qquad k_BT_F=\varepsilon_F \qquad (4)

where: $n=N/V$ — number density; $\varepsilon_F$ — Fermi energy, equal to $\mu(0)$ ; $T_F$ — Fermi temperature, the single scale that decides whether a gas is "cold" ( $T\ll T_F$ , degenerate) or "hot" ( $T\gg T_F$ , classical). For $g_s=2$ the bracket $6\pi^2 n/g_s$ becomes $3\pi^2 n$ . Now the payoff. Feeding the Sommerfeld expansion of Sommerfeld Expansion into the constraint (2) and solving order by order gives the low-temperature behaviour of $\mu$ — a quadratic decline whose coefficient is a pure number.

\mu(T)=\varepsilon_F\left[1-\frac{\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2-\frac{\pi^4}{80}\left(\frac{k_BT}{\varepsilon_F}\right)^4+\cdots\right] \qquad (5)

where: $k_BT/\varepsilon_F$ — the degeneracy parameter, tiny in metals. The leading correction is negative: $\mu$ drops below $\varepsilon_F$ . The coefficient $\pi^2/12$ is not fitted — it falls out of the algebra, as the derivation below shows. Push to high temperature and the constraint (2) gives the opposite limit — the classical Maxwell–Boltzmann chemical potential, which is large and negative.

\mu(T)\to k_BT\ln\!\left(\frac{n\lambda_T^3}{g_s}\right),\qquad \lambda_T=\sqrt{\frac{2\pi\hbar^2}{mk_BT}} \qquad (6)

where: $\lambda_T$ — the thermal de Broglie wavelength (owned by Thermal de Broglie Wavelength); $n\lambda_T^3$ — the phase-space density, $\ll 1$ in the classical regime, so the logarithm is negative and $\mu<0$ . Equations (5) and (6) are the two ends of the single curve in the hero figure.

Derivation AQuadratic decrease of the chemical potential via the Sommerfeld expansion

We want to prove the boxed low-temperature law $\mu(T)=\varepsilon_F[1-\frac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$ and, in particular, pin down the pure number $\pi^2/12$ . The strategy is to hold the density fixed — that is the only physical input — and demand that the number integral (2) return the same $N$ at temperature $T$ that it returned at $T=0$ . The Sommerfeld expansion, which we borrow from Sommerfeld Expansion rather than re-derive, turns that integral into a controlled series in $(k_BT/\mu)^2$ ; matching the two sides order by order and inverting gives $\mu(T)$ . Watch the sign of the correction — it is the whole point, and it comes from the fact that the density of states rises with energy.

Step 1. Write down the Sommerfeld formula. For any smooth weight $H(\varepsilon)$ , the expansion states $\int_0^\infty H(\varepsilon)f(\varepsilon)\,d\varepsilon = \int_0^\mu H\,d\varepsilon + \frac{\pi^2}{6}(k_BT)^2\,H'(\mu)+\cdots$ , valid when $k_BT\ll\mu$ . Why: only the thin shell of width $k_BT$ around $\mu$ deviates from the $T=0$ step, and the leading correction is proportional to the slope $H'(\mu)$ there.

Step 2. Apply it with the weight $H=g\propto\sqrt\varepsilon$ . The zeroth term is $\int_0^\mu C\sqrt\varepsilon\,d\varepsilon=\frac{2}{3}C\mu^{3/2}$ , and the correction uses $g'(\mu)=\frac{C}{2}\mu^{-1/2}$ , giving $N=\frac{2}{3}C\mu^{3/2}+\frac{\pi^2}{6}(k_BT)^2\frac{C}{2}\mu^{-1/2}$ . Why: the number is just the density of states weighted by occupation, so $g$ is the weight and its slope at $\mu$ carries the temperature correction.

Step 3. Factor out the leading term. Pulling $\frac{2}{3}C\mu^{3/2}$ to the front, $N=\frac{2}{3}C\mu^{3/2}\big[1+\frac{\pi^2}{8}(k_BT/\mu)^2\big]$ . Why: the ratio of the two terms is $\frac{\pi^2}{6}\cdot\frac{1}{2}/\frac{2}{3}=\frac{\pi^2}{8}$ — this inner coefficient is checked symbolically in the verify script.

