The treatment of infinitesimal quantities Please be advised that my question is different from some of the existing threads like this one.
I have long been convinced that if we are to question the value of something which we ultimately are going to take derivative of, then second order quantities are of no importance to us. For example, in the extremisation of an action. Moreover, in perturbation theory we are entitled to ignore higher order terms, the exact order of which we decide based on the precision we would like to have.
I am less convinced however, that this particular treatment of translation operation by J.J Sakurai in page 40 of his book Modern Quantum Mechanics, third edition is correct:
Suppose for one particular position state $|x\rangle$ we define a translation operator $\mathcal{X(dx)}:\mathcal{H}\to\mathcal{H}$ such that its effect on state $|x\rangle$ is $$\mathcal{X}(dx)|x\rangle:=|x+dx\rangle$$
Among many properties we would like such translation operator to have, one of them is the conservation of probability which demands $\mathcal{X}^\dagger(dx)\mathcal{X}(dx)=id$. Asserted by J.J. Sakurai, the translation operator is of the form
$$\mathcal{X}(dx)=id-iA\cdot dx$$ for some Hermitian operator $A$. Calculation shows that
$$\mathcal{X}^\dagger(dx)\mathcal{X}(dx)=id-i(A-A^\dagger)\cdot dx+O(dx^2)$$where the second order terms do not have zero coefficients. So why are we allowed to ignore these second order terms in this specific case?
 A: Strictly speaking you are right, but at the sacrifice to the relaxed good-faith spirit of the introduction
involved in the book!
Indeed, to satisfy your exceptionally literalist meaning, you'd need the full translation group element to be
$$\mathcal{X}(dx)={\mathbb I}-iA\cdot dx+ O(dx^2),$$ for  a Hermitian Lie group algebra element  $A$.
The reader is invited to implicitly ignore/blur-out all such higher order infinitesimals,  which, complacently, the authors assume the readers have already internalized at the sight of shift variables written as dx.
In practice/reality, of course, a reader should appreciate quite soon that this is seat-of-the-pants shorthand for
$$\mathcal{X}(dx)= \exp(-iA\cdot dx),$$
the standard Lagrange translation operator, exponentiation of the gradient operator. Take extra care to appreciate the correctness of the signs involved!
A: If you wish, you can write $\mathcal X(\mathrm dx) = \mathbb I - iA\mathrm dx - B \mathrm dx^2+\mathcal O(\mathrm dx^3)$
$$\implies \mathcal X(\mathrm dx)^\dagger \mathcal X(\mathrm dx) = \mathbb I - i(A-A^\dagger)\mathrm dx - (B+B^\dagger - A^\dagger A) \mathrm dx^2 + \mathcal O(\mathrm dx^3)$$
$$\implies A = A^\dagger, \qquad  B + B^\dagger = A^2$$
The additional requirement that $\mathcal X(\alpha) \mathcal X(\beta)= \mathcal X(\alpha+\beta)$ yields
$$\mathcal X(\alpha)\mathcal X(\beta) = (\mathbb I -iA\alpha - B\alpha^2 + \mathcal O(\mathrm dx^3))(\mathbb I-iA\beta - B\beta^2+\mathcal O(\mathrm dx^3)) $$
$$= \mathbb I - iA(\alpha+\beta)-B(\alpha^2+\beta^2) -A^2 \alpha\beta + \mathcal O(\mathrm dx^3)$$
$$ \implies A^2 = 2B$$
and so
$$\mathcal X(\mathrm dx) = \mathbb I - iA \mathrm dx - \frac{1}{2}A^2 \mathrm dx^2 + \mathcal O(\mathrm dx^3)$$

This argument may be continued to arbitrarily high orders, but it is not necessary to do so.  The facts that that $\mathcal X(0) = \mathbb I$  and that $\mathcal X(\alpha)\mathcal X(\beta) = \mathcal X(\alpha+\beta)$ for all $\alpha,\beta$ are a remarkably strong requirement. In particular, they tells us that if
$$-iA := \lim_{\epsilon\rightarrow 0}  \frac{\mathcal X(\epsilon)-\mathbb I}{\epsilon} \equiv \mathcal X'(0)$$
then it follows that $\mathcal X^{(n)}(0) = \big(-iA\big)^n$, which is proved easily by induction.  As a result, if we write $\mathcal X(\mathrm dx)$ as a power series we obtain
$$\mathcal X(\mathrm dx) = \sum_{n} \frac{\big(-iA\big)^n}{n!} \mathrm dx^n = \mathbb I - iA\mathrm dx - \frac{A^2}{2}\mathrm dx^2  + i\frac{A^3}{3!} \mathrm dx^3 + \ldots$$
In a rough sense, the fact that $\mathcal X(\alpha+\beta) = \mathcal X(\alpha)\mathcal X(\beta)$ tells us that the first-order behavior uniquely determines the higher order behavior. This is precisely analogous to the properties of the more familiar $\mathbb R$-valued exponential function, of which $\mathcal X$ may be seen as an operator-valued generalization.
In the interest of rigor, note that the formalization of this is given by Stone's theorem, the proof of which does not actually require the existence of a convergent power series (which is good, because the power series written above has zero radius of convergence). The spirit of the theorem, however, remains more or less the same as the argument given here.
