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  "body": "\n## Definition\n\nLet $d \\ge 1$, let $\\hbar, m \u003e 0$ and $g \\in \\R$ be constants, and let $V \\colon \\R^d \\to \\R$ be a given function, the external potential.\nThe *Gross–Pitaevskii equation* is the equation\n$$i \\hbar \\, \\partial_t \\psi = - \\frac{\\hbar^2}{2m} \\Delta \\psi + V(x) \\psi + g \\abs{\\psi}^2 \\psi$$\nfor an unknown function $\\psi \\colon \\R \\times \\R^d \\to \\C$, $(t, x) \\mapsto \\psi(t, x)$, where $\\Delta$ is the [[Laplacian|0E0B6RA8K0]] in $x$.\nThe constant $g$ measures the strength of the interaction, repulsive for $g \u003e 0$ and attractive for $g \u003c 0$.\n\n## Remarks\n\nFor $V = 0$ it is, after scaling, the cubic [[nonlinear Schrödinger equation|GEJHRXC176]].\nA standing wave $\\psi(t, x) = e^{-i\\mu t/\\hbar} \\phi(x)$ solves it exactly when $\\phi$ solves the stationary equation $\\mu \\phi = -\\tfrac{\\hbar^2}{2m} \\Delta \\phi + V \\phi + g \\abs{\\phi}^2 \\phi$, and $\\mu$ is the chemical potential.\n\nExample: for $V = 0$ and $n \u003e 0$, the constant state $\\phi = \\sqrt{n}$ solves the stationary equation with $\\mu = g n$, since $\\Delta \\phi = 0$ and $g \\abs{\\phi}^2 \\phi = g n \\phi$.\n\nNon-example: if $V$ is not constant, then no nonzero constant $\\phi$ solves the stationary equation, because $\\mu \\phi = (V(x) + g \\abs{\\phi}^2) \\phi$ would force $V$ to be constant.\n\nThe ground state minimises the Gross–Pitaevskii energy $\\int \\big( \\tfrac{\\hbar^2}{2m} \\abs{\\nabla \\phi}^2 + V \\abs{\\phi}^2 + \\tfrac{g}{2} \\abs{\\phi}^4 \\big) dx$ under the constraint $\\int \\abs{\\phi}^2 \\, dx = N$, the number of particles.\n",
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