I read the wikipedia article on the Lindblad operator, but I still don't understand what this operator is supposed to describe. I therefore considered setting up an example in order to get the idea.

So let $H$ be the Hamiltonian of the first $n$ states of the hydrogen atom and $V(t)$ an interaction due to an external electric field. Now, I consider the evolution of density matrices in this finite-dimensional space by $$i\rho'(t) = [H-V(t), \rho(t)]+ L(\rho(t)).$$ My question is then: What exactly is the physical meaning behind this additional term $L$?

I mean the equation: $i\rho'(t) = [H-V(t), \rho(t)]$ would define the propagation of an initial distribution of states in this $n$-dimensional space of my reduced hydrogen atom under the influence of the electric field. But what am I exactly modeling if I additionally have this $L$ term there ?

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    $\begingroup$ You are modelling the environmental conditions such as dephasing and relaxation $\endgroup$
    – TanMath
    Aug 4, 2015 at 18:31
  • $\begingroup$ @TanMath sorry, but this is far too short for me. I am really looking for a thorough explanation that outlines an experimental situation and describes the differences to the model without this term. $\endgroup$
    – user167575
    Aug 4, 2015 at 18:38
  • $\begingroup$ Would it help to say that in your specific case the $L$ term describes things like spontaneous emission? I'm trying to figure out how much you already know so I know where to start in an answer. $\endgroup$
    – DanielSank
    Aug 4, 2015 at 18:57
  • $\begingroup$ @DanielSank yes, I read that, as spontaneous emission can normally be only described in terms of QED but not QM (at least this is what I heard). I would appreciate it, if you could adapt your answer to "my Hamiltonian", as you are apparently willing to do. $\endgroup$
    – user167575
    Aug 4, 2015 at 19:18
  • $\begingroup$ @DanielSank in especially, it would be interesting for me to understand, why the Lindblad operator describes this and not thermal effects or whatever. $\endgroup$
    – user167575
    Aug 4, 2015 at 19:20

3 Answers 3


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:

  1. It is still linear dynamics, in terms of $\rho$.
  2. 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.
  3. It leaves $\dot\rho$ Hermitian, which is important because only Hermitian operators have all-real eigenvalues.
  4. 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.

  • 2
    $\begingroup$ "which ensure the positivity of the Lindbladian" Actually, it's a generator of a completely positive, not just positive map. There are no criteria on $A$ as the Lindblad form of the generator is sufficient to guarantee that the map is completely positive and trace preserving. $\endgroup$
    – Bubble
    Aug 7, 2015 at 2:43
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    $\begingroup$ In fact, if we restrict ourselves to time local and Markovian master equations then Lindblad form is also necessary for the Liouvillian to be a generator of a CP trace preserving map. $\endgroup$
    – Bubble
    Aug 7, 2015 at 2:50
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    $\begingroup$ Small nitpick: point 3 is not completely correct. Namely, the statement "only Hermitian operators have all-real eigenvalues" is incorrect. For example, the matrix $\begin{bmatrix}1 & -1 \\ 0 & -1\end{bmatrix}$ is non-Hermitian, but possesses a real spectrum $\{1,-1\}$. For those who are interested, reality of the spectrum is guaranteed for diagonalizable matrix when it is pseudo-Hermitian with respect to a positive definite metric (see Ali Mostafazadeh's work on pseudo-Hermiticity for more information). $\endgroup$ Dec 17, 2019 at 19:09
  • $\begingroup$ Is it known if the Lindbladian is a trace preserving positive opetaror? $\endgroup$
    – Marion
    Nov 27, 2021 at 11:38

Let's suppose you don't have this operator, but you only have the self-adjoint Hamiltonian part. This means you have the usual Schrödinger equation (or Liouville equation since it's for the density matrix)

$$ i \dot{\rho}=[H,\rho] $$

and the solution will be $\rho(t)=e^{iHt}\rho(0)e^{-iHt}$, hence the solution will evolve according to the (strongly continuous) unitary group associated to your Hamiltonian. In other words, at any time your density matrix is just a unitary conjugation of the density matrix you started out with. In particular, its spectrum will never change. This means (for example) that if you start with a pure state (rank-one projection), you'll always stay in a pure state.

The behaviour we have just described is that of a closed system. Now, suppose we don't have a closed system, but we have an environment somewhere that is interacting with our system. In particular, you might want to consider a bath and see effects such as thermalisation (which implies that the spectrum of $\rho$ needs to change!). You can of course try to model bath + system with Hamiltonians and then solve the whole system, but you can also trace out the bath directly and have a look at the evolution equation for the remaining system you are interested in.

Under quite general assumptions (time-homogeneity, Markovianity), the equation you'll get is the Lindblad equation. The Hamiltonian part is the Hamiltonian part of your system and the additional terms (which is the Lindblad operator in many cases) describe the effects of the environment. In other words, if you can empirically guess the right form of the Lindblad-superoperator, you don't need to worry that you cannot model the whole closed system (or forget part of it). This is essentially what is known as the paradigm of "open quantum systems".

In short: The Lindblad operator describes the interaction of your system with an environment and it models the effects of the environment on the model. This could be anything. For example, it could be that you are interested in describing the double slit experiment with electrons. The Hamiltonian part would be the electrons and the slit, but you really don't know all the other particles around, you don't have a quantum description of the photon source (or it's complicated) and you don't have a quantum description of your measurement procedure, etc. But all these things interact with your measurement, so you might want to take them into consideration (e.g. if you want to see decoherence). The upshot of the above is that their influence on your system is precisely given by the Lindblad operator.

How to find the Lindbladian in a particular situation? This is a completely different story of which I don't know very much about...

  • 1
    $\begingroup$ the thing is that I don't understand which interaction I can capture with this guy? Somebody suggested in the comments that it might be spontaneous emission. So is it really anything I want? $\endgroup$
    – user167575
    Aug 4, 2015 at 19:20
  • 1
    $\begingroup$ @user167575 Yes, it's anything you want, as long as that thing has no time correlations. $\endgroup$
    – DanielSank
    Aug 4, 2015 at 19:47
  • $\begingroup$ @DanielSank then why did you say that it's spontaneous emission in my exmaple, or was this just a possible application, as this is a external effect in qm that would not occur in a closed system? $\endgroup$
    – user167575
    Aug 4, 2015 at 19:50
  • $\begingroup$ @user167575 I said "things like spontaneous emission" :-) It is, as you guess, just an example of a (relevant) process which happens because the Hamiltonian ignores some degrees of freedom (in this case the electromagnetic field). Note that you can only use the master equation form if we assume that the electromagnetic environment has no memory; i.e. that radiation emitted from the atom never comes back. $\endgroup$
    – DanielSank
    Aug 4, 2015 at 19:54
  • $\begingroup$ When you say "How to find the Lindbladian in a particular situation?", do you mean how to find the collapse operator? $\endgroup$
    – TanMath
    Aug 5, 2015 at 2:06

One detail

First thing I would like to note is that the operator you are talking about is called the Lindblad superoperator. A superoperator is like an operator that acts on other linear operators (in this case, the density matrix).

Lindblad Equation

What you have written is known as the Lindblad Equation. The Lindblad Equation is one example of the many equations used to described the dynamics of open quantum systems. The Lindblad Equation has the general form of $$\dot \rho = -i[H, \rho] + [ \sum \gamma (A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A)]$$ where the part in brackets is known as the Lindblad superoperator. Therefore, this equation can be rewritten as: $$\dot \rho= -i[H, \rho] + L(\rho)$$

More details

The Lindblad superoperator models the environmental conditions that make up the open quantum system such as dephasing and relaxation. The operator $A$ is known as the collapse operator and it is important for deciding what the Lindblad superoperator describes. This operator is through which the environment couples to the system. Different collapse operators describe different aspects of the environment.

$\gamma$ is an important constant that usually describes dephasing rate, rephasing rate, relaxation rate, etc. It is basically a corresponding rate for the coulping of the environment to the system. It also is important to the master equation. Note that whenever this constant is equal to zero, then we get the quantum Liovillian equation for a closed system without any environmental effects.

One last thing I would like to add is that you can add as many Lindblad superoperators to the commutator to describe for different environmental conditions. One can describe for rephasing, another for dephasing, etc. It all depends on the environment of the quantum system.

In summary, the Lindblad superoperator models an environmental coupling to the system. Without it, you get a model for a closed system with no environmental effects. That's why the L term is important.

More resources

If you want to learn more, you should look into open quantum systems. Here is a link to start with. This link was also helpful to me when I was studying open quantum systems. And here is the original article by Lindblad about the Lindblad Equation

  • 1
    $\begingroup$ The second link is down. Could you please post the name of that lecture? I find most treatments of the subject very confusing, and it's great to have begginer-level articles. The first article you mention made me understand it a little after weeks of trying to understand tons of articles. $\endgroup$
    – lytex
    Jan 11, 2019 at 8:46
  • $\begingroup$ What do you mean with "rephasing"? As far as I am aware there is no such process that can be modelled with the Lindblad equation. Do you have any references mentioning this process? $\endgroup$ Jan 7, 2022 at 12:19

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