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While deriving the equilibrium lattice constants we use expressions for potential like Lennard-Jones potential which have 6th and 12th order terms or Madelung energy for ionic crystals.

While deriving the properties of phonons like dispersion relation , we use a form of potential energy that is quadratic i.e., the assumption which we make when we write equations of motion for atoms whose motion depends linearly on the distance between nearest neighbors .

Why the inconsistency?

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2 Answers 2

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Phonons are collective modes in solids and a general derivation is needed independent of particular lattice constants to first order.

Lattice constants define individual solids.

Are you aware of the harmonic oscillator approximation? All symmetric potentials have as a first term in their expansion the quadratic, thus the harmonic oscillator approximation covers the first order in the expansion for a lot of potential functions with only a constant to qualify the particular potential. In this way a general formulation for phonons will not depend on the particular lattice to first order.

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  • $\begingroup$ So , what you mean is that phonons are collective modes which are small displacements from the equilibrium positions and their motion can be approximated using the harmonic oscillator motion , while for deriving the lattice constants it depends on the specifics of the lattice ? But then for phonons with high energies , wouldn't the displacements of atoms that are oscillating be large enough to warrant a usage of specific potentials ? $\endgroup$
    – Sajid
    Commented Oct 2, 2014 at 6:23
  • $\begingroup$ Yes, thats the gist of it. Two different needs $\endgroup$
    – anna v
    Commented Oct 2, 2014 at 6:23
  • $\begingroup$ for the "large enough to warrant the usage of specific potentials" I think that the problem would become too complicated. energies comparable to lattice transitions would disrupt the framework of Phonon approximation.phonon spectra are in milieletronvolts $\endgroup$
    – anna v
    Commented Oct 2, 2014 at 13:53
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You can use either potential for either purpose; it's just that some potentials are better for the different purposes.

The reason is that these are empirical potentials; their constants are tweaked to work for a certain purpose. For example, if you're looking at the phonon band structure, you want $\omega\left(\vec{q}\right)$ to be as accurate as possible. However, it will not be exact. If you're looking at the phonon group velocities, you want $\nabla\omega\left(\vec{q}\right)$ to be as accurate as possible. This will usually not happen with the same empirical potential, but if you had the exact potential, it wouldn't be an issue.

Moreover, if you want a good empirical potential to get the phonon dispersion relation, you don't have to use quadratic potentials; that's often used as a teaching example.

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