Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 84 (Quantum Fluids)

Treat $\psi$ and its complex conjugate as independent variables in the functional derivative. Integrating the gradient variation by parts, with vanishing boundary terms, givesThe factors follow from ${\partial }_{{\psi }^{\ast }}|\psi {|}^4=2|\psi {|}^2\psi$ and ${\partial }_{{\psi }^{\ast }}|\psi {|}^6=3|\psi {|}^4\psi$. Thus the cubic-quintic Gross–Pitaevskii equation is

At fixed particle number $N_p=\int |\psi {|}^{2}d\mathbf{x}$, an equilibrium wavefunction is a constrained critical point of the energy functional. Introduce the chemical potential $\mu$ as the Lagrange multiplier for $N_p$, and impose $\delta (E-\mu N_p)=0$. The resulting stationary cubic-quintic Gross–Pitaevskii equation isThe time-dependent equilibrium has $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$, so its number density is stationary even though its quantum phase changes uniformly. The specified particle number fixes $\mu$ together with the boundary conditions. In an infinite uniform bulk, particle number is understood through a finite volume or a prescribed bulk density.

For a nonzero uniform wavefunction, the Laplacian vanishes. Dividing the stationary cubic-quintic Gross–Pitaevskii equation by ${\psi }_0$ yieldsEquivalently, the interaction energy density is $e(n)=V_0{n}^2/2+W_0{n}^3/3$ and its derivative $e^{\prime }(n_0)$ is the chemical potential. Positive local compressibility requires ${\mu }^{\prime }(n_0)=V_0+2W_0{n}_0>0$. This stability condition is distinct from the sign of $\mu$ itself.

For the displayed scaling, assume $\mu >0$ and $n_0>0$. SetSubstituting in the stationary cubic-quintic Gross–Pitaevskii equation and dividing by $\mu \sqrt{n_0}$ givesThe uniform-state relation proves $\alpha +\beta =1$, so the real bulk amplitude is one; its constant quantum phase can be chosen to make $\tilde{\psi }\to 1$ in the zero-winding sector. The unit of distance $\ell$ is the kinetic-to-chemical-potential scale, reducing to the stated healing length convention for purely repulsive two-body coupling.

The positive-$\mu$ assumption is necessary for the printed sign convention. If $\mu <0$, the real length is $\hslash /\sqrt{2m|\mu |}$, and the dimensionless linear term is $-1$, with interaction coefficients divided by $|\mu |$; the displayed equation with linear term $+1$ does not follow. At $\mu =0$ this scaling is undefined. These are limitations of this particular nondimensionalization, not a claim that all cubic-quintic bulk states have positive chemical potential.

A straight quantum vortex with integer winding number $\mathcal{N}$ has $\tilde{\psi }=R(r)e^{i\mathcal{N}\theta }$, independent of $z$. Here $r$ is measured in the unit of distance from the preceding part. The polar Laplacian supplies the centrifugal term, givingFor $\mathcal{N}\ne 0$, regularity requires $R\sim c{r}^{|\mathcal{N}|}$ near the axis; the bulk condition is $R\to 1$. The full wavefunction tends to $e^{i\mathcal{N}\theta }$ rather than one. This is the constant far-field phase excludes net vortex winding distinction: the equation remains applicable to a vortex, but the nonwinding boundary condition must be interpreted as an amplitude condition.

Let $F(R)=R-\alpha R^3-\beta R^5$. Since $\alpha +\beta =1$, one has $F(1)=0$ and $F^{\prime }(1)=1-3\alpha -5\beta =-2(1+\beta )$. Insert $R=1-p/r^2+\cdots$. Its derivative terms start at order $r^{-4}$, whileCancellation of the order-$r^{-2}$ terms proves the cubic-quintic quantum vortex tailFor a stable nondegenerate bulk state in this scaling, $1+\beta =n_0(V_0+2W_0{n}_0)/\mu >0$. At $\beta =-1$ the amplitude restoring derivative vanishes, and an inverse-square perturbation cannot cancel the centrifugal term, so the stated expansion fails. For zero winding the uniform solution has $p=0$.

To first order, $|1+u+iv{|}^2=1+2u$ and $(1-|\psi {|}^2)\psi =-2u$. Equating real and imaginary parts of the Gross–Pitaevskii equation therefore givesDifferentiate the second equation in time and use the first to eliminate $v$:For a Fourier mode with $k=|\mathbf{k}|$, the dispersion relation isThis is the normalized Bogoliubov spectrum. At small wavenumber the positive branch has sound speed ${\lim }_{k\to 0}{\omega }_{+}(k)/k=1/\sqrt{2}$; at large wavenumber it approaches the free-particle quadratic dispersion. The linear frequencies are real, confirming stability of the repulsive uniform bulk.

Introduce the translating coordinate $Z=z-Ut$ and put $\psi (\mathbf{x},t)=\Psi (x,y,Z)$. Then ${\psi }_t=-U{\Psi }_Z$. Substitution in the Gross–Pitaevskii equation gives the stationary Gross–Pitaevskii solitary wave equationThe complex bulk field is fixed by $\Psi \to 1$ far from the localized disturbance. Spatial derivatives in this equation are taken in the translating frame. There is no additional uniform temporal phase here because the normalization has already made the background $\psi =1$ stationary.

Use decaying variations with the same complex bulk field one, so that integration by parts produces no variation boundary term. The functional derivatives of energy and renormalized momentum of a condensate areFor the second identity, variation of the two momentum terms and integration of the ${\partial }_z\delta {\psi }^{\ast }$ term each supply $(2i{)}^{-1}{\psi }_z$, so together they give $-i{\psi }_z$. It follows thatis exactly the stationary Gross–Pitaevskii solitary wave equation. Thus each solitary wave is a constrained energy critical point with multiplier $U$.

Let $s$ parametrize a differentiable family of these waves. Apply the critical-point identity to the variation ${\partial }_s\psi$, keeping the multiplier fixed at its value for that member of the family. Including the complex conjugate variation givesHence the energy–momentum slope of a solitary wave iswherever $dp/ds\ne 0$. At a momentum turning point the parametrized identity $dE/ds=Udp/ds$ remains the correct statement; a globally single-valued energy as a function of momentum is not needed. Differentiating $E-U(s)p$ as a composite function would include an extra $-p{U}^{\prime }$ term, so that composite derivative must not be set to zero.

Split the solitary-wave energy into longitudinal, transverse and interaction contributions,Consider the longitudinal dilation ${\psi }_{\lambda }(x,y,z)=\psi (x,y,\lambda z)$ for $\lambda >0$. Changing variables to $Z=\lambda z$ givesThe renormalized momentum of a condensate is unchanged: its single $z$ derivative contributes $\lambda$, cancelling the measure factor ${\lambda }^{-1}$. Since the wave is a critical point of $E-Up$, differentiating this admissible scaling variation at $\lambda =1$ givesTherefore the longitudinal scaling identity for a condensate solitary wave yieldsThis directional Pohozaev identity assumes the finite energy and sufficient decay to justify the dilation variation; it can equivalently be obtained with cutoffs followed by a limit. It is an identity for solitary-wave critical points, not for arbitrary localized fields.

Write $g=V_0$ for the effective coupling and keep $N$ for the number obtained by the stated two-dimensional normalization. A repulsive interaction, $g>0$, permits the Thomas–Fermi approximation for a condensate: density-gradient kinetic energy is small compared with the trapping and interaction terms over most of the cloud. The stationary Gross–Pitaevskii equation then givesand $n_{\mathrm{TF}}=0$ outside the cloud, up to the narrow edge region. HereNormalize the parabolic number density directly:Consequently the two-dimensional Thomas–Fermi condensate parameters areWith $a_{\mathrm{ho}}=\sqrt{\hslash /(m\omega )}$ and central healing length $\xi =\hslash /\sqrt{2m\mu }$, the bulk approximation requiresIndeed $R/\xi =2\mu /(\hslash \omega )\gg 1$. In the coupling convention printed in the paper, formal substitution of $g=4\pi {\hslash }^{2}a/m$ givesThis is the requested relationship in that convention; $\omega$ sets the physical size and chemical potential, while the interaction-strength ratio to $\hslash \omega$ is independent of $\omega$.

