I'm currently calculating the dynamics of the density operator for a 3-level system. To do this, I've set up a Hamiltonian as:
\begin{align}
H =
\left(
\begin{array}{ccc}
\hbar \delta _q & \hbar \tilde{\Omega }_q & 0 \\
\hbar \Omega _q & 0 & \hbar \Omega _d \\
0 & \hbar \tilde{\Omega }_d & \hbar \delta _d \\
\end{array}
\right)
\end{align}
For each of the spontaneous decay channels, the lindblad form of the decay is:
\begin{align}
D_\sigma[\rho]= (\sigma \rho \sigma^{\dagger} – \frac{1}{2}(\sigma^\dagger \sigma \rho + \rho \sigma^\dagger \sigma)
\end{align}
Then overall, the time evolution can be cast into the following form:
\begin{align}
\dot{\rho}= \frac{i}{\hbar}[\rho, H] + \Gamma_{12} D_{\sigma_{12}}[\rho] + \Gamma_{13} D_{\sigma_{13}}[\rho] + \Gamma_{23} D_{\sigma_{23}}[\rho] = \mathcal{L}[\rho]
\end{align}
With $\mathcal{L}$ being a linear map acting on the operator $\rho$.
I now try to find the steady state solution of the system, by simply finding the operator $\rho$ that $\mathcal{L}$ computes to 0 – If you want, the kernel of the system.
I do so by using a computer algebra system, and perform the calculation symbolically.
Now, when I find the 0-space, what usually happens is that the solutions I get have complex diagonal elements, which means that my solution is not hermitian.
Of course, solutions of eigenvalue equations are just solutions up to a complex constant, but by that alone I am not able to cure the non-hermicity in this case.
Which makes me wonder: Should I even expect to get a solution that qualifies as a density matrix? (Self-Adjoint, and trace 1), by finding the kernel of the Liouvillian? I know that the Liouvillian is trace and hermiticity preserving, but does that also play a role in my calculation here, where I didn't pick a density-operator $\rho(0)$ for a start?
Best Answer
I pondered a bit, and I think I have the answer to the question:
If $\rho$ is part of the Kernel, that is
\begin{align} \mathcal{L}[\rho]= 0 \end{align}
Then \begin{align} \mathcal{L}[\rho]^{\dagger}=0 \\ =(\frac{i}{\hbar}[\rho, H])^{\dagger} - \sum_i \Gamma_i ( \sigma_i \rho \sigma_i^{\dagger} - \frac{1}{2} ( \rho \sigma_i^{\dagger} \sigma + \sigma_i^{\dagger} \sigma \rho))^{\dagger} \\ =(\frac{i}{\hbar}[\rho^{\dagger}, H]) - \sum_i \Gamma_i ( \sigma_i \rho^{\dagger} \sigma_i^{\dagger} - \frac{1}{2} (\sigma_i^{\dagger} \sigma \rho^{\dagger} + \rho^{\dagger} \sigma_i^{\dagger} \sigma)) \\ \end{align} because $H$ is hermitean. Then \begin{align} = \mathcal{L}[\rho^{\dagger}]= 0 \end{align} That means, when $\rho$ is in the kernel, then $\rho^{\dagger}$ is as well, so for every element $\rho$ of the kernel, $\rho + \rho^{\dagger}$ is as well in the kernel, and qualifies (up to normalization of the trace} as a density operator.