I studied a disordered state (FCC solid solution) and used a special quasirandom structure supercell with 32 atoms to represent the system. After relaxing it thoroughly with VASP, I used phonopy for the harmonic phonon calculation. Since the DFT supercell itself was big enough (32 atoms and ~6.5 angstroms along all three lattice vectors), I used a $1\times1\times1$ supercell for the phonopy calculation. I read somewhere that using a $1\times1\times1$ supercell for phonon calculations is incorrect (no matter how big the cell is). I couldn't understand why. Is that actually the case, and if so, why?
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asked May 14, 2021 at 8:10
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1 Answer
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If you have a system without periodicity like your disordered solid solution, then you should use a $1\times1\times1$ $\mathbf{q}$-point grid for a phonon calculation (equivalent to a $1\times1\times1$ supercell). Using a larger supercell will introduce an artificial periodicity in the system.
Having said this, it is still really important to converge with respect to system size in disordered systems. But what this means is that you should consider generating special quasirandom structures with more atoms in the quasirandom cell, and then for those doing again a $1\times1\times1$ phonon calculation.
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$\begingroup$ Thank you for your answer ProfM. We get to choose a q-mesh as an input to the code. I couldn't understand the equivalence of a $1\times1\times1$ q grid and a $1\times1\times1$ supercell. Could you explain that a bit more? $\endgroup$ Commented May 14, 2021 at 14:32
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2$\begingroup$ In general, a calculation in a primitive cell using a $n_1\times n_2\times n_3$ $\mathbf{q}$-point mesh is equivalent to a calculation in a supercell of size $n_1\times n_2\times n_3$ using the $\Gamma$ point only. This is because a particular $\mathbf{q}$-point describes a phonon that corresponds to an atomic distortion whose wavelength fits in the corresponding supercell. $\endgroup$– ProfMCommented May 14, 2021 at 15:44
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$\begingroup$ Understood the first part but couldn't understand the "atomic distortion whose wavelength fits in the corresponding supercell" part. $\endgroup$ Commented May 16, 2021 at 17:51
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1$\begingroup$ In 1D for simplicity, a phonon of a wave vector $q$ has wavelength $\lambda=2\pi/q$. Roughly speaking, this means that the real space distortion of the atoms has a periodicity of wavelength $\lambda$. This means that you need a supercell of that size to capture the full atomic displacements in real space. $\endgroup$– ProfMCommented May 16, 2021 at 18:34
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1$\begingroup$ Phonons depend on the forces that atoms experience when other atoms move, and these are typically short range. This means that the matrix of force constants typically converges with a relatively small "coarse" $\mathbf{q}$-point grid (equivalent to a relatively small supercell). From this, you can then Fourier interpolate the dynamical matrix to any $\mathbf{q}$-point to access phonons of any wavelength. This is what is typically done to, for example, plot a "continuous" phonon dispersion. $\endgroup$– ProfMCommented May 16, 2021 at 21:41