User:EverettYou/Renormalization

Regularization

Fourier Transform (Real to Reciprocal)

In general, the $d$-dimentional Fourier transformation of an isotropic function $f(r)$ is defined as

${\mathcal {F}}_{d}[f(r)]\equiv \int _{\mathbb {R} ^{d}}f(r)e^{-\mathrm {i} {\boldsymbol {p}}\cdot {\boldsymbol {r}}}\mathrm {d} ^{d}{\boldsymbol {r}}={\frac {2\pi ^{d/2}}{\Gamma ({\tfrac {d}{2}})}}\int _{0}^{\infty }f(r)r^{d-1}\;{}_{0}F_{1}{\big (}{\tfrac {d}{2}},-({\tfrac {pr}{2}})^{2}{\big )}\mathrm {d} r,$

where ${}_{0}F_{1}$ is a generalized hypergeometric function.

Derivation of the Fourier transform formula

To derive this formula, we first note that the integrand has an SO(2) rotational symmetry about the axis along the direction of ${\boldsymbol {p}}$. Let $\theta$ be the angle between ${\boldsymbol {p}}$ and ${\boldsymbol {r}}$, we have ${\boldsymbol {p}}\cdot {\boldsymbol {r}}=pr\cos \theta$, and the volume element can be written as

$\mathrm {d} ^{d}{\boldsymbol {r}}=S_{d-2}(r\sin \theta )r\mathrm {d\theta } \mathrm {d} r={\frac {2\pi ^{(d-1)/2}}{\Gamma ({\tfrac {d-1}{2}})}}\,r^{d-1}\mathrm {d} r\,\sin ^{d-2}\theta \mathrm {d} \theta ,$

where $S_{n}(R)$ denotes the area of hypersphere $S^{n}$ of radius $R$. So

${\mathcal {F}}_{d}[f(r)]={\frac {2\pi ^{(d-1)/2}}{\Gamma ({\tfrac {d-1}{2}})}}\int _{0}^{\infty }f(r)r^{d-1}\mathrm {d} r\int _{0}^{\pi }e^{-ipr\cos \theta }\sin ^{d-2}\theta \mathrm {d} \theta$

The integral over $\theta$ can be carried out first by the variable substitution $x=\cos \theta$, as

$\int _{0}^{\pi }e^{-ipr\cos \theta }\sin ^{d-2}\theta \mathrm {d} \theta =\int _{-1}^{1}e^{-iprx}(1-x^{2})^{\tfrac {d-3}{2}}\mathrm {d} x={\frac {\pi ^{1/2}\Gamma ({\tfrac {d-1}{2}})}{\Gamma ({\tfrac {d}{2}})}}{}_{0}F_{1}{\big (}{\tfrac {d}{2}},-({\tfrac {pr}{2}})^{2}{\big )}.$

Plugging in the above result, we obtain the formula for the Fourier transform of power functions.

When the Gamma function $\Gamma ({\tfrac {d+n}{2}})$ is not singular, i.e. $d+n\neq 0,-2,-4,-6,\cdots$, we have the following results:

${\mathcal {F}}_{d}[r^{n}]={\frac {2^{d+n}\pi ^{d/2}\Gamma ({\frac {d+n}{2}})}{\Gamma (-{\frac {n}{2}})}}{\frac {1}{p^{d+n}}},$

${\mathcal {F}}_{d}[r^{n}\ln r]={\frac {2^{d+n}\pi ^{d/2}\Gamma ({\frac {d+n}{2}})}{\Gamma (-{\frac {n}{2}})}}{\frac {1}{p^{d+n}}}{\frac {1}{2}}{\big (}\psi ({\tfrac {d+n}{2}})+\psi (-{\tfrac {n}{2}})-\ln({\tfrac {p^{2}}{4}}){\big )},$

where $\Gamma (\cdot )$ and $\psi (\cdot )$ are the gamma function and the digamma function (0th polygamma function) respectively. When such condition is violated, the Fourier transform falls back to the hypergeometric function integral.

