Lattice vibration

Harmonic phonon

Hamiltonian and commutation relations:

\[ H = \sum_{ l \kappa \mu } \frac{ p_{\mu}(l\kappa)^{2} }{ 2 M_{\kappa} } + \frac{1}{2} \sum_{ l \kappa \mu } \sum_{ l' \kappa' \mu' } \Phi_{ \mu \mu' }(l\kappa, l'\kappa') u_{\mu}(l\kappa) u_{\mu'}(l'\kappa')\]

Dynamical matrix 1

\[ D_{\mu \mu'}(\kappa\kappa'; \mathbf{q}) = \frac{1}{ \sqrt{ M_{\kappa} M_{\kappa'} } } \sum_{l'} \Phi_{ \mu \mu' }(0\kappa, l'\kappa') e^{i \mathbf{q} \cdot ( \mathbf{r}(l'\kappa') - \mathbf{r}(0\kappa) )}\]

We denote \([\mathbf{D}(\mathbf{q})]_{\kappa\mu, \kappa'\mu'} = D_{\mu \mu'}(\kappa\kappa'; \mathbf{q})\), then

\[\begin{split} \mathbf{D}(-\mathbf{q}) &= \mathbf{D}(\mathbf{q})^{\ast} \\ \mathbf{D}(\mathbf{q})^{\dagger} &= \mathbf{D}(\mathbf{q}) \quad \mbox{(Hermite)}.\end{split}\]

Here we use \(\Phi_{ \mu' \mu }(0\kappa', -l\kappa) = \Phi_{ \mu \mu' }(0\kappa, l'\kappa')\).

Let normalized eigenvector of \(\mathbf{D}(\mathbf{q})\) as \([\mathbf{e}(\mathbf{q}\nu)]_{\kappa\mu} = e_{\mu}(\kappa; \mathbf{q}\nu)\) with

\[\begin{split} \mathbf{D}(\mathbf{q}) \mathbf{e}(\mathbf{q}\nu) &= \omega_{\mathbf{q}\nu}^{2} \mathbf{e}(\mathbf{q}\nu) \quad (\nu = 1, \dots, 3N) \\ \omega_{-\mathbf{q}\nu}^{2} &= \omega_{\mathbf{q}\nu}^{2}.\end{split}\]

We can choose as

\[ e_{\mu}(\kappa; -\mathbf{q}\nu) = e_{\mu}(\kappa; \mathbf{q}\nu)^{\ast}.\]

Later we denote \(q = (\mathbf{q}, \nu)\) and \(-q = (-\mathbf{q}, \nu)\).

Normal coordinates

\[\begin{split} u_{\mu}(l\kappa) &=: \frac{1}{\sqrt{ L^{3} M_{\kappa} }} \sum_{q} Q_{q} e_{\mu}(\kappa; q) e^{i \mathbf{q} \cdot \mathbf{r}(l\kappa)} \\ p_{\mu}(l\kappa) &=: \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{q} P_{q} e_{\mu}(\kappa; -q) e^{-i \mathbf{q} \cdot \mathbf{r}(l\kappa)} \\ Q_{q} &= \sum_{ l\kappa\mu } \sqrt{ \frac{M_{\kappa}}{L^{3}} } u_{\mu}(l\kappa) e_{\mu}(\kappa; -q) e^{-i \mathbf{q} \cdot \mathbf{r}(l\kappa)} \\ P_{q} &= \sum_{ l\kappa\mu } \frac{1}{\sqrt{ L^{3}M_{\kappa} }} p_{\mu}(l\kappa) e_{\mu}(\kappa; q) e^{i \mathbf{q} \cdot \mathbf{r}(l\kappa)} \\ Q_{-q} &= Q_{q}^{\ast} \\ P_{-q} &= P_{q}^{\ast} \\ H &= \frac{1}{2} \sum_{q} \left( P_{q} P_{-q} + \omega_{q}^{2} Q_{q} Q_{-q} \right) \\\end{split}\]

