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We found the following classical crystal Hamiltonian in terms of the normal modes
{$$ \begin{equation}{\cal H} =\frac 1 {2}\sum_{\mathbf ks} [|\mathbf P_{\mathbf ks}|^2 +\omega^2_{\mathbf ks}|\mathbf Q_{\mathbf ks}|^2]\end{equation}$$}
Let us define the following operators
{$$a_{\mathbf ks} = \frac 1{\sqrt {2 \hbar \omega_{\mathbf ks}}}(\omega_{\mathbf ks}Q_{\mathbf ks}+iP_{\mathbf ks})\qquad a^\dagger_{\mathbf ks} = \frac 1{\sqrt {2 \hbar \omega_{\mathbf ks}}}(\omega_{\mathbf ks}Q_{\mathbf k}-iP_{\mathbf ks})$$}
It can be checked that they obey
{$$[a_{\mathbf ks},a^\dagger_{\mathbf k^\prime s^\prime}]=\delta_{\mathbf k\mathbf k^\prime}\delta_{ss^\prime}\qquad [a^\dagger_{\mathbf ks},a^\dagger_{\mathbf ks}] = [a_{\mathbf ks},a_{\mathbf ks}]=0$$}
Their definition implies that
{$$Q_{\mathbf ks} = \sqrt{\frac \hbar {2\omega_{\mathbf ks}}}(a^\dagger_{\mathbf ks} +a_{\mathbf ks})\qquad
P_{\mathbf ks} = i\sqrt{\frac {\hbar\omega_{\mathbf ks}} 2} (a^\dagger_{\mathbf ks} - a_{\mathbf ks})$$}
and, by substitution in Eq. (1)
{$${\cal H} = \sum_{\mathbf ks}\hbar \omega_{\mathbf ks}\left(a_{\mathbf ks}a^\dagger_{\mathbf ks}+\frac 1 2\right)$$}
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