Series expansion of unitary operators in terms of other operators I am reading lecture notes on local gauge invariance, part of Prof. Ethan Neil's course on Quantum Mechanics at the University of Colorado.
There, he writes about introducing a so-called comparator $U(x_1,x_2),\; x_{1,2} \in \mathbb{R}^4$ that transforms as
$$U(x_1,x_2) \to e^{i\theta(x_1)}U(x_1,x_2)e^{-i\theta(x_2)}$$
in order for (e.g.) the Schrödinger equation to be invariant under the local gauge transformation
$$\psi(x) \to e^{i \theta(x)} \psi(x) $$
of the wave function $\psi(x)\; (x \in \mathbb{R}^4$).
He explains why the comparator needs to be unitary and then states that

Since we have a unitary operator with a small parameter $\epsilon$, we
can write it as a series expansion in terms of another operator
$U(x+\epsilon,x) = 1 + i \frac{e}{\hbar c} \epsilon A_x(x) + \mathcal{O}(\epsilon^2)$

where $A(x)$ is the other operator and obviously corresponds to the vector potential.
My question is regarding the cited statement: how does he conclude that such an expansion in terms of another operator exists? I suppose there is a theorem for unitary operators he is having in mind but I am not sure.
 A: First of all, what is the meaning of $U(x_1, x_2)$? In classical physics, it would be a complex number with magnitude 1 such that the phase difference between $\psi(x_1)$ and $U(x_1, x_2) \psi(x_2)$ is constant under gauge transformations (since $U$ itself also transforms so as to cancel out the difference).
(In a nonabelian gauge theory, we can have irreducible representations with dimension greater than 1. In such cases, the value of $U$ would not be a complex number, but some matrix.)
In a quantum gauge theory, $\psi$ is operator-valued so $U$ must also be operator-valued.
In order for the limit to exist in the expression for the covariant derivative, $U(x + \epsilon, x)$ must approach the identity (1) as $\epsilon$ approaches zero. So we define $U$ such that $U(x, x) = 1$ for all $x$.
An assumption is made that $U(x_1, x_2)$ is an analytic function of $x_1$ and $x_2$, so it can be expanded in a Taylor series:
$$U(x + \epsilon, x) = U(x, x) + \epsilon (\partial_1 U)(x, x) + \frac{\epsilon^2}{2} (\partial_1 \partial_1 U)(x, x) + \ldots$$
(Here, $\partial_1$ denotes the partial derivative with respect to the first argument. In spatial dimension greater than 1, the expansion becomes $U(x, x) + \epsilon \cdot (\nabla_1 U)(x, x) + \ldots$)
The author has given the name $A$ to the first derivative appearing in the above expansion, with appropriate scaling such that they could write $U(x + \epsilon, x) = 1 + i\frac{e}{\hbar c} \epsilon A(x) + O(\epsilon^2)$. This has been done because it turns out that $A$ is the electromagnetic vector potential.
A: The general principle at work here is made precise by Stone's theorem on one-parameter unitary groups.
In short, if $U(\epsilon)$ is a strongly continuous$^\dagger$ family of unitary operators parameterized by a continuous variable $\epsilon$ which obey$^\ddagger$ $U(\epsilon)U(\epsilon')=U(\epsilon+\epsilon')$, then there exists a (possibly unbounded) self-adjoint operator $A$ given by
$$A\psi := \lim_{\epsilon\rightarrow 0} \frac{U(\epsilon)\psi - \psi}{i\epsilon}$$
whose domain of definition is all $\psi$ in the Hilbert space such that this limit exists, and which we often call the infinitesimal generator of the family $U(\epsilon)$.  Formally, one can understand $A$ (or rather, $iA$) as the operator-valued derivative of $U$ evaluated at $0$, and write $U(\epsilon)=e^{i\epsilon A}$.
In practice, one can often determine $A$ by expanding $U$ in a formal power series about $\epsilon=0$ and then reading off the coefficent of $\epsilon$.  The actual convergence of this power series is not guaranteed over the full Hilbert space, but by looking at the linear term we can read off the "formula" for $A$. For example, consider the family of operators $U(\epsilon)$ defined by
$$\big(U(\epsilon)\psi\big)(x) = e^{i\epsilon x} \psi(x)$$
It's not difficult to show that it satisfies the conditions for Stone's theorem to apply.  Expanding it in a formal power series to linear order yields the following expression:
$$\big(U(\epsilon)\psi\big)(x)= e^{i\epsilon x}\psi(x) \simeq \psi(x) + i\epsilon x \psi(x) \overset{!}{=}(1+i\epsilon A)\psi(x)$$
Comparing both sides, we see that $\big(A\psi\big)(x)= x \psi(x)$, which we identify as the position operator (which I will now call $X$).  Observe that the formal power series
$$U(\epsilon) = \mathbf 1 + i\epsilon X  - \frac{\epsilon^2}{2}X^2 + \ldots$$
actually converges on a very small subset of $L^2(\mathbb R)$ (at minimum, we must have that $X^n\psi\in L^2(\mathbb R)$ for all $n$); nevertheless, by reading off the linear term, we were able to determine the infinitesimal generator of $U(\epsilon)$. A similar exercise leads to the observation that the infinitesimal generator of the spatial displacement operators $\big(T(\epsilon)\psi\big)(x) = \psi(x-\epsilon)$ is the momentum operator (and in fact, this is often taken to be the definition of the momentum operator).

$^\dagger$A strongly continuous family of operators is one which satisfies the expression $\lim_{t\rightarrow t_0} \Vert A_t\psi - A_{t_0}\psi \Vert\rightarrow 0$ for all $t_0\in \mathbb R$ and $\psi$ in the Hilbert space. For reference, this can be contrasted with a weakly continuous family of operators, which need only satisfy $\lim_{t\rightarrow t_0} \langle \phi,A_t\psi\rangle = \langle\phi,A_{t_0}\psi\rangle$ for all $t_0\in \mathbb R$ and $\phi,\psi$ in the Hilbert space.  The former implies the latter, but the latter does not imply the former.
$^\ddagger$Note that this automatically implies that $U(0)=\mathbf 1$
