# What does the term $\mathcal O(\epsilon^2)$ mean?

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.

• $O(\epsilon^2)$ is big O notation from asymptotic analysis. Why would you think otherwise? – catalogue_number Aug 6 '19 at 6:34

$$\mathcal O(\epsilon^2)$$ 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.
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.