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Expand an interacting Bose field about a macroscopically occupied condensate and retain terms quadratic in its fluctuations. A Bogoliubov transformation diagonalizes the stable quadratic Hamiltonian. For contact repulsion this gives the Bogoliubov quasiparticle dispersion $E_k=\sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$, a linear sound mode at small momentum and particle-like behavior at large momentum. Mixing creation and annihilation operators also produces quantum depletion of a Bose condensate. The approximation requires controlled fluctuations; it does not give an exact strongly interacting liquid or the full fluctuation-dominated critical region.

For a weakly depleted homogeneous Bose gas with repulsive contact interaction, set $g=uN/V$ and ${ϵ}_k={\hslash }^2{k}^2/(2m)$. A zero-mode coherent state times a two-mode squeezed vacuum for each opposite-momentum pair has mean particle number $N=N_0+{\sum }_{k\ne 0}{\sinh }^2{\theta }_k$. Substitution of $N_0=N-\sum {\sinh }^2{\theta }_k$ into the leading condensate interaction energy, followed by a consistent quadratic truncation, gives the displayed variational energy. Omitting this substitution loses the subtraction embodied in the number-conserving Bogoliubov quadratic Hamiltonian.

For each mode set $A_k={ϵ}_k+g$ and $E_k=\sqrt{A_k^2-g^2}$. If $A_k>|g|$, choose $\tanh (2{\theta }_k^{\ast })=g/A_k$. The mode contribution equals $\frac{1}{2}{E}_k\cosh [2({\theta }_k-{\theta }_k^{\ast })]-A_k/2$, proving a unique minimum $(E_k-A_k)/2$. At fixed ultraviolet cutoff this lowers the energy relative to the uncorrelated condensate. The approximation requires small quantum depletion of a Bose condensate; an unregulated three-dimensional ground-state energy also requires renormalization of a three-dimensional dilute-gas contact coupling.

For the normalized Hartree product state $|N:\varphi ⟩$, applying the bosonic annihilation operator gives $\psi (y)|N:\varphi ⟩=\sqrt{N}\varphi (y)|N-1:\varphi ⟩$. Taking the inner product proves the displayed factorization of the one-body density matrix. Its diagonal is the total number density in this approximation. For a real positive orbital, $G^{(1)}(0,r)=\sqrt{n(0)n(r)}$. Restoring the kinetic energy in a Gross–Pitaevskii equation can smooth a Thomas–Fermi approximation for a condensate edge while retaining this factorization. Describing quantum depletion of a Bose condensate additionally requires noncondensate mode occupations and their contribution to the one-body density matrix.

The coherent state in the zero mode has mean occupation number $N_0$. The other occupation numbers follow from part (c), giving exactlyImpose the fixed mean particle number $⟨\hat{N}⟩=N$ by choosingHere $D$ is the quantum depletion of a Bose condensate. The approximation requires $0\le D\ll N$, so $N_0$ remains positive and macroscopically large. The specified coherent state and multimode squeezed vacuum are not exact eigenstates of the total number operator; $N$ in this variational construction is their prescribed mean particle number.

Write $P={\sum }_{\mathbf{k}\ne 0}{s}_k{c}_k$. Substituting into the retained part of the interaction energy before truncating givesThe last two terms quadratic in $D,P$ have no explicit large-$N$ factor when the Bogoliubov transformation parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the number-conserving Bogoliubov quadratic Hamiltonian. Thus, defining the physical number density correctly as $\rho =N/V=N/L^3$,Replacing $N_0$ by $N$ at the outset would give the wrong coefficient $2{\sinh }^2{\theta }_k$: expanding the leading condensate interaction energy supplies the necessary subtraction.

