# Is the Heisenberg picture of an open-system very different than that of a closed one?

For a closed system the time evolution (in the Heisenberg picture) of an operator $A$ is given by

$$A(t) = U^{\dagger}(t)AU(t)$$

with $U^{\dagger} U = 1\!\!1$, so that for some other operator $C$ we have: $$C(t) = (AB)(t) = A(t)B(t)$$

However for an open system, the time evolution of an operator is given by:

$$A(t) = \sum_{\alpha,\beta} W_{\alpha,\beta}^{\dagger}(t)\,A\,W_{\alpha,\beta}(t)$$

Where the Krauss operators $W$ satisfy $\sum_{\alpha,\beta} W_{\alpha,\beta}^{\dagger}(t) W_{\alpha,\beta}(t) = 1\!\!1$. So that in general it seems that $C(t) = (AB)(t) \neq A(t)B(t)$, since:

$$C(t) = \sum_{\alpha,\beta} W_{\alpha,\beta}^{\dagger}(t)\,A B\,W_{\alpha,\beta}(t) \neq \sum_{\alpha,\beta,\gamma,\delta} W_{\alpha,\beta}^{\dagger}(t)\,A\,W_{\alpha,\beta}(t) W_{\gamma,\delta}^{\dagger}(t)\,B\,W_{\gamma,\delta}(t) = A(t)B(t)$$

Is there some property that I miss of these $W$ operators, or is this Heisenberg picture so different than that of a closed system? For example the commutator equality $[x(t),p(t)] = i$ does not seem to hold in general.

• Perhaps one can try to take a step back by making the environment explicit. In the full system, prior to tracing out the environment, the Heisenberg picture should still work. – flippiefanus Aug 16 '18 at 12:25

Indeed, for a product operator $\hat{C} = \hat{A}\hat{B}$, it is not true that $\hat{C}(t) = \hat{A}(t) \hat{B}(t)$ for a general (i.e. non-unitary) evolution in the Heisenberg picture. It is instructive to consider the simple example of a harmonic oscillator equilibrating with a thermal bath. This is described by a Lindblad equation $$\dot{\rho} = -i[\omega \hat{a}^\dagger\hat{a},\rho] + \gamma \mathcal{D}[\hat{a}] \rho + \gamma {\rm e}^{-\beta \omega} \mathcal{D}[\hat{a}^\dagger]\rho,$$ where $[\hat{a},\hat{a}^\dagger]=1$, $\omega$ is the oscillator frequency, $\gamma$ is the damping rate, $\beta$ is the inverse temperature and $\mathcal{D}[\hat{L}]\rho = \hat{L}\rho\hat{L}^\dagger - \tfrac{1}{2}\{\hat{L}^\dagger\hat{L},\rho\}$. In the Heisenberg picture, the solution for the ladder operators is $$\hat{a}(t) = {\rm e}^{-i \omega t - \bar{\gamma}t/2}\hat{a}(0),$$ where $\bar{\gamma} = \gamma(1-{\rm e}^{-\beta \omega})$. This expresses the fact that initial oscillations (which are caused by initial coherences in the energy eigenbasis) should decay to zero in the thermal steady state. Now consider the evolution of $\hat{n} = \hat{a}^\dagger\hat{a}$. If it were true that $\hat{n}(t) = \hat{a}^\dagger(t)\hat{a}(t)$, then we would have $\hat{n}(t) = {\rm e}^{-\bar{\gamma}t}\hat{n}(0),$ which is obviously wrong since it would mean no excitations at any temperature in the thermal steady state. In fact, we have $$\hat{n}(t) = {\rm e}^{-\bar{\gamma}t}(\hat{n}(0)-n_\beta) + n_\beta,$$ where $n_\beta = ({\rm e}^{\beta \omega}-1)^{-1}$ is the equilibrium excitation number.
The point of the example is that the relation $\hat{C}(t) = \hat{A}(t) \hat{B}(t)$ places rigid constraints on the relationship between coherences and populations. This constraint holds true for unitary dynamics, which preserves the purity of states. However, many physically relevant situations involve initial coherence decaying to zero while the associated populations do not, in such a way that the purity decreases. Such non-unitary dynamics therefor cannot obey $\hat{C}(t) = \hat{A}(t) \hat{B}(t)$ in general.