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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.

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.

Put $A_k={ϵ}_k+n{V}_k$ and $B_k=n{V}_k$, and assume the homogeneous condensate is stable, so $A_k>|B_k|$ for nonzero modes. A real Bogoliubov transformation isThe canonical condition preserves the bosonic canonical commutation relations. Cancellation of pair terms requires $2A_k{u}_k{v}_k=B_k(u_k^2+v_k^2)$. ThereforeThe sign of $v_k$ follows that of $B_k$. Normal ordering the transformed operators gives the bosonic Bogoliubov zero-point shift:Rearranging the last two terms is exactly the requested $-\frac{1}{2}\sum A_k+\sum E_k({\alpha }_k^{†}{\alpha }_k+1/2)$ form. Each momentum appears in the sums, so opposite-momentum pairs are not counted again separately.

The Bogoliubov quasiparticle dispersion is linear at low momentum when $V_k\to V_0>0$:It is a gapless phonon of the condensate, unlike the free-particle quadratic spectrum. For a contact repulsion $V_k=g>0$, the healing length $\xi =\hslash /\sqrt{2mng}$ controls crossover to the particle regime $E_k\simeq {ϵ}_k+ng$. A momentum-dependent potential can modify the higher-energy curve; the sketch uses the contact case and does not assume every interaction has that entire shape.

For each opposite-momentum pair use the Bogoliubov transformationEquivalently $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, with $\tanh (2{\theta }_k)=B_k/A_k$. The identity $u_k^2-v_k^2=1$ preserves the canonical commutation relation. Vanishing anomalous terms require $B_k(u_k^2+v_k^2)-2A_k{u}_k{v}_k=0$, and hence the Bogoliubov quasiparticle dispersion isSubstitution and normal ordering giveThis is the requested form after putting each $E_k/2$ inside the oscillator term. By momentum-pair counting in bosonic diagonalization, all sums cover the full nonzero-momentum set; no further factor of two is needed. The bosonic Bogoliubov zero-point shift combines the two vacuum sums before taking a continuum limit.

If $V_k\to V_0>0$ smoothly, the low-energy excitations are gapless phonons:For contact repulsion, the healing length $\xi =\hslash /\sqrt{2mn{V}_0}$ separates this regime from the particle-like regime $E_k\simeq {ϵ}_k+n{V}_0$. Stability requires ${ϵ}_k+2n{V}_k>0$ for every nonzero mode. A general momentum-dependent interaction need not share the contact model's entire curve. An unregularized contact model also makes the vacuum-energy shift ultraviolet divergent; a physical finite-range interaction or scattering-length renormalization is needed for that absolute ground-state energy.

For a dilute three-dimensional weakly interacting Bose gas with repulsive short-range scattering, use a contact coupling $g=4\pi {\hslash }^2{a}_s/m>0$ and $n{a}_s^3\ll 1$. In imaginary time ${\tau }_E$, the grand-canonical action isThe uniform saddle has $n_0=\mu /g$. Its real-time equation is the Gross–Pitaevskii equation; quantum depletion means condensate and total density need not coincide exactly.

Writing $\psi =\sqrt{n_0}+\delta \psi$ and retaining quadratic fluctuations gives a Nambu kernel with diagonal ${ϵ}_k+g{n}_0$, anomalous coupling $g{n}_0$ and ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The Bogoliubov quasiparticle dispersion isPhase and density fluctuations form the gapless sound mode associated with broken global $U(1)$ symmetry. At large momentum the spectrum crosses towards particle behaviour; the healing length is $\xi =\hslash /\sqrt{2mg{n}_0}$. The depletion integral $\int d^{3}k{v}_k^2/(2\pi {)}^3$ is finite in three dimensions and gives $(n-n_0)/n\simeq 8\sqrt{n{a}_s^3}/(3\sqrt{\pi })$ to leading order. The contact coupling denotes a scattering-length-renormalized interaction; naive ultraviolet zero-point integrals require renormalization.

The condensate phase gives superfluid velocity ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ and quantized circulation $\oint {\mathbf{v}}_s\cdot d\mathbf{l}=2\pi \hslash \ell /m$. The Landau criterion, $v_c={\min }_k{E}_k/(\hslash k)$, gives $c_s$ for this homogeneous Bogoliubov spectrum. Boundaries and vortex nucleation can cause dissipation below that value, so it is not a universal measured critical speed. Condensation and superfluid stiffness are related but distinct; dimensional fluctuations can remove true condensate order while retaining superfluid responses.

