mbd_hamiltonian Module

Form and solve either a real or a complex MBD Hamiltonian.

The real-typed version is equivalent to $\mathbf q=0$.

$$\begin{gathered} E_\text{MBD}(\mathbf q)=\frac12\operatorname{Tr}\big(\sqrt{\mathbf Q(\mathbf q)}\big)- 3\sum_i\frac{\omega_i}2,\qquad \mathbf Q_{ij}(\mathbf q)=\omega_i^2\delta_{ij}\mathbf I+ \omega_i\omega_j\sqrt{\alpha_{0,i}\alpha_{0,j}}\mathbf T_{ij}(\mathbf q) \\ \mathbf Q(\mathbf q)\equiv \mathbf C(\mathbf q)\boldsymbol\Lambda(\mathbf q)\mathbf C(\mathbf q)^\dagger,\qquad \boldsymbol\Lambda(\mathbf q) \equiv\operatorname{diag}(\{\tilde\omega_i(\mathbf q)^2\}),\qquad \operatorname{Tr}\big(\sqrt{\mathbf Q(\mathbf q)}\big) =\sum_i\tilde\omega_i(\mathbf q) \end{gathered}$$

$$\begin{aligned} \partial E_\text{MBD}&=\frac14\operatorname{Tr}\big( \mathbf C\boldsymbol\Lambda^{-\frac12}\mathbf C^\dagger \partial\mathbf Q \big)- 3\sum_i\frac{\partial\omega_i}2 \\ \frac{\partial E_\text{MBD}}{\partial X_i}&= \operatorname{Re}\frac12\sum_{p\zeta}( \mathbf C\boldsymbol\Lambda^{-\frac12}\mathbf C^\dagger )_{i\zeta,p} \frac{\partial Q_{p,i\zeta}}{\partial X_i}- \frac32\frac{\partial\omega_i}{\partial X_i} \end{aligned}$$

$$\begin{aligned} \partial\mathbf Q_{ij}=& 2\delta_{ij}\omega_i\partial\omega_i\mathbf I+ \omega_i\omega_j\sqrt{\alpha_{0,i}\alpha_{0,j}}\mathbf T_{ij}\left( \frac{\partial\omega_i}{\omega_i}+ \frac{\partial\omega_j}{\omega_j}+ \frac12\frac{\partial\alpha_{0,i}}{\alpha_{0,i}}+ \frac12\frac{\partial\alpha_{0,j}}{\alpha_{0,j}} \right) \\ &+\omega_i\omega_j\sqrt{\alpha_{0,i}\alpha_{0,j}} \partial\mathbf T_{ij} \end{aligned}$$