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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.

Frederik Ravn Klausen
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    What is preventing you from just taking your favourite many body system and pick some $V$? – jacob1729 Dec 18 '20 at 14:54
  • The difficulty is that I am not convinced that there is a steady state operator which is trace-class. Suppose for example that you pick each $V_j$ self-adjoint. In that case we know that in a finite dimensional system (i.e. ($l^2(\lbrack - L , L \rbrack)$) the steady state is $1/n \cdot 1_n $ proportional to the identity. It is also true that $\mathcal{L}(1) = 0$ for the infinite case, but there the identity is not trace-class anymore and hence it does not correspond to a state. – Frederik Ravn Klausen Dec 18 '20 at 14:59

2 Answers2

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"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).

Goffredo_Gretzky
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  • Thanks, it was very nice that you pointed this out. With infinite volume I mean that the physical system is infinitely large (spatially). For example one could consider the infinite volume limit of a spin-chain say in the single particle sector. – Frederik Ravn Klausen Dec 18 '20 at 14:33
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    Oh, I see, that's an interesting point. We can say that, anyway, we can take any system spread all over an infinite volume, couple it to a thermal environment at zero temperature, and we still get the ground state of the system as stationary state, as in the example of my answer. – Goffredo_Gretzky Dec 18 '20 at 15:05
  • We have to be careful, however, while dealing with open systems of infinite dimension: most of the theorems about open quantum systems hold only for finite-dimensional systems. For instance, the theorem stating that there exists at least one steady state is valid only for finite-dimensional Hilbert spaces. However, even in the infinite case we can often restrict ourselves to a certain scenario where the dynamics is confined to a finite subset of the Hilbert space (in the example of the answer, we consider only the first $n$ bosonic levels). This would not be valid anymore for pure absorption! – Goffredo_Gretzky Dec 18 '20 at 15:10
  • Yes, that is a very good example. However, it is not clear to me how to couple to a thermal environment at zero temperature using the Lindblad form. – Frederik Ravn Klausen Dec 20 '20 at 17:06
  • And yes, we have to be very carefull with those infinite dimensional systems indeed. Do you have any references on what could happen? – Frederik Ravn Klausen Dec 20 '20 at 17:06
  • I have added to the answer a concrete example of how to derive the zero-temperature GKLS equation of a chain of harmonic oscillators that can be extended till infinity. – Goffredo_Gretzky Dec 21 '20 at 11:29
  • As for the problem of infinite-dimension Hilbert spaces, I think the pathological scenarios where there is no steady state have not been really studied yet. However, a simple example may be this: consider the master equation of my answer (even for a single harmonic oscillator), but take the adjoint jump operators (i.e. you have terms like $a^\dagger \rho a-\ldots$). This would correspond to the case of a thermal bath pumping energy into the system, without any emission. There exists no steady state, since the oscillator energy is always increasing toward $\infty$. – Goffredo_Gretzky Dec 21 '20 at 11:32
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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.

Frederik Ravn Klausen
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