density operator - Is it possible to go from the Master Equation formalism to Heisenberg-Langevin equations

If I have derived a master equation (e.g. in the Lindblad form) and solved for the density matrix, $\rho(t)$ I can get the mean value of an operator, A as:

$ = \mathrm{Tr}A\rho $.

But this is just the mean value. Suppose I want an equation for the operator in the Heisenberg picture

$ A_H (t) $

For this I can derive quantum-Langevin equations (aka. Heisenberg-Langevin equations see e.g. API p.340ff) which gives this in terms of the Hamiltonian for the full system including the reservoir OR in terms of quantum Langevin forces.

Is there a way to go from the density matrix in the Schrödinger picture to the operator in the Heisenberg picture (in the case of an open system like described above)?

API: Cohen-Tannoudji et al., Atom Photon Interactions, 1998 & 2004.

Answer

For a linear master equation as the one described by the Lindbladian there is no problem in passing from the Schrödinger to the Heisenberg picture, apart from an eventual loss of regularity wrt time.

This is because there is a duality chain between the "set of normal states" $V_*$ (to be precise the normal states are positive elements with norm one of this predual space), the von Neumann algebra of observables $V$ and the states $V^*$ (again the positive elements with norm one): $$V^*=(V)^*=(V_*)^{**}\;,$$ where the $*$ operation stands for the topological dual.

So, suppose you are given a (strongly-continuous) semigroup (Lindblad semigroup) $(e^{\,\cdot\,\mathscr{L}})_*: \mathbb{R}_+\times V_*\to V_*$, that to a time $t\geq 0$ and a normal state (density matrix) $\rho\in V_*$ associates another normal state $(e^{\,t\,\mathscr{L}})_*\,\rho$.

By the aforementioned map we can define the evolution $e^{\,\cdot\,\mathscr{L}}: \mathbb{R}_+\times V\to V$ on observables $V$ easily by duality: $$\forall A\in V,\rho\in V_*, t\geq 0,\quad \rho(e^{\,t\,\mathscr{L}}A)=(e^{\,t\,\mathscr{L}})_*\,\rho(A)\; .$$

The semigroup $e^{\,\cdot\,\mathscr{L}}$ is called the dual semigroup of $(e^{\,\cdot\,\mathscr{L}})_*$. The only care that has to be taken is that this dual semigroup is, in general, less regular than the original one: if the original one is strongly continuous, the dual is a priori only weakly continuous. There is however a subspace $V^{\dagger}$ of the observables where the restriction of the dual semigroup is again strongly continuous. Nevertheless you have defined your Heisenberg evolution, starting from the Schrödinger one (and you can do the same thing to obtain the evolution for general states as well).

Remark. Here I have considered the Lindblad generator $(\mathscr{L})_*$ as a linear operator on the states (density matrices); its explicit form is the usual $$(\mathscr{L})_*\rho= -i[H,\rho]+\frac{1}{2}\sum_j\Bigl([U_j\rho,U_j^{\dagger}]+[U_j,\rho U_j^{\dagger}]\Bigr)\; .$$ The generator for the observables semigroup $\mathscr{L}$ (acting in $V^\dagger$ where it can be defined) is simply $$\mathscr{L}X= i[H,X]+\sum_j\Bigl(U_j^\dagger X U_j -\tfrac{1}{2}\{U^\dagger_jU_j,X\}\Bigr)\; .$$