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  "body": "\n## Definition\n\nLet $M = \\begin{pmatrix} a \u0026 b \\\\ c \u0026 d \\end{pmatrix}$ be a real matrix with $ad - bc = 1$, an element of the [[special linear group|0V9PMN0RBH]] $SL(2, \\R)$, and assume $b \\ne 0$.\nFix a square root $\\sqrt{1/(ib)}$ of the complex number $1/(ib)$.\nThe *linear canonical transformation* with matrix $M$ sends an integrable function $f \\colon \\R \\to \\C$ to the function\n$$L_M f(u) = \\sqrt{\\frac{1}{i b}}\\; e^{i\\pi \\frac{d}{b} u^2} \\int_\\R e^{-2\\pi i \\frac{u t}{b}}\\, e^{i\\pi \\frac{a}{b} t^2} f(t)\\, dt, \\qquad u \\in \\R.$$\n\n## Remarks\n\nThe map $L_M$ is the composite of a multiplication by the chirp $e^{i\\pi a t^2 / b}$, a [[Fourier transform|DPVM34ADZV]] rescaled by $1/b$, a multiplication by $e^{i\\pi d u^2 / b}$, and the factor $\\sqrt{1/(ib)}$ of modulus $\\abs{b}^{-1/2}$.\nThe rescaled Fourier transform multiplies the $L^2$ norm by $\\abs{b}^{1/2}$, the factor divides it by the same amount, and the chirps do not change it.\nSo $L_M$ preserves the $L^2$ norm, and $L_M$ extends to a [[unitary operator|MM0EJ2GDE4]] of $L^2(\\R)$, the [[$L^2$ space|SASP14MA1Z]].\nFor $b = 0$ the transform is defined by $L_M f(u) = \\sqrt{d}\\, e^{i\\pi c d u^2} f(du)$, a scaling followed by a chirp multiplication.\nFor two matrices one has $L_{M_1} L_{M_2} = \\pm L_{M_1 M_2}$; the sign ambiguity comes from the choice of square roots.\n\nExample: $M = \\begin{pmatrix} 0 \u0026 1 \\\\ -1 \u0026 0 \\end{pmatrix}$, which has determinant $0 \\cdot 0 - 1 \\cdot (-1) = 1$.\nHere $a = d = 0$ and $b = 1$, so $L_M f(u) = \\sqrt{1/i} \\int_\\R e^{-2\\pi i u t} f(t)\\, dt$.\nWith $\\sqrt{1/i} = e^{-i\\pi/4}$, this is $e^{-i\\pi/4} \\hat f(u)$, the Fourier transform up to a constant phase.\nThe matrices $\\begin{pmatrix} \\cos\\alpha \u0026 \\sin\\alpha \\\\ -\\sin\\alpha \u0026 \\cos\\alpha \\end{pmatrix}$ give the fractional Fourier transforms, in the normalization with the kernel $e^{-2\\pi i u t}$, up to a constant phase.\n\nNon-example: $M = \\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 2 \\end{pmatrix}$ has determinant $2$, so it is not in $SL(2, \\R)$ and has no linear canonical transformation.\nThe formula above does not use $c$, and with $a = 1$, $b = 1$, $d = 2$ it gives the transform of $\\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 2 \\end{pmatrix}$, the only matrix of determinant $1$ with these entries $a$, $b$, $d$.\n",
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