# Proof of differentiate form of dynamical semigroups

I am studying some basics of the pure mathematical background for open quantum systems from Angel Rivas`s book which is "Open quantum systems, an introduction".

Here is a theorem (Page 6, Theorem 1.3.1) that sounds really interesting:

If $$T_{t}$$ forms a uniformly continuous one-parameter semigroup, then the map $$t \mapsto T_{t}$$ is differentiable, and the derivative of $$T_{t}$$ is given by: $$\frac{\mathrm{d} T_{t}}{\mathrm{d} t}=LT_{t}$$ with $$L = \frac{\mathrm{d} T_{t}}{\mathrm{d} t}\mid _{t=0}$$.

Proof: Since $$T_{t}$$ is uniformly continuous on $$t$$, the function $$V_{(t)}$$ defined by: $$V_{(t)}=\int_{0}^{t} T_s\, \mathrm ds, \qquad t\geq 0$$ is differentiable with $$\frac{\mathrm{d} V_{(t)}}{\mathrm{d} t}=T_{t}$$. In particular, $$\lim_{t\rightarrow 0}\frac{V_{(t)}}{t}=\lim_{t\rightarrow 0}\frac{V_{(t)} - V_{(0)}}{t} = \frac{\mathrm{d} T_{t}}{\mathrm{d} t}\mid _{t=0}=T_{0}=\mathbb{I}$$ this implies that there exist some $$t_{0} > 0$$ small enough such that $$V_{(t_{0})}$$ is invertible.

I literally do not understand the term "some" (why we should not say "every") here and why the above equation implies that the $$t_{0}$$ should be small enough.

Since $$\lim_{t\to 0} \frac{V(t)}{t} = \mathbb I$$, we know for small $$t$$ $$V(t) = t\, \mathbb I + o(t) . \tag{1}$$ That is, $$V(t)$$ is asymptotically close to the invertible operator $$t\, \mathbb I$$ for small $$t$$, and since the set of invertible operators is open in the space of linear operators$$^1$$, there must be a $$t_0$$ so that $$V(t_0)$$ is invertible itself.

However, there is no reason to assume that $$V(t)$$ is invertible for every $$t>0$$, since we do not know much about $$V(t)$$ except for the asymptotic behavior at small $$t$$.

[1] Intuitively, this statement means: if an operator $$T$$ is invertible, and another operator $$S$$ is "close enough" to $$T$$, then $$S$$ is also invertible. I think Rivas is appealing to this intuitive argument; it is however not fully rigorous. We can make it rigorous by recalling the following fact:

If an operator $$T$$ is invertible, and $$|| S - T || \leq || T^{-1} ||^{-1}$$, then $$S$$ is invertible.

Here, $$|| \cdot ||$$ denotes the operator norm. In our application, $$T$$ corresponds to $$t\, \mathbb I$$ and $$S$$ corresponds to $$V(t)$$. Using $$|| \mathbb I || = 1$$, we learn that $$V(t)$$ is invertible if

$$|| V(t) - t\, \mathbb I || \leq t \tag{2}$$

holds (and $$t>0$$), and Eq. (1) implies that (2) holds for small enough $$t$$.

• Thank you for your consideration. Could you please explain this " the set of invertible operators is open in the space of linear operators" more? Commented Jan 5, 2023 at 4:45
• @S.NavidElyasi I have added an explanation Commented Jan 5, 2023 at 5:17