Is the choice of a lattice for describing a crystal arbritary ( within the constraint of system symmetries)?

It seems to me that in order to correctly depict the crystal momentum of modes, it is essential to choose only a particular lattice. In other words, band diagrams obtained between one lattice and another, connected through zone-folding, do not provide an accurate picture in terms of crystal momentum.

In order to illustrate this, I will consider a 1D monoatomic chain of atoms and describe their longitudinal phonon modes through two lattices as shown below:

This uses a lattice with a 1-atom basis:

enter image description here

This uses a lattice with a 2-atom basis: enter image description here

The problem is, that if we use the second lattice, and use zone-folding to represent all our modes, the dynamically unapproachable phonons at the Zone edge when we use the first lattice, suddenly become susceptible to Raman scattering.

  • 1
    $\begingroup$ Your handwritten text is not very easy to read. It'd be better if you converted most of it into real text, and formatted the rest (that's undetachable from the images) with some digital font. $\endgroup$
    – Ruslan
    Commented Apr 24, 2020 at 7:07

1 Answer 1


Of course, the choice of lattice you use is just an intermediate step in a calculation, and has no bearing on the final results. But a bad choice of lattice might make the result harder to see.

Suppose that the scattering for any kind of particle whatsoever responds to a potential $\rho(\mathbf{x})$. Under the Born approximation, the amplitude to scatter a particle and change its momentum by $\mathbf{k}$ is proportional to $$\widetilde{\rho}(\mathbf{k}) = \int d\mathbf{x} \, e^{i \mathbf{k} \cdot \mathbf{x}} \rho(\mathbf{x}).$$ For phonons, Bloch's theorem tells us that $$\rho(\mathbf{x}) = u(\mathbf{x}) e^{i \mathbf{k}' \cdot \mathbf{x}}$$ where $\mathbf{k}'$ is the phonon's crystal momentum, and $u(\mathbf{x})$ is periodic. Because of this periodicity, it's useful to do the integral in two stages. We decompose it into an integral over a "unit cell" $V$, and a sum over such cells. This gives $$\widetilde{\rho}(\mathbf{k}) = \sum_{\text{cells at } \mathbf{x}_i} e^{i (\mathbf{k} + \mathbf{k}') \cdot \mathbf{x}_i} S(\mathbf{k}), \quad S(\mathbf{k}, \mathbf{k}') = \int_V d\mathbf{x} \, e^{i(\mathbf{k} + \mathbf{k}') \cdot \mathbf{x}} u(\mathbf{x})$$ where the integral $S(\mathbf{k}, \mathbf{k}')$ is called a form factor. Performing the sum, $$\widetilde{\rho}(\mathbf{k}) \propto S(\mathbf{k}, \mathbf{k}') \sum_{\mathbf{G}} \delta(\mathbf{k} + \mathbf{k}' - \mathbf{G})$$ where the sum is over reciprocal lattice vectors $\mathbf{G}$.

When you change the choice of unit cell, you just change what goes into the form factor, and what goes into the sum, but the overall result is of course the same. In your first example, the scattering amplitude off the indicated optical phonons vanishes because the sum vanishes -- that is, whenever the frequencies match, none of the delta functions in the sum are nonzero. In the second example, it vanishes because the form factor vanishes: at the point you mark, it has the form $+1 - 1 = 0$. Both options are valid.

  • $\begingroup$ I encourage the OP to draw out the distorted lattice from the phonon according to the two schemes then read this answer by knzhou. $\endgroup$
    – KF Gauss
    Commented Apr 27, 2020 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.