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For $i{\psi }_t=-D{\nabla }^2\psi +[V(n)+i\Gamma (n)]\psi$, with real $D,V,\Gamma$ and $\psi =\sqrt{n}{e}^{iS}$, the continuity equation isMultiply the equation by ${\psi }^{\ast }$ and subtract its complex conjugate multiplied by $\psi$ to obtain this identity; real interaction and potential terms cancel. Thus the current velocity is $\mathbf{j}/n=2D\nabla S$. With dimensional kinetic operator $-{\hslash }^2{\nabla }^2/(2m)$ and left side $i\hslash {\psi }_t$, this gives the physical superfluid velocity $(\hslash /m)\nabla S$. In normalized equations with $D=1/2$ or $D=1$, the current velocity is respectively $\nabla S$ or $2\nabla S$. The distinction fixes the radial-flow slope of a driven quantum vortex.

If a continuous nonzero complex field tends uniformly to one on large circles, its values on a sufficiently large circle lie in a disk about one that excludes zero. That disk is contractible, so the field's winding number on that circle is zero. A lone unit-charge quantum vortex therefore cannot satisfy the literal boundary $\psi \to 1$. For a vortex it is the wave amplitude, rather than the entire complex field, that can approach one; $f(r)e^{i\theta }$ gives a simple illustration of direction-dependent far-field complex argument.

For a cubic-quintic Gross–Pitaevskii equation with normalized bulk amplitude one, a straight quantum vortex of integer winding number $\mathcal{N}$ has radial equation $R^{\prime \prime }+R^{\prime }/r-{\mathcal{N}}^{2}R/r^2+(1-\alpha R^2-\beta R^4)R=0$. With $\alpha +\beta =1$, the derivative of the nonlinear term at $R=1$ is $-2(1+\beta )$. Balancing the order-$r^{-2}$ centrifugal term with the bulk amplitude restoring term proves the displayed tail for a stable nondegenerate bulk state, $1+\beta >0$. A vortex's amplitude tends to one; its full complex field tends to $e^{i\mathcal{N}\theta }$, not a direction-independent constant. At $\beta =-1$ the stated inverse-square expansion cannot cancel the centrifugal term.

To isolate finite size from quantum vortex penetration, keep the same weak-field London equations but replace the half-space by an infinite slab $-d/2\le z\le d/2$. Put equal tangential magnetic fields $B_0\hat{\mathbf{x}}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed current density areThe hyperbolic cosine profile has equal boundary values, while the electric currents on the two sides have opposite signs. In particular,For the stipulated thickness $d=2{\lambda }_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a Meissner state. For $d\gg {\lambda }_L$, the center magnetic field is exponentially small and the mean is about $2{\lambda }_L{B}_0/d$, reflecting the two surface layers. For $d\ll {\lambda }_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their demagnetizing field and edge geometry accounted for; the slab calculation only isolates the effect of thickness.

For the normalized conservative equation $-2i{\psi }_t={\nabla }^2\psi +(1-|\psi {|}^2)\psi$, a localized disturbance translating at speed $v$ obeys ${\nabla }^2\psi +(1-|\psi {|}^2)\psi -2iv{\psi }_x=0$ in translating coordinates. Its complex-field boundary is $\psi \to 1$. This permits zero-net-winding vortex pairs, but excludes a lone quantum vortex. The usual subsonic far field has $\psi -1=O(r^{-1})$ in two dimensions.

In healing-length units a straight quantum vortex obeys $f^{\prime \prime }+f^{\prime }/r-{\mathcal{N}}^{2}f/r^2+(1-f^2)f=0$ and $f\to 1$. Inserting $f=1-a/r^2+\cdots$ makes its order-$r^{-2}$ balance $-{\mathcal{N}}^2+2a=0$, proving the tail. The physical healing length convention must be stated; the deficit of density $1-f^2$ has twice the leading coefficient of the amplitude deficit $1-f$.