Creation and annihilation operators

\[\begin{split} A_{q} &:= \frac{1}{\sqrt{2\hbar\omega_{q}}} \left( \omega_{q} Q_{q} + i P_{-q} \right) \\ A_{q}^{\dagger} &:= \frac{1}{\sqrt{2\hbar\omega_{q}}} \left( \omega_{q} Q_{-q} - i P_{q} \right) \\ Q_{q} &= \sqrt{\frac{\hbar}{2\omega_{q}}} \left( A_{q} + A_{-q}^{\dagger} \right) \\ P_{q} &= i \sqrt{\frac{\hbar\omega_{q}}{2}} \left( A_{-q}^{\dagger} - A_{q} \right) \\ \left[ A_{q}, A_{q'}^{\dagger} \right] &= \delta_{qq'} \\ \left[ A_{q}, A_{q'} \right] &= 0 \\ \left[ A_{q}^{\dagger}, A_{q'}^{\dagger} \right] &= 0 \\ H &= \frac{1}{2} \sum_{q} \hbar \omega_{q} \left( A_{q}^{\dagger}A_{q} + \frac{1}{2} \right)\end{split}\]

In canonical ensemble:

\[\begin{split} \langle A_{q}^{\dagger}A_{q} \rangle_{\beta} &= \frac{1}{ e^{\beta \hbar \omega_{q}} - 1 } \\ \langle |Q_{q}|^{2} \rangle_{\beta} &= \frac{\hbar}{2\omega_{q}} \left( 2 \langle A_{q}^{\dagger}A_{q} \rangle_{\beta} + 1 \right)\end{split}\]

Action on displacements

We define left group action for positions \(\mathbf{r}(l\kappa) = \mathbf{r}(l) + \mathbf{r}(0\kappa)\) by \(g = (\mathbf{R}_{g}, \mathbf{\tau}_{g}) \in \mathcal{G}\) as

\[ g \mathbf{r}(l\kappa) := \mathbf{R}_{g} \mathbf{r}(l\kappa) + \mathbf{\tau}_{g}.\]

We denote that site \(\mathbf{r}(0, \kappa)\) is transformed to \(\mathbf{r}(0, g \kappa ) + \mathbf{h}_{g}(\kappa)\) by symmetry operation \(g\). Then,

\[\begin{split} g \mathbf{r}(\mathbf{l}, \kappa) &= \mathbf{R}_{g} \mathbf{r}(l) + \mathbf{r}(0, g\kappa) + \mathbf{h}_{g}(\kappa) \\\end{split}\]

We define left group action for displacement at \(\mathbf{r}(l\kappa)\) as 2

\[ g u_{\mu}(\mathbf{r}(l\kappa)) := \sum_{\nu} [\mathbf{R}_{g}]_{\mu\nu} u_{\nu}(g \mathbf{r}(l\kappa)).\]

Consider Fourier transformation of \(\mathbf{u}(\mathbf{r}(l\kappa))\)

\[\begin{split} \mathbf{u}(\kappa; \mathbf{q}) &:= \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{l} \mathbf{u}(\mathbf{r}(l\kappa)) e^{ i \mathbf{q} \cdot \mathbf{r}(l) } \\ \mathbf{u}(\mathbf{r}(l\kappa)) &= \frac{1}{\sqrt{M_{\kappa}L^{3}}} \sum_{\mathbf{q}} \mathbf{u}(\kappa; \mathbf{q}) e^{ -i \mathbf{q} \cdot \mathbf{r}(l) } \\ g u_{\mu}(\kappa; \mathbf{q}) &= \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{l} g u_{\mu}(\mathbf{r}(l\kappa)) e^{i \mathbf{q}\cdot \mathbf{r}(l)} \\ &= \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{l}\sum_{\nu} R_{g,\mu\nu} u_{\nu}\left( \mathbf{R}_{g}\mathbf{r}(l) + \mathbf{r}(0, g\kappa) + \mathbf{h}_{g}(\kappa) \right) e^{i \mathbf{q}\cdot \mathbf{r}(l)} \\ &\quad (\mathbf{r}(l') := \mathbf{R}_{g}\mathbf{r}(l) + \mathbf{h}_{g}(\kappa) ) \\ &= \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{l'}\sum_{\nu} R_{g,\mu\nu} u_{\nu}\left( \mathbf{r}(l', g\kappa) \right) e^{i \mathbf{q}\cdot \mathbf{R}_{g}^{-1}(\mathbf{r}(l') - \mathbf{h}_{\kappa} ) } \\ &= \sqrt{\frac{M_{\kappa}}{L^{3}}} \sum_{l'}\sum_{\nu} R_{g,\mu\nu} u_{\nu}\left( \mathbf{r}(l', g\kappa) \right) e^{i \mathbf{R}_{g}\mathbf{q}\cdot (\mathbf{r}(l') - \mathbf{h}_{\kappa} ) } \quad (\because \mathbf{R}_{g} \in O(3)) \\ &= \sum_{\kappa'\mu'} u_{\mu'}(\kappa'; \mathbf{R}_{g} \mathbf{q} ) \Gamma_{\kappa'\mu'; \kappa\mu}^{\mathbf{q}}(g) \quad (\because M_{g\kappa} = M_{\kappa}),\end{split}\]