The integral can be formally carried out by Mathematica with the option GenerateConditions -> False. Following is a table of Fourier transform of $r^{n}$ and $r^{n}\ln r$ in several lowest dimensional spaces. Only leading contributions are kept.

$d=$	1	2	3	4	5
$r^{-10}$	$-{\frac {\pi p^{9}}{362880}}$	$-{\frac {\pi p^{8}\ln p}{73728}}$	${\frac {\pi ^{2}p^{7}}{20160}}$	${\frac {\pi ^{2}p^{6}\ln p}{4608}}$	$-{\frac {\pi ^{3}p^{5}}{1440}}$
$r^{-9}$	$-{\frac {p^{8}\ln p}{20160}}$	${\frac {2\pi p^{7}}{11025}}$	${\frac {\pi p^{6}\ln p}{1260}}$	$-{\frac {4\pi ^{2}p^{5}}{1575}}$	$-{\frac {1}{105}}\pi ^{2}p^{4}\ln p$
$r^{-8}$	${\frac {\pi p^{7}}{5040}}$	${\frac {\pi p^{6}\ln p}{1152}}$	$-{\frac {1}{360}}\pi ^{2}p^{5}$	$-{\frac {1}{96}}\pi ^{2}p^{4}\ln p$	${\frac {\pi ^{3}p^{3}}{36}}$
$r^{-7}$	${\frac {1}{360}}p^{6}\ln p$	$-{\frac {2\pi p^{5}}{225}}$	$-{\frac {1}{30}}\pi p^{4}\ln p$	${\frac {4\pi ^{2}p^{3}}{45}}$	${\frac {4}{15}}\pi ^{2}p^{2}\ln p$
$r^{-6}$	$-{\frac {\pi p^{5}}{120}}$	$-{\frac {1}{32}}\pi p^{4}\ln p$	${\frac {\pi ^{2}p^{3}}{12}}$	${\frac {1}{4}}\pi ^{2}p^{2}\ln p$	$-{\frac {\pi ^{3}p}{2}}$
$r^{-5}$	$-{\frac {1}{12}}p^{4}\ln p$	${\frac {2\pi p^{3}}{9}}$	${\frac {2}{3}}\pi p^{2}\ln p$	$-{\frac {4\pi ^{2}p}{3}}$	$-{\frac {8}{3}}\pi ^{2}\ln p$
$r^{-4}$	${\frac {\pi p^{3}}{6}}$	${\frac {1}{2}}\pi p^{2}\ln p$	$-\pi ^{2}p$	$-2\pi ^{2}\ln p$	${\frac {2\pi ^{3}}{p}}$
$r^{-3}$	$p^{2}\ln p$	$-2\pi p$	$-4\pi \ln p$	${\frac {4\pi ^{2}}{p}}$	${\frac {8\pi ^{2}}{p^{2}}}$
$r^{-2}$	$-\pi p$	$-2\pi \ln p$	${\frac {2\pi ^{2}}{p}}$	${\frac {4\pi ^{2}}{p^{2}}}$	${\frac {4\pi ^{3}}{p^{3}}}$
$r^{-1}$	$-2\ln p$	${\frac {2\pi }{p}}$	${\frac {4\pi }{p^{2}}}$	${\frac {4\pi ^{2}}{p^{3}}}$	${\frac {16\pi ^{2}}{p^{4}}}$
$\ln r$	$-{\frac {\pi }{p}}$	$-{\frac {2\pi }{p^{2}}}$	$-{\frac {2\pi ^{2}}{p^{3}}}$	$-{\frac {8\pi ^{2}}{p^{4}}}$	$-{\frac {12\pi ^{3}}{p^{5}}}$
$r$	$-{\frac {2}{p^{2}}}$	$-{\frac {2\pi }{p^{3}}}$	$-{\frac {8\pi }{p^{4}}}$	$-{\frac {12\pi ^{2}}{p^{5}}}$	$-{\frac {64\pi ^{2}}{p^{6}}}$
$r^{3}$	${\frac {12}{p^{4}}}$	${\frac {18\pi }{p^{5}}}$	${\frac {96\pi }{p^{6}}}$	${\frac {180\pi ^{2}}{p^{7}}}$	${\frac {1152\pi ^{2}}{p^{8}}}$
$r^{5}$	$-{\frac {240}{p^{6}}}$	$-{\frac {450\pi }{p^{7}}}$	$-{\frac {2880\pi }{p^{8}}}$	$-{\frac {6300\pi ^{2}}{p^{9}}}$	$-{\frac {46080\pi ^{2}}{p^{10}}}$
$r^{7}$	${\frac {10080}{p^{8}}}$	${\frac {22050\pi }{p^{9}}}$	${\frac {161280\pi }{p^{10}}}$	${\frac {396900\pi ^{2}}{p^{11}}}$	${\frac {3225600\pi ^{2}}{p^{12}}}$
