[Fourier transform and its application study notes] 16. Continue last content, crystal imaging

X-ray Crystal Imaging

1) x-ray was found in 1895 Yurunchin (Roentgen), with a wavelength of $10^{-8}$ centimeters, commonly used to measure the visible wavelength of the method will be due to its wavelength is too small to measure.

2) Crystal (crystals), the atomic structure of the crystal conforms to a certain law-atoms arranged in an orderly manner into a lattice. Laue (Laue) made a series of famous experiments in 1912 to study the nature of crystals using X-ray diffraction experiments.

Raul assumes:

1) x-ray is a wave, so it can be diffracted

2) crystals can act as suitable diffraction gratings, i.e. crystals have lattice atoms (lattice atomic)-periodic atomic structures, which can be compared with the wavelength of X-ray

In one-dimensional cases, a one-dimensional crystal is a straight line formed by the arrangement of atoms and other spaces.

The line is infinitely prolonged, there are countless atoms in the line, and the electron density distribution of the crystals is to be studied in the experiment--it can be understood as the X-ray transmittance of the crystals. The electron density of the whole crystal is the periodic arrangement of the electron density of a single atom. The density of a single atom is recorded as $\rho$, which is periodically

The atomic density of the entire one-dimensional crystal is:

$\begin{align*}
\rho_p (x)
&=\sum_{k=-\infty}^{\infty}\rho (X-KP) \ \
&=\sum_{k=-\infty}^{\infty} \rho (x) *\delta (X-KP) \qquad (Shift property \ of \ \delta) \ \
&=\rho (x) *\sum_{-\infty}^{\infty}\delta (X-KP)
\end{align*}$

According to the conclusion of our last lesson, the diffraction fringes should be determined by the Fourier transform $\mathcal{f}\rho_p$ with the equation.

New symbol $ш$ (Read sha)

We introduced the new symbol $ш$ to make

$ш_p (x) = \displaystyle{\sum_{k=-\infty}^{\infty}\delta (X-KP)}$

Since the symbol $ш$ is similar to the equal spacing pulse symbol, this symbol is introduced, $ш_p (x) $ represents an infinite number of $\delta$, and the spacing of each $\delta$ is $p$

So

$\rho_p (x) =\rho (x) *ш_p (x) $

$\mathcal{f}\rho_p=\mathcal{f} (\rho (x) *ш_p (x)) = (\mathcal{f}\rho) (\mathcal{f}ш_p) $

The $\rho$ is determined by the crystalline properties, and the object we need to study is $ш_p$. So, does $ш_p$ make sense?

We know that $\delta$ as a distributed meaningful,<\delta,\varphi> represents the value of the test function from the $0$ point $\varphi (0) $, then <Ш_p,\varphi> represents the $\delta$ at each time Value is taken.

When $p=1$

$<ш,\varphi> = \displaystyle{\sum_{k=-\infty}^{\infty}\varphi (k)}$

Since $\varphi$ is a fast-descending function, $\displaystyle{\sum_{k=-\infty}^{\infty}\varphi (k)}$ is convergent, that is, $<ш,\varphi>$ is meaningful, then $<\ The mathcal{f}ш,\varphi>$ also make sense.

$<\mathcal{f}ш,\varphi> = <Ш,\mathcal{F}\varphi> = \displaystyle{\sum_{k=-\infty}^{\infty}\mathcal{f}\ Varphi (k)}$

According to the previous solution, we can write

$\mathcal{f}ш= \displaystyle{\sum_{k=-\infty}^{\infty} \mathcal{f}\delta_k = \sum_{k=-\infty}^{\infty}e^{-2\pi ikx} }$

But this form is not the value we are ultimately asking for, we need to introduce other methods to solve it.

Poisson summation formula (the Poisson sum Formula)

Note: Since the derivation of the Poisson summation formula in the course is somewhat unclear, we are using the derivation method on the wiki.

$ш$ is a pulse function with a period of $1$, i.e.

$ш (t) = \displaystyle{\sum_{k=-\infty}^{\infty}\delta (t-k)}$

Decomposition of $ш$ into the form of Fourier series

$\begin{align*}
Ш (t)
& = \sum_{k=-\infty}^{\infty}c_ke^{2\pi ikt}\\
& = \sum_{k=-\infty}^{\infty}\left (\int_0^1ш (t) e^{-2\pi Ikt}dt \right) e^{2\pi ikt}\\
& = \sum_{k=-\infty}^{\infty} \left (\int_{0}^1 \delta (t) e^{-2\pi Ikt}dt \right) e^{2\pi ikt}\\
& = \sum_{k=-\infty}^{\infty}e^{-2\pi I0t}e^{2\pi ikt}\\
& = \sum_{k=-\infty}^{\infty}1\cdot E^{2\pi ikt}\\
& = \sum_{k=-\infty}^{\infty}e^{2\pi Ikt}
\end{align*}$

