Why does the Lindblad master equation in Heisenberg picture not preserve commutation relations? From the book of Breuer and Petruccione [1], the Lindblad master equation for open quantum systems reads $\mathrm{d} \rho (t) / \mathrm{d}t = \mathcal{L} \rho(t)$, where $\mathcal{L}$ is the Lindblad superoperator. In analogy to closed quantum systems, it can be written in the Heisenberg picture for any operator $A_H(t)$, and reads
$$ \frac{\mathrm{d}}{\mathrm{d}t} A_H(t) = \mathcal{L}^\dagger A_H(t)$$
In particular, for the annihilation operator $a$, and assuming the Lindblad operator $\mathcal{L}\rho = \kappa \mathcal{D}[a]\rho = \kappa \left(a\rho a^\dagger - \{ a^\dagger a, \rho \} / 2\right)$ which represents single-photon decay, then the adjoint master equation on $a$ reads
$$ \frac{\mathrm{d}}{\mathrm{d}t}a(t) = \kappa\mathcal{D}^\dagger[a] a(t) = - \frac{\kappa}{2} a(t).$$
This differential equation is easily solved, and yields $a(t) = a(0) \exp(-\kappa t / 2)$. From this, we can evaluate the commutation relation, which reads
$$ [a(t), a^\dagger(t)] = \exp(-\kappa t) [a(0), a^\dagger(0)] = \exp(-\kappa t)$$
This seems to contradict the usual formula $[a, a^\dagger] = 1$ (assuming $\hbar = 1$). How can I resolve this contradiction?
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[1] H.-P. Breuer and F. Petruccione, The theory of open quantum systems (Oxford University Press, 2002).
 A: First of all, there is a priori no reason to assume that $a(t)$ and $a^\dagger(t)$ satisfy the canonical commutation relation for $t>0$.
But that is probably not a very satisfying answer to your question, so let me motivate why operators generally cannot maintain their commutation relations at times $t>0$.
The Lindblad equation describes relaxation to equilibrium, and therefore the system state in the Schrödinger picture generically* converges towards a unique steady state $\rho_\infty$ at long times, $\rho(t) \to \rho_\infty$, independent of the initial state $\rho(0) = \rho_0$.
A direct consequence is that, if $A$ is some operator, the expectation value $\operatorname{tr}[A\, \rho(t)]$ converges to a number $A_\infty$ independent of $\rho_0$.
Translating that to the Heisenberg picture, we find
$$ \lim_{t \to \infty} \operatorname{tr}[A(t)\, \rho_0] = \lim_{t \to \infty} \operatorname{tr}[A\, \rho(t)] = A_\infty , $$
which is only possible if
$$ \lim_{t \to \infty} A(t) = A_\infty\, \mathbb I . $$
Here, $\mathbb I$ is the identity operator.
Operators in the Heisenberg picture must therefore converge to a multiple of the identity, and their commutator will go to zero.
* Not every Lindblad equation has a unique steady state, but almost all Lindblad equations that we as Physicists are interested in do. There are mathematical criteria for that. In any case, it is not important for the argument above whether all Lindblad equations are like that.
