Semi-group in quantum open systems

Question

In the literature of Open Quantum System, one often comes across the following ($t_2>t_1,>0$):

Semi-group property of a map: $A(t_1+t_2,0) = A(t_2,0) A(t_1,0)$.

What does this mean physically, and why the name semi-group?

Answer 1

I realize that this is a bit late for a reply. However, just in case you never found encountered a response to the question regarding the physicality of such semigroups I would like to add a brief note.

The term semigroup is not a very loaded word. If you are familiar with group theory then you will notice that this is like a group. However, a semigroup does not have inverses! In the context of quantum opensystems, processes such as decoherence are ireversible processes and therefore an element $ A(-s, 0 )$,$s>0$ is not physical because it would imply that entropy increasing processes such as decoherence are reversible, which is not the case.

I'll reiterate here with some equations. The semigroup $\Lambda:=\{A(t^{'},0)\}_{t^{'}\geq 0}$ summarizes the dynamics of a quantum open system with initial state $\rho(0)$. i.e. $\rho(t) = A(t,0)\rho(0).$ If we allowed the term $A(-t,0)$ to be part of the allowed dynamics then we could do the following

$$A(-t,0) \rho(t) =A(-t,0)A(t,0)\rho(0) = A(0,0) \rho(0) = \rho(0).$$

To understand why such, reversing elements should not be around let us quickly look at how such semigroups come about.

Consider a compound quantum system with associated $Hilbert$ space $H_{S}\otimes H_{E}$, $S$ for system and $E$ for environment. The state of the system may be taken to be $\rho(0) = \rho_{S}(0)\otimes\rho_{E}(0) \in B(H_{S}\otimes H_{E})$ at $t=0$, $B(H_{S}\otimes H_{E})$ is the space of bounded linear operators of the $Hilbert$ space $H_{S}\otimes H_{E}$and we assume $\rho(0)_{S}$ and $\rho(0)_{E}$ to be pure states.At time $t =0$ $\rho(0)$ is a pure state and therefore has Von Neumann entropy zero, i.e. $S(\rho(0))$. However, this is for the total system and environment grouping. Let us zoom into the system by doing a partial trace over the environment. $$\rho(0)_{S} = Tr_{E}(\rho(0)).$$

It turns out that $S\rho(0)_{S} = 0$ as well from the definition of the Von Neumann entropy.

Now, when we do a UNITARY evolution to the entire compound system, i.e. S and E together and then partial trace the environment we get the NON-UNITARY evolution of system $S$. i.e.

$$\rho(t)_{S} = Tr_{E}[U\rho_{S}(0)\otimes \rho_{E}(0)U^{\dagger}]. (1)$$

$U\in B(H_{S}\otimes H_{E})$.

This wacky partial trace object reduces to an element of a semigroup of completely positive maps for a fixed t>0. It can be shown that in general $S(\rho_{S}(t))>0$. I included the environment and the partial trace just to appreciate that there is a loss of "information", and to appreciate that it is going to our selected environment. Now, $S(\rho_{S}(t))>0$ means an increas in entropy which is an irreversible process. Irreversible means that inverse dynamics for some element $A(t,0)$ is not physcial.

Sorry about any typos, I wrote quite fast.