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Expand a weakly depleted Bose gas around its macroscopically occupied mode. Replacing a condensate operator by an amplitude of order $\sqrt{N}$ gives quadratic, cubic and quartic fluctuation interaction scales proportional to $N{N}_{\mathrm{ex}}$, $\sqrt{N}{N}_{\mathrm{ex}}^{3/2}$ and $N_{\mathrm{ex}}^2$, with the same interaction-volume factor. Thus the depletion fraction supplies formal amplitude counting. This is not a uniform bound on infrared modes or a substitute for checking the interaction regime. At fixed particle number, expansion of the condensate interaction must precede the final quadratic truncation; the resulting cancellation is the number-conserving Bogoliubov quadratic Hamiltonian.

The static density susceptibility is the linear density response to a chemical-potential perturbation. In the quadratic uniform contact Bose gas, ${\chi }_{\rho }(q)=[g+q^2/(4m{\rho }_0){]}^{-1}$. Its inverse is a density stiffness. At zero wavevector $\partial \rho /\partial \mu =1/g$ at mean field, while the isothermal compressibility includes the additional factor $1/{\rho }^2$.

At fixed temperature, the continuum excited population of a three-dimensional ideal Bose gas reaches a finite maximum as the chemical potential approaches the ground energy. Adding particles beyond that maximum creates macroscopic ground occupation. Thus $N^{\prime }=\min (N,N_c)$ and $N_0=\max (0,N-N_c)$ to leading thermodynamic-limit order. Finite systems retain nonmacroscopic ground occupation below the nominal threshold.

Interpret the box as a hard-wall cube. Its one-particle levels are ${ϵ}_{n_x{n}_y{n}_z}={\pi }^2{\hslash }^2(n_x^2+n_y^2+n_z^2)/(2m{L}^2)$ for positive integers, giving $E_0=3{\pi }^2{\hslash }^2/(2m{L}^2)$. Counting lattice points in a positive octant and differentiating the large-radius count gives the supplied continuum density of states. Expressing its prefactor through $E_0$ yieldsHere $\mathcal{A}$ denotes the dimensionless constant requested in this question, to distinguish it from the unrelated hyperfine constant of question 1. Periodic boundary conditions would have zero ground energy and would not permit this parametrization by a nonzero $E_0$.

Separate the ground-state occupation from excited particles. The finite-box ground-energy convention in Bose condensation makes the chemical potential approach the ground energy from below as that occupation becomes macroscopic. In the thermodynamic continuum approximation one shifts the energy zero to the ground state and sets the limiting chemical potential to zero in the excited-state integral. The difference between the exact shifted density and its $\sqrt{ϵ}$ bulk form gives finite-size corrections; at the transition $k_B{T}_c/E_0\sim N^{2/3}\gg 1$. ThusThe identity follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating each term. The excited states have a finite saturation capacity. Setting that capacity equal to $N$ determines condensation:Equivalently $k_B{T}_c=(2\pi {\hslash }^2/m)[n/\zeta (3/2){]}^{2/3}$, with $n=N/L^3$.

At fixed $N$, below $T_c$ the chemical potential is pinned to its limiting value andAbove $T_c$, $N^{\prime }\simeq N$ and there is no macroscopic $N_0$; the chemical potential adjusts rather than allowing the saturation expression to exceed the available particles. At fixed $T$, put $N_c=N_{\max }^{\prime }(T)$. For $N<N_c$, $N^{\prime }\simeq N$ and $N_0$ is not macroscopic; for $N>N_c$, $N^{\prime }=N_c$ saturates while $N_0=N-N_c$ grows linearly. This is particle saturation in a homogeneous Bose gas; the sharp threshold is a thermodynamic-limit approximation to finite-box occupations.

The interaction scale compares differently with confinement and transition energies:Thus the stated range makes interactions dominate the ground-state confinement energy while remaining small on the thermal scale controlling $T_c$. This is an interaction-dominated ground state with nearly ideal condensation temperature. It also gives the dilute-gas parameter $n{a}^3=N(a/L{)}^3\ll 1$. A contact model has $g=4\pi {\hslash }^{2}a/m$ and $\mu \simeq gn$; the question's $E_I$ suppresses order-one factors, and the uniform ground energy per particle is $gn/2$, rather than exactly $gn$. Weak interactions can shift $T_c$ slightly, but their dimensionless strength $a{n}^{1/3}\ll 1$ makes that a small relative effect. In a hard-wall box, boundary healing layers do not overturn bulk interaction dominance when their thickness is small compared with $L$.

At $k_{B}T\ll \mu$, thermally occupied Bogoliubov modes have $E_k\simeq \hslash c_{s}k$, with $c_s=\sqrt{\mu /m}$. There is one acoustic branch for this spinless gas. Its thermal energy determines the phonon heat capacity of a dilute Bose gas:The zero-temperature interaction energy does not itself contribute to this leading temperature derivative at fixed $N,V$. This bulk power law also requires $k_{B}T\gg \hslash c_s/L$ so that the phonon spectrum may be treated as continuous; a strictly finite box eventually has discrete-level activated behaviour.

For the noninteracting condensed Bose gas, the ideal Bose-gas heat capacity below condensation follows by integrating $ϵg(ϵ)f(ϵ)$:Its limiting value at $T_c^{-}$ is about $1.926N{k}_B$. At very low temperature the interacting phonon curve is suppressed relative to the ideal curve, has slope three on logarithmic axes, and crosses over towards particle-like $T^{3/2}$ behaviour around $k_{B}T\sim \mu$. Since $\mu \ll k_B{T}_c$, most of the interval below $T_c$ is approximately ideal at leading weak-coupling order. The figure marks a qualitative extrapolation near the transition: a zero-temperature Bogoliubov spectrum does not determine precise interacting critical heat capacity or its small transition shift. The illustrative fixed-$\mu$ interpolation is not claimed as a controlled critical-region calculation.

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 a partially condensed equilibrium ideal Bose gas in a fixed three-dimensional box, the energy scales as $T^{5/2}$. Unbiased random particle removal keeps mean energy per particle fixed, so re-equilibration gives $T\propto N^{2/5}$ until condensation disappears. Removal of an excited particle can be followed by partial replenishment from the zero-energy condensate. This differs from selective evaporative cooling of atoms, which preferentially removes high-energy atoms and reduces mean energy per particle.

With the ground-state energy set to zero, the chemical potential of an ideal Bose gas cannot exceed zero. The limiting population of excited states as it approaches zero is the saturation particle number, provided the integral converges. Additional particles occupy the ground state macroscopically. For a three-dimensional uniform box, $N_c=\zeta (3/2)V/{\lambda }^3$ with thermal wavelength $\lambda =h/\sqrt{2\pi m{k}_{B}T}$.