Infinite Representations of $SO(2)$ 
So I want to show that a possible (infinite dimensional) representation of the generator of $\mathfrak{so}(2)$ is this: 
  $$ T = i\bigg(y\frac{\partial}{\partial x} -x \frac{\partial}{\partial y} \bigg). $$ 

Attempt: 


*

*$\mathfrak{so}(2)$ is one-dimensional (because $SO(2)$ is one dimensional, & the tangent space to a manifold has the same dimension as the manifold) subgroup so it has one generator. If $g(\phi) \in SO(2)$ is connected to the identity, where $\phi$ is the parameter: 
$$ g(\phi) = 1 + iT\phi \quad : \ T = -i\frac{\partial  g(\phi)}{\partial \phi}  $$ 

*Apparently the group action of $g$ on $\psi(\mathbf{x}) = \langle \mathbf{x}| \psi \rangle$ is 
$ g\psi = \psi(R^{-1}\mathbf{x})$ 


*

*How do we prove this? Perhaps could argue, if $ |\psi'\rangle$  is the rotated state and $\mathbf{x'}$ the rotated position: 
$$ \langle \mathbf{r'}|\psi'\rangle = \langle \mathbf{r} | \psi \rangle \implies \psi'(\mathbf{r'}) = \psi(\mathbf{r})\implies \psi'(\mathbf{r}) = \psi(R^{-1}\mathbf{r}) $$ 
Where $R^{-1}$ is the inverse rotation matrix.Then define $g\psi = \psi'(\mathbf{r})$ - but what is the difference between $g$ and $R$? $R$ is the fundamental representation of the group element $g$? 

*This proof is completely useless for fields although Srednicki would seem to imply a generalization of this result??


*Assuming 2 fine: 
 $$ g(\phi)\psi(\mathbf{x}) = (1+iT\phi)\psi(\mathbf{x}) = \psi(R^{-1}\mathbf{x})$$ 
$$ R^{-1} = \begin{pmatrix} cos\phi & -sin\phi \\ sin\phi & cos\phi \\ \end{pmatrix}  \implies \psi(R^{-1}\mathbf{x}) = \psi\begin{pmatrix} x-\phi y \\ y+ \phi x \end{pmatrix} $$ 