where

(1)\[\begin{split} \Gamma_{\kappa'\mu'; \kappa\mu}^{\mathbf{q}}(g) &:= \exp \left( -i \mathbf{R}_{g} \mathbf{q} \cdot \mathbf{h}_{g}(\kappa) \right) [\mathbf{R}_{g}]_{\mu'\mu} \delta_{ g\kappa, \kappa' } \\ &:= \exp \left( -2 \pi i \mathbf{R}_{g, f}^{\top} \mathbf{q}_{f} \cdot \mathbf{h}_{g, f}(\kappa) \right) [\mathbf{R}_{g}]_{\mu'\mu} \delta_{ g\kappa, \kappa' } \\ &\quad ( \mathbf{R}_{g, f} := \mathbf{A}^{-1} \mathbf{R}_{g} \mathbf{A}, \mathbf{q} =: 2\pi \mathbf{A}^{-\top} \mathbf{q}_{f}, \mathbf{v} =: \mathbf{A} \mathbf{v}_{f} )\end{split}\]

Equation (1) is essentially the same with Eq. (2.37) of [MV68].

We write matrix representation \([\mathbf{\Gamma}^{\mathbf{q}}(g)]_{ \kappa'\mu'; \kappa\mu } := \Gamma_{\kappa'\mu'; \kappa\mu}^{\mathbf{q}}(g)\). Then,

\[\begin{split} \left[ \mathbf{\Gamma}^{ \mathbf{q}}(gg') \right]_{ \kappa'\mu', \kappa\mu } &= \delta_{gg'\kappa, \kappa'} R_{gg', \mu'\mu} \exp \left( -i \mathbf{R}_{gg'}\mathbf{q} \cdot \mathbf{h}_{gg'}(\kappa) \right) \\ &= \sum_{ \kappa''\nu } \delta_{g\kappa'', \kappa'} \delta_{g'\kappa, \kappa''} R_{g, \mu'\nu} R_{g', \nu\mu} \exp \left( -i \mathbf{R}_{g'}\mathbf{q} \cdot \mathbf{h}_{g'}(\kappa) \right) \exp \left( -i \mathbf{R}_{g}\mathbf{R}_{g'}\mathbf{q} \cdot \mathbf{h}_{g}(g\kappa') \right) \\ &= \left[ \mathbf{\Gamma}^{ \mathbf{R}_{g'} \mathbf{q}}(g) \mathbf{\Gamma}^{ \mathbf{q}}(g') \right]_{ \kappa'\mu', \kappa\mu } \\ \mathbf{\Gamma}^{\mathbf{q}}(g)^{\dagger} \mathbf{\Gamma}^{ \mathbf{q}}(g) &= \mathbf{1} \quad \mbox{(Unitary)} \\ \Gamma^{\mathbf{q}}((E, \mathbf{t}))_{ \kappa'\mu', \kappa\mu } &= \exp \left( -i \mathbf{q} \cdot \mathbf{t} \right) \delta_{\mu'\mu} \delta_{ \kappa, \kappa' }.\end{split}\]

