I mentioned the idea of *K*-space in post 22.22 and introduced the vector ** K** in post 19.20; in post 22.12 I described the Fourier transform as a function of

**. The purpose of this post is to bring all these ideas together.**

*K*In the left-hand picture above, a ray is scattered by a particle at O. Let’s think about the ray scattered, through an angle 2*ϕ* by the particle. Following post 19.20, I will define the incident direction by a unit vector *k*_{o} and the scattered direction by the unit vector ** k**. The change in direction of the ray is given by (

**–**

*k*

*k*_{o}). I am now going to define the vector

**by**

*K** K* = (2π/

*λ*)(

*–*

**k**

**k**_{0}) (1)

as shown in the right-hand picture above. Here *λ* represents the wavelength of the ray, for the reasons given in post 22.14. The picture below shows that, for all scattering directions, the origin of the ** K** vectors is at O’ that lies in the direction of

**a distance 2π/**

*k**λ*from O, when the direction of the incident ray doesn’t change. Of course, scattered rays and, therefore,

**vectors are not confined to the plane of the picture, which is a two-dimensional representation of something that happens in three dimensions. In post 19.20, we saw that the modulus of**

*K***is given by**

*K**K* = (4π/*λ*)sin(*ϕ*/2).

The vectors ** K** define a space, called

*K*-space, whose origin is at O’. If we multiply our definition of

**, in equation 1, by 2π the resulting vector defines what is sometimes called**

*K**reciprocal space*.

In posts 22.12 and 22.14, we saw that the rays scattered by an object (its diffraction pattern) could be represented, in amplitude and phase by the Fourier transform, *F*, of the density of scattering matter, *ρ*, in the object. *F* is a function of ** K** and is defined by

where the limits of integration are over the volume of the scattering object. Here ** r** is a vector defining positions in the scattering object so that

*ρ*is a function of

**. We define the origin of the**

*r***vectors to be at O. Then the vectors**

*r***are in**

*r**real space*, whose origin is at O.

Let’s explore the relationship between *K*-space and real space in more detail.

From the definition of ** K**, in equation 1, all possible

**vectors define a sphere of radius 2π/**

*K**λ*, whose centre is at O, the origin of real space, as shown in the picture above (where the

**vectors are shown in red). The picture shows a circular section of this sphere because it is a two-dimensional representation of what happens. This is because the points of the red arrows are all a distance of (2π/**

*K**λ*) from O. Equation 2 implies that

*K*-space has an infinite extent but the picture shows that, for a fixed position of the scattering object, we can only explore the region of

*K*-space that lies on the surface of this sphere – called the

*Ewald sphere*after the German physicist Paul Ewald (1888-1985).

We can explore more of *K*-space by rotating the object in real space – leading to an equal rotation, about a parallel axis, in *K*-space. Then further regions of *K*-space pass through the Ewald sphere – providing more information about the object. But we can only explore regions of *K*-space for which *K* < 2π/*λ* which limits the resolution of the information we can obtain with a given wavelength – if we want to obtain higher resolution information, we must extend the region of *K*-space that we can explore by increasing 2π/*λ, *that is by decreasing* λ*. This is the conclusion we reached, by a different route, in post 22.13.

The ideas in the previous two paragraphs are important when we use x-ray diffraction to investigate the positions of atoms in molecules because the radius of curvature of the Ewald sphere is appreciable because the spacing between atoms is comparable to *λ*. If we use electrons of wavelength 3 pm (see post 22.13), the radius of curvature is much greater so that, for scattering angles not approaching 90^{o}, the Ewald sphere approximates to a plane section through *K*-space. Then the diffraction pattern we record is almost a plane section through *K*-space.

What information does a plane section of *K*-space provide? Calculating the inverse Fourier transform of equation 2 gives

We can write the dot product ** r.K** as

** r.K** =

*xK*+

_{x}*yK*+

_{y}*zK*

_{z}where *x*, *y* and *z* are orthogonal Cartesian coordinates in real space and *K _{x}*,

*K*and

_{y}*K*, are the components of

_{z}**in those directions. Now let’s suppose that our plane section is perpendicular to the**

*K**z*-axis, so that

** r.K** =

*xK*+

_{x}*yK*.

_{y}Now

Note that *ρ*(*x*,*y*) is the projection of *ρ*(** r**) on the

*x*,

*y*plane. So a section of

*K*-space provides information about a projection in real space.

When we think about diffraction, it is often convenient to think about real space, the space occupied by the scattering object, and *K*-space, the space occupied by its diffraction pattern.

Related posts

19.20 Diffraction

22.12 Diffraction, Fourier transforms and image formation

22.14 X-ray diffraction

22.21 An ideal crystal

22.22 Fourier transform of a one-dimensional lattice

Follow-up posts