Past exam of the atomic and Optical Physics course of the University of Cambridge / 2016 / Paper 1 / 3 / Solution

Write ${ϵ}_k={\hslash }^2{k}^2/(2m)$, $E_k=\hslash \omega$, and take $\mu >0$. The Bogoliubov spectrum is gapless and positive for $k\ne 0$. Its small- and large-momentum limits areThe low-energy excitations are collective sound modes; high-energy excitations approach particles with a mean-field energy offset. With healing length $\xi =\hslash /\sqrt{2m\mu }$ and $q=|k|\xi$,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 $E_k-\hslash \mathbf{k}\cdot \mathbf{v}\le 0$. For this isotropic spectrum,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 givesHere $\rho$ 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: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 $L_z=-i\hslash {\partial }_{\theta }$ gives angular momentum $n\hslash$ per atom and $Nn\hslash$ in total for the uniform ring. The superfluid velocity isThe interaction chemical potential is ${\mu }_0=gN/(2\pi R)$; a flowing uniform state also has the constant kinetic contribution ${\hslash }^2{n}^2/(2m{R}^2)$ to its total chemical potential. Its speed of sound is set by the interaction term. Applying $|v|<\sqrt{{\mu }_0/m}$ givesFor 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,When both halves are occupied, normalization fixesand thereforeThese 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 $\pi R$, andAt 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: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 requiresThe 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,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 givesThe winding condition and equal lengths requireThus the left-hand sound-speed test isThe right-hand condition is only $n^2<3m{R}^2{\mu }_0/{\hslash }^2$, so it is less restrictive.

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 $m{v}^2/2$ 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. 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.