The printed definition $\rho =N/L$ is an error. In three dimensions the contact interaction coupling $u$ has units of energy times volume, so $uN/L$ does not even have units of energy. There is a direct counterexample with no depletion: when ${\theta }_k=0$, the specified coherent state has exactly $⟨H_{\mathrm{int}}⟩=u{N}^2/(2L^3)$. In fixed length and energy units, $L=2$, $u=1$ and $N=100$ give $625$, whereas the printed $u\rho N/2$ with $\rho =N/L$ gives $2500$. The requested formula and the minimization in part (e) hold with $\rho =N/L^3$.

Put ${ϵ}_k={\hslash }^2{k}^2/(2m)>0$ for $\mathbf{k}\ne 0$ and $g=u\rho$, using $\rho =N/L^3$. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, $u>0$, each ${ϵ}_k+g>g>0$. Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution ${\theta }_k^{\ast }$, $\cosh 2{\theta }_k^{\ast }=({ϵ}_k+g)/E_k$ and $\sinh 2{\theta }_k^{\ast }=g/E_k$. HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from $\theta =0$ isIndeed $E_k<{ϵ}_k+g$. More directly, for small positive $\theta$,The pair vacuum expectation value lowers the interaction energy at first order in $\theta$, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when ${ϵ}_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-{ϵ}_k$ in some mode, $f_k(\theta )={ϵ}_k\sinh (2\theta )/2$ tends to $-\infty$ as $\theta \to -\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large $k$, $E_k-{ϵ}_k-g=-g^2/(2{ϵ}_k)+O({ϵ}_k^{-2})$, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.

In the ground state of the diagonal Hamiltonian, ${\alpha }_{\mathbf{k}}|0_B⟩=0$. The Bogoliubov transformation therefore gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Summing all nonzero momenta proves the quantum depletion of a Bose condensate:The nonzero depletion arises from correlated pairs in the interacting quasiparticle vacuum, not thermally populated quasiparticles.

The printed expression is the zero-temperature result. At a nonzero temperature the complete leading-order expression instead hasThe additional term is thermal depletion and cannot be discarded merely because the gas is condensed.

In the Bogoliubov quasiparticle vacuum, ${\alpha }_{\mathbf{k}}|0_{\alpha }⟩=0$, but the inverse Bogoliubov transformation gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Consequently the quantum depletion of a Bose condensate isThis expression assumes zero temperature. At nonzero temperature, thermal Bogoliubov quasiparticles add to the depletion:Thus the printed formula is the quantum contribution, or a low-temperature approximation when the thermal contribution is negligible. It is nonnegative for a stable quadratic Hamiltonian and vanishes in the noninteracting zero-temperature limit.

A complex order parameter captures both the amplitude and phase rigidity of a superfluid. A local isotropic Ginzburg-Landau theory writes its free-energy functional, to fourth order in amplitude and second order in gradients, asThe uniform minimum has $\Psi =0$ for $\alpha >0$ and $|{\Psi }_0{|}^2=-\alpha /\beta$ for $\alpha <0$. Its phase remains arbitrary under a global $U(1)$ transformation. Minimizing with respect to ${\Psi }^{\ast }$ gives $-K{\nabla }^2\Psi +\alpha \Psi +\beta |\Psi {|}^2\Psi =0$. The phenomenological description specifies the allowed structures and coefficients; microscopic input is needed to identify the normalization of $\Psi$ and calculate those coefficients. A static functional by itself does not fix the dynamical equation or mode frequencies.

For a dilute three-dimensional Bose gas with short-range repulsion and particle mass $m_B$, the microscopic low-energy Hamiltonian isThe interaction parameter is expressed through the low-energy s-wave scattering length $a_s$, rather than an unrenormalized bare delta potential. In a symmetry-selected description, $\Psi =⟨\hat{\psi }⟩$ and $|\Psi {|}^2=n_0$ is condensate density. Equivalently, fixed-number off-diagonal long-range order identifies the macroscopically occupied natural orbital and its amplitude without a nonzero one-point expectation.

