Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 84 / 4

Here equilibrium means a time-independent number density and current, with a common temporal quantum phase. Write $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }\Psi (\mathbf{x})$ with real $\mu$, since an imaginary temporal frequency would change the density. The stationary pumped Gross–Pitaevskii equation isThe chemical potential is the common phase-rotation frequency times $\hslash$; this open system has no conserved particle-number constraint fixing it as in the conservative problem. Local gain and loss need not cancel separately in an inhomogeneous equilibrium, because a stationary condensate number current can transport particles between gain and loss regions.

Define the physical velocity potential by $\psi =\sqrt{n}\exp (im\varphi /\hslash )$, so the superfluid velocity is $\mathbf{v}=\nabla \varphi$. The condensate number current isMultiplying the pumped Gross–Pitaevskii equation by ${\psi }^{\ast }$ and subtracting its complex conjugate gives the density equation. Equivalently, substituting the amplitude and phase and equating imaginary parts yieldsEquating real parts gives the other gain-loss Madelung equations relation,The last term is the quantum potential, retained wherever $n>0$. Taking a gradient gives the equivalent velocity equationin an irrotational region. In equilibrium $n_t=0$ and $\varphi (\mathbf{x},t)=\Phi (\mathbf{x})-\mu t/m$, soThis convention distinguishes the physical velocity potential from the dimensionless quantum phase $m\varphi /\hslash$; using the latter instead would change the divergence prefactor.

For a positive uniform number density $n_{\ast }$ and constant superfluid velocity $v\hat{\mathbf{x}}$, current divergence vanishes. The stationary density equation therefore imposes $(\gamma -\Gamma n_{\ast })n_{\ast }=0$, giving $n_{\ast }=\gamma /\Gamma$. The quantum potential vanishes, so the phase equation sets $\mu =m{v}^2/2+V_0\gamma /\Gamma$. Thus the uniform-current pumped polariton condensate isDirect substitution verifies the equation: the kinetic term gives $m{v}^2/2$, the interaction gives $V_0\gamma /\Gamma$, and the gain-loss term is zero. A constant overall phase is arbitrary.

With the stated $\gamma ,\Gamma >0$, this nonzero plane wave exists for any real $v$ and real interaction coefficient $V_0$ when $\mu$ is determined by the solution. If a particular real chemical potential is prescribed instead, the existence condition isEquality gives the uniform condensate at rest. A nonzero density generally requires $\gamma /\Gamma >0$; a zero nonlinear-loss coefficient with positive pumping would not allow this nonzero stationary balance. The vacuum is a separate zero solution, unstable to small perturbations for positive pump.

Existence should not be confused with stability. For completeness, linearize in the comoving frame with $\psi ={\psi }_{\ast }[1+u+iw]$, and let ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The linear real-amplitude and phase equations givesoFor $V_0\ge 0$ both growth rates have nonpositive real parts, including the neutral uniform-phase mode. For $V_0<0$, sufficiently small nonzero $k$ makes the constant term negative, so one growth rate is positive: the plane wave still exists but is unstable in an infinite system. Uniform drift adds only a Doppler phase to these growth rates in this model.

Let $N={\int }_{S}ndS$ be the particle number per unit axial length, and consider the axially homogeneous equilibrium implicit in a cross-sectional description. Integrating the stationary gain-loss Madelung equations over $S$ and applying the divergence theorem givesThe wall condition $n=0$ means the lateral condensate number current vanishes. This is also directly apparent from $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$ for a regular Dirichlet wavefunction, without assigning a phase at its zeros. Consequently the cylindrical pump-loss balance isIntegrated particle creation equals integrated nonlinear loss; there is no assumption that $n=q$ pointwise. In fact the Cauchy-Schwarz inequality gives $N^2\le |S|{\int }_S{n}^2=|S|qN$, so a nonzero equilibrium obeys $N\le q|S|$.

The axial condition is needed for a slice-by-slice statement. More generally, put $J(z)={\int }_{S}n{\partial }_z\Phi dS$. Lateral no-flux and stationary density giveThus the requested equality also holds whenever the integrated axial current is constant. If axially inhomogeneous equilibria with nonzero axial flux divergence are allowed, the lateral boundary condition alone proves this more general balance, rather than the stated equality at every cross-section.