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
\[ D_{\mu \mu'}(\kappa\kappa'; \mathbf{q})
= \frac{1}{ \sqrt{ M_{\kappa} M_{\kappa'} } } \sum_{l'} \Phi_{ \mu \mu' }(0\kappa, l'\kappa') e^{i \mathbf{q} \cdot ( \mathbf{r}(l'\kappa') - \mathbf{r}(0\kappa) )}\]
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
\[ g u_{\mu}(\mathbf{r}(l\kappa))
:= \sum_{\nu} [\mathbf{R}_{g}]_{\mu\nu} u_{\nu}(g \mathbf{r}(l\kappa)).\]
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
\[ \mathbf{\Phi}(\mathbf{R}_{g} \mathbf{q})
= \mathbf{\Gamma}^{\mathbf{q}}(g) \mathbf{\Phi}(\mathbf{q}) \mathbf{\Gamma}^{\mathbf{q}}(g)^{\dagger}.\]
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
\[ \tilde{\Gamma}^{\mathbf{q}\alpha s}(h)_{\lambda'\lambda}
:= \sum_{\sigma\nu\nu'} c(\mathbf{q}\alpha s\lambda')_{\sigma\nu'}^{\ast} \Gamma^{\mathbf{q}\alpha}(h)_{\nu'\nu} c(\mathbf{q}\alpha s\lambda)_{\sigma\nu}
\quad (h \in \mathcal{G}^{\mathbf{q}}).\]
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 \(S^{2 d_{\alpha} - 2}\).