Often, instead of $\mathbf{R}^{3n}/S_n$, you may want to resolve the singularity. Let me explain a toy model where that resolution appears naturally.

Consider $n$ identical particles on $\mathbf{C}$ with the configuration space $M^n=(\mathbf{C}^n-\Delta)/S_n$. You can think of this space as the space of unordered eigenvalues of $n\times n$ matrices over $\mathbf{C}$, i.e. $M^n=Mat_n(\mathbf{C})^{diag}/GL_n(\mathbf{C})$, where $Mat_n(\mathbf{C})^{diag}$ is the space of diagonalizable matrices with distinct eigenvalues and $GL_n(\mathbf{C})$ acts by conjugation.

A heuristic argument (which is precise when the action of $G$ is nice) shows that $T^*(M/G)\cong T^*M//G$, where $//$ is the Hamiltonian reduction. In my case, there is a well-known compactification of $T^*M^n$ called the Calogero-Moser space $C_n$ obtained by the Hamiltonian reduction of $T^* Mat_n(\mathbf{C})$ along some orbit.

Cotangent bundles have natural quantizations (functions are replaced by differential operators on the base and the Hilbert space is just $L^2$ functions on the base), and the quantization of the Calogero-Moser space $C_n$ is obtained by a procedure called the quantum Hamiltonian reduction from the quantization of $T^*Mat_n(\mathbf{C})$.

For a reference, see Etingof's lectures http://arxiv.org/abs/math/0606233v4. In particular, see proposition 2.6. Note, that he is more precise than I am, and so considers the action by $PGL_n(\mathbf{C})$ since it does not have any center.

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