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Consider a U(1) Chern Simons theory on a torus $\mathbb{T}$: \begin{align} L &= \frac{k}{4\pi} \int_{\mathbb{T}} a \partial a \end{align} where a is some U(1) gauge field, $k\in\mathbb{Z}$ and we used the short hand notation $a \partial a \equiv \epsilon^{\mu \nu \lambda} a_\mu \partial_\nu a_\lambda$.

Consider Wilson Loops of the form \begin{equation} W(C) = \mathcal{P} e^{i \oint_C a \cdot dl}. \end{equation} Here $\mathcal{P}$ denotes path ordering and $C$ denotes some closed loop on the torus $\mathbb{T}$.

Consider the two non-contractible loops on $\mathbb{T}$ denoted by $\mathcal{a}$ and $\mathcal{b}$. (For a picture see: http://share.pdfonline.com/7e91df64f6e84f43bff166c6911972d6/torus_a_b.htm)

THe ground state manifold of of the theory is $|k|$-fold degenerate.

Consider a basis that consists of wrapping "quasiparticles" around the b loop of the torus: $\left| n \right\rangle$ with n=0...$|k|-1$. Then the Wilson loop operators act as \begin{align} W(b)|n \rangle &= |n + 1 \text{ mod } |k| \rangle, \nonumber \\ W(a) |n \rangle &= e^{2\pi i n /k} |n \rangle. \end{align}.

What is the reason for this? I assume it must somehow be related to the explicit construction of the ground state manifold?

Since I am working myself into Chern Simons theory at the moment I would also be happy for some advice on readable literature with a focus on condensed matter problems.

Best regards.

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2 Answers 2

There are two ways to quantize a gauge theory.

1) First quantize all degrees of freedom and then reduce the gauge freedom by imposing conditions on the quantum Hilbert space.

2) First reduce the gauge redundancy then quantize the reduced phase space.

For realistic gauge theories in $4D$, there are problems in implementing either of these methods. The first method would require gauge fixing and the Faddeev-Popov procedure, but actually there is no good gauge fixing due to the Gribov ambiguity. The second method would result a very complicated non-flat infinite dimensional phase space with singularities.

In the case of the pure $3D$ Chern-Simons however, the second method results a finite dimensional quantizable phase space. In the Abelian (and also Non-Abelian) Chern-Simons theory on the torus, this construction can be performed explicitly.

Recall that the phase space of a quantum theory can be identified with the space of its classical solutions. In the case of the Chern Simons theory the classical solution satisfy:

$$\mathbf{F}=0$$

($\mathbf{F}$ is the gauge field strength). In flat space, this condition would mean that the solution is a pure gauge and the removal of the gauge redundancy would mean that the phase space is one point. However on the torus there is another solution which is not a pure gauge: The constant solution, because on one hand it satisfies the classical equation of motion and on the other hand it is normalizable. Thus a classical solution on the torus would have the general form:

$$\mathbf{A(x)}=\frac{\mathbf{\alpha}}{2\pi}+\mathbf{\nabla}{\phi(\mathbf{x})}$$

Where $\mathbf{\alpha}$ is a constant.

Remembering that the Poisson brackets obtained from the classical Abelian Chern Simons Lagrangian read:

$$\{A_i, A_j \} = \frac{2 \pi i \epsilon_{ij}}{k} \delta^2(\mathbf{x}-\mathbf{y})$$

And observing that

$$\mathbf{\alpha} = \oint\mathbf{A}$$

(The integration is over a single turn). Then by integrating the Poisson bracket equation over its two circle generators, we obtain the following Poisson brackets for $\alpha$

$$\{\alpha_i, \alpha_j \} = \frac{2 \pi i \epsilon_{ij}}{k} $$

$\alpha$ defines a two dimensional phase space. Quantizing this space would mean to find functions on the two dimensional phase space satisfying these Poisson brackets. In this case this can be accomplished by inspection:

$$ \alpha_i = p_i - \frac{2 \pi i \epsilon_{ij}}{k} q_i$$

This is just the momentum operator in the presence of a constant magnetic field $$\mathbf{B} = \frac{2 \pi }{k} \mathbf{\hat{z}}$$ in the $z$-direction. ("Magnetic translation operator"). Thus the quantization of the Wilson loop along an integral number of loops:

$$ W(\mathbf{l}) = e^{i \oint_{\mathbf{l}}\mathbf{A}} = e^{i \mathbf{\alpha}. \mathbf{l}}$$

(i.e., this time we do not restrict the integration to a single turn:) $\mathbf{l} = 2 \pi (m, n)$ where $m$ and $n$ are integers.

It is not hard using the canonical commutation relations (Weyl algebra) to obtain:

$$ W(\mathbf{l}_1) W(\mathbf{l}_2) = e^{i \mathbf{B} .( \mathbf{l}_1 \times \mathbf{l}_2)} W(\mathbf{l}_2) W(\mathbf{l}_1) $$