An experimental physicist who coauthored a 2000 study of quantum vortex formation in a stirred rubidium-87 Bose-Einstein condensate. A focused laser beam stirred the gas inside a magnetic trap. Above a threshold rotation frequency, vortices appeared; faster stirring produced multiple vortices. Measuring the single-vortex lifetime after stirring stopped made rotational excitations a controlled probe of a dilute quantum fluid.^[ref]

The Landau criterion asks whether creating an excitation can lower a moving superfluid's energy. For a weakly coupled, heavy obstacle moving through a uniform liquid initially at zero temperature, let $ϵ(p)>0$ be the excitation energy in the fluid rest frame, with momentum magnitude $p$. Transferring momentum $\mathbf{p}$ from an obstacle moving at velocity $\mathbf{v}$ supplies, to leading order in its negligible recoil, energy $\mathbf{v}\cdot \mathbf{p}$. Emission requiresFor an isotropic dispersion relation, the easiest direction is parallel to $\mathbf{v}$, giving the necessary kinematic conditionAn infimum includes cases in which the limiting slope occurs only as $p\to 0$. A finite-mass obstacle of mass $M$ instead loses $\mathbf{v}\cdot \mathbf{p}-p^2/(2M)$, so the corresponding emission inequality includes the extra positive recoil term $p^2/(2M)$ on the left. This is an energy and momentum balance, not a calculation of an emission rate.

The Landau criterion excludes spontaneous emission into the specified spectrum below its threshold. Above it, kinematic permission alone does not guarantee strong coupling or rapid decay. It is not a universal observed flow threshold: walls, channel geometry and quantum vortex nucleation can enable other decay routes, and finite temperature permits scattering from existing excitations. Using only bulk phonons and rotons leaves those processes outside the model.

Landau's 1941 energy and momentum argument used a roton approximation different from the finite-momentum minimum illustrated below.^[ref]

Landau received his Nobel Prize in Physics in 1962 while hospitalized in Moscow after a car accident.^[ref]

For a deliberately simple Bose-Einstein condensation comparison, stipulate an ideal free-particle excitation law $ϵ=p^2/(2m)$. Then $ϵ/p=p/(2m)$ has infimum zero. Macroscopic one-particle occupation therefore supplies no positive emission threshold by this argument alone. By contrast a linear phonon law $ϵ=cp$ yields $v_L=c$. The dispersion relation and the collective response contain information that the occupation criterion alone does not supply; Bose-Einstein condensation and thermodynamic superfluidity must be tested separately.

To check the picture algebraically, on the stipulated roton branch,Thus tangency occurs at $q_L=\sqrt{4.3}$, slightly above the roton energy minimum. At an energy minimum $e^{\prime }=0$, but a tangent through the origin needs $e^{\prime }=e/q>0$. Minimizing energy and minimizing the energy-to-momentum ratio are different operations. This geometric construction makes the criterion useful even when the full dispersion relation is supplied numerically rather than by a simple formula.

For a bulk type-II superconductor with negligible demagnetizing field, raising the applied magnetic field intensity past the lower critical value $H_{c1}$ favors a superconducting mixed state containing quantum vortex lines; superconductivity survives until the upper critical value $H_{c2}$. The magnetic field extends around each vortex over a penetration scale, not solely inside its core. Thus a nonzero average magnetic flux density can coexist with superconductivity. Below $H_{c1}$, a type-II superconductor also supports a Meissner state; a vortex-free state above the equilibrium entry field may instead be metastable. ^[ref]

Real preparation histories may leave quantum vortex lines held in place by material inhomogeneities, retaining magnetic flux and producing hysteresis. This flux pinning restricts access to equilibrium: it is different from the arbitrary frozen interior magnetic field of the ideal perfect conductor model. Harvard's vortex-imaging work illustrates strong pinning to bulk disorder in a type-II superconductor. ^[ref] Separately, flux conservation in a superconducting loop and fluxoid and flux quantization in a superconductor explain why a hole can retain magnetic flux. Neither bore retention, vortex penetration nor weak screening in a thin sample should be silently identified with the equilibrium Planar London screening calculation.

For a centered quantum vortex in a two-dimensional Thomas–Fermi condensate, use an intermediate radius ${\xi }_{\mathrm{core}}\ll L\ll R$. The approximately uniform core contributes $\pi n_c{\hslash }^2{\mathcal{N}}^2[\log (L/\xi )+L_{0|\mathcal{N}|}]/m$. The outer superfluid velocity is $\hslash \mathcal{N}/(mr)$, so its kinetic contribution is $(\pi {\hslash }^2{\mathcal{N}}^2/m){\int }_L^R{n}_c(1-r^2/R^2)dr/r$. Integration cancels $L$ and produces the constant $-1/2$. The displayed result is the matched small-core approximation at fixed particle number, not a formula for arbitrarily large winding at fixed cloud size. The angular momentum is exactly $L_z=N\hslash \mathcal{N}$ for a normalized axisymmetric single-winding state.

