Fourier transform總結(jié)
[TOC]
總的分類
根據(jù)變量的連續(xù)與否，F(xiàn).T.總共有三種衷戈。
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完全離散
$\sum_l e^{-i\frac{2\pi}{N}k'l} e^{i\frac{2\pi}{N}kl}=N\delta_{kk'}$
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部分離散
$\int_{-\infty}^{\infty} e^{-ik'x} e^{ikx} \mathrmvwvozxox = L \delta_{kk'}$
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完全連續(xù)
$\int_{-\infty}^{\infty} e^{-ikx} e^{ik'x} \mathrms7n7afhx = 2\pi \delta (x-x')$
部分離散
動(dòng)量本征函數(shù)
$\frac{\hbar}{i} \frac{\mathrmqiwhh71}{\mathrm2zslagex} \psi = p \psi \Longrightarrow \psi = e^{i\frac{p}{\hbar}x}$
歸一化
$\langle e^{ikx} \mid e^{ikx'} \rangle = \int_{-\infty}^{\infty} e^{-ikx} e^{ikx'} \mathrmtycbmzgk = \int_{-\infty}^{\infty} e^{ik(x'-x)} \mathrmc3kiqoqk = 2\pi \delta(x-x')$
周期為$L$
$f(x)$是周期函數(shù)
$f(x+L) = f(x)$
將$x$從$2\pi$拉伸到$L$
$x \rightarrow x \cdot \frac{2\pi}{L}$
相應(yīng)歸一化到$2L$
$\langle e^{ikx \cdot \frac{2\pi}{L}} \mid e^{ikx' \cdot \frac{2\pi}{L}} \rangle = \int_{-\infty}^{\infty} e^{-ikx \cdot \frac{2\pi}{L}} e^{ikx' \cdot \frac{2\pi}{L}} \mathrmfaptskuk = L \delta(x-x')$
Fourier展開系數(shù)由下式求得
$f(x)= \frac{1}{L} \sum_k \mid e^{ikx \cdot \frac{2\pi}{L}} \rangle \langle e^{ikx \cdot \frac{2\pi}{L}} \mid f(x) \rangle = \sum_k e^{ikx \cdot \frac{2\pi}{L}} C_k $
$C_n = \frac{1}{L} \int_{-\infty}^{\infty} e^{-ikx \cdot \frac{2\pi}{x}} \mathrmgwgzk7rx$
完全連續(xù)
當(dāng)周期$L \rightarrow \infty?$時(shí)，$\frac{2\pi}{L} \rightarrow 0?$防泵，指數(shù)上的$k\frac{2\pi}{L}?$由原來(lái)離散的取值$\frac{2\pi}{L}, 2\frac{2\pi}{L}, 3\frac{2\pi}{L}, \cdots?$變成連續(xù)的變量励背，記為新的$k?$祖乳，且$\mathrmishaegtk = \frac{2\pi}{L}?$。
則Fourier展開變?yōu)?/p>
$f(x)= \lim_{L\rightarrow \infty}\frac{1}{2\pi} \frac{2\pi}{L} \sum_k \mid e^{ikx \cdot \frac{2\pi}{L}} \rangle \langle e^{ikx \cdot \frac{2\pi}{L}} \mid f(x) \rangle = \int {-\infty}^{\infty} \mathrmbbbmmd2k \frac{1}{2\pi} \mid e^{ikx \cdot \frac{2\pi}{L}} \rangle \langle e^{ikx \cdot \frac{2\pi}{L}} \mid f(x) \rangle = \frac{1}{2\pi} \int{-\infty}^{\infty} \mathrmufj3tqik \int_{-\infty}^{\infty} \mathrmjfyuu76x' e^{ikx} \cdot e^{-ikx’} f(x‘)$
也就是
$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathrmpww8k26k \int_{-\infty}^{\infty} \mathrmdkvkdqdx' e^{ik(x-x')} f(x')$
完全離散
類似群論中的不可約表示力喷。
$\sum_l e^{-i\frac{2\pi}{N}k'l} e^{i\frac{2\pi}{N}kl}=N\delta_{kk'}$
