|
For simplicity we consider a Bravais lattice. The crystal Hamiltonian is
{$$\begin{equation} \sum _{\mathbf n} \frac {P_{\mathbf n}^2} {2M} + \frac 1 2 \sum_{\mathbf n \mathbf m} \mathbf u_{\mathbf n}\cdot C_{\mathbf n\mathbf m}\cdot u_{\mathbf m}\end{equation}$$}
It corresponds to a system of classical Newton equations where the forces are the opposite of the derivatives of the second term, the potential energy
{$$ M\ddot u_{\mathbf n j} = -\sum_{\mathbf n \mathbf m} C_{\mathbf n j\mathbf m l} u_{\mathbf n l}$$}
We want to identify the collective excitations of the atoms that correspond to the energy dispersion, i.e. the normal modes, and to express the Hamiltonian in terms of these modes. By Bloch theorem each component of the solution that obeys the periodicity of the lattice must be of the form
{$$u_{\mathbf nj}(\mathbf k,t) = v_{\mathbf k j} e^{i\mathbf k\cdot \mathbf R_{\mathbf n }} e^ {-i\omega t}$$}
where {$ \mathbf k = \frac {n_a} {N_a}\mathbf a^* +\frac {n_b} {N_b} \mathbf b^* +\frac {n_c} {N_c}\mathbf c^*$}
Substituting this solution in the Newton equations the left member is {$-\omega^2 u_{\mathbf n j}$} and the following dynamical matrix appears in the right member
{$$D_{jl}(\mathbf k) =\sum_{\mathbf m} \frac {C_{\mathbf n j, \mathbf m l}} {M} e^{i\mathbf k \cdot (\mathbf R_{\mathbf m}-\mathbf R_{\mathbf n})}$$}
so that the system of equations become
{$$\omega^2 \, v_{\mathbf kj} = \sum_ l D_{jl}(\mathbf k) \,\, v_{\mathbf k l} $$}
The solution is given by the following determinant equation
{$$\begin{equation}|D_{jl}(\mathbf k )-\omega^2 \delta_{jl}|=0\end{equation}$$}
with eigenvalues {$\omega_{\mathbf kj}$} and unit eigenvectors {$\hat v_{\mathbf kj}$}, identifying the three directions in which atoms collectively oscillate with a specific frequency, the eigenvalue of the determinant equation. These are the two transverse and the longitudinal modes.
{
The solutions yield the unitary displacements of atom {$\mathbf n$} in the direction of the eigenvector {$\hat v_{\mathbf k j}$} due to the {$\mathbf k$} eigenvectors as {$ \mathbf u_{\mathbf n}(\mathbf k, t) = \hat v_{\mathbf kj} e^{i\mathbf k\cdot \mathbf R_{\mathbf n} -i\omega(\mathbf k) t}$}. The general displacement of atom {$\mathbf n$} may be expressed in terms of these unitary displacements as
{$$\mathbf u_{\mathbf n}(t) = \frac 1 {\sqrt{ NM}}\sum_{\mathbf kj}\underbrace{A_{\mathbf kj} e^{-i\omega_{\mathbf kj} t}}_{ Q_{\mathbf kj}}\hat v_{\mathbf kj}\,e^{i\mathbf k\cdot \mathbf R_{\mathbf n} }$$}
The conjugated moments of these displacement is
{$$\mathbf P_{\mathbf n} = M\dot {\mathbf u}_{\mathbf n}(t) =-i\sqrt{\frac M N}\sum_{\mathbf kj}\underbrace{\dot Q_{\mathbf kj}}_{ P_{\mathbf kj}} \hat v_{\mathbf kj}\,e^{i\mathbf k\cdot\mathbf R_n}$$}