The Meissner effect is the equilibrium expulsion of a sufficiently weak, static magnetic field from the bulk of a material exhibiting superconductivity. Expulsion includes a magnetic field that was already present before cooling: zero electrical resistance alone does not predict this. The Field-cooled Meissner experiment tests that distinction. A finite London penetration depth means that the magnetic field and screening electric current occupy a surface layer, rather than disappearing at a mathematical boundary.^[ref]

This makes superconductivity useful for shielding weak magnetic fields, and makes its magnetic response an independent test of the phase. The striking result is that the same screened equilibrium can be reached by two different preparation histories. The assumptions matter: Finite-thickness Meissner screening can leave an appreciable magnetic field throughout a small sample, and the Limits of Meissner screening distinguish quantum vortex penetration and retained magnetic flux from this equilibrium state.

A Meissner state is an equilibrium state of superconductivity without quantum vortex penetration, with screening electric currents responding to the applied magnetic field. In a solid, simply connected sample much thicker than the London penetration depth, the deep interior magnetic field is approximately zero. This is a statement about a phase and a geometry, not a rule that every sample exhibiting superconductivity always has $\mathbf{B}=0$. In particular, the Meissner state exists for sufficiently weak magnetic fields in both type-I superconductors and type-II superconductors. ^[ref]

For a straight unit-winding quantum vortex, the real radial amplitude satisfiesIn the frame translating with the ring, put $Z=z-ut$ and $\Phi =e^{i\theta }(R+\varphi )$, where $\varphi$ is independent of the angular coordinate. The cylindrical Laplacian has the exact identityAfter inserting this into the translating Gross–Pitaevskii equation, subtract the straight-vortex equation. Because $R$ is real,and direct expansion of the nonlinear difference givesThus the exact ring-on-line perturbation equation, including its quadratic and cubic terms, isThe two appearances of $R|\varphi {|}^2$ are responsible for the coefficient $2$ multiplying ${\varphi }^{\ast }$ inside the bracket. No small-amplitude approximation is needed for this identity.

Use the stationary quantum vortex fieldwith no dependence on the coordinate along the vortex line. The polar Laplacian contributes $R^{\prime \prime }+R^{\prime }/r-{\mathcal{N}}^{2}R/r^2$. ConsequentlyA regular vortex with nonzero integer winding number has $R\sim C{r}^{|\mathcal{N}|}$ near the origin. The angular phase gives the superfluid velocity $\hslash \mathcal{N}/(mr)$ outside the core. For $s\ne 0$, the exterior amplitude responds to the periodically varying potential; it is not generally a constant bulk value. The radial potential here is not a Cartesian translation-invariant optical lattice, so the Cartesian Bloch theorem is not an exact theorem for its full radial equation.

A nonzero radial lattice prevents a unique universal curve without values of the depth, period, chemical potential, coupling and winding number. An explicit example of the requested profile is shown below for $\mathcal{N}=1$, repulsive coupling, $d=4{\ell }_0$ and $s{E}_R/\mu =0.6$; since $E_R/\mu ={\pi }^2/16$, this corresponds to $s\simeq 0.973$.

The plotted dimensionless amplitude is $f=R/\sqrt{\mu /U}$. The generator solves the nonlinear radial equationby finite differences and damped Newton iteration; $r$ here is in healing length units. It imposes regularity $f(0)=0$ and a zero radial derivative at $r=32$, well outside the plotted interval $0\le r\le 20$. The exterior boundary is at a potential minimum, and its placement is checked by an extended-domain calculation.

The quantum vortex core still forces the amplitude to vanish at the origin. Farther out, amplitude minima occur near potential maxima, where repulsion and trapping reduce the local equilibrium density, and amplitude maxima occur near the low-potential annuli. Gradient energy smooths the modulation, so the profile does not exactly follow the local estimate $f_{\mathrm{TF}}=\sqrt{[1-V/\mu {]}_{+}}$. It approaches a periodically modulated exterior background rather than the homogeneous limit one; the dashed zero-lattice curve is included solely as a reference. At sufficiently deep lattice barriers the density troughs can become much more pronounced than in this example.

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 $p=0$.

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.

The axial angular momentum per unit volume is $mn(r)r{u}_{\theta }(r)=\hslash n(r)$ for a singly quantized positive quantum vortex. Thus every particle in the prescribed shell contributes $\hslash$:This yieldsThis is $\hslash$ times the shell particle number. For a wall-truncated cloud with $R=R_0<R_r$, the corresponding result is $L_z=\pi H_0\hslash n_0[R^2-a_0^2-(R^4-a_0^4)/(2R_r^2)]$. Reversing the quantized circulation reverses $L_z$ without changing the flow kinetic energy. For winding number $q$, $L_z$ is multiplied by $q$.

Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.