= Solution
Write $\epsilon_k=\hbar^2k^2/(2m)$, $E_k=\hbar\omega$, and take $\mu>0$. The <Bogoliubov spectrum> is gapless and positive for $k\ne0$. Its small- and large-momentum limits are
$$
\boxed{E_k\simeq\hbar c|k|,\qquad c=\sqrt{\mu/m}\quad(\epsilon_k\ll\mu),}
$$
$$
E_k=\epsilon_k+\mu+O(\mu^2/\epsilon_k)\qquad(\epsilon_k\gg\mu).
$$
The low-energy excitations are collective sound modes; high-energy excitations approach particles with a mean-field <energy> offset. With <healing length> $\xi=\hbar/\sqrt{2m\mu}$ and $q=|k|\xi$,
$$
\frac{E_k}{\mu}=\sqrt{q^2(q^2+2)}.
$$
The crossover occurs at $q$ of order one, not at an additional <energy> gap.
The <critical velocity of a homogeneous condensate> follows because the <Landau criterion> compares an excitation <energy> with the <energy> that flow can release by transferring its <momentum>. Creating a mode is energetically possible when $E_k-\hbar\mathbf k\cdot\mathbf v\leq0$. For this isotropic spectrum,
$$
v_c=\inf_{k>0}\frac{E_k}{\hbar k}
=\inf_{k>0}\sqrt{\frac{\mu}{m}+\frac{\hbar^2k^2}{4m^2}}
=\boxed{\sqrt{\frac{\mu}{m}}}.
$$
The infimum is the long-wavelength <speed of sound>. This tests energetic stability against the stated excitations; it is not a calculation of a dissipation rate or a universal vortex-nucleation threshold.
For an external potential varying slowly on the healing scale, the <Thomas–Fermi approximation for a condensate> neglects density-gradient <kinetic energy> in the stationary <Gross–Pitaevskii equation>. In occupied regions it gives
$$
g\rho(\mathbf r)\simeq\mu-V(\mathbf r),\qquad
c(\mathbf r)^2=\frac{g\rho(\mathbf r)}m,\qquad
\boxed{v_c(\mathbf r)\simeq\sqrt{\frac{\mu-V(\mathbf r)}m}.}
$$
Here $\rho$ is the appropriate local <number density>; in the ring it is a linear density and $g$ is the one-dimensional coupling. This is the <local-density sound speed of a condensate>. The formula applies where $\mu>V$, not in an empty region where its square root would be imaginary.
Normalize the condensate order parameter to the atom number:
$$
N=\int_0^{2\pi}|\psi(\theta)|^2R\,d\theta,\qquad
\boxed{\psi_0=\sqrt{\frac{N}{2\pi R}}}
$$
up to an arbitrary constant phase. A <wavefunction> normalized to one instead has amplitude $1/\sqrt{2\pi R}$, with $N|\psi|^2$ then providing density. Acting with $L_z=-i\hbar\partial_\theta$ gives \b[<angular momentum> $n\hbar$ per atom and $Nn\hbar$ in total] for the uniform ring. The <superfluid velocity> is
$$
v=\frac{\hbar}{mR}\partial_\theta\arg\psi=\frac{n\hbar}{mR}.
$$
The interaction <chemical potential> is $\mu_0=gN/(2\pi R)$; a flowing uniform state also has the constant kinetic contribution $\hbar^2n^2/(2mR^2)$ to its total <chemical potential>. Its <speed of sound> is set by the interaction term. Applying $|v|<\sqrt{\mu_0/m}$ gives
$$
\boxed{n^2<\frac{mR^2\mu_0}{\hbar^2}.}
$$
For a <uniform ring condensate>, this is the continuum <phonon> criterion used here. Quantization of the available excitation <wavelengths> in a very small ring can modify a finite-size threshold.