$r^{-9/2}$	${\frac {16}{105}}{\sqrt {2\pi }}p^{7/2}$	${\frac {\pi p^{5/2}\Gamma \left(-{\frac {5}{4}}\right)}{4{\sqrt {2}}\Gamma ({\tfrac {9}{4}})}}$	$-{\frac {16}{15}}{\sqrt {2}}\pi ^{3/2}p^{3/2}$	${\frac {\pi ^{2}{\sqrt {p}}\Gamma \left(-{\frac {1}{4}}\right)}{{\sqrt {2}}\Gamma ({\tfrac {9}{4}})}}$	${\frac {16{\sqrt {2}}\pi ^{5/2}}{5{\sqrt {p}}}}$
$r^{-7/2}$	${\frac {8}{15}}{\sqrt {2\pi }}p^{5/2}$	${\frac {\pi p^{3/2}\Gamma \left(-{\frac {3}{4}}\right)}{2{\sqrt {2}}\Gamma ({\tfrac {7}{4}})}}$	$-{\frac {8}{3}}{\sqrt {2}}\pi ^{3/2}{\sqrt {p}}$	${\frac {{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {1}{4}})}{{\sqrt {p}}\Gamma ({\tfrac {7}{4}})}}$	${\frac {8{\sqrt {2}}\pi ^{5/2}}{3p^{3/2}}}$
$r^{-5/2}$	$-{\frac {4}{3}}{\sqrt {2\pi }}p^{3/2}$	${\frac {\pi {\sqrt {p}}\Gamma \left(-{\frac {1}{4}}\right)}{{\sqrt {2}}\Gamma ({\tfrac {5}{4}})}}$	${\frac {4{\sqrt {2}}\pi ^{3/2}}{\sqrt {p}}}$	${\frac {2{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {3}{4}})}{p^{3/2}\Gamma ({\tfrac {5}{4}})}}$	${\frac {4{\sqrt {2}}\pi ^{5/2}}{p^{5/2}}}$
$r^{-3/2}$	$-2{\sqrt {2\pi }}{\sqrt {p}}$	${\frac {{\sqrt {2}}\pi \Gamma ({\tfrac {1}{4}})}{{\sqrt {p}}\Gamma ({\tfrac {3}{4}})}}$	${\frac {2{\sqrt {2}}\pi ^{3/2}}{p^{3/2}}}$	${\frac {{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {1}{4}})}{p^{5/2}\Gamma ({\tfrac {3}{4}})}}$	${\frac {6{\sqrt {2}}\pi ^{5/2}}{p^{7/2}}}$
$r^{-1/2}$	${\frac {\sqrt {2\pi }}{\sqrt {p}}}$	${\frac {\pi \Gamma ({\tfrac {3}{4}})}{{\sqrt {2}}p^{3/2}\Gamma ({\tfrac {5}{4}})}}$	${\frac {{\sqrt {2}}\pi ^{3/2}}{p^{5/2}}}$	${\frac {2{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {7}{4}})}{p^{7/2}\Gamma ({\tfrac {5}{4}})}}$	${\frac {5{\sqrt {2}}\pi ^{5/2}}{p^{9/2}}}$
$r^{1/2}$	$-{\frac {\sqrt {\frac {\pi }{2}}}{p^{3/2}}}$	${\frac {{\sqrt {2}}\pi \Gamma ({\tfrac {1}{4}})}{p^{5/2}\Gamma \left(-{\frac {1}{4}}\right)}}$	$-{\frac {3\pi ^{3/2}}{{\sqrt {2}}p^{7/2}}}$	${\frac {16{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {9}{4}})}{p^{9/2}\Gamma \left(-{\frac {1}{4}}\right)}}$	$-{\frac {21\pi ^{5/2}}{{\sqrt {2}}p^{11/2}}}$
$r^{3/2}$	$-{\frac {3{\sqrt {\frac {\pi }{2}}}}{2p^{5/2}}}$	${\frac {8{\sqrt {2}}\pi \Gamma ({\tfrac {7}{4}})}{p^{7/2}\Gamma \left(-{\frac {3}{4}}\right)}}$	$-{\frac {15\pi ^{3/2}}{2{\sqrt {2}}p^{9/2}}}$	${\frac {32{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {11}{4}})}{p^{11/2}\Gamma \left(-{\frac {3}{4}}\right)}}$	$-{\frac {135\pi ^{5/2}}{2{\sqrt {2}}p^{13/2}}}$
$r^{5/2}$	${\frac {15{\sqrt {\frac {\pi }{2}}}}{4p^{7/2}}}$	${\frac {16{\sqrt {2}}\pi \Gamma ({\tfrac {9}{4}})}{p^{9/2}\Gamma \left(-{\frac {5}{4}}\right)}}$	${\frac {105\pi ^{3/2}}{4{\sqrt {2}}p^{11/2}}}$	${\frac {64{\sqrt {2}}\pi ^{2}\Gamma ({\tfrac {13}{4}})}{p^{13/2}\Gamma \left(-{\frac {5}{4}}\right)}}$	${\frac {1155\pi ^{5/2}}{4{\sqrt {2}}p^{15/2}}}$