Fourier transformation of force constants

\[\begin{split} \Phi_{\mu\mu'}(\kappa\kappa'; \mathbf{q}) &:= \frac{1}{\sqrt{M_{\kappa}M_{\kappa'}}} \sum_{l'} \Phi_{\mu\mu'}(0\kappa; l'\kappa') e^{ i \mathbf{q} \cdot \mathbf{r}(l') } \\ \sum_{ l l' } \Phi_{ \mu \mu' }(l\kappa, l'\kappa') u_{\mu}(l\kappa) u_{\mu'}(l'\kappa') &= \sum_{\mathbf{q}} \Phi_{\mu\mu'}(\kappa\kappa'; \mathbf{q}) u_{\mu}(\kappa; \mathbf{q}) u_{\mu'}(\kappa'; -\mathbf{q}) \\\end{split}\]

The condition that potential energy is invariant under symmetry operations is rewritten as 3

\[ \mathbf{\Phi}(\mathbf{R}_{g} \mathbf{q}) = \mathbf{\Gamma}^{\mathbf{q}}(g) \mathbf{\Phi}(\mathbf{q}) \mathbf{\Gamma}^{\mathbf{q}}(g)^{\dagger}.\]

Small representation of \(\mathcal{G}^{\mathbf{q}}\)

For \(h, h' \in \mathcal{G}^{\mathbf{q}}\),

\[\begin{split} \mathbf{\Gamma}^{ \mathbf{q}}(h) \mathbf{\Gamma}^{ \mathbf{q}}(h') &= \mathbf{\Gamma}^{ \mathbf{q}}(hh') \\ \mathbf{\Phi}(\mathbf{q}) &= \mathbf{\Gamma}^{\mathbf{q}}(h) \mathbf{\Phi}(\mathbf{q}) \mathbf{\Gamma}^{\mathbf{q}}(h)^{\dagger} \quad (\forall h \in \mathcal{G}^{\mathbf{q}}).\end{split}\]

We can introduce projective representation \(\overline{\Gamma}^{ \mathbf{q}}\),

\[\begin{split} \mathbf{\Gamma}^{ \mathbf{q}}(h) &=: e^{ -i \mathbf{q} \cdot \mathbf{v}_{h} } \overline{\mathbf{\Gamma}}^{ \mathbf{q}}(h) \\ \overline{\mathbf{\Gamma}}^{ \mathbf{q}}(h) \overline{\mathbf{\Gamma}}^{ \mathbf{q}}(h') &= e^{ -i \mathbf{q} \cdot ( \mathbf{R}_{h} \mathbf{v}_{h'} - \mathbf{v}_{h'} ) } \overline{\mathbf{\Gamma}}^{ \mathbf{q}}(hh') \\ \overline{\mathbf{\Gamma}}^{ \mathbf{q}}((E, \mathbf{t})) &= \mathbf{1}\end{split}\]

Thus, we only need to consider projective representation \(\overline{\mathbf{\Gamma}}^{ \mathbf{q}}\) for little co-group \(\overline{\mathcal{G}}^{\mathbf{q}} \simeq \mathcal{G}^{\mathbf{q}} / \mathcal{T}\). The decomposition of the projective representation

\[\begin{split} \overline{\Gamma}^{\mathbf{q}} &= \sum_{\alpha} \sum_{\sigma} \overline{\Gamma}^{\mathbf{q}\alpha\sigma} \\\end{split}\]

can be performed with spgrep, where \(\alpha\) represent irrep and \(\sigma = 1,\dots, m_{\alpha}\) distinguish equivalent irreps to \(\alpha\). The corresponding small representation of \(\mathcal{G}^{\mathbf{q}}\) is obtained by \(\mathbf{\Gamma}^{ \mathbf{q}\omega}(h) := e^{ -i \mathbf{q} \cdot \mathbf{v}_{h} } \overline{\mathbf{\Gamma}}^{ \mathbf{q}\omega }(h)\). We call orthonormal basis vectors \(f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu)\) forming irrep \(\Gamma^{\mathbf{q}\alpha}\) as modified eigenvectors:

\[\begin{split} h f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu) &= \sum_{\nu'} f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu') \Gamma^{\mathbf{q}\alpha}(h)_{\nu',\nu} \quad (h \in \mathcal{G}^{\mathbf{q}}, \nu = 1, \dots, d_{\alpha}) \\ \sum_{\kappa\mu} f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu)^{\ast} f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu') &= \delta_{\nu\nu'} \quad (\nu, \nu' = 1, \dots, d_{\alpha})\end{split}\]