For the nonrotating equilibrium with a raised left half, let $\rho_L$ and $\rho_R$ be the bulk linear densities. Within the same <Thomas–Fermi approximation for a condensate>,
$$
g\rho_L=\mu-V_0,\qquad g\rho_R=\mu,\qquad
N=\pi R(\rho_L+\rho_R).
$$
When both halves are occupied, normalization fixes
$$
\boxed{\mu=\mu_0+\frac{V_0}{2},\quad
N_L=\frac N2\left(1-\frac{V_0}{2\mu_0}\right),\quad
N_R=\frac N2\left(1+\frac{V_0}{2\mu_0}\right),}
$$
and therefore
$$
\boxed{\frac{N_L}{N_R}=\frac{2\mu_0-V_0}{2\mu_0+V_0}.}
$$
These expressions require $V_0<2\mu_0$. Beyond that threshold, extending them would give a negative left density and is unphysical. Instead the left bulk empties, all $N$ atoms occupy a segment of length $\pi R$, and
$$
\boxed{N_L=0,\quad N_R=N,\quad \mu=\frac{Ng}{\pi R}=2\mu_0
\qquad(V_0>2\mu_0).}
$$
At equality the left bulk just reaches zero density. <kinetic energy> rounds the interface and produces tails beyond the <Thomas–Fermi approximation for a condensate>, so an exactly empty half is the idealized bulk result.
A nonconstant $V(\theta)$ breaks <rotational symmetry>:
$$
[H,L_z]=i\hbar\,\partial_\theta V\ne0.
$$
Consequently <angular momentum> is no longer a good <quantum number>, and an eigenfunction need not be $e^{in\theta}$ times a constant amplitude. Nevertheless, while the order parameter is nonzero around the whole ring, single-valuedness requires
$$
\boxed{\oint d\phi=2\pi n,\qquad
\oint v\,ds=\frac{2\pi n\hbar}{m}.}
$$
The integer is a <phase winding of a condensate>, not an assertion that the inhomogeneous state remains an angular-momentum <eigenstate>. A <superfluid phase slip> can change it if the amplitude vanishes locally.
Conservation of particles gives $\partial_t\rho+\partial_s(\rho v)=0$. For steady density the one-dimensional particle current $j=\rho v$ is constant. Since the two halves have equal lengths,
$$
\rho_Lv_L=\rho_Rv_R,\qquad
\boxed{\frac{v_L}{v_R}=\frac{\rho_R}{\rho_L}=\frac{N_R}{N_L}.}
$$
The <current bottleneck in an inhomogeneous condensate> occurs because a lower-density region has faster flow at the same current; it also has a smaller <speed of sound>. This makes the raised half the limiting region.
For $V_0=2\mu_0/3$ the equilibrium approximation gives
$$
\mu=\frac43\mu_0,\quad g\rho_L=\frac23\mu_0,\quad
g\rho_R=\frac43\mu_0,\quad v_L=2v_R.
$$
The winding condition and equal lengths require
$$
\pi R(v_L+v_R)=\frac{2\pi n\hbar}{m},\qquad
v_R=\frac{2n\hbar}{3mR},\quad v_L=\frac{4n\hbar}{3mR}.
$$
Thus the left-hand sound-speed test is
$$
\frac{16n^2\hbar^2}{9m^2R^2}<\frac{2\mu_0}{3m}
\quad\Longleftrightarrow\quad
\boxed{n^2<\frac38\frac{mR^2\mu_0}{\hbar^2}.}
$$
The right-hand condition is only $n^2<3mR^2\mu_0/\hbar^2$, so it is less restrictive.
\Image[/past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2016/paper-1-bogoliubov-ring.png]
{title=<phonon> stability and the density bottleneck in a split ring}
{description=Original dimensionless dispersion and piecewise bulk density/velocity profiles for the specified $V_0=2\mu_0/3$. The ring is unwrapped into its angular coordinate; no source diagram is reproduced. Narrow transition regions are omitted in the bulk approximation.}
This <local sound-speed estimate for ring-flow stability> uses the <step-potential equilibrium in a ring condensate> densities together with the local sound-speed test, as suggested by the preceding approximations. A quantitatively self-consistent finite-current solution also includes $mv^2/2$ and the <quantum pressure of a condensate> in the stationary <Gross–Pitaevskii equation>, allowing flow to change the density. Smooth steps suppress reflection but do not alone remove those terms. \b[The $3/8$ result is the stated local-density estimate; it should not be promoted to an exact stability theorem for every finite interacting ring.]
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