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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.

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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 $t$.

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

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