A neutral superfluid's broken global $U(1)$ phase symmetry produces a Goldstone mode. Coupling phase to density fluctuations generally gives a long-wavelength phonon with $E(p)\simeq cp$, rather than the quadratic dispersion of a noninteracting condensate. In a dilute interacting Bose gas, the Bogoliubov quasiparticle dispersion isFor a heavy weak obstacle moving at speed $v$, emitting an excitation of momentum $\mathbf{p}$ changes the available energy by $E(p)-\mathbf{v}\cdot \mathbf{p}$. The Landau criterion therefore givesThe dilute-gas spectrum has $v_L=c$. In liquid helium, a roton minimum can make the minimum ratio smaller than the sound speed; the slope of a line from the origin tangent to the full dispersion determines the kinematic bound. A neutral paired Fermi superfluid also has a fermionic pair-breaking bound of order $\Delta /p_F$ in weak coupling. The smallest relevant branch bound must be considered. Actual flow can fail earlier through vortex nucleation, surface defects or other instabilities: the excitation-energy criterion is not a complete dynamical guarantee of dissipation-free flow.

A neutral superfluid can select a phase of its complex order parameter, breaking a global $U(1)$ symmetry. A spatially uniform change of that phase costs no energy. Slowly varying changes therefore produce a gapless Goldstone mode. For a dilute weakly repulsive Bose gas, the Bogoliubov quasiparticle dispersion isso $E_k\simeq \hslash c_{s}k$ at long wavelengths, with $c_s=\sqrt{g_B{n}_0/m_B}$. Density and phase fluctuations are canonically coupled components of this sound branch, rather than two independent positive-frequency modes. Superfluid helium and dilute atomic condensates supply neutral examples, although the dilute-gas formula is not quantitatively valid for strongly interacting liquid helium.

In a superconductor, the pair field is charged and its phase couples to the electromagnetic gauge field. Only combinations such as $\nabla \theta -q\mathbf{A}/\hslash$ are physically meaningful. The Anderson-Higgs mechanism in a superconductor produces electromagnetic screening and removes the neutral sound mode as an independent gapless charged-density oscillation. In a homogeneous three-dimensional continuum, long-range Coulomb forces give a longitudinal plasma gap, approximately ${\omega }^2={\omega }_p^2+c_s^2{k}^2$ at small $k$. The static transverse response gives the Meissner effect andwhen $n_s$ counts superconducting electrons of effective mass $m$. Counting pairs instead uses density $n_s/2$, charge $-2e$ and mass $2m$, giving the same stiffness. The condensate amplitude may also have a gapped Higgs mode, but that amplitude oscillation is distinct from the electromagnetic Higgs mechanism. Local gauge transformations remain a redundancy; this is not spontaneous breaking of an observable local symmetry in the same sense as the neutral global phase symmetry.

For independent $\chi ,{\chi }^{\ast }$, the derivatives needed in the ${\chi }^{\ast }$ Euler-Lagrange field equation areConsequently the two conjugate field equations areWrite $\chi ={\chi }_1+i{\chi }_2$ with real components, and put $A={\hslash }^2/(2M)$. Separating real and imaginary parts givesFor each real field take the real part of a complex plane wave amplitude, ${\chi }_j=\mathrm{Re}[q_j{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. With ${ϵ}_k={\hslash }^2|\mathbf{k}{|}^2/(2M)$ the two equations becomeA nonzero normal mode requires their determinant to vanish, soThe positive-frequency Bogoliubov quasiparticle dispersion is thereforeThere is also the negative-frequency conjugate solution, as required for real component fields. Since $\mu >0$, ${\omega }^2\ge 0$, consistent with the stability of the chosen uniform ground state.

Expand a repulsive Bose-Einstein condensate around the real uniform saddle ${\psi }_0=\sqrt{\mu /g}$ as $\psi ={\psi }_0+\chi$. After removing a constant and a temporal boundary term, the quadratic Lagrangian density is the displayed expression. Its equation $i\hslash \dot{\chi }=-{\hslash }^2{\nabla }^2\chi /(2M)+\mu (\chi +{\chi }^{\ast })$ mixes a fluctuation and its complex conjugate. Writing $\chi ={\chi }_1+i{\chi }_2$ gives canonically paired amplitude and phase fluctuations and one positive Bogoliubov quasiparticle dispersion branch, $\hslash \omega =\sqrt{{ϵ}_k({ϵ}_k+2\mu )}$ with ${ϵ}_k={\hslash }^2{k}^2/(2M)$. The phase direction is a Goldstone mode; its zero uniform restoring force does not produce a second independent positive-frequency amplitude branch in this first-order-in-time theory.