How does the reciprocal lattice takes into account the basis of a crystal structure?

Question

I am reading about solid state physics, and I think I got right the concept of crystal lattice. We first define a Bravais lattice as the set of vectors spanned by $\{\vec{a}_1,\vec{a}_2,\vec{a}_3 \}$ over the integers, i.e. vector of the form $$R=n_1\vec{a}_1+n_2\vec{a}_2+n_3\vec{a}_3 $$ Now, we need to define the atoms that constitute the crystal lattice. We may arrange a point in the Bravais to be the origin and take the minimum cell and define some points within, i.e. $$r=x_1\vec{a}_1+x_2\vec{a}_2+x_3\vec{a}_3 ;\quad x_i\in [0,1] \text{ for } i\in\{1,2,3 \}$$ Given a Bravais Lattice and a basis, we call the whole structure the crystal lattice. Now we further define a reciprocal space, which is like an scaled dual in some way such that it is the lattice spanned over the integers of $\{\vec{b}_1,\vec{b}_2,\vec{b}_3 \}$ such that $$\langle b_i,a_j\rangle =2\pi \delta_{ij} $$ So, it is said that x-ray diffraction is a kind of map of the reciprocal lattice, and hence of the crystal lattice, but the basis is never taken into account for the construction of the reciprocal lattice. How can we know the structure (the complete crystal lattice). One big problem that I see in this is that we can have the same crystal lattice described by two different Bravais lattices (and of course two different basis). How does this work? Is there something that I am not understanding?

Answer 1

Your understanding that the reciprocal lattice is only defined for a Bravais lattice is correct. When you have a so-called lattice with a basis, you would calculate the reciprocal lattice primitive vectors using only the primitive vectors of the underlying Bravais lattice as:

\begin{aligned}\mathbf {b} _{1}&=2\pi {\frac {\mathbf {a} _{2}\times \mathbf {a} _{3}}{\mathbf {a} _{1}\cdot \left(\mathbf {a} _{2}\times \mathbf {a} _{3}\right)}}\\\mathbf {b} _{2}&=2\pi {\frac {\mathbf {a} _{3}\times \mathbf {a} _{1}}{\mathbf {a} _{2}\cdot \left(\mathbf {a} _{3}\times \mathbf {a} _{1}\right)}}\\\mathbf {b} _{3}&=2\pi {\frac {\mathbf {a} _{1}\times \mathbf {a} _{2}}{\mathbf {a} _{3}\cdot \left(\mathbf {a} _{1}\times \mathbf {a} _{2}\right)}}\end{aligned}

Indeed, this does not take into account the effect of the basis. That, however, plays a role in X-ray diffraction experiments and affects which peaks are actually measured. The resultant intensity is determined by how the waves diffracted from each of the constituent atoms of the basis interfere. For instance, consider the body-centered cubic lattice as a simple cubic lattice with a basis at $(0,0)$ and $(\frac{a}{2}, \frac{a}{2}, \frac{a}{2})$. It's possible that at some incident angles diffracted waves from these two atoms interfere destructively and you don't see an intensity peak even though you would expect it from a simple cubic reciprocal lattice. This is taken into account by the geometrical structure factor $F_{hkl}$, defined as: $$F_{hk\ell }=\sum _{j=1}^{N}f_{j}\mathrm {e} ^{[-2\pi i(hx_{j}+ky_{j}+\ell z_{j})]}$$
where $(hkl)$ indicates the scattering plane, the index $j$ is summed over each atom in the basis with coordinates $(x_{j},y_{j},z_{j})$ and scattering factor $f_{j}$. This sum can be zero for certain values for $h,k,l$ and in such cases, no diffraction peak is seen even though it is expected from the reciprocal lattice structure of the underlying Bravais lattice.

(I've ignored that BCC is itself a Bravais lattice, for the purpose of this example. Of course, you could calculate the BCC reciprocal lattice primitive vectors using BCC primitive vectors and you'd get the right reciprocal lattice. XRD peak intensities thus determined will be most accurate, but simple structure factor calculations for SC with a basis would at least tell you which peaks won't be observed at all.)

Chapter 6 of the book Solid State Physics by Ashcroft and Mermin is a good reference for X-ray diffraction, and the concept of structure factor in particular.

Answer 2

The reciprocal lattice definitely takes into account the basis for the Bravais Lattice. Take for example a 3-dimensional lattice with basis $\{ \vec{a}_1, \vec{a}_2, \vec{a}_3 \}$. The reciprocal basis is then

Which clearly depends on the basis vectors.

Every lattice has a reciprocal lattice. In some ways, the reciprocal lattice is like the Fourier transform of the basis. For a specific set of Fourier basis functions (in this case the exponential series), it is unique.

In response to comment:

There was a clarification, the word "basis" in the question refers to the structure of the crystal itself in terms of the atoms within the crystal.

This is taken into account when you actually try to work out for example the diffraction of Xrays and such. There is a term called the structure factor that makes use of the specific position of atoms within your lattice. The "map" of the reciprocal lattice that you would obtain then also depends also on the position of atoms within the lattice cells.

For example. if you have a simple cubic lattice, you may also have atoms in the center of the cell, at fractional basis positions. These are taken into account when you try to determine scattering angles for example.

However, this becomes a much more general question, and is much more involved. A resource I used when I was studying this was the Solid State Physics textbook by Kittel, Chapter 2 page 41 in the 8th edition if you would like to check.

Answer 3

Reciprocal lattice simply does not take into account the basis of a crystal structure.

The lattice is an array of geometrical points in the space, whereas the structure is the convolution of the lattice with a basis (consituted of one or more atoms). So the lattice is not a real object, the crystal structure is a real (ideally) object that you can analyze using a diffractometer.

What you analyze by diffraction is the crystal structure, not purely the lattice.

A lattice unit cell in the reciprocal space corresponds to a Brillouin zone.