-I now want to taylor expand this, but I'm not sure how?? If this were a simple multi variable  functions $f: \mathbb{R} \rightarrow \mathbb{R} $  this would be obvious - but it's not. 
-I'm also slightly concerned how I have ended up representing one generator as an infinite number of objects? - i.e this representation of $T$ is a member of $L^{2}(\mathbb{R}^{3})$, labelled by the tuple of parameters $(x,y)$ 
 A: Alright, this is a new answer. First off, I'll fix some notation. It will be slightly different from the one in your question, but I believe it will help.
We will denote an element of $SO(2)$ by $R(\phi)$. To speak of "the" generator we need to choose a specific parametrization of $SO(2)$; it looks like the one we're using here is
$$R(\phi) = \begin{pmatrix}
\cos \phi & \sin \phi \\
- \sin \phi & \cos \phi \end{pmatrix}.$$
This is the same as what you called $g(\phi)$, so let's not use two names for the same thing. The generator (i.e., basis element) of the algebra is
$$T = -i R'(0) = \begin{pmatrix}
0 & -i \\
i & 0 \end{pmatrix},$$
so that we have $R(\phi) = e^{i \phi T}$. The Lie algebra is a one-dimensional real vector space, isomorphic to $\mathbb{R}$. We would normally be interested in the commutators of elements, but since this is one dimensional this is pretty trivial: for any elements $A = \alpha T$ and $B = \beta T$ in the Lie algebra, we have $[A,B] = [\alpha T, \beta T] = 0$.
Next we have the vector space $V$ of smooth complex functions on $\mathbb{R}^2$, and we want to look at both group and algebra representations. A group representation is a function $\rho_G$ that takes an element of $SO(2)$ and returns a linear operator on $V$, such that $\rho_G(R_1) \rho_G(R_2) = \rho_G(R_1 R_2)$. There is a canonical representation on the space of functions, given by $(\rho_G(R)\psi)(\mathbf{x}) \equiv \psi(R^{-1} \mathbf{x})$. The inverse is necessary to make this obey the group composition law. Next, a Lie algebra representation is a function $\rho_A$ which takes an element $X$ of the algebra and returns a linear operator on $V$, such that $[\rho_A(X_1), \rho_A(X_2)] = \rho_A([X_1,X_2])$. We are given a representation defined by $\rho_A(T) = i \left(y \partial_x - x \partial_y \right)$, with $\rho_A(X)$ defined for all $X$ by linearity, since $\{T\}$ is a basis for the algebra. Showing that this is in fact a representation is easy because the commutators are always trivial. If our algebra wasn't one dimensional we would have some work to do.
The hard part here is showing that the group and algebra representations correspond to each other. In other words, we need to show that the relation $R(\phi) = e^{i \phi T}$ is also true for the representations; that is, $\rho_G(e^{i\phi T}) = e^{i \phi \rho_A(T)}$, or
$$\psi(e^{-i \phi T} \mathbf{x}) = (e^{i \phi \rho_A(T)} \psi) (\mathbf{x})$$
(remember that the left hand side is by definition the representation $\rho_G$). Now, the great thing about this is that because everything is an exponential function of $\phi$, we only need to check the equalities to first order in $\phi$, since the properties of the exponential guarantee that they will then be true for all $\phi$. We then have that, to first order,
$$\psi(e^{-i \phi T} \mathbf{x}) \approx \psi(\mathbf{x} - i \phi T \mathbf{x}) = \psi \begin{pmatrix} x - \phi y \\ y + \phi x \end{pmatrix}$$
and
$$(e^{i \phi \rho_A(T)} \psi) (\mathbf{x}) \approx (1 - \phi(y \partial_x + x \partial_y)) \psi(\mathbf{x}).$$
To Taylor expand the first expression, simply use the fact that $\psi(\mathbf{x} + \mathbf{d}) \approx \psi(\mathbf{x}) + \nabla \psi \cdot \mathbf{d}$. You will find that it is indeed equal to the second expression.
I want to highlight something, though. If you just want to show that the given differential operator gives a Lie algebra representation, all of this is unnecessary. Since the Lie algebra is one dimensional, you can represent the matrix $T$ by anything and the commutator relations will be automatically satisfied, since they're trivial.
A: I can't claim I understood the flow of your question to be helpful, but here are two basic facts manifest whenever you are looking at abelian rotations:


*

*In polar coordinates, r,θ, you just have $T=-i\frac{\partial}{\partial \theta}$, a mere plane rotation obviously commuting with itself. It is a one-dimensional representation. (The Lie group itself is an infinite-dimensional group, but that only counts the infinite $\phi$ angles.)


$$ g(\phi) = 1 + i\phi T\quad +O(\phi^2)= 1+\phi\frac{\partial}{\partial\theta}+O(\phi^2) \\
 T = -i\frac{\partial g(\phi)}{\partial \phi} |
_{\phi=0} ~~.$$ The rotation merely translates the angle θ of your representation by the parameter $\phi$. 


*

*So, explicitly,
$$ g(\phi)\psi(\mathbf{x})  \sim (1+iT\phi)~\psi( e^{i\theta}r)=\left(1+\phi\frac{\partial}{\partial\theta}\right )\psi( e^{i\theta}r)=\psi( e^{i\theta}r)+i\phi~ r e^{i\theta} \psi'( e^{i\theta}r).$$ 


This is but the infinitesimal version of your final expression, $\psi( e^{i(\theta+\phi)}r)=\psi(x-\phi y, y+\phi x)\sim (1+\phi(x\partial_y-y\partial_x))\psi(x,y)$.
Two refs: WP  and Woit 2017  (Quantum Theory, Groups and Representations--An Introduction).