We can subdivide eigenvectors further by decomposing \(\Gamma^{\mathbf{q}}\) into irreps,

\[\begin{split} \left[ F^{\mathbf{q}\alpha\sigma} \right]_{\kappa\mu, \nu} &:= f_{\mu}(\kappa; \mathbf{q}\alpha\sigma\nu) \\ \mathbf{F}^{\mathbf{q}\alpha\sigma \dagger} \mathbf{\Gamma}^{\mathbf{q}}(h) \mathbf{F}^{\mathbf{q}\alpha\sigma} &= \mathbf{\Gamma}^{\mathbf{q}\alpha}(h) \quad (h \in \mathcal{G}^{\mathbf{q}}) \\\end{split}\]

Solve dynamical matrix w.r.t. modified eigenvectors

Block-diagonalize fourier transformed force constants:

\[\begin{split} \mathbf{F}^{\mathbf{q}\alpha} &:= \left( \mathbf{F}^{\mathbf{q}\alpha 1} \dots \mathbf{F}^{\mathbf{q}\alpha m_{\alpha}} \right) \quad \in \mathbb{C}^{3N \times m_{\alpha}d_{\alpha}} \\ \mathbf{\Phi}(\mathbf{q}\alpha) &:= \mathbf{F}^{\mathbf{q}\alpha \dagger} \mathbf{\Phi}(\mathbf{q}) \mathbf{F}^{\mathbf{q}\alpha} \quad \in \mathbb{C}^{ m_{\alpha}d_{\alpha} \times m_{\alpha}d_{\alpha} }, \\\end{split}\]

where \(\mathbf{\Phi}(\mathbf{q}\alpha)\) is hermitian.

Diagonalize \(\mathbf{\Phi}(\mathbf{q}\alpha)\)

\[\begin{split} \sum_{\sigma' \nu'} \Phi(\mathbf{q}\alpha)_{\sigma\nu, \sigma'\nu'} c(\mathbf{q} \alpha s\lambda)_{\sigma' \nu'} &= \omega_{\alpha s}^{2} c(\mathbf{q} \alpha s\lambda)_{\sigma \nu} \quad (s = 1, \dots, m_{\alpha}, \lambda = 1, \dots, d_{\alpha}) \\ \sum_{\sigma\nu} c(\mathbf{q}\alpha s\lambda)_{\sigma\nu}^{\ast} c(\mathbf{q}\alpha s'\lambda')_{\sigma\nu} &= \delta_{s s'} \delta_{\lambda \lambda'} \\ \sum_{s\lambda} c(\mathbf{q}\alpha s\lambda)_{\sigma\nu}^{\ast} c(\mathbf{q}\alpha s\lambda)_{\sigma'\nu'} &= \delta_{\sigma \sigma'} \delta_{\nu \nu'} \quad (\sigma = 1, \dots, m_{\alpha}, \nu = 1, \dots, d_{\alpha}) \\\end{split}\]

where \(s\) labels real eigenvalues \(\omega_{\alpha s}^{2}\) and \(\lambda\) labels degenerated eigenvectors. Here, we assume each equivalent irrep \(\Gamma^{\mathbf{q}\alpha \sigma}\) gives a different eigenvalue. Also, we choice eigenvectors \(\mathbf{c}(\mathbf{q} \alpha s\lambda) := \{ c(\mathbf{q}\alpha s\lambda)_{\sigma\nu} \}_{\sigma\nu}\) are mutually orthogonal even within degenerated eigenvalues.

