= Solution
Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with <spin degeneracy> two. Their effective densities of states are
$$
N_c=2\left(\frac{m_ck_BT}{2\pi\hbar^2}\right)^{3/2},\qquad N_v=2\left(\frac{m_hk_BT}{2\pi\hbar^2}\right)^{3/2}.
$$
For the <semiconductor mass-action law>, integrating the square-root <density of states> against the Boltzmann tail gives $n=N_c e^{(\mu-E_c)/(k_BT)}$ and $p=N_v e^{(E_v-\mu)/(k_BT)}$. Multiplication cancels the <chemical potential>, giving
$$
\boxed{np=A(k_BT)^3e^{-E_g/(k_BT)},\qquad A=\frac{4(m_cm_h)^{3/2}}{(2\pi\hbar^2)^3}.}
$$
If there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi\hbar^2)^3$. No numerical mass or valley multiplicity has been invented.
When $\mu=E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive <temperature>. Define the <Normalized complete Fermi-Dirac integral>
$$
F_{1/2}(\eta)=\frac2{\sqrt\pi}\int_0^\infty\frac{\sqrt x}{1+e^{x-\eta}}\,dx,\qquad f_0=F_{1/2}(0)=(1-2^{-1/2})\zeta(3/2)\simeq0.765147.
$$
For the <carrier densities with chemical potential pinned to a band edge>, with the two specified equal masses, put $N(T)=2[mk_BT/(2\pi\hbar^2)]^{3/2}$. Direct Fermi-Dirac integration then gives
$$
\boxed{n(T)=N(T)f_0,\qquad p(T)=N(T)F_{1/2}[-E_g/(k_BT)].}
$$
Thus $n\propto T^{3/2}$ exactly within the parabolic model. At low <temperature>, $p\simeq N(T)e^{-E_g/(k_BT)}$ is thermally activated; at $k_BT\gg E_g$, $p/n\to1$ from below. Both vanish at zero <temperature> and increase at fixed <chemical potential>. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.
For the <Pauli reduction of the pinned-edge mass-action product>, the exact product relative to the classical law is
$$
\frac{np}{N(T)^2e^{-E_g/(k_BT)}}=f_0e^{E_g/(k_BT)}F_{1/2}[-E_g/(k_BT)]<1.
$$
It approaches $f_0$ at low <temperature> and $f_0^2\simeq0.58545$ in the formal high-<temperature> limit. The low-<temperature> activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-<temperature> limit here refers to the given unbounded parabolic bands, not a claim that real <semiconductor> band structures remain unchanged at arbitrarily high <temperature>.<Quantum Gases by David Tong>[Sections 3.1 and 3.6.1–3.6.4: density of states, Fermi occupation and Sommerfeld expansion]{ref}
Finally, for a fixed positive $\mu$, the zero-<temperature> Fermi sphere has radius $k_\mu=\sqrt{2m\mu}/\hbar$, so
$$
\boxed{n(0)=\frac{k_\mu^3}{3\pi^2}=\frac1{3\pi^2}\left(\frac{2m\mu}{\hbar^2}\right)^{3/2},\qquad n(T)=N(T)F_{1/2}[\mu/(k_BT)].}
$$
The low-<temperature> expansion at \b[fixed $\mu$] gives the <fixed-chemical-potential Sommerfeld density increase>:
$$
\boxed{n(T)=n(0)\left[1+\frac{\pi^2}{8}\left(\frac{k_BT}{\mu}\right)^2+O\!\left(\frac{k_BT}{\mu}\right)^4\right].}
$$
The <monotonic density at fixed chemical potential in a square-root density of states> initially rises quadratically from $n(0)$. For $k_BT\gg\mu$, its leading behaviour crosses to $f_0N(T)\propto T^{3/2}$. The fixed <chemical potential> is what allows the particle number to change.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2015/paper-1-pinned-carrier-densities.png]
{title=Pinning the chemical potential to or above a band edge changes carrier populations and the mass-action product}
{source=paper-1-pinned-carrier-densities.py}
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