Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 1

The Hamiltonian functional isTreating the field and its complex conjugate as independent variables, and integrating the gradient term by parts with periodic, decaying or fixed-field boundary conditions, givesThus $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$ is the Gross–Pitaevskii equation. The factor $1/2$ in the interaction energy avoids counting each pair twice.

At prescribed particle number, minimize $H$ subject to $N=\int |\psi {|}^2{d}^{3}x$. Introducing the chemical potential as a Lagrange multiplier, the variation of $H-\mu N$ vanishes at a ground-state profile ${\psi }_0$, provingThe corresponding time-dependent field is $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$. A solution of this stationary equation is only a candidate ground state; the ground state is the minimizer among all fields satisfying the normalization. For the uniform repulsive gas in a periodic volume, the gradient energy is nonnegative and $\int n^2{d}^{3}x\ge N^2/\mathcal{V}$, so the constant-density, constant-phase state is indeed the minimum. The uniform-background solutions below assume $U>0$ and a positive background density.

Let the uniform background number density be $n_{\infty }>0$ and set $\mu =U{n}_{\infty }$. The subtraction of this chemical potential requires a phase rotation as well as a rescaling:Substitution into the physical Gross–Pitaevskii equation givesConsequently the requested coefficients follow fromThe length unit is the specified healing length. Removing tildes produces the paper's dimensionless equation; in particular the coefficient of its time derivative fixes $t_0$, not its reciprocal. Without the phase rotation the linear $+\Psi$ term would not appear.

The dimensionless ground state is $\Psi =e^{i\chi }$ for any constant phase $\chi$, conventionally $\Psi =1$. Its subtracted energy is proportional to $\int [|\nabla \Psi {|}^2+(1-|\Psi {|}^2{)}^2/2]d^{3}x$, which is nonnegative and vanishes for this state. In physical variables it has density $n_{\infty }$ and phase rotating at $\mu /\hslash$.

Use translating coordinates $Z=z-vt$ and a profile $\psi (\mathbf{x},t)=\Phi (x,y,Z)$; the background condensate remains at rest in the laboratory. Then ${\psi }_t=-v{\Phi }_Z$, and the Gross–Pitaevskii solitary wave equation isThis is a coordinate translation of a traveling solution, rather than an additional phase boost to change the background flow.

Set $\Phi =1+f+ig$ with small real $f,g$. Since $|\Phi {|}^2=1+2f+O(f^2+g^2)$, the real and imaginary linearized equations areFor a longitudinal sinusoid, choose $f=A\cos (kZ)$ and $g=B\sin (kZ)$. Their coefficients obeyA nonzero solution therefore requires $4v^2=k^2+2$, giving the positive-direction phase velocityEquivalently the time-dependent linear Bogoliubov spectrum is ${\omega }^2=k^2(k^2+2)/4$ and $v=\omega /k$. Its long-wavelength sound speed is $1/\sqrt{2}$, which becomes $\sqrt{\mu /m}$ in physical units using ${\ell }_0/t_0=\sqrt{2\mu /m}$. For a general wavevector $\mathbf{k}$, the stationary resonance is $4v^2{k}_z^2=k^2(k^2+2)$; specifying only its magnitude does not determine the velocity unless its direction is also fixed. This sinusoidal dispersion describes linear radiation about the background, not an amplitude-independent dispersion law for every nonlinear solitary wave; localized subsonic waves do not require a real oscillatory far-field wavenumber.

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.