Diffraction and $k$-space

Regarding diffraction I am a little bit lost reading about reciprocal space and the space of $k$'s. As I understand it the Fourier relationship between a wavepacket $\Psi(\vec r,t)$ and the complex weighting factors of each constituent plane wave $A(\vec k)$ is given by: $$\Psi(\vec r,t)=\frac{1}{\sqrt{2\pi}}\int ^{\infty}_{-\infty}A(\vec k)e^{i(\vec k\vec r-\omega t)}d\vec k$$ demonstrating a sort of linear superposition of reflected plane waves from a diffraction grating (or crystal lattice).Further by Parseval's theorem the intensity of this reflected packet is given by: $$\int^{\infty}_{-\infty}\big|\Psi(\vec r,t)\big|^2d\vec r=\int^{\infty}_{-\infty}\big|A(\vec k)\big|^2d\vec k$$

However I am not really sure how this relates to the other sort of understanding of $k$ space. That is to say the space that can give us meaningful information about crystal lattices and unit cells. Are they the same spaces?

Would this mean therefore that the intensity/position of the diffraction spots can be related to the structure of the solid's lattice. If so how can we understand distributions in terms of the Fourier relationship above?

I understand there have been several questions so far regarding the reciprocal k-space however so far I have not found anything that helps me particularly grasp this aspect of diffraction.

As you can see I am quiet confused in this matter and would greatly appreciate some help!

• You are on the right track. Indeed, the intensity/position of the diffraction spots are directly related to the structure of the lattice along with the scattering properties of the individual atoms on the different lattice sites. Nov 20, 2014 at 15:00
• Thank you for your comment. It is this relation that I am not too sure I understand! How does the spacing of the lattice effect the resulting diffraction image? Nov 20, 2014 at 20:17

My understanding is that A(k) is the space of wave vectors, however we know these only differ by direction not magnitude for the diffracted wavepacket (Rayleigh scattering). Therefore the function A(k) indicates the "spread" of the directionality of the wave vectors. So a narrow distribution of A(k) indicates a larger intensity $|\Psi(x)|^2$ and therefore a wider diffraction grating (or indeed lattice spacing).
The mathematical reasoning behind this is simple: $\Psi(x)$ and A(k) are Fourier conjugates.