General form, properties
A Lindblad form $$\dot \rho = -i[\eta, \rho] + A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A$$ has three important properties:
- It is still linear dynamics, in terms of $\rho$.
- It is trace-free regardless of the trace of $\rho$. This means that the total sum of the eigenvalues, which starts out as 1, does not change.
- It leaves $\dot\rho$ Hermitian, which is important because only Hermitian operators have all-real eigenvalues.
- There are usually some simple criteria on $\hat A$ that I don't really remember any more, which ensure the positivity of the Lindbladian, so it doesn't ever take the positive eigenvalues to negative ones.
Its general physical meaning is therefore "nonunitary dynamics which can nonetheless be modeled without slaughtering our state matrix."
One interpretation you can always reach for
If you are not pleased with that definition, the most common nonunitary process in quantum mechanics is measurement, so let me show you how you can interpret any Lindblad form as a continuous quantum measurement. This is a common[1] [2] way to consider taking some $\rho$ with otherwise unitary dynamics and coupling it with a continuous measurement of the system.
A simple measurement looks like this: we bring some qubit with energy Hamiltonian $\epsilon ~c^\dagger c$ to the system, put it in its ground state $|0\rangle\langle 0|$, which we can write as $c c^\dagger$ for short. We'll assume that whatever the qubit is made out of commutes with all of the operators etc. which properly act on the "system" $\rho$. The qubit then couples to the system with some interaction term $\hat v^\dagger c^\dagger + \hat v c$, extremely generic.
During a time $dt/2$ the system will then evolve like $$\rho ~cc^\dagger \mapsto \rho ~ c c^\dagger - i~\frac{dt}2\left([\eta, \rho] ~ c c^\dagger + \hat v^\dagger \rho ~ c^\dagger - ~\rho ~\hat v~ c\right) $$
The only problem here is that these latter terms are still a little "in the past"; so let's evolve each of those terms $\rho c$ and $\rho c^\dagger$ forward another $dt/2$ to find a second-order effect:$$\begin{align}
\rho ~c^\dagger \mapsto& \rho ~ c^\dagger - i~{dt\over 2}\left([\eta, \rho] ~ c^\dagger + \hat v \rho ~ c c^\dagger - ~\rho ~\hat v~ c^\dagger c\right)\\
\rho ~c \mapsto& \rho ~ c - i~{dt\over 2}\left([\eta, \rho] ~ c + \hat v^\dagger \rho ~ c^\dagger c - \rho ~\hat v^\dagger~ c c^\dagger\right) \end{align}$$
We then measure it in the qubit's $|0\rangle, |1\rangle$ basis and discard the measurement. This collapses the qubit to either $|0\rangle$ or $|1\rangle$ and therefore the $|0\rangle\langle 1| = c$ and $|1\rangle\langle 0| = c^\dagger$ terms of the density matrix, so let's look only at the $c c^\dagger$ and $c^\dagger c$ terms:
$$ \rho ~cc^\dagger \mapsto \rho ~ c c^\dagger - i~ dt [\eta, \rho] ~ c c^\dagger -\frac{dt^2}4 \left( \hat v^\dagger (\hat v \rho ~ c c^\dagger - ~\rho ~\hat v~ c^\dagger c) - ~(\hat v^\dagger \rho ~ c^\dagger c - \rho ~\hat v^\dagger~ c c^\dagger)~\hat v\right) $$We see that the $\hat v^\dagger \rho \hat v$ terms correspond to $c^\dagger c$ and apparently collapse the overall system somewhat infinitesimally, something like $|\psi\rangle \mapsto |\psi\rangle + \sqrt{dt} v^\dagger |\psi\rangle.$ Usually textbooks/papers say by fiat that "we're measuring $\sqrt{dt} \hat v^\dagger $" or so; this is the real interpretation: asymptotically strong coupling that does not grow as rapidly as the measurement interval that we're applying it over, so that we get state-lengthening due to the Quantum Zeno effect.
By "tracing over" the qubit, which is what you do when you want to get the effective density matrix of the system for all Hermitian system-operators generating expectation values, and defining $A = \sqrt{dt/2} ~ \hat v^\dagger$, this physical process corresponds to the first equation I wrote. It is therefore the limit of a system which is coupled to a qubit which you measure every time frame $dt$, which is coupled to the system by an interaction Hamiltonian $(A c^\dagger + A^\dagger c)/\sqrt{dt/2}.$
Other resources
You can often derive very similar expressions when, say, you weakly couple your system to an infinite bath of bosons, since those can also cause constant decoherence in a similar fashion. If you want some examples, Wiseman's textbook Quantum Measurement and Control may be up your alley. (I think it for example had a lasing cavity which naturally tended towards that expression where $A$ was just the annihilator of the bosons in the cavity, which explains that they come to a coherent state.) If you don't have it in your library, this arXiv paper, also linked above, covers much of the same ground. The buzzword is "quantum trajectories", which also covers simulations of quantum systems when you add measurements.