If the Fourier transform is UV regularized, all $\ln p$ should be understood as $\ln(p/\Lambda )$. Results that do not contain $\ln p$ are not affected by regularization at the leading order.

Inverse Fourier Transform (Reciprocal to Real)

The $d$-dimensional inverse Fourier transform is defined as

${\mathcal {F}}_{d}^{-1}[f(p)]\equiv {\frac {1}{(2\pi )^{d}}}\int _{\mathbb {R} ^{d}}f(p)e^{\mathrm {i} {\boldsymbol {p}}\cdot {\boldsymbol {r}}}\mathrm {d} ^{d}{\boldsymbol {p}}$.

Due to the momentum-position duality, the inverse Fourier transform can be obtained by taking the Fourier transform result, exchanging $p$ and $r$, and dividing by $(2\pi )^{d}$.

${\mathcal {F}}_{d}^{-1}[f(p)]={\frac {1}{(2\pi )^{d}}}{\mathcal {F}}_{d}[f(r)]/.r\to p$.

Heat Kernel Regularization

In quantum field theory applications, it is often desired that the momentum integral is cut off at a UV scale $\Lambda$. The heat kernel regularization suppresses the UV contribution in the momentum integral by an envelope function $e^{-p/\Lambda }$. The UV regularized inverse Fourier transform is defined as

${\mathcal {F}}_{d}^{-1}[f(p);\Lambda ]\equiv {\frac {1}{(2\pi )^{d}}}\int _{\mathbb {R} ^{d}}f(p)e^{-p/\Lambda }e^{\mathrm {i} {\boldsymbol {p}}\cdot {\boldsymbol {r}}}\mathrm {d} ^{d}{\boldsymbol {p}}$.

Since the UV cutoff $\Lambda$ is expected to be large, following an $1/\Lambda$ expansion $e^{-p/\Lambda }=1-p/\Lambda +\cdots$, the regularized inverse Fourier transform can be calculated order-by-order as

${\mathcal {F}}_{d}^{-1}[f(p);\Lambda ]={\mathcal {F}}_{d}^{-1}[f(p)]-{\mathcal {F}}_{d}^{-1}[f(p)p]/\Lambda +\cdots$.