Step 4. Impose number conservation. At $T=0$ , $N=\frac{2}{3}C\varepsilon_F^{3/2}$ . Equating, $\varepsilon_F^{3/2}=\mu^{3/2}[1+\frac{\pi^2}{8}(k_BT/\mu)^2]$ , so $\mu=\varepsilon_F[1+\frac{\pi^2}{8}(k_BT/\mu)^2]^{-2/3}$ . Why: the density did not change, so the number integral cannot either — that single constraint fixes $\mu(T)$ .

Step 5. Invert to leading order. Replace $\mu\to\varepsilon_F$ inside the small correction (consistent to this order) and expand with $(1+x)^{-2/3}\approx 1-\frac{2}{3}x$ where $x=\frac{\pi^2}{8}(k_BT/\varepsilon_F)^2$ . The coefficient is $-\frac{2}{3}\cdot\frac{\pi^2}{8}=-\frac{\pi^2}{12}$ , and the boxed result stands. Why: because $g$ rises with energy, the correction subtracts — a density of states that fell with energy would flip the sign and $\mu$ would rise. *Verified: the $-\pi^2/12$ coefficient is reproduced symbolically from the binomial inversion in verify_chemical-potential-fermi-gas-temperature.py, matching Pathria & Beale Eq. (8.1.34) and Ashcroft & Mermin Eq. (2.77).*

Go deeper · Derivation B · 1 minThe classical limit: why the chemical potential goes large and negative

Derivation BThe classical limit: why the chemical potential goes large and negative

Here we prove the opposite end of the curve — that at high temperature $\mu\to k_BT\ln(n\lambda_T^3/g_s)$ , the familiar Maxwell–Boltzmann chemical potential, which is large and negative for a dilute gas. The physical idea is that when the phase-space density $n\lambda_T^3$ is small, every single-particle state is nearly empty, the " $+1$ " in the Fermi–Dirac denominator becomes irrelevant, and Fermi statistics collapses onto Boltzmann statistics. The mathematical work is one Gaussian-type integral; the interesting part is checking self-consistency — that the very approximation we make forces $\mu$ to be negative, which is exactly what "dilute" requires.

Step 1. Take the small-occupation approximation. When $f\ll 1$ everywhere, $e^{(\varepsilon-\mu)/k_BT}\gg 1$ and $f(\varepsilon)\approx e^{-(\varepsilon-\mu)/k_BT}=z\,e^{-\varepsilon/k_BT}$ with fugacity $z=e^{\mu/k_BT}$ . Why: dropping the $+1$ is legitimate only if occupations are small, which we must confirm at the end.

Step 2. Do the number integral. Substituting into (2), $N=z\,C\int_0^\infty\sqrt\varepsilon\,e^{-\varepsilon/k_BT}\,d\varepsilon$ . The integral is standard: $\int_0^\infty\sqrt\varepsilon\,e^{-\varepsilon/k_BT}\,d\varepsilon=\Gamma(\tfrac32)(k_BT)^{3/2}=\frac{\sqrt\pi}{2}(k_BT)^{3/2}$ . Why: it is a gamma function, $\Gamma(3/2)=\sqrt\pi/2$ , verified symbolically in the script.

Step 3. Collect constants into $\lambda_T$ . Assembling the prefactors gives $n=g_s\,z/\lambda_T^3$ with $\lambda_T=\sqrt{2\pi\hbar^2/mk_BT}$ . Why: the messy combination of $C$ , $\Gamma(3/2)$ , and $(k_BT)^{3/2}$ is exactly what packages into the thermal wavelength — the same $\lambda_T$ that governs quantum concentration.