Evaluating $\hat{H}-\mu \hat{N}$ in a coherent-field condensate variational state gives the Ginzburg-Landau functional of a dilute Bose gas, with the mean-field microscopic identificationsIts stationary variational equation is the Gross–Pitaevskii equation. The time-dependent variational action additionally contains $i\hslash {\Psi }^{\ast }{\partial }_t\Psi$, givingThis term expresses density-phase conjugacy. It is the missing microscopic dynamical input; a purely dissipative relaxation law would describe a different approximation.

Bogoliubov theory writes $\hat{\psi }=\Psi +\delta \hat{\psi }$ and retains quadratic fluctuations. In a uniform condensate, $\mu =g_B{n}_0$ and the fluctuation matrix isIts eigenvalues are $\pm \sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$. The positive branch is the gapless sound mode, continuously crossing over to particle-like excitations at larger $k$. The same result follows by linearizing the time-dependent Gross–Pitaevskii equation about its uniform solution. Thus the microscopic condensate amplitude, scattering interaction and mass determine the mean-field static coefficients and the collective mode. Quadratic quantum fluctuations also produce quantum depletion of a Bose condensate, which a pure coherent-state product omits.

These identifications are controlled for a dilute weakly interacting gas away from a fluctuation-dominated critical region. They are not a universal statement that the exact finite-temperature GL coefficient is always $-\mu$ or that its quartic coefficient never receives corrections. Integrating out noncondensed and short-wavelength modes produces effective temperature-dependent coefficients and can change their near-critical behavior. The phase rigidity described by $K$ and the thermodynamic order-parameter expansion remain the shared organizing structure.

For BCS theory, the microscopic order parameter is instead an anomalous pair amplitude, commonly normalized as an energy gap $\Delta \propto ⟨{\hat{\psi }}_{\downarrow }{\hat{\psi }}_{\uparrow }⟩$. It is not an ordinary single-electron field expectation or automatically the square root of a pair number density. Integrating out fermions and expanding in small $\Delta$ and slow gradients near $T_c$ gives the microscopic derivation of Ginzburg-Landau theory. The BCS pairing-field effective action supplies its coefficients through normal-state pair-response functions and thermal fermion sums.

For example, a clean isotropic three-dimensional weak-coupling s-wave model, with $N(0)$ the density of states per spin at the Fermi energy and $\Delta$ measured in energy units, giveswith microscopic Ginzburg-Landau coefficients from BCS theoryThe numerical factors arise from fermionic Matsubara sums and the isotropic velocity average; they depend on the stated gap and density-of-states conventions. Rescaling the order parameter rescales all three coefficients. The expansion applies near the transition with sufficiently slow spatial variation; disorder, anisotropic pairing and strong coupling alter the coefficients or require additional components. Microscopic derivation also fixes the pair charge in the covariant gradient, producing the superconducting screening current and the Meissner effect.^[ref]

The neutral Bose and charged Fermi theories therefore share a complex order parameter and a phase-stiffness functional, while deriving that field from different microscopic objects. The Bogoliubov transformation mixes particle and hole fluctuations of the bosonic condensate in one case and paired fermionic excitations in the other. The static GL limit organizes long-distance behavior; the full microscopic theories retain excitation spectra and temperature dependence beyond that local near-transition expansion.

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 condensateThermal 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 factorsThe 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 givesSince $E_{\mathbf{k}}^2-{\xi }_{\mathbf{k}}^2={\Delta }^2$,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, $n_{\mathbf{k}\sigma }=[1-(\xi /E)\tanh (E/(2k_{B}T))]/2$ in the same mean-field model.

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 isThe integral is $[\xi -\sqrt{{\xi }^2+{\Delta }^2}{]}_0^{\infty }/2=\Delta /2$. Therefore the BCS population above the normal-state Fermi surface isThe 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 $T_c\simeq 4.2\mathrm{K}$ and the weak-coupling scale ${\Delta }_0\simeq 1.76k_B{T}_c\simeq 0.64\mathrm{meV}$, with a representative metallic $\mu \simeq 7\mathrm{eV}$. These input approximations give $\Delta N/N\sim 7\times 10^{-5}$, 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 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.