Fourier transform 总结

2018-09-07

-physics

分类

根据变量的连续与否,傅立叶变换有以下几种情况:

$$ \begin{align*} \sum_{l = 1}^N e^{-i k'l} e^{i kl}=N\delta_{k,k'}, \quad l = 1, 2, \cdots N, \quad k = 0, \frac{2\pi}{N}, 2 \frac{2\pi}{N}, \cdots (N-1)\frac{2\pi}{N} \end{align*} $$ $$ \begin{align*} \int_{0}^{L} e^{-ik'x} e^{ikx} \mathrm{d}x = L \delta_{k, k'}, \quad k = 0, \pm\frac{2\pi}{L}, \pm 2\frac{2\pi}{L}, \cdots \end{align*} $$ $$ \begin{align*} \int_{-\infty}^{\infty} e^{-ikx} e^{ik'x} \mathrm{d}x = 2\pi \delta (k-k') \end{align*} $$

上式的等号在分布意义下成立。

例子

实空间离散

周期性边界条件下的紧束缚模型

$$ H = -t \sum_{j=1}^{N} \left( c_j^{\dagger} c_{j+1} + c_{j+1}^{\dagger} c_j \right), \quad c_{N + 1} = c_1 $$

具有平移对称性,本征态是平面波

$$ \begin{align} a_k^{\dagger} = \sum_j c_j^{\dagger} \langle j| k\rangle =\frac{1}{\sqrt{N}} \sum_j c_j^{\dagger} e^{i j k},\quad k = 0, \frac{2\pi}{N}, 2\frac{2\pi}{N}, \cdots (N - 1)\frac{2\pi}{N} \end{align} $$

所以

$$ \begin{align} H =& -t \sum_{j = 1}^N \sum_{k_1, k_2} \frac{1}{N} \left( a_{k_1}^\dagger a_{k_2} e^{-i j(k_1 - k_2)} e^{i k_2} + a_{k_2}^\dagger a_{k_1} e^{-i j(k_2 - k_1)} e^{-ik_2} \right) \\ =& -t \sum_k 2\cos k a_{k}^\dagger a_{k} \end{align} $$

实空间有限尺寸

尺寸为 L 的周期性边界条件的一维系统(箱归一化)

动量本征态 $\psi_k(x) = \langle x | k\rangle$ 是周期函数

$$ \begin{align*} \psi_k(x+L) = \psi_k(x) = \frac{1}{\sqrt{L}}e^{ikx}, \quad k = 0, \pm \frac{2\pi}{L}, \pm 2\frac{2\pi}{L}, \pm 3\frac{2\pi}{L}, \cdots \end{align*} $$

动量本征态正交归一

$$ \begin{align*} \langle k | k'\rangle = \int_{0}^{L} e^{-ikx} e^{ik'x} \mathrm{d}x = L \delta(k-k') \end{align*} $$

实空间无穷大

一维无穷空间中的动量本征函数

$$ \begin{align*} -i \frac{\mathrm{d}}{\mathrm{d}x} \langle x|k\rangle = k \langle x| k\rangle \quad \Longrightarrow \quad \psi_k(x) = \langle x| k\rangle = \frac{1}{\sqrt{2\pi}} e^{i kx} \end{align*} $$

正交归一

$$ \begin{align*} \langle k | k' \rangle = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-ikx} e^{ik'x} \mathrm{d}x =\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i(k'-k)x} \mathrm{d}x = \delta(k'-k) \end{align*} $$

#Fourier transform #数学