Lattice constant in plane wave ansatz

Physics Q&A
1

Dear staff,

I have a question because I see it being different in exercises, but if there are multiple atoms in a unit cell, for example 2 and the lattice constant is a: do I use a/2 in the plane wave ansatz or a? I thought a because it made more sense to me because then the wave would not be in the “other” atoms, if that makes sense. So could you also give a brief explanation of why this constant is chosen?

Kind regards,
Fenna

2

You are correct, the plan wave ansatz must have the periodicity of the crystal. Where do you see a different ansatz being used?

3

Thank you! I saw it on extra exercises online but they were maybe incorrect then :slight_smile:

4

Um, which ones? Do you have a link?

5

I saw it here: [https://unlcms.unl.edu/cas/physics/tsymbal/teaching/SSP-927/Section%2005_Lattice_Vibrations.pdf

at page 24, " One of the possible choices of the primitive basis is: ~a1 = a~x,~a2 = a~y,~a3 =
1
2
a(~x + ~y + ~z), however the nearest neighbors are at R~ =
a
2
(±x ± y ± z). Altogether 8
neighbors each at distance √
3a/2. We obtain
E~k = Ea + h(0) + 8W cos(akx/2) cos(aky/2) cos(akz/2) , (91)
where W = h(
√
3a/2)−h(0)I(
√
3a/2). "

"

6

but I think maybe then the difference is that when you use different bases, that you can use the same wavefunction, instead of different ones, so that the periodicity is smaller, maybe?

7

Haha, I know the author of the notes :sweat_smile:

There are indeed two plane wave ansatzes that are used. One is where the phase factor only contains the coordinates of the unit cell regardless of the specific atom position, so \exp[i \bm{k}(n \bm{a}_1 + m \bm{a}_2 + k\bm{a}_3)]. This is what we use in the course.

Another option is the ansatz where the phase factor has coordinates of the specific atom \exp[i \bm{k}(n \bm{a}_1 + m \bm{a}_2 + k\bm{a}_3 + \bm{r}_i)] (note the added \bm{r}_i). That’s used in the lecture notes you found.

Both approaches are correct, and both have their advantages. Our approach is more commonly used because it’s simpler:

  • There are fewer quantities that enter the tight binding equations (the coordinates of the atoms drop out), so there’s less room for errors.
  • As a result the Hamiltonian H(k) is periodic within the Brillouin zone (the other approach only gives the energies that are periodic).

The other approach is useful for questions where specific atomic positions matter: polarization within a unit cell, etc. Also all the hopping terms get phase factors that are proportional to the distance between atoms.

8

Ah I see! This clarifies a lot. Thank you :smiley: