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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?

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  • $\begingroup$ The identity is always in the kernel which, up to renormalization, qualifies as a valid density matrix. $\endgroup$
    – LPZ
    Commented May 6, 2023 at 17:59
  • $\begingroup$ @LPZ this however doesn't tell me wether every linear independent element of the Kernel does, up to renormalization, qualify as a valdi density matrix. $\endgroup$ Commented May 6, 2023 at 19:26
  • $\begingroup$ As you’ve said yourself, there are elements of the kernel that are non valid density matrices even after renormalisation. If I understand correctly, your original question is whether there are density matrices in the kernel. The answer is yes just by this simple example. $\endgroup$
    – LPZ
    Commented May 6, 2023 at 22:29
  • $\begingroup$ @LPZ The identity matrix is not always in the kernel. It is always a left eigenvector, that is, an eigenvector of $\mathcal L^\dagger$, but only a right eigenvector if the channel is unital. $\endgroup$
    – Noiralef
    Commented May 7, 2023 at 0:31
  • $\begingroup$ @Noiralef aren't eigenvectors of $\mathcal{L}$ what I'm interested in, in that case? If I shove all this super-operator terminology asside, it's just a linear map that can have a kernel - no? So why should I be interested in $\mathcal{L}^{\dagger}$? $\endgroup$ Commented May 7, 2023 at 11:12

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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.

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  • $\begingroup$ Actually, you’d need to also check positivity of $\rho$ for it to be a density matrix (or else you could have negative probabilities). This is where knowing that the identity is in the kernel comes in handy. By convex combination, you can construct a valid density matrix. $\endgroup$
    – LPZ
    Commented May 6, 2023 at 22:27

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