Unitary Representation of $\text{SO}(3)$ in Position Representation

Question

Let $R\in\text{SO}(3)$ be an arbitrary rotation, and let $U_R$ be the unitary representation of $R$ on some Hilbert space $\mathcal H$. To me, the defining property of $U_R$ is how it conjugates the position operator $\textbf X$: $$U_R^{-1}\textbf X U_R=R\textbf X$$ And in position space, I would expect that: $$\langle{\textbf x}|U_R|{\psi}\rangle=\langle{R^{-1}\textbf x}|\psi\rangle$$ for arbitrary $\textbf x\in\textbf R^3$ and $|\psi\rangle\in\mathcal H$ but I'm not sure how to actually prove this. Any hints or advice would be appreciated.

Answer 1

The eigenvector $|x\rangle$ of the position operator $\hat X$ associated to the eigenvalue $x$ being defined by $$\hat X|x\rangle=x|x\rangle$$ the defining property of $U_R$ gives $$U_R^{-1}\hat XU_R|x\rangle=R\hat X|x\rangle=Rx|x\rangle$$ which leads to $$\hat XU_R|x\rangle=RxU_R|x\rangle$$ since $Rx$ is a number and not an operator. As a consequence, $U_R|x\rangle$ is the eigenvector of the position operator $\hat X$ associated to the eigenvalue $Rx$. Since $U_R$ is unitary, it is normalized and $$U_R|x\rangle=|Rx\rangle$$ Acting on the left of the defining property of $U_R$ with $\langle x|$ or taking the adjoint of the latter relation gives $$\langle x|U_R^+=\langle x|U_R^{-1}=\langle Rx|$$ so that $$\langle x|U_R=\langle R^{-1}x|$$ Finally, $$\langle x|U_R|\psi\rangle=\langle R^{-1}x|\psi\rangle$$