There is a dimensional convention to make explicit. A physical three-dimensional scattering length $a$ makes $4\pi {\hslash }^{2}a/m$ a three-dimensional contact coupling, whereas a genuinely two-dimensional particle-number normalization requires coupling units of energy times area. For an axial factor $\chi (z)$ normalized to one, the effective two-dimensional contact coupling is $g_{2D}=(4\pi {\hslash }^{2}a/m)\int |\chi {|}^{4}dz$. For a Gaussian of width $a_z$ this is $4\pi {\hslash }^{2}a/(m\sqrt{2\pi }{a}_z)$, and the physical TF condition becomes $Na/a_z\gg 1$ up to numerical constants. Alternatively the printed coupling can be retained if $N$ is a line density. The paper supplies no axial scale, so its $Na$ expressions require that effective-unit or line-density convention; the formulas in terms of $g$ above are unambiguous. The approximation fails inside vortex cores and at the cloud edge, and it is not a repulsive TF ground-state construction for $g\le 0$.

Use the central number density $n_c$, radius $R$ and central healing length $\xi$ from the preceding part. The vortex energy means the excess over the no-vortex ground state at fixed particle number. Choose an intermediate matching radius $\xi \ll L\ll R$. Inside $L$, the trapped number density differs little from $n_c$, so the resolved uniform vortex core givesOutside the core, the singly quantized superfluid velocity is $u_{\theta }=\hslash /(mr)$. Its leading flow kinetic energy in the parabolic background is thereforeAdding the regions cancels the arbitrary matching radius. Sending $L/R\to 0$ while $\xi /L\to 0$ gives the matched vortex energy in a parabolic condensateThe $-1/2$ comes from the decrease of background number density away from the centre. It would be missed by treating the whole trap as uniform. To this accuracy, using the unperturbed TF density in the outer region is appropriate at fixed $N$: the first variation of trapping plus interaction energy is $\mu \delta N$ and therefore vanishes after the small normalization adjustment. The resolved core energy supplies the finite core correction $L_{01}$, while remaining cloud-edge and density-relaxation corrections vanish in the small-core TF limit. This is an asymptotic energy estimate, not an exact formula at finite $R/\xi$.

For an axisymmetric centred unit quantum vortex, write $\psi =f(r)e^{i\theta }$, including its radial density depletion in $f$. The axial angular momentum operator is ${\hat{L}}_z=-i\hslash {\partial }_{\theta }$, so every particle in this single-winding state has eigenvalue $\hslash$. HenceThis is exact at the specified particle number and does not require neglecting the core or using the parabolic TF profile.

Let $\mathcal{N}\in \mathbb{Z}$ be the vortex winding number, distinct from the particle number $N$. Outside its core, the superfluid velocity is $u_{\theta }=\hslash \mathcal{N}/(mr)$. The outer flow energy is therefore multiplied by ${\mathcal{N}}^2$, while the resolved core contribution contains the appropriate constant $L_{0|\mathcal{N}|}$. The same matching integration givesThe core equation depends on ${\mathcal{N}}^2$, so reversing circulation leaves this energy unchanged. On the other hand, $-i\hslash {\partial }_{\theta }[f(r)e^{i\mathcal{N}\theta }]=\hslash \mathcal{N}f(r)e^{i\mathcal{N}\theta }$, giving the exact resultNegative winding reverses angular momentum; zero winding has zero excess vortex energy and zero angular momentum. The energy approximation applies to each fixed nonzero winding in the TF limit. More generally its core radius, of order $|\mathcal{N}|\xi$, must be much smaller than $R$ so that a matching interval exists. It is not valid uniformly for arbitrary growing $|\mathcal{N}|$ at fixed cloud size. The leading quadratic energy cost also explains why multiply quantized vortices can lower their energy by separating into singly quantized vortices; this energy comparison does not by itself establish a particular dynamical splitting rate.

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.