Quantum systems with steady states: does a system always approach the steady state as it evolves?

Question

Say we have a quantum system whose dynamics results in there being a steady state. For example, it may be described by a Linblad master equation with several opposing dissipators. It is obvious that given that enough time has passed, the steady-state will be reached. However, I am interested in the process by which this happens.

I find the concept of "approaching the steady state" itself rather vague since it is not straightforward to put a geometric interpretation to the evolution of a generally mixed quantum state. Nevertheless, there are distance measures which can be used to gauge how close the state is to the steady state, for example the trace distance. In this case, my question may ask whether the trace distance between the state $\rho(t)$ at any given time and the steady state $\rho_\mathrm{ss}$ monotonically decreases as the system evolves—the trace distance monotonically decreasing is what I mean by "always approaches" in the question title.

I'd also love to welcome any other form of interpretation here as it is honestly confusing for me.

Answer 1

Yes, the time-evolving state $\rho(t)$ monotonically approaches the steady state $\rho_{\rm ss}$ in the sense of trace distance. This also holds for any other distance measure (e.g. relative entropy) that is monotonic under completely positive trace preserving (CPTP) maps. Monotonicity of the trace distance $D(\rho_1,\rho_2) = \frac{1}{2}||\rho_1-\rho_2||_1$ means that $$ D(\mathcal{M}\rho_1,\mathcal{M}\rho_2) \leq D(\rho_1,\rho_2) ,$$ for any pair of states $\rho_{1,2}$ and any CPTP map $\mathcal{M}$. Now consider the CPTP map $\mathcal{M} = e^{\mathcal{L}t}$, which is the time evolution operator generated by the Lindblad master equation $\dot{\rho} = \mathcal{L}\rho$. The solution of the Lindblad equation is simply $\rho(t) = e^{\mathcal{L}t}\rho(0)$. Since the steady state obeys $e^{\mathcal{L}t}\rho_{\rm ss} = \rho_{\rm ss}$ by definition, one has $$ D\big(\rho(t),\rho_{\rm ss}\big) = D\big(e^{\mathcal{L}t}\rho(0), e^{\mathcal{L}t}\rho_{\rm ss}\big) \leq D\big(\rho(0),\rho_{\rm ss}\big).$$ Since the Lindblad equation generates a CPTP evolution for all $t$, the above inequality proves monotonic decrease of trace distance to the steady state.

The intuition here is that relaxation to a unique stationary state implies loss of memory of the initial conditions over time. Therefore, any initial preparation should become less distinguishable from the steady state as time increases. Note the even stronger implication, $$D\big(\rho_1(t),\rho_2(t)\big) \leq D\big(\rho_1(0),\rho_2(0)\big),$$ implying a loss of distinguishability between any two initial preparations $\rho_{1,2}(0)$ over time.

As an aside, applying similar reasoning to the relative entropy, $$D(\rho_1||\rho_2) = {\rm tr}(\rho_1\ln \rho_1) - {\rm tr}(\rho_1\ln \rho_2), $$ gives rise to the Spohn definition of entropy production, $\sigma = -(d/dt) D(\rho(t)||\rho_{\rm ss})\geq 0$.