When irrep \(\alpha\) appears more than once (\(m_{\alpha} > 1\)), irrep formed by eigenvectors is no longer same as \(\Gamma^{\mathbf{q}\alpha}\) [MV68]:

\[\begin{split} \tilde{\mathbf{f}}(\mathbf{q}\alpha s \lambda) &:= \sum_{\sigma\nu} \mathbf{f}(\mathbf{q}\alpha \sigma \nu) c(\mathbf{q}\alpha s\lambda)_{\sigma \nu} \\ h \tilde{\mathbf{f}}(\mathbf{q}\alpha s \lambda) &= \sum_{\lambda'} \tilde{\mathbf{f}}(\mathbf{q}\alpha s\lambda') \tilde{\Gamma}^{\mathbf{q}\alpha s}(h)_{\lambda'\lambda},\end{split}\]

where 4

\[ \tilde{\Gamma}^{\mathbf{q}\alpha s}(h)_{\lambda'\lambda} := \sum_{\sigma\nu\nu'} c(\mathbf{q}\alpha s\lambda')_{\sigma\nu'}^{\ast} \Gamma^{\mathbf{q}\alpha}(h)_{\nu'\nu} c(\mathbf{q}\alpha s\lambda)_{\sigma\nu} \quad (h \in \mathcal{G}^{\mathbf{q}}).\]

Now go back to the other convention of dynamical matrix:

\[\begin{split} e_{\mu}(\kappa; \mathbf{q}\alpha s \lambda) &:= e^{ -i\mathbf{q} \cdot \mathbf{r}(0\kappa) } \tilde{f}_{\mu}(\kappa; \mathbf{q}\alpha s \lambda) \\ D_{\mu\mu'}(\kappa\kappa'; \mathbf{q}) &= e^{ i \mathbf{q} \cdot \left( \mathbf{r}(0\kappa') - \mathbf{r}(0\kappa) \right) } \Phi_{\mu\mu'}(\kappa\kappa'; \mathbf{q}) \\ \sum_{ \kappa'\mu' } D_{\mu\mu'}(\kappa\kappa'; \mathbf{q}) e_{\mu'}(\kappa'; \mathbf{q}\alpha s\lambda) &= \omega_{\alpha s}^{2} e_{\mu}(\kappa; \mathbf{q}\alpha s \lambda) \\ h \mathbf{e}(\mathbf{q}\alpha s \lambda) &= \sum_{\lambda'} \mathbf{e}(\mathbf{q}\alpha s'\lambda') \tilde{\Gamma}^{\mathbf{q}\alpha s}(h)_{\lambda'\lambda} \quad (h \in \mathcal{G}^{\mathbf{q}}).\end{split}\]

Modulation

Modulation associated with qpoint \(\mathbf{q}\) and frequency \(\mathbf{\omega}_{\mathbf{q}}\):

\(\mathbf{q} \neq \mathbf{0}\) case:

\[\begin{split} u^{( \mathbf{q} \alpha s)}_{\mu}(l\kappa) &= \frac{1}{\sqrt{L^{3}M_{\kappa}}} \sum_{ \lambda } \left( Q^{ (\mathbf{q} \alpha s) }_{\lambda} e_{\mu}(\kappa; \mathbf{q} \alpha s \lambda) e^{ i \mathbf{q} \cdot \mathbf{r}(l\kappa) } + \mathrm{c.c.} \right) \\ Q^{ (\mathbf{q} \alpha s) }_{\lambda} &\in \mathbb{C} \\\end{split}\]

\(\mathbf{q} = \mathbf{0}\) case:

\[\begin{split} u^{( \mathbf{0} \alpha s)}_{\mu}(l\kappa) &= \frac{1}{\sqrt{L^{3}M_{\kappa}}} \sum_{ \lambda } Q^{ (\mathbf{0} \alpha s) }_{\lambda} e_{\mu}(\kappa; \mathbf{0} \alpha s \lambda ) \\ Q^{ (\mathbf{0} \alpha s) }_{\lambda} &\in \mathbb{R} \\\end{split}\]

From unitary arbitrariness of \(\tilde{\Gamma}^{\mathbf{q}\alpha s}\), we can choose as \(\mathrm{Im}\, Q^{ (\mathbf{q} \alpha s) }_{\lambda=1} = 0\). Thus, sampling of \(\mathbf{Q}^{ (\mathbf{q} \alpha s) }\) is attributed to sampling points from unit sphere \(S^{2 d_{\alpha} - 2}\).

