# Semi-group in quantum open systems

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?

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.

• What is (mildly) annoying is that the semigroup in question is I'm pretty sure actually a monoid since it has an identity element $A(0)$. I wonder if at some point in the past this terminology wasn't quite fixed. Aug 10, 2021 at 11:36
• The terminology "Dynamical Semigroups" is often utilized, with an emphasis on the "Dynamical" part. I think some people just thought the words dynamical and monoids where contradictory since the identity element dos not correspond to any dynamics. Nevertheless, the identity element corresponds to the value $t=0$ where $t$ is often taken to be time. In the end I see no reasoning why we are not calling these structures dynamical Monoids. :) It is similar to the tradition of calling the dirac distributions dirac functions I guess. Aug 10, 2021 at 23:01