We want to rewrite the Hamiltonian in terms of the components at wave vector {$\mathbf k$} in direction {$j$}, {$\ Q_{\mathbf kj}$}, called the normal modes, and their conjugated moment {$ P_{\mathbf kj}$}. By substituting both {$\mathbf u_{\mathbf n}$} and {$\mathbf P_{\mathbf n}$} in Eq. (1) we get
{$$ \begin{align*}{\cal H} &=\sum_{\mathbf k \mathbf k^\prime } \underbrace{\frac 1 N\sum_{\mathbf n}e^{i(\mathbf k+\mathbf k^\prime)\mathbf R_{\mathbf n}}}_{\delta(\mathbf k +\mathbf k^\prime)} \frac 1 {2}
\left[\sum_j P_{\mathbf k j} P_{\mathbf k^\prime j} \right.\\&+ \sum_{jl} v_{\mathbf k j}
\underbrace{\sum_{\mathbf m } \frac {C_{\mathbf n j \mathbf m l}} M e^{i\mathbf k^\prime\cdot(\mathbf R_{\mathbf m}-\mathbf R_{\mathbf n}) } }_{D_{j l}(\mathbf k^\prime)} v_{\mathbf k^\prime l}
Q_{\mathbf kj}Q_{\mathbf k^\prime l}\Bigg]\end{align*}$$}
Notice that in the second term we used
{$$ e^{i(\mathbf k \cdot \mathbf R_n+\mathbf k^\prime \cdot \mathbf R_m)} = e^{i(\mathbf k+\mathbf k^\prime)\cdot\mathbf R_{\mathbf n}} e^{i\mathbf k^\prime\cdot (\mathbf R_{\mathbf m}-\mathbf R_{\mathbf n})}$$}
The delta function in the sum over {$\mathbf k^\prime$} yields {$\mathbf k^\prime=-\mathbf k$}. Consider further the second term. Since the sum over {$j,l$} is equivalent to the trace of {$D_{jl}(\mathbf k)$}, the trace is invariant for choice of basis, i.e. it can be taken in the diagonal form, {$v^2_{\mathbf kj}=1$} and {$\omega_{-\mathbf kj}=\omega_{\mathbf kj}$}, by Eq. (2) this second term becomes {$M^2\omega^2_{\mathbf kj}Q_{\mathbf kj} Q_{-\mathbf kj} $}. Furthermore the displacements are real, therefore {$Q_{-\mathbf kj}= Q^*_{\mathbf kj}$}, and, as a consequence {$P_{-\mathbf kj}=P^*_{\mathbf kj}$}. Finally we obtain
{$$ {\cal H} =\frac 1 2\sum_{\mathbf kj} [| P_{\mathbf kj}|^2 +\omega^2_{\mathbf kj} | Q_{\mathbf kj}|^2]$$}
The result can be extended to the equivalent dynamical matrix determinant equation that is obtained for crystal lattice with a basis of {$\tau=1\cdots r$} atoms. The only subtle point is that we define local versions of the eigenvectors
{$$v_{\mathbf k \tau j} = v_{\mathbf k j}e^{i\mathbf k\cdot \mathbf r_\tau } $$}
with
{$$\begin{equation}|D_{j \tau l \tau^\prime}(\mathbf k )-\omega^2 \delta_{jl}\delta_{\tau \tau^\prime}|=0\end{equation}$$}
Here we get {$3r$} solutions for each {$\mathbf k$} vector, with eigenvalues {$\omega_{\mathbf ks}$} {$s=1\cdots 3r$} and eigenvectors {$v_{\mathbf k s}$}. Hence for each eigenvector we have {$r$}
{$$v_{\mathbf k \tau s} = v_{\mathbf k s}e^{i\mathbf k\cdot \mathbf r_\tau } $$}
distinct local vectors, proportional to one another, one each atom of the basis. The normal coordinates and their conjugated moments are
{$$Q_{\mathbf ks}= A_{\mathbf ks} e^{-i\omega_{\mathbf ks} t}\qquad P_{\mathbf ks}= \dot Q_{\mathbf ks} $$}
and the Hamiltonian in this case reads
{$$ {\cal H} =\frac 1 {2}\sum_{\mathbf ks} [| P_{\mathbf ks}|^2 +\omega^2_{\mathbf ks} | Q_{\mathbf ks}|^2]$$}
| 