Electron density as a Fourier transform of the structure factor

The electron densities of atoms in a crystal are Fourier transforms of the structure factor.

In the article on scattering factor, we have restricted the electron density ρ to a lattice point, which is too simplistic. We have also described the scattering factor (eq27) as

$df=\rho e^{i\phi}dV$

or in its integrated form

$f=\int \rho e^{i\phi}dV\; \; \; \; \; \; \; (40)$

If we now extend the distribution of ρ through the unit cell, eq40 becomes the structure factor:

$F_{hkl}=\int \rho 'e^{i\phi}dV$

where ρ’ = ρ(xyz) is the electron density at coordinates xyz in the unit cell and $\phi=2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c} )$ , i.e.

$F_{hkl}=\int \rho(xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c} )}dV\; \; \; \; \; \; \; (41)$

Let dV be an infinitesimal volume of the unit cell with edges dx, dy, dz that are parallel to the unit cell axes of a, b, c (volume of unit cell is V). The ratio of dV/V must be equal to (dxdydz)/abc and we can rewrite eq41 as

$F_{hkl}=\frac{V}{abc}\int_{0}^{a}\int_{0}^{b}\int_{0}^{c}\rho (xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz\; \; \; \; \; \; \; (42)$

A Fourier series is an expansion series used to represent a periodic function and is given by:

$f(x)=\sum_{n=0}^{\infty }[A_ncos(nx)+B_nsin(nx)]$

or by its complex form

$f(x)=\sum_{n=-\infty }^{\infty }c_ne^{i\frac{2\pi nx}{a}}$

Due to the repetitive arrangement of atoms in a crystal, the electron density in a crystal is also periodic and can be expressed as a Fourier series:

$\rho (x)=\sum_{n=-\infty }^{\infty }c_ne^{i\frac{2\pi nx}{a}}\; \; \; \; \; \; \; (43)$

In three dimensions, eq43 becomes

$\rho (xyz)=\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}e^{i2\pi(\frac{nx}{a}+\frac{ my}{b}+\frac{ oz}{c})}\; \; \; \; \; \; \; (44)$

Substitute eq44 in eq42

$F_{hkl}=\frac{V}{abc}\int_{0}^{a}\int_{0}^{b}\int_{0}^{c}\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}e^{i2\pi(\frac{nx}{a}+\frac{my}{b}+\frac{oz}{c})} e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz$

$F_{hkl}=\frac{V}{abc}\sum_{n=-\infty }^{\infty }\sum_{m=-\infty }^{\infty }\sum_{o=-\infty }^{\infty }c_{nmo}\int_{0}^{a}e^{i2\pi\frac{x}{a}(n+h)}dx\int_{0}^{b}e^{i2\pi\frac{y}{b}(m+k)}dy\int_{0}^{c}e^{i2\pi\frac{z}{c}(o+l)}dz\; \; \; \; (45)$

If n ≠ –h,

$\int_{0}^{a}e^{i2\pi\frac{x}{a}(n+h)}dx=\int_{0}^{a}[cos2\pi\frac{x}{a}(n+h)+isin2\pi\frac{x}{a}(n+h)]dx$

$=\int_{0}^{a}cos2\pi\frac{x}{a}(n+h)dx=0$

which makes F_hkl = 0.

Similarly, F_hkl = 0 if m ≠ -k or o ≠ -l. Therefore, the surviving term in the triple summation in eq45 corresponds to the case of n = –h, m = –k, o = –l, giving

$F_{hkl}=\frac{V}{abc}c_{-h,-k,-l}\int_{0}^{a}dx\int_{0}^{b}dy\int_{0}^{c}dz=Vc_{-h,-k,-l}$

$c_{-h,-k,-l}=\frac{F_{hkl}}{V}\; \; \; \; \; \; \; (46)$

When n = –h, m = –k, o = –l, eq44 becomes

$\rho (xyz)=\sum_{-h=-\infty }^{\infty }\sum_{-k=-\infty }^{\infty }\sum_{-l=-\infty }^{\infty } c_{-h,-k,-l}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}$

Since $\sum_{-h=-\infty }^{\infty }\sum_{-k=-\infty }^{\infty }\sum_{-l=-\infty }^{\infty }=\sum_{h=\infty }^{-\infty }\sum_{k=\infty }^{-\infty }\sum_{l=\infty }^{-\infty }=\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }$

$\rho (xyz)=\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }c_{-h,-k,-l}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}\; \; \; \; \; \; \; (47)$

Substitute eq46 in eq47

$\rho (xyz)=\frac{1}{V}\sum_{h=-\infty }^{\infty }\sum_{k=-\infty }^{\infty }\sum_{l=-\infty }^{\infty }F_{hkl}e^{-i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}\; \; \; \; \; \; \; (48)$

Since the limits of the integrals in eq42 can be changed to run from $-\infty$ to $\infty$ without affecting the values of the integrals because $\rho(xyz)$ is zero outside the crystal, we have

$F_{hkl}=\frac{V}{abc}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\rho (xyz)e^{i2\pi(\frac{hx}{a}+\frac{ky}{b}+\frac{lz}{c})}dxdydz\; \; \; \; \; \; \; (48a)$

Eq48 and eq48a are a Fourier transform pair. As mentioned in the previous article, the intensity of a diffraction signal is proportional to the square of the magnitude of the three-dimensional structure factor, i.e. $I \propto \left | F_{hkl} \right |^2$ . If we know the value of F_hkl (which in principle is the square root of the intensity of a peak from an X-ray diffraction experiment $\sqrt{\left | F_{hkl} \right |^2}=\left | F_{hkl} \right |$ ) and having indexed the plane contributing to this intensity peak (i.e. knowing the h, k, l values), we can determine ρ(xyz) using a mathematical software. The solution to ρ(xyz) is an electron density map that elucidates bond lengths and bond angles of the compound. However, a problem called the phase problem arises.

Electron density as a Fourier transform of the structure factor

next article: phase problem

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