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  Geometric quantization of identical particles

+ 12 like - 0 dislike


It is well known that the quantum mechanics of $n$ identical particles living on $\mathbb{R}^3$ can be obtained from the geometric quantization of the cotangent bundle of the manifold $M^n = \frac{\mathbb{R}^{3n}-\Delta}{S_n}$, where $\Delta$ is the set of coincidences and $S_n$ is the permutation group of $n$ elements acting naturally on the individual copies of $\mathbb{R}^3$, please see, for example, Souriau: Structure of dynamical systems. $T^*M^n$ is multiply connected with $\pi_1(T^*M^n) = S_n$. Given the canonical symplectic structure on $T^*M^n$,the set of inequivalent quantizations has a one to one correspondence to the set of character representations of the fundamental group $\mathrm{Hom}(\pi_1(M^n), U(1))= \mathbb{Z}_2$ corresponding to the identity and the parity characters. These quantizations correspond exactly to the pre-quantization of bosons and fermions. The boson and fermion Fock spaces modeled on $\mathrm{L}^2(R^3)$ emerge as the quantization of Hilbert spaces corresponding to these two possibilities.

Many authors pointed out that the removal of the coincidence set from the configuration space may seem not to be physically well motivated. The standard reasoning for this choice is that without the removal, the configuration space becomes an orbifold rather than a manifold. Some authors indicate also that without the removal, the configuration space is simply connected thus does allow only Bose quantization (Please, see for example the reprinted article by Y.S. Wu in Fractional statistics and anyon superconductivity By Frank Wilczek.

My question:

Are there any known treatments or results of the problem of geometric quantization of the configuration space as an orbifold (without the removal of the coincidence set), in terms of orbifold line bundles, etc.? Partial results or special cases are welcome. Does this quantization allow the possibility of Fermi statistics?

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asked Nov 8, 2011 in Theoretical Physics by David Bar Moshe (4,355 points) [ no revision ]
Although I don't know the answer to the question, and definitely am no expert on geometric quantization, it's worth noting that some of the state-spaces of classical dynamics are endowed with a symplectic structure that does *not* arise from a potential (per Sec. 18.36 of Souriau), the most prominent example of which is the Bloch sphere. In principle these "qudit" dynamical systems can be (must be?) quantized without reference to any tangent bundle, action principle, or Lagrangian; indeed for large-scale simulation purposes such non-geometric quantization is a standard procedure.

This post has been migrated from (A51.SE)
A tiny remark: if you choose to think of the configuration space (without the removal) as an orbifold, then it is not simply connected: its orbifold fundamental group is $S_n$.

This post has been migrated from (A51.SE)
@Dan Thanks, your remark is very helpful. I guess that the orbifold fundamental group should be the relevant object to the orbifold quantization.

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2 Answers

+ 5 like - 0 dislike

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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answered Nov 8, 2011 by Pavel Safronov (1,120 points) [ no revision ]
Thank you for your answer and reference

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+ 2 like - 0 dislike

Such quantization produces the same results as quantization with the coincidence set removed. Here is why. Consider the canonical way to perform geometric quantization of $T^*M$, namely:

  • Line bundle is trivial
  • Connection is given by the canonical 1-form $p\ dx$ in physical notation
  • Lagrangian foliation has fibres of $T^*M$ for leaves

This yields quantum mechanics in the position picture

Now, take $X = T^*M^n / S_n$ to be the multi-particle orbifold phase space. Line bundles on $X$ are just $S_n$-equivariant line bundles on $T^*M^n$. Hence we can take the same line bundle with the action of S_n either trivial or multiplied by the sign of the permutation. All the other structures are S_n invariant and therefore descend to $X$. Clearly, the result is either bosons or fermions depending on the chosen line bundle (representation of $S_n$)

EDIT: It seems that there is no analogue of this result for anyonic statistics with dim M = 2. This is because for dim M = 2, the phase space of distinguishable hardcore (incidence removed) particles already has a non-trivial fundamental group.

This post has been migrated from (A51.SE)
answered Nov 12, 2011 by Squark (1,725 points) [ no revision ]
I think $(T^*\mathbf{R}^{3n})/S_n$ is just one of many ways to "complete" $T^*(\mathbf{R}^{3n}-\Delta)/S_n$. Also, I do not understand what you mean by "replace $S_n$ by $B_n$": what kind of action of $B_n$ on $\mathbf{R}^{2n}$ do you have in mind? In David's notation it arises as $\pi_1(M^n)$ in dimension 2, and so will not be seen in the orbifold picture.

This post has been migrated from (A51.SE)
@Squark Thank you for your answer

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What I wrote about dimension 2 was nonsense. See my edit.

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