= Solution
For a single quantum state, the equilibrium occupations are
$$
\boxed{\bar n_F(\epsilon)=\frac1{e^{(\epsilon-\mu)/(k_BT)}+1},\qquad
\bar n_B(\epsilon)=\frac1{e^{(\epsilon-\mu)/(k_BT)}-1}.}
$$
Both reduce to $e^{(\mu-\epsilon)/(k_BT)}$ when $\epsilon-\mu\gg k_BT$.
For one three-dimensional quadratic branch, the <density of states> per unit <volume> is
$$
g(\epsilon)=\frac1{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt\epsilon.
$$
With $\lambda_T=\sqrt{2\pi\hbar^2/(mk_BT)}$, changing integration variable to $x=\epsilon/(k_BT)$ gives
$$
\boxed{n_{\rm ex}(\mu=0)=\frac{\zeta(3/2)}{\lambda_T^3}\propto T^{3/2},\qquad
n_{\rm classical}=\frac{e^{\mu/(k_BT)}}{\lambda_T^3}.}
$$
A degeneracy factor multiplies both <densities> if additional equivalent branches are present. The first expression is the noncondensed <density>: a macroscopically occupied <ground state> must be counted separately.
<Bose-Einstein condensation> occurs when the <excited states> can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total <density>,
$$
\boxed{T_c=\frac{2\pi\hbar^2}{mk_B}
\left[\frac{n}{\zeta(3/2)}\right]^{2/3}\propto n^{2/3}.}
$$
For <field-induced triplon condensation>, write $\Delta=\nu n-\eta(H-H_g)$, so $\epsilon(k)=\hbar^2k^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic <chemical potential> is zero, the Bose distribution requires $\Delta\ge0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.
For the <triplon condensation field scaling>, the critical <density> at onset is $n_T=\zeta(3/2)/\lambda_T^3$. Therefore
$$
\boxed{H_c(T)-H_g=\frac{\nu}{\eta}\zeta(3/2)
\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}\propto T^{3/2}.}
$$
The low-temperature <phase> boundary rises from $(T,H)=(0,H_g)$; the condensed <phase> lies on the higher-field side.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2014/paper-1-tsp2014-triplons.png]
{title=Original shifted-parabola triplon dispersions and low-temperature field–temperature phase diagram; model choices are dimensionless and illustrative}
For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the <condensate density in a shifted-parabola triplon model>, added <condensate density> keeps the minimum at zero: $\nu(n_0+n_T)=\eta(H-H_g)$. At fixed <temperature> this gives
$$
\boxed{n_0=\frac{\eta}{\nu}[H-H_c(T)]\quad(H>H_c),}
$$
with $n_0=0$ below onset. At zero <temperature> it becomes $n_0=\eta(H-H_g)/\nu$ within that model.
This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed <phase>. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{\epsilon_k^0(\epsilon_k^0+2g n_0)}$, with $\epsilon_k^0=\hbar^2k^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. \b[The gap closes and the <condensate density> grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model.] A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.<Bose-Einstein Condensation of Dilute Magnons in TlCuCl3 by Tetsuro Nikuni and others>[arXiv:cond-mat/9908118v1, equations (4)–(9): normal Hartree-Fock shift, T^(3/2) onset and gapless condensed spectrum]{ref}
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