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:
1. $\mathfrak{so}(2)$ 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: $$ g(\phi) = 1 + iT\phi \quad : \ T = -i\frac{\partial g(\phi)}{\partial \phi} $$
2. 364
Apparently the group action of $g$ on $\psi(\mathbf{x}) = \langle \mathbf{x}| \psi \rangle$ is $ g\psi = \psi(R^{-1}\mathbf{x})$
3. 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 $g$?
4. This proof is completely useless for fields although <PERSON> would seem to imply a generalization of this result??
5. Assuming 2 fine: $$ g(\phi)\psi(\mathbf{x}) = (1+iT\psi(\mathbf{x})\phi) = \psi(R^{-1}\mathbf{x})$$ $$ R^{-1} = \begin{pmatrix} cos\phi & -sin\phi \ sin\phi & cos\phi \ \end{pmatrix} \implies \psi(R^{-1}\mathbf{x}) = \psi\begin{pmatrix} x-\phi y \ y+ \phi x \end{pmatrix} $$
-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 $(x,y)$ 804