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Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

In the two-fluid model of superfluidity, the superfluid fraction is ${\rho }_s/\rho$, where ${\rho }_s$ is the superfluid mass density and $\rho$ the total mass density. For a one-component fluid of particles of the same mass this also equals $n_s/n$, using the superfluid number density $n_s$ and total number density $n$. It measures the superfluid response and is distinct from the condensate fraction, which measures occupation of a single-particle state. Interactions can make the two fractions very different.

The renormalized phase stiffness has this discontinuity at the Berezinskii–Kosterlitz–Thouless transition. The algebraic correlation function has exponent $1/4$ at the transition from the ordered side.

The stiffness here is an energy in two dimensions, not an areal mass density. For a neutral superfluid of particles of mass $m$, let $n_s$ be the superfluid number density per unit area. Substituting the superfluid velocity ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ into the kinetic energy gives ${\rho }_s={\hslash }^2{n}_s/m$. Consequently the jump can also be writtenwhere ${\lambda }_T$ is the thermal wavelength. It is the large-distance superfluid number density, rather than the total density or a locally inferred condensate density, that enters this relation.

Within the vortex-pair renormalization flow, $y$ changes from irrelevant to relevant at $\pi K_R=2$, with $K_R$ the large-distance dimensionless phase stiffness. Combining this with the spin wave result $\eta =1/(2\pi K_R)$ gives $\eta =1/4$. Above the transition the homogeneous equilibrium phase stiffness vanishes at infinite observation scale. A discontinuous response coefficient is compatible with a diverging correlation length; it does not imply a latent heat.^[ref]