Step 4. Solve for $\mu$ and check self-consistency. Then $z=n\lambda_T^3/g_s$ , so $\mu=k_BT\ln(n\lambda_T^3/g_s)$ . Because the classical regime means $n\lambda_T^3\ll 1$ , the logarithm is negative and $\mu<0$ — confirming that occupations really were small. Why: the approximation is self-validating; a positive $\mu$ would have signalled degeneracy and invalidated dropping the $+1$ . *Verified: $\Gamma(3/2)=\sqrt\pi/2$ and the Gaussian integral $\int\sqrt\varepsilon\,e^{-\varepsilon/k_BT}d\varepsilon=(\sqrt\pi/2)(k_BT)^{3/2}$ are checked symbolically, and $\mu<0$ for $n\lambda_T^3\ll1$ numerically, in the verify script; matches Reif §9.10 and Pathria Eq. (8.1.4).*

Limiting cases

The chemical potential interpolates between two clean regimes set by the single dimensionless ratio $k_BT/\varepsilon_F$ .

Case	Result	Meaning
Zero temperature (fully degenerate)	$\mu(0)=\varepsilon_F$	Sharp Fermi surface; $f$ is a step at $\varepsilon_F$ . The reference value for all corrections.
Low temperature $T\ll T_F$	$\mu(T)\simeq\varepsilon_F[1-\frac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$	Quadratic decrease from the Sommerfeld expansion; the dominant behaviour in every metal.
Crossover $T\approx 0.99\,T_F$	$\mu\approx 0$	The chemical potential passes through zero just below the Fermi temperature; degeneracy is being lost.
High temperature $T\gg T_F$ (classical)	$\mu(T)\simeq k_BT\ln(n\lambda_T^3/g_s)<0$	Maxwell–Boltzmann limit; the exclusion-driven behaviour smoothly merges into the ideal-gas result with $\mu$ large and negative.

Common trapThe chemical potential equals the Fermi energy at all temperatures

It does not — $\mu=\varepsilon_F$ holds only at $T=0$ . The Fermi energy is a fixed property of the density (it never changes as you warm the gas), while $\mu(T)$ moves, and above roughly the Fermi temperature it even goes negative. The reason the two are conflated for metals is purely quantitative: room temperature is $\sim 10^{-2}\,T_F$ there, so the correction is one part in $10^5$ . Two further traps: the half-filled energy is $\mu$ , not $\varepsilon_F$ ( $f(\mu)=\tfrac12$ always); and a negative chemical potential is not unphysical — it is the signature of a classical gas, appearing in every Sackur–Tetrode calculation you have done.

Worked examples

1 · The chemical potential of copper at room temperature (confidence builder)

We compute how far $\mu$ has slid from $\varepsilon_F$ in a real metal — and see why nobody bothers to distinguish them.

Step 1. Copper's conduction electrons have $\varepsilon_F=7.0$ eV. The Fermi temperature is $T_F=\varepsilon_F/k_B=(7.0\ \text{eV})/(8.617\times10^{-5}\ \text{eV/K})=\mathbf{8.12\times10^{4}\ K}$ — over eighty thousand kelvin.

Step 2. At $T=300$ K the degeneracy parameter is $k_BT/\varepsilon_F=(8.617\times10^{-5}\times300)/7.0=0.02585/7.0=\mathbf{3.69\times10^{-3}}$ . Room temperature is a tiny fraction of $T_F$ .

Step 3. The Sommerfeld shift is $\varepsilon_F-\mu=\varepsilon_F\frac{\pi^2}{12}(k_BT/\varepsilon_F)^2=7.0\times0.8225\times(3.69\times10^{-3})^2=\mathbf{7.85\times10^{-5}\ eV}$ .

Step 4. As a fraction, $(\varepsilon_F-\mu)/\varepsilon_F=1.12\times10^{-5}$ — about one part in $10^5$ . So $\mu=6.9999\ \text{eV}$ : identical to $\varepsilon_F$ to five figures. This is why "Fermi energy" and "chemical potential" are used interchangeably for metals. The distinction only matters when you compute the tiny derivative $\partial\mu/\partial T$ , which drives the electronic heat capacity.

2 · Where does the chemical potential cross zero for a cold atomic gas? (exam level)

Ultracold $^{40}$ K fermions are cooled below $T_F$ ; we locate the temperature at which $\mu$ changes sign, and contrast the answer with the naive guess $T=T_F$ .

