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In my solid state book, when deriving the dispersion relation of a diatomic one-dimensional chain, the author makes the following Ansatz for the oscillation of the first (x) and seond (y) atom in the chain:

$\delta x_n = A_x e^{i \omega t - ikna}$
$\delta y_n = A_y e^{i \omega t - ikna}$
where n is the number of the current unit cell, a is the lattice constant and the $\delta$ quantities are the deviations of the atom positions from their rest position. These are subsequently plugged into the equations of motion which yields an Eigenvalue problem for the frequency.

My question is now the following: Within each unit cell, both atoms have different positions. Why are we not including those positions in the exponent (for example $\delta x_n = A_x e^{i \omega t - ik(na + x_{unit})}$ where $x_{unit}$ is the distance of the rest position of x from the reference point in a unit cell) and instead pretend that both atoms within a unit cell are in phase and only differ by their amplitude?

At first I thought this distance could be neglected compared to $na$ but this should not be the case since $x_{unit}$ is in the same order of magnitude as the lattice constant and should therefore make the same difference as increasing the number of lattice constants by 1.

Is there a physical reason for this assumption? Or are we just assuming in-phase oscillations to make the calculations easier and it magically still works?

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1 Answer 1

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The amplitudes $A_x$ and $A_y$ are allowed to be complex, so any constant phase factor like $e^{ikx_{cell}}$ can be absorbed into them without any loss of generality. The phase difference can then be extracted by computing $\mathrm{arg}(A_y/A_x)$.

This phase difference generally depends on $\omega$ and $ka$, but e.g. in the $ka\rightarrow 0$ limit of the acoustic branch, the atoms oscillate in-phase whereas in the same limit of the optical branch, the atoms oscillate out-of-phase.

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  • $\begingroup$ Does that mean that the exact sub structure of a unit cell does not at all affect the dispersion relation, just the fact that there are two atoms and/or two types of bonds? $\endgroup$
    – Takitoli
    Commented Dec 9, 2022 at 21:46
  • $\begingroup$ @Takitoli In a realistic model, the unit cell's substructure will be related to the nature (strength, length, etc) of the bonds between the atoms, so it's not exactly true that the substructure is irrelevant to the dispersion relation. But all other things being equal (which again, they generally won't be), the equilibrium positions of the atoms within the unit cell simply contribute an overall phase as per the solution to your exercise. $\endgroup$
    – J. Murray
    Commented Dec 9, 2022 at 22:02

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