Example of an infinite volume Lindblad system What is an explicit example of a Lindbladian
\begin{align*}
L(\rho) = - i \lbrack H_A, \rho \rbrack + G \sum_{j} V_j \rho V_j^* - \frac{1}{2}(V_j^* V_j \rho  + \rho V_j^* V_j)
\end{align*}
acting on the space of trace class operators on some Hilbert space $\mathcal{H}$ such that there is a trace-class operator $\rho_\infty$ with $L(\rho_\infty ) = 0$.
EDIT: As pointed out in an answer I did not specify what I mean by infinite volume. I mean that the Hilbert space is for example $l^2(\mathbb{Z})$ with the standard orthonormal basis and that the state $\rho_\infty$ should be an element of the trace-class operators on $l^2(\mathbb{Z})$. I.e. that the system is not just a single mode, but physically extended.
 A: "Infinite volume Lindblad system" doesn't make much sense to me. Apparently, what you are asking for is an example of a Lindblad system with steady state $\rho_{\infty}$: if $L(\rho_{\infty})=0$, then this state is invariant under the action of the dynamical semigroup $\phi(t)=\exp L t$ driving the open system dynamics.
The simplest example we may think of is a single bosonic mode $a$ with frequency $\omega$ (e.g. a monochromatic laser in the quantum regime with very few photons) immersed in the electromagnetic field at zero temperature. The open system dynamics is then given by the following master equation [1]:
$$
L(\rho)=-i[H_S,\rho]+\gamma_0 \left(a \rho a^\dagger-\frac{1}{2}\{a^\dagger a,\rho\}\right),
$$
where $H_S=\hbar\omega a^\dagger a$ and $\gamma_0$ is the standard emission coefficient. You can easily verify that $\rho_\infty=|0\rangle\!\langle 0|$ is the unique steady state of the open dynamics; physically, this means that the bosonic mode is decaying toward the vacuum state.

As for an open quantum system spread over an infinite volume, we may consider a chain of $2n+1$ harmonic oscillators with bosonic annihilation operators $\{a_j\}_{j=-n}^n$, and then let $n$ goes to infinity. The system Hamiltonian is $H_S=\sum_{j=-n}^n \hbar\omega_j a_j^\dagger a_j$, and we consider a single common bosonic bath acting all over the chain. The bath Hamiltonian is $H_B=\sum_k \hbar\Omega_k b_k^\dagger b_k$, and we take it at zero temperature, i.e. the state of the bath is $\rho_B=\bigotimes_k |0\rangle_k\langle 0|$. We connect the harmonic oscillators and the thermal bath through the interaction Hamiltonian $H_I=\sum_{k} g_k \left(\sum_{j=-n}^n a_j b_k^\dagger +h.c.\right)$. Following the standard derivation of the master equation in the Markovian limit [1], you can see that the GKLS equation driving the dynamics of the state of the system of harmonic oscillators $\rho_S$ reads:
$$
L(\rho_S)=-i[H_S,\rho_S]+\sum_{j=-n}^n\sum_{j'=-n}^n\gamma_0 \left( a_j \rho a_{j'}^\dagger-\frac{1}{2}\{a_{j'}^\dagger a_j,\rho\}\right).
$$
Once again, you can easily verify that $\rho_{\infty}=\bigotimes_{j=-n}^n |0\rangle_j\langle 0|$ (the ground state of the system Hamiltonian) is the steady state of the dynamics. This is because the action of the thermal bath (expressed by the collective jump operator $\sum_{j=-n}^n a_j$, i.e. annihilation of all the oscillator modes) consists in completely absorbing the energy of the system.
[1] H.-P. Breuer and F. Petruccione, The theory of open quantum systems (Oxford University Press, 2002).
A: In the following paper https://arxiv.org/abs/2206.09879 (work I did in connection to asking the question) several examples of infinite volume Lindblad systems are discussed.
In the single particle sector, consider the Hilbert space $l^2(\mathbb{Z})$ and some space of operators on that Hilbert space, e.g. trace-class, Hilbert-Schimdt, compact or bounded operators. Then consider the Lindbladian $\mathcal{L}$ defined on one of these space, for example with a nearest neighbor hopping Hamiltonian
$ H = \sum_{k \in \mathbb{Z}}  \mid k \rangle \langle k+1\mid + \mid k+1\rangle \langle k \mid $ and Lindblad operators $V_k = \mid k \rangle \langle k\mid$ for each $k \in \mathbb{Z}$.
Since the Lindbladian is a super operator a lot of mathematical complications arise. I.e. one needs to think very carefully about which space of operators the Lindbladian acts on and many properties that are true in finite dimensions are not true in infinite dimensions.