Step 1. The naive Sommerfeld formula (5), extrapolated, would predict $\mu=0$ where $1-\frac{\pi^2}{12}(T/T_F)^2=0$ , i.e. $T/T_F=\sqrt{12/\pi^2}=\mathbf{1.10}$ . But equation (5) is a low-temperature expansion and is not trustworthy near $T_F$ — this is only an order-of-magnitude guide.

Step 2. Solving the exact number constraint (2) numerically — setting $\mu=0$ and finding the temperature at which the full Fermi integral returns the fixed density — gives $T=\mathbf{0.989\,T_F}$ . The crossing sits just below the Fermi temperature, not above it.

Step 3. Interpret the number. At $T=0.989\,T_F$ the gas is right at the boundary between degenerate and classical: $\mu$ is passing through zero on its way to the large-negative classical branch. For a $^{40}$ K gas with $T_F\approx 1\ \mu$ K, this crossover happens near $\mathbf{0.99\ \mu K}$ — squarely in the range these experiments reach, which is why cold-atom labs can map the entire $\mu(T)$ curve of the figure directly through density profiles.

Step 4. Sanity check against copper: at $T_F\sim 8\times10^4$ K, no metal ever reaches $0.99\,T_F$ (it would vaporise a thousand times over), so in a metal $\mu$ never leaves the flat degenerate region — another reason the sign change is a cold-atom, not a solid-state, phenomenon.

Origins · 1 minFermi temperature and how this was discovered

Fermi temperature

E_F/k_B ≈ 10⁴–10⁵ K in metals · the degeneracy scale

The Fermi temperature $T_F = \varepsilon_F/k_B$ is the single scale that decides everything in this article: the chemical potential barely moves from $\varepsilon_F$ while $T \ll T_F$ , and slides appreciably — eventually going negative into the classical branch — only once $T$ approaches $T_F$ . For an ordinary metal $\varepsilon_F \approx 7\ \mathrm{eV}$ gives $T_F \approx 8\times10^{4}\ \mathrm{K}$ , so at room temperature $T/T_F \approx 0.004$ and the electron gas is profoundly degenerate — the metal melts and boils long before its electrons notice a temperature. This enormous scale is exactly why Arnold Sommerfeld's 1927–28 free-electron theory of metals succeeded where the classical Drude model failed: equipartition wrongly demanded that a mole of electrons contribute $\tfrac32 R$ to the heat capacity, but because $T \ll T_F$ only the thin shell within $\sim k_BT$ of the Fermi surface can be excited (the rest are Pauli-blocked), and the Sommerfeld expansion in the small parameter $k_BT/\varepsilon_F = T/T_F$ extracts that shell's contribution order by order.

History

1900

— Drude models metals as a classical electron gas; explains conduction but predicts a huge electronic heat capacity that is not observed.

1925

— Pauli states the exclusion principle, the microscopic reason fermions fill states one at a time.

1926

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

1927

— Sommerfeld applies Fermi–Dirac statistics to the metallic electron gas, resolving the heat-capacity paradox; the full paper appears 1928.

1936

— Sommerfeld and Bethe's handbook article codifies the low-temperature expansion now taught as the Sommerfeld expansion.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 8 · Kardar, Statistical Physics of Particles, Ch. 7 · Reif, Fundamentals of Statistical and Thermal Physics, §9.10, §9.16 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 7 · Landau & Lifshitz, Statistical Physics Part 1, §57–58 · Ashcroft & Mermin, Solid State Physics, Ch. 2.

Provenance. All equations are standard. Verifiers should reproduce: the Sommerfeld coefficient $\pi^2/12$ in $\mu(T)$ (Pathria & Beale 3rd ed. Eq. 8.1.34; Ashcroft & Mermin Eq. 2.77); the density-of-states prefactor $C=\frac{g_sV}{4\pi^2}(2m/\hbar^2)^{3/2}$ and the $\varepsilon_F$ formula (Kardar Ch. 7); the classical limit $\mu=k_BT\ln(n\lambda_T^3/g_s)$ with $\lambda_T=\sqrt{2\pi\hbar^2/mk_BT}$ (Reif §9.10); the half-filling identity $f(\mu)=\tfrac12$ ; and the $\mu=0$ crossover at $T\approx0.989\,T_F$ from the exact number integral (verify_chemical-potential-fermi-gas-temperature.py, 19 checks).

