Lindblad equation unitary term (in practice) The Lindblad equation is given by
$$
\dot\rho=-{i\over\hbar}[H,\rho]+\sum_{i = 1}^{N^2-1} \gamma_i\left(L_i\rho L_i^\dagger -\frac{1}{2} \left\{L_i^\dagger L_i, \rho\right\} \right),
$$
where $H$ is not the usual Hamiltonian for the closed system but rather is given by some complicated combination of the system and system-bath interactions. However, in practice, I have gathered that people usually just set $H$ to be simply the system Hamiltonian. For instance, for the case of a harmonic oscillator coupled to a heat bath I have seen that people set $H=\omega a^{\dagger}a$.
From a physical point of view this makes complete sense to me: sure you carefully came up with a complicated equation but in essence, all it is saying is that $\rho$ mainly evolves unitarily by some operator (call it $H$) and non-unitary dynamics are taken care of by some other operators.( I.e. We could have guessed this equation to some degree by just physical considerations).
Now, if you decouple the system from the bath ($\gamma_i=0$) then $H$ really is the system’s Hamiltonian so it makes complete sense to me when people do things like $H=\omega a^{\dagger}a$. However, in general this is not correct (just look at the horrible formula by which $H$ is actually given) so my question is:
When is it okay, or at least a good approximation, to set $H$ equal to the system’s Hamiltonian?
Furthermore, since I can more or less choose the $L_i$'s in my model, is there a particular class of systems/$L_i$’s that admit this choice of $H$?
Thanks in advance!
 A: You may check "the theory of open quantum systems" by Prof. Petruccione, especially section 3.3 microscopic derivations. When a system is coupled to an environment, the dynamics of the total system+environment state will be governed by Schrödinger equation with the total Hamiltonian including system Hamiltonian $H_s$, environment Hamiltonian $H_e$, and interaction Hamiltonian between system and environment, $H_I$. When the environmental coupling spectrum, called spectral density that is fully characterised by the environment and interaction Hamiltonians, is broad enough and the overall coupling strength between system and environment is weak enough, the dynamics of the open quantum system can be described by a Lindblad equation using a second order perturbation theory with several approximations (Born, Markov and secular). In this case, the Lindblad equation in the Schrödinger picture will be described by
$$
\frac{\mathrm{d}\rho_s(t)}{\mathrm{d}t} = -i[H_s+H_{\text{Lamb}},\rho_s(t)] + L[\rho_s(t)],
$$
where $\rho_s(t)$ is the reduced density matrix of the open quantum system, obtained by partial trace over environmental degrees of freedom, $H_s$ is the system Hamiltonian you mentioned, and $H_{Lamb}$ is the Lamb shift term describing the energy-level shift of the eigenstates of the system Hamiltonian due to environmental couplings, meaning that the energies of system eigenstates will be modified by environment. In addition to the Lamb shift, the environment will destroy the coherence of the system density matrix, described by the dissipator $L[\rho_s(t)]$ you mentioned. The decoherence rates $\gamma_i$ and noise operators $L_i$ in your equations will be determined by the properties of system eigenstates and environmental spectral densities where the latter is fully characterised by environment and interaction Hamiltonians. In summary, when there is a model Hamiltonian for system and environment, the dynamics of an open quantum system can be described by a Lindblad equation when certain conditions should be satisfied (Born, Markov, secular approximations). The decoherence rates $\gamma_i$ and noise operators $L_i$ you mentioned will be determined by the second order perturbation theory. For more details, you may check the book by Prof. Petruccione. I hope this information will be helpful!
A: As was suggested by user296595, Breuer & Petruccione book chapter 3.3 Microscopic derivation is what you need.
I suppose, you are interested whether a harmonic oscillator, if interacting with a bath, will remain harmonic.
Here is the general summary of relevant formulas and definitions with application to a harmonic oscillator at the end. Looking at the equation (3.141), the contribution from the thermostat to Hamiltonian in the interaction picture is
$$
H_{LS}=\sum_{\omega}\sum_{\alpha\beta}S_{\alpha\beta}(\omega)A_\alpha(\omega)A^\dagger_\alpha(\omega),
$$
where $S_{\alpha\beta}(\omega)$ is a function dependent on the thermostat, and $A_\alpha(\omega)$ are some operators, derived from the complete interaction operator of the system with thermostat. If the full interaction operator is expressed as (Eq. 3.119 in the book)
$$
H_I=\sum_{\alpha} A_\alpha\otimes B_{\alpha}
$$
then $A_\alpha(\omega)$ is defined as
$$
A_\alpha(\omega)=\sum_{\epsilon'-\epsilon=\omega} \Pi(\epsilon)A_\alpha\Pi(\epsilon'),
$$
where $\Pi(\epsilon)$ is a projection operator onto the state with the energy $\epsilon$.
Edit: As one can see, if $H_I\sim\varepsilon$ then $H_{LS}\sim\varepsilon^2$.
Applying this machinery to a harmonic oscillator, you can derive after some work, that the oscillator remains harmonic if and only if the interaction terms $A_\alpha$ contain only $a$ and $a^{\dagger}$, i.e. your bath interacts linearly with the oscillator.