References

MV68(1,2)

A. A. MARADUDIN and S. H. VOSKO. Symmetry properties of the normal vibrations of a crystal. Rev. Mod. Phys., 40:1–37, Jan 1968. URL: https://link.aps.org/doi/10.1103/RevModPhys.40.1, doi:10.1103/RevModPhys.40.1.


1

This is the same phase convention with phonopy. There is the other formulation for defining dynamical matrix as

\[\Psi_{\mu \mu'}(\kappa\kappa'; \mathbf{q}) = \frac{1}{ \sqrt{ M_{\kappa} M_{\kappa'} } } \sum_{l'} \Phi_{ \mu \mu' }(0\kappa, l'\kappa') e^{i \mathbf{q} \cdot \mathbf{r}(l')}.\]
2

This definition actually satisfies the condition of left group action

\[\begin{split} \left[ g \left( g' \mathbf{u}(\mathbf{r}(l\kappa)) \right) \right]_{\mu} &= \left[ g \left\{ \sum_{\nu} R_{g', \mu'\nu} u_{\nu}( g'\mathbf{r}(l\kappa) ) \right\}_{\mu'} \right]_{\mu} \\ &= \sum_{ \mu'\nu } R_{g, \mu\mu'} R_{g', \mu'\nu} u_{\nu}( gg'\mathbf{r}(l\kappa) ) \\ &= \sum_{ \nu } R_{gg', \mu\nu} u_{\nu}( gg'\mathbf{r}(l\kappa) ) \\ &= \left[ (gg') \mathbf{u}(\mathbf{r}(l\kappa)) \right]_{\mu}.\end{split}\]
3

The derivation is as follows:

\[\begin{split}\sum_{ l\kappa\mu l'\kappa'\mu' } \Phi_{ \mu \mu' }(l\kappa, l'\kappa') u_{\mu}(l\kappa) u_{\mu'}(l'\kappa') &= \sum_{ l\kappa\mu l'\kappa'\mu' } \Phi_{ \mu \mu' }(l\kappa, l'\kappa') gu_{\mu}(l\kappa) gu_{\mu'}(l'\kappa') \\ &= \dots = \sum_{ } \left[ \mathbf{\Gamma}^{\mathbf{q}}(g) \mathbf{\Phi}(\mathbf{q}) \mathbf{\Gamma}^{\mathbf{q}}(g)^{\dagger} \right]_{\kappa\mu, \kappa'\mu'} u_{\mu}(\kappa; \mathbf{R}_{g}\mathbf{q}) u_{\mu'}(\kappa'; -\mathbf{R}_{g}\mathbf{q}).\end{split}\]
4

The derivation is cumbersome (to me). Let \(\mathbf{C}^{\mathbf{q}\alpha s} = ( \mathbf{c}(\mathbf{q}\alpha s 1), \dots, \mathbf{c}(\mathbf{q}\alpha s d_{\alpha}) )\).

\[\begin{split}\tilde{\mathbf{\Gamma}}^{\alpha s}(h) &:= \mathbf{C}^{\mathbf{q}\alpha s} \begin{pmatrix} \mathbf{\Gamma}^{\mathbf{q}\alpha}(h) & & \\ & \ddots & \\ & & \mathbf{\Gamma}^{\mathbf{q}\alpha}(h) \\ \end{pmatrix} \mathbf{C}^{\mathbf{q}\alpha s \dagger} \\ \tilde{\Gamma}^{\alpha s}(h)_{\lambda' \lambda} &= \sum_{\sigma' \nu' \sigma \nu} c(\mathbf{q}\alpha s \lambda')_{\sigma'\nu'}^{\ast} \delta_{\sigma \sigma'} \Gamma^{\mathbf{q}\alpha}(h)_{\nu'\nu} c(\mathbf{q}\alpha s \lambda)_{\sigma\nu} \\ &= \sum_{\sigma \nu' \nu} c(\mathbf{q}\alpha s \lambda')_{\sigma\nu'}^{\ast} \Gamma^{\mathbf{q}\alpha}(h)_{\nu'\nu} c(\mathbf{q}\alpha s \lambda)_{\sigma\nu} \\\end{split}\]