Builds on: Quantum Statistics: Occupation Numbers · Density of States for Free Particles · Fermi Energy, Momentum and Fermi Temperature · Degenerate Fermi Gas at Zero Temperature · Sommerfeld Expansion Leads to: Electronic Heat Capacity of Metals · Degeneracy Pressure of Fermions · Bose-Einstein Condensation in the Ideal Gas

If μ is barely different from ε_F in a metal, why bother tracking μ(T) at all?

Because the physics you care about lives in the derivative, not the value. The electronic heat capacity, the thermopower, and the temperature dependence of many transport coefficients all depend on how $\mu$ (and the occupation near it) changes with $T$ , and those changes — though tiny in absolute terms — are the entire thermal response of the electron gas. The value of $\mu$ being frozen at $\varepsilon_F$ is exactly what lets you use $\varepsilon_F$ as a fixed reference while computing the small dynamical corrections around it. So the near-constancy is a feature, not a reason to ignore the motion.

Is the chemical potential the same as the Fermi level in a semiconductor?

Yes — "Fermi level" is the condensed-matter name for the chemical potential of the electrons, and it obeys exactly the same number constraint. The reason it feels different is that in a semiconductor the density of states has a gap, so $\mu$ can sit inside the forbidden gap where there are no states at all. It still marks the energy at which the occupation would be one half if a state existed there, and it still slides with temperature — toward midgap as intrinsic carriers are thermally generated. Same quantity, more dramatic motion because of the gap.

Why does the sign of the μ shift depend on the shape of g(ε)?

The shift comes from the asymmetry of thermal smearing. Warming promotes electrons from just below $\mu$ to just above it. If there were the same number of states on each side, the promotions would balance and $\mu$ would not move. But in 3D, $g(\varepsilon)\propto\sqrt\varepsilon$ means there are more states just above $\mu$ than just below, so more electrons leave than the region can spare — to conserve $N$ , $\mu$ must drop so that the filled region shrinks slightly. In a hypothetical band where $g$ decreased with energy, the imbalance would reverse and $\mu$ would rise. The sign is a direct readout of the local slope of $g$ .

Does μ ever exceed ε_F for the 3D free Fermi gas?

No — for the ordinary 3D free-particle density of states, $\mu(T)\le\varepsilon_F$ for all $T\ge0$ , with equality only at $T=0$ . The Sommerfeld correction is strictly negative and the exact curve stays below $\varepsilon_F$ all the way down to the classical branch. You can engineer $\mu>\varepsilon_F$ in systems with an unusual density of states that falls with energy (for instance near certain band features), but that is a property of the band structure, not of Fermi statistics itself.

How can the chemical potential be negative — doesn't that mean it costs negative energy to add a particle?

It does, and that is perfectly physical for a classical gas. The chemical potential is the free-energy cost of adding a particle at fixed $T$ and $V$ : $\mu=(\partial F/\partial N)_{T,V}$ . Adding a particle raises the energy but also raises the entropy (more ways to arrange the gas), and at high temperature the entropy term $-T\,\Delta S$ wins, making the net free-energy cost negative. So $\mu<0$ simply says that, thermodynamically, a dilute hot gas is "happy" to accept more particles. Every classical ideal gas has $\mu<0$ — the Sackur–Tetrode expression makes this explicit.

Why is f(μ) = 1/2 exactly, at every temperature?

Set $\varepsilon=\mu$ in the Fermi function: the exponent $(\varepsilon-\mu)/k_BT$ becomes zero, so $f=1/(e^0+1)=1/(1+1)=1/2$ . There is no temperature dependence because the exponent vanishes identically at that point, independent of how large $k_BT$ is. This is the cleanest operational definition of the chemical potential for a Fermi gas: it is the energy at which a single-particle state is exactly half-likely to be occupied. It is not the top of the filled region (that is $\varepsilon_F$ , and only at $T=0$ ); it is the centre of the thermal smear.