So

$\displaystyle{\sum_{k=-\infty}^{\infty}e^{2\pi Ikt} = \sum_{k=-\infty}^{\infty}\delta (t-k)}$

Based on this condition, we make the following deduction

$\begin{align*}
\sum_{k=-\infty}^{\infty}\mathcal{f}f (k)
&=\sum_{k=-\infty}^{\infty}\left (\int_{-\infty}^{\infty}f (x) e^{-2\pi ikx}dx \right) \ \
&=\int_{-\infty}^{\infty}f (x) \left (\sum_{k=-\infty}^{\infty}e^{-2\pi ikx} \right) dx\\
&=\int_{-\infty}^{\infty}f (x) \left (\sum_{k=-\infty}^{\infty}e^{2\pi ikx} \right) dx \ \
&=\int_{-\infty}^{\infty}f (x) \left (\sum_{k=-\infty}^{\infty}\delta (x-k) \right) dx\\
&=\sum_{k=-\infty}^{\infty}\left (\int_{-\infty}^{\infty}f (x) \delta (x-k) dx \right) \ \
&=\sum_{k=-\infty}^{\infty}f (k)
\end{align*}$

That

$\displaystyle{\sum_{k=-\infty}^{\infty}\mathcal{f}f (k) = \sum_{k=-\infty}^{\infty}f (k)}$

This equation is the Poisson summation formula.

Fourier transform of $ш$

According to the Poisson summation formula, the $\mathcal{f}ш$ solution process is as follows

$\begin{align*}
<\mathcal{F}Ш,\varphi>
&=<a,\mathcal{f}\varphi>\\
&=\sum_{k=-\infty}^{\infty}\mathcal{f}\varphi (k) \ \
&=\sum_{k=-\infty}^{\infty}\varphi (k) \qquad (the\ poisson\ sum\ Formula) \ \
&=<Ш,\varphi>
\end{align*}$

So

$\mathcal{f}ш=ш$

Fourier transform of $ш_p$

First, convert the $ш_p$ into a $ш$ form.

$\begin{align*}
Ш_p
&=\sum_{k=-\infty}^{\infty}\delta (X-KP) \ \
&=\sum_{k=-\infty}^{\infty}\delta (P (\frac{x}{p}-k) \ \
&=\sum_{k=-\infty}^{\infty}\frac{1}{p}\delta (\frac{x}{p}-k) \qquad (\delta\ scaling\ property) \ \
&=\frac{1}{p}\sum_{k=-\infty}^{\infty}\delta (\frac{x}{p}-k) \ \
&=\frac{1}{p}ш (\frac{x}{p})
\end{align*}$

Fourier transform the $ш_p$

$\begin{align*}
\mathcal{f}ш_p
&=\frac{1}{p}\mathcal{f} (ш (\frac{x}{p})) \ \
&=\frac{1}{p}\cdot P (\mathcal{f}ш) (px) \qquad (fourier\ scaling\ theorem) \ \
&=ш (px) \ \
&=\sum_{k=-\infty}^{\infty}\delta (px-k) \ \
&=\frac{1}{p}\sum_{k=-\infty}^{\infty}\delta (x-\frac{k}{p}) \ \
&=\FRAC{1}{P}Ш_{\FRAC{1}{P}}
\end{align*}$

So

$\mathcal{f}ш_p = \frac{1}{p}ш_{\frac{1}{p}}$

Crystal Imaging

We already know that the atomic density of one-dimensional crystals is

$\rho_p (x) = \displaystyle{\rho (x) *\sum_{-\infty}^{\infty}\delta (X-KP) = \rho (x) *ш_p}$

The imaging of a crystal depends on the Fourier transform of its atomic density.

$\begin{align*}
\mathcal{f}\rho_p
&=\MATHCAL{F} (\rho*ш_p) \ \
&=\MATHCAL{F} (\rho) \mathcal{f} (ш_p) \ \
&=\MATHCAL{F} (\rho) (\frac{1}{p}ш_{\frac{1}{p}}) \ \
&=\mathcal{f}\rho (x) \left (\frac{1}{p}\sum_{k=-\infty}^{\infty}\delta (x-\frac{k}{p}) \right) \ \
&=\frac{1}{p}\sum_{k=-\infty}^{\infty}\mathcal{f}\rho (x) \delta (x-\frac{k}{p}) \ \
&=\frac{1}{p}\sum_{k=-\infty}^{\infty}\mathcal{f}\rho (\frac{k}{p}) \delta (X-\frac{k}{p}) \qquad (\delta\ Sampling\ property)
\end{align*}$

We can see that the image is sampled as $\mathcal{f}\rho$ at various points on the pitch of the $\frac{1}{p}$

The atomic spacing of the crystals is $p$, and the space after the image is $\frac{1}{p}$, so the diffraction imaging spacing of the crystals and the atomic spacing of the crystals are reciprocal.

[Fourier transform and its application study notes] 16. Continue last content, crystal imaging