Phonons: Theory: Difference between revisions

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(Created page with "To understand them we start by looking at the Taylor expansion of the total energy (<math>E</math>) around the set of equilibrium positions of the nuclei (<math>\{\mathbf{R}^0...")
 
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\sum_{J\beta} D_{I\alpha J\beta} (\mathbf{q})
\sum_{J\beta} D_{I\alpha J\beta} (\mathbf{q})
\xi^{\mu }_{J\beta}(\mathbf{q}) =
\xi^{\mu }_{J\beta}(\mathbf{q}) =
\omega(\mathbf{q})^2 \xi^{\mu }_{I\alpha}(\mathbf{q})
\omega^\mu(\mathbf{q})^2 \xi^{\mu }_{I\alpha}(\mathbf{q})
</math>
</math>


with the dynamical matrix
with
:<math>
:<math>
D_{I\alpha J\beta} (\mathbf{q}) =  
D_{I\alpha J\beta} (\mathbf{q}) =  
\frac{1}{\sqrt{M_I M_J}} C_{I\alpha J\beta} e^{i\mathbf{q} \cdot (\mathbf{R}_J-\mathbf{R}_I)}
\frac{1}{\sqrt{M_I M_J}} C_{I\alpha J\beta} e^{i\mathbf{q} \cdot (\mathbf{R}_J-\mathbf{R}_I)}
</math>
</math>
the dynamical matrix in the harmonic approximation.
Now by solving the eigenvalue problem above we can obtain the phonon modes
<math>\xi^{\mu }_{I\alpha}(\mathbf{q})</math> and frequencies
<math>\omega^\mu(\mathbf{q})^2</math> at any arbitrary '''q''' point.

Revision as of 09:45, 1 August 2022

To understand them we start by looking at the Taylor expansion of the total energy () around the set of equilibrium positions of the nuclei ()

where the positions of the nuclei. The first term in the expansion corresponds to the forces

,

and the second to the second-order force-constants


We can define a variable that corresponds to the displacement of the atoms with respect to the equilibrium position which leads to

If we take the system to be in equilibrium, the forces on the atoms are zero and then the Hamiltonian of the system is

and the equation of motion

Using the following ansatz

where are the phonon mode eigenvectors. Replacing we obtain the following eigenvalue problem

with

the dynamical matrix in the harmonic approximation. Now by solving the eigenvalue problem above we can obtain the phonon modes and frequencies at any arbitrary q point.