Stefano Maffezzoli Felis edited Energies.tex  over 8 years ago

Commit id: f15d9d4b6477f5aea9594f9cba93505c393e20fc

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Each block can be easily diagonalized and one gets eigentates and eigenenergies of the total Hamiltonian  \begin{equation}  \begin{align}  &E^{1/2}_N=\epsilon_N\mp \sqrt{(2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^2})^2+(\epislon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)-\epislon_1\; \sqrt{(2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^2})^2+(\epsilon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)-\epsilon_1\;  e^{-2\Gamma_1^2}L_N(4\Gamma_1^2))^2}\\\nonumber &E^{3/4}_N=\epsilon_N\mp \sqrt{(2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^2})^2+(\epislon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)+\epislon_1\; \sqrt{(2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^2})^2+(\epsilon_2\; e^{-2\Gamma_2^2}L_N(4\Gamma_2^2)+\epsilon_1\;  e^{-2\Gamma_1^2}L_N(4\Gamma_1^2))^2}\\ \end{align}  \end{equation}