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  "body": "\n## Definition\n\nLet $\\psi_n(x) = (2^n n! \\sqrt{\\pi})^{-1/2} H_n(x) e^{-x^2/2}$, $n \\ge 0$, be the normalized [[Hermite functions|52VS07H6GK]]; they form an orthonormal basis of $L^2(\\R)$, the [[$L^2$ space|SASP14MA1Z]].\nLet $\\alpha \\in \\R$.\nThe *fractional Fourier transform* of angle $\\alpha$ is the [[unitary operator|MM0EJ2GDE4]] $\\mathcal{F}_\\alpha$ on $L^2(\\R)$ given by\n$$\\mathcal{F}_\\alpha \\Big( \\sum_{n \\ge 0} c_n \\psi_n \\Big) = \\sum_{n \\ge 0} e^{-i n \\alpha}\\, c_n \\psi_n$$\nfor every square-summable sequence $(c_n)$.\n\n## Remarks\n\nLet $\\mathcal{F} f(\\xi) = \\frac{1}{\\sqrt{2\\pi}} \\int_\\R f(x) e^{-i x \\xi}\\, dx$ be the unitary [[Fourier transform|DPVM34ADZV]].\nEach $\\psi_n$ satisfies $\\mathcal{F} \\psi_n = (-i)^n \\psi_n$, so $\\mathcal{F}_{\\pi/2} = \\mathcal{F}$.\nAlso $\\mathcal{F}_0 = \\id$, $\\mathcal{F}_\\pi f(x) = f(-x)$, $\\mathcal{F}_\\alpha \\mathcal{F}_\\beta = \\mathcal{F}_{\\alpha + \\beta}$, and $\\mathcal{F}_\\alpha$ depends only on $\\alpha$ modulo $2\\pi$.\nFor $0 \u003c \\alpha \u003c \\pi$ it is an integral operator: for $f$ integrable and square-integrable,\n$$\\mathcal{F}_\\alpha f(\\xi) = \\sqrt{\\frac{1 - i \\cot\\alpha}{2\\pi}} \\int_\\R \\exp\\Big( \\frac{i}{2} \\cot\\alpha\\, (x^2 + \\xi^2) - \\frac{i\\, x \\xi}{\\sin\\alpha} \\Big) f(x)\\, dx,$$\nwith the principal branch of the square root; at $\\alpha = \\pi/2$ this is the formula for $\\mathcal{F}$.\nSome authors use the order $a = 2\\alpha/\\pi$ instead of the angle, so that $\\mathcal{F}$ has order $1$.\n\nExample: $f = \\psi_0 = \\pi^{-1/4} e^{-x^2/2}$.\nIts only coefficient is $c_0 = 1$, and $e^{-i \\cdot 0 \\cdot \\alpha} = 1$, so $\\mathcal{F}_\\alpha \\psi_0 = \\psi_0$ for every $\\alpha$.\n\nNon-example: the transform $f \\mapsto \\int_\\R f(x) e^{-i x \\xi}\\, dx$ without the factor $\\frac{1}{\\sqrt{2\\pi}}$ is not $\\mathcal{F}_{\\pi/2}$.\nIt sends $\\psi_n$ to $\\sqrt{2\\pi}\\, (-i)^n \\psi_n$, so it multiplies the coefficient $c_n$ by a number of modulus $\\sqrt{2\\pi} \\ne 1$, and it is not unitary.\n",
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