= Solution
<Condensate depletion> means particles outside a macroscopically occupied bosonic single-particle mode. Interactions can deplete that mode even at zero <temperature>: the <Bogoliubov transformation> mixes creation and <annihilation operators>, so its <quasiparticle> vacuum contains particles at nonzero <momentum>. A weak homogeneous three-dimensional gas with repulsive scattering length $a_s$ has leading <quantum depletion of a Bose condensate>
$$
\frac{n-n_0}{n}=\frac8{3\sqrt\pi}\sqrt{na_s^3},\qquad na_s^3\ll1.
$$
Thermal <quasiparticles> add another contribution. Depletion and normal-fluid density are distinct: a Galilean-invariant fluid can have quantum depletion while retaining its full superfluid <mass> response at zero <temperature>.
For the fermionic pairing problem, mean-field decoupling of the <reduced BCS pairing Hamiltonian> introduces the anomalous amplitude $\Delta$. Each paired Nambu block has diagonal entries $\xi_{\mathbf k},-\xi_{\mathbf k}$ and off-diagonal pairing entries of magnitude $\Delta$. A phase convention with negative off-diagonal entries gives eigenvalues $\pm E_{\mathbf k}$, where $E_{\mathbf k}=\sqrt{\xi_{\mathbf k}^2+\Delta^2}$. Its canonical Bogoliubov rotation yields the <BCS coherence factors>
$$
u_{\mathbf k}=\frac1{\sqrt{1+s_{\mathbf k}^2}},\qquad
v_{\mathbf k}=\frac{s_{\mathbf k}}{\sqrt{1+s_{\mathbf k}^2}},\qquad
\frac{v_{\mathbf k}}{u_{\mathbf k}}=\frac{E_{\mathbf k}-\xi_{\mathbf k}}\Delta.
$$
The partner transformation has the complementary sign required by fermionic anticommutation. At zero <temperature> all positive-energy <BCS quasiparticle> modes are empty. Expanding the supplied <electron> operator therefore gives
$$
\langle a^\dagger_{\mathbf k\sigma}a_{\mathbf k\sigma}\rangle
=u_{\mathbf k}^2\langle\alpha^\dagger\alpha\rangle
+v_{\mathbf k}^2\langle\alpha\alpha^\dagger\rangle=v_{\mathbf k}^2.
$$
Since $E_{\mathbf k}^2-\xi_{\mathbf k}^2=\Delta^2$,
$$
\boxed{n_{\mathbf k\sigma}=\frac{s_{\mathbf k}^2}{1+s_{\mathbf k}^2}
=\frac12\left(1-\frac{\xi_{\mathbf k}}{\sqrt{\xi_{\mathbf k}^2+\Delta^2}}\right).}
$$
The <electron> occupation is close to one far below the normal-state <Fermi surface>, equals one half at $\xi=0$, and decays toward zero above it. Pairing smooths the Fermi step over an <energy> range of order $\Delta$. The positive-energy <quasiparticle> occupation is instead zero everywhere at this <temperature>; the <electron> sea and the <quasiparticle> vacuum are different descriptions of the same state. At finite <temperature>, $n_{\mathbf k\sigma}=[1-(\xi/E)\tanh(E/(2k_BT))]/2$ in the same mean-field model.
\Image[/past-exam-of-the-superconductivity-and-quantum-coherence-course-of-the-university-of-cambridge/2015/paper-1-electron-quasiparticle-occupation.png]
{title=Zero quasiparticle occupation coexists with electron population above the normal-state Fermi surface}
{description=Original isotropic BCS calculation. The left panel uses an illustrative gap-to-chemical-potential ratio of 0.1; it is not measured mercury or cuprate data.}
For a three-dimensional parabolic normal band, the total <density of states> including both spins is $D(\mu)=3N/(2\mu)$. In the weak-gap limit, replace it by this constant across the narrow pairing region and use $\mu\simeq E_F$. The number of <electrons> above the normal-state <Fermi surface> is
$$
\Delta N\simeq D(\mu)\int_0^\infty\frac12\left(1-\frac\xi{\sqrt{\xi^2+\Delta^2}}\right)d\xi.
$$
The integral is $[\xi-\sqrt{\xi^2+\Delta^2}]_0^\infty/2=\Delta/2$. Therefore the <BCS population above the normal-state Fermi surface> is
$$
\boxed{\frac{\Delta N}{N}\simeq\frac{3\Delta}{4\mu}.}
$$
The constant-density-of-states approximation and a small gap relative to the <chemical potential> are essential. This is the question's explicitly defined fermionic population measure; it is not the fraction of <electrons> outside a single bosonic condensate orbital.
For an order-of-magnitude mercury estimate, use $T_c\simeq4.2\,\mathrm K$ and the weak-coupling scale $\Delta_0\simeq1.76k_BT_c\simeq0.64\,\mathrm{meV}$, with a representative metallic $\mu\simeq7\,\mathrm{eV}$. These input approximations give \b[$\Delta N/N\sim7\times10^{-5}$, of order $10^{-4}$]. The use of a weak-coupling gap is an estimate, not an assertion of mercury's exact measured gap.
For a copper oxide, an illustrative gap $20\,\mathrm{meV}$ and effective chemical-potential scale $0.5\,\mathrm{eV}$ give \b[about three per cent] in this formula. Choosing $0.2$–$1\,\mathrm{eV}$ instead gives roughly $7.5$–$1.5$ per cent. These are explicit example inputs: the paper specifies no compound, doping or band parameters from which a unique number follows. A real cuprate's anisotropic gap, quasi-two-dimensional band and strong correlations require a band-resolved occupation integral, so the reliable contrast is the much larger gap-to-electronic-energy scale, rather than a universal cuprate percentage.
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