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Stefano Maffezzoli Felis edited Energies.tex
about 8 years ago
Commit id: 852800425e5e69650c122f6bb0d27e90755ef415
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...
the eigenstates of this Hamiltonian are:
\begin{equation}\label{eqn:basis}
|N_{nm}nm>=\ddag{\gamma_{nm}}|N>|n>|m> |N_{nm}nm>=\ddag{\omega_0^{-1}\gamma_{nm}}|N>|n>|m>
\end{equation}
with $N\in\mathbb{N}$, $n,m=\{+,-\}$ and $\gamma_{nm}=n\Gamma_1+m\Gamma_2$ $\in\mathbb{R}$ are the eigenvalues of the operator $\hat{\gamma}$ on the 2-qubits basis $\{|nm>\}=\{|++>,|+->,|-+>,|-->\}$.
Negletting the constant terms the Hamiltonian (\ref{eqn:DHD}) becomes
\begin{equation}
H_{2q}=\hbar\omega_0 D^{\dagger}(\hat{\gamma})\adag a D(\hat{\gamma}) + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z -
2\hbar\Gamma_1\Gamma_2\sigma^{(1)}_x\otimes\sigma^{(2)}_x 2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^{-1}}\sigma^{(1)}_x\otimes\sigma^{(2)}_x
\end{equation}
Calculate now the matrix's elements in the basis of the eigenstates (\ref{eqn:basis}) in order to find later the eigenenergies of the system
...
&=-\epsilon_{12}(nm)\delta_{sn}\delta_{tm} \nonumber
\end{align}
\end{equation}
where $\bar{n}$ is the negation of $n$, $H_1=\hbar\omega_0 D^{\dagger}(\hat{\gamma})\adag a D(\hat{\gamma})$, $H_2=\hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z$, $H_3=-
2\hbar\Gamma_1\Gamma_2\sigma^{(1)}_x\otimes\sigma^{(2)}_x$, 2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^{-1}}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$, $\epsilon_N=\hbar\omega_0 N$, $\epsilon_1=\hbar \frac{\omega_1}{2}$, $\epsilon_2= \hbar \frac{\omega_2}{2}$ and
$\epsilon_{12}=2\hbar\Gamma_1\Gamma_2$. $\epsilon_{12}=2\hbar\frac{\Gamma_1\Gamma_2}{\omega_0^{-1}}$. One can easily identify two diagonal terms and an off-diagonal one.
The overlap between two displaced number states $|N_{nm}>$ is
\begin{equation}\label{eqn:Lag}