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  "body": "\n## Statement\n\nLet $f \\in L^2(\\R)$, and let $\\hat f$ be its Fourier transform on $L^2(\\R)$, normalized so that $\\hat f(\\xi) = \\int_\\R f(x) e^{-2\\pi i x \\xi}\\, dx$ for integrable $f$.\nThen for all real numbers $a$ and $b$,\n$$\\left( \\int_\\R (x - a)^2 \\abs{f(x)}^2\\, dx \\right) \\left( \\int_\\R (\\xi - b)^2 \\abs{\\hat f(\\xi)}^2\\, d\\xi \\right) \\ge \\frac{\\norm{f}_2^4}{16 \\pi^2}.$$\nEquality holds if and only if $f(x) = C e^{2\\pi i b x} e^{-c (x - a)^2}$ for some $C \\in \\C$ and $c \u003e 0$.\n\n## Remarks\n\nFor a Schwartz function $f$ and $a = b = 0$, integrate by parts: $\\int \\abs{f}^2\\, dx = -\\int x\\, \\frac{d}{dx}\\abs{f(x)}^2\\, dx = -2 \\operatorname{Re} \\int x f(x) \\overline{f'(x)}\\, dx$.\nThe [[Cauchy–Schwarz inequality|0RFWW56Y22]] bounds this by $2 \\norm{x f}_2 \\norm{f'}_2$.\nBy the [[differentiation rule|HHNR31W3WD]] and the [[Plancherel theorem|HPZHRZ0502]], $\\norm{f'}_2 = 2\\pi \\norm{\\xi \\hat f}_2$.\nSquaring gives the inequality; translation in $x$ and modulation by $e^{2\\pi i b x}$ give general $a$ and $b$, and a density argument gives all $f$ in [[$L^2(\\R)$|SASP14MA1Z]].\nEquality in Cauchy–Schwarz forces $f' = -2c x f$, whose solutions are Gaussians.\n\nIf one of the two integrals is infinite and $f \\ne 0$, then the other is positive, so the left side is $+\\infty$ and the inequality holds.\n\nExample: for $f(x) = e^{-\\pi x^2}$, one has $\\hat f = f$, $\\norm{f}_2^2 = 2^{-1/2}$ and $\\int x^2 e^{-2\\pi x^2}\\, dx = \\frac{1}{4\\pi} 2^{-1/2}$, so the left side equals $\\frac{1}{16\\pi^2} \\cdot \\frac12 = \\frac{\\norm{f}_2^4}{16\\pi^2}$.\n\n## Explanations\n\n### L1 human\n\nIf $\\abs{f}^2$ is a probability density, its spread around a point $a$ and the spread of $\\abs{\\hat f}^2$ around a point $b$ cannot both be small.\nTheir product is at least $1/(16\\pi^2)$.\nIn quantum mechanics, position and momentum are related by this Fourier transform, and the inequality becomes Heisenberg's uncertainty principle.\n\n### L2 human\n\nThe proof is a commutator argument: the operators of multiplication by $x$ and of differentiation do not commute, and Cauchy–Schwarz turns this into a lower bound.\nThe same scheme gives the Robertson uncertainty relation for any two self-adjoint operators.\nThe inequality also follows from the entropic uncertainty principle, since among densities with a given variance the Gaussian has the largest entropy.\nIn $\\R^n$ the analogous bound is $\\norm{\\abs{x} f}_2 \\norm{\\abs{\\xi} \\hat f}_2 \\ge \\frac{n}{4\\pi} \\norm{f}_2^2$.\n",
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