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Recently I started to read about solid state physics, and as an introduction to analyzing the problem of electrons in crystals with periodic potential, I started to learn about Bravais lattice.

I've read that sometimes, in order to describe a complex Bravais lattice as a simple Bravais lattice, such as simple cubic lattice, we may want to choose a few points of the original lattice, which would form one point of the new simplified lattice.

For example, we can describe a BCC lattice as a simple cubic lattice, where each point in the simple cubic lattice represents 2 points of the original BCC lattice.

My questions are:

1.Given some general Bravais lattice, is there a mathematical way to determine if this given Bravais lattice can be described as a simple Bravais lattice, say simple cubic lattice?

2.Assume we know that a given Bravais lattice can be simplified and can be described as a simple cubic lattice, is there a mathematical way to determine how many points of the original lattice would be in the basis of the new simplified lattice?

  1. Is the choice of the basis unique? For example, I've read that we can describe a FCC lattice as a simple cubic lattice with a basis consists of 4 points of the original lattice, how can we know for sure that with a basis that consists 3 points of the original lattice we can't describe the lattice? How can we know that there is no basis with, say 6 points of the original lattice that can be used to describe FCC lattice as a simple cubic lattice?

If indeed there is a mathematical way to do the things I described here, it would be very helpful if someone can demonstrate it by describing FCC lattice as a simple cubic lattice.

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  1. Bravais lattices aren't that many, just 14 in 3-D, so there's not much variability and you can easily check whether you can describe it as a simpler Bravais lattice. The point is: is there an underlying simpler Bravais lattice? If the answer is yes, then you can describe your lattice as a simpler one with basis. This holds, in general for every lattice, even non-Bravais ones. In a more formal way, the question would be: is there a subset of lattice points $\{\boldsymbol{R}\}$ which form a simpler Bravais lattice?
  2. Again, the simplest way would be to look at the crystal and take the basis so to fill the whole crystal. The number of atoms in a basis should however give you the correct number of atoms per unit cell. The FCC has a density of 4 atoms. If you describe it as a simple cubic with basis, each point should weigh 4 so that $4\times8/8=4$.
  3. There is some kind of freedom in the choice of the basis, but I don't think you have freedom in the number of atoms in the basis (once you have fixed the simpler Bravais lattice) because the density should be the same. You can, however, choose the direction of the basis vectors, just like you can with any Bravais lattice.

I hope this answer is what you looked for.

EDIT: I came up with a somewhat more formal way to evaluate the number of atoms in a basis, though it's basically what I said above. The density of atoms can be evaluated as $1/V_{WS}$, where $V_{WS}$ is the (primitive) volume of the Wigner-Seitz cell. The Wigner-Seitz cell always contains one lattice point by definition, so calling $V_{WS,2}$ the volume of the W-S cell of the simpler lattice and $N$ the number of atoms in the basis you should have: \begin{equation} \frac{N}{V_{WS,2}}=\frac{1}{V_{WS}} \end{equation}

In the case of an FCC describes as a SC with basis $V_{WS}=a^3/4$ while $V_{WS}=a^3$, so you need $N=4$.

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  • $\begingroup$ So whenever we want to describe a Bravais lattice with a simpler lattice with a basis, the number of the atoms in the basis should be the number of atoms per unit cell? $\endgroup$
    – FreeZe
    Commented Mar 1, 2021 at 0:06
  • $\begingroup$ It should be the number of atoms per unit cell of the original lattice divided by the number of atoms per unit cell of the simpler cell I think. That's so that the number of atoms in the lattice with basis is the number of atoms per (the simpler) unit cell times the number of atoms in a basis. I can't think of a formal proof right now but it checks out for all the lattices I've tried. If the simpler lattice is the simple cubic then yes, the number of atoms in the basis should be the number of atoms per unit cell. $\endgroup$ Commented Mar 1, 2021 at 8:50
  • $\begingroup$ I edited the answer $\endgroup$ Commented Mar 1, 2021 at 9:09
  • $\begingroup$ Best regards from Sicily.....:-) $\endgroup$
    – Sebastiano
    Commented Mar 4, 2021 at 22:27
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    $\begingroup$ @Sebastiano Ciao! $\endgroup$ Commented Mar 4, 2021 at 22:46

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