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In the highest upvoted answer to Where does the $i$ come from in the Schrödinger equation? the author writes the following equation:

$$ U^\dagger U=(\mathbb I+\epsilon^* A^\dagger)(\mathbb I+\epsilon A)=I+\epsilon^*A^\dagger+\epsilon A+\mathcal O(\epsilon^2) $$

What does that last term mean? It looks a lot like the big O notation from asymptotic analysis.

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    $\begingroup$ $O(\epsilon^2)$ is big O notation from asymptotic analysis. Why would you think otherwise? $\endgroup$ – catalogue_number Aug 6 at 6:34
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$\mathcal O(\epsilon^2)$ just means of order $\epsilon^2$ i.e. terms that are proportional to $\epsilon^2$.

The point is that is $\epsilon$ is much less than unity then $\epsilon^2 \ll \epsilon$, so terms proportional to $\epsilon^2$ and higher powers can be ignored.

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Yes, it is the big $O$ notation, but here we are dealing with operators and not numbers, so that a further interpretation needs. It means that there is a constant $C>0$ such that $$ ||U^\dagger U - (I+\epsilon^*A^\dagger+\epsilon A) || \leq C|\epsilon|^2$$ if $|\epsilon |\leq E$ for some constant $E$. Actually, regirously speaking, the initial statement is false if $A$ is not bounded. In that case the correct statement has a more delicate form related to the strong-operator topology instead of the norm topology.

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