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  "body": "\n## Statement\n\nLet $n \\ge 1$ and $a \u003e 0$, and let $f(x) = e^{-\\pi a \\abs{x}^2}$ for $x \\in \\R^n$.\nLet $\\hat f(\\xi) = \\int_{\\R^n} f(x) e^{-2\\pi i x \\cdot \\xi}\\, dx$ be its [[Fourier transform|DPVM34ADZV]].\nThen\n$$\\hat f(\\xi) = a^{-n/2}\\, e^{-\\pi \\abs{\\xi}^2 / a} \\quad \\text{for every } \\xi \\in \\R^n.$$\nIn particular, for $a = 1$ the function $e^{-\\pi \\abs{x}^2}$ is its own Fourier transform.\n\n## Remarks\n\nSince $e^{-\\pi a \\abs{x}^2} = \\prod_j e^{-\\pi a x_j^2}$, the integral splits into $n$ one-dimensional integrals, so it suffices to take $n = 1$.\nFor $n = 1$ and $a = 1$, put $g = \\hat f$.\nBy the [[rules for differentiation and multiplication|HHNR31W3WD]], the identity $f'(x) = -2\\pi x f(x)$ becomes $g'(\\xi) = -2\\pi \\xi g(\\xi)$.\nSo $g(\\xi) = g(0) e^{-\\pi \\xi^2}$, and $g(0) = \\int_\\R e^{-\\pi x^2}\\, dx = 1$ is the Gaussian integral.\nThe case of general $a$ follows by the change of variables $x \\mapsto x / \\sqrt{a}$.\n\nThe statement shows a trade-off: a narrow Gaussian (large $a$) has a wide transform (variance proportional to $a$).\nFor every $a \u003e 0$ the product of the two widths is as small as possible, so every Gaussian is an equality case of the Heisenberg inequality.\n\n## Explanations\n\n### L1 human\n\nThe bell curve keeps its shape under the Fourier transform.\nSqueezing the bell in space spreads it out in frequency by the same factor, and the reverse.\nThe standard bell $e^{-\\pi x^2}$ is exactly fixed.\n\n### L2 human\n\nThis computation is the base case for many proofs.\nThe Fourier inversion theorem, the Plancherel theorem and the solution of the heat equation all use the Gaussian as an approximate identity whose transform is known.\nIt also gives the characteristic function of the normal distribution after the change of variables $\\xi = -t / (2\\pi)$.\nGaussians are the extremal functions in the Babenko–Beckner and the Heisenberg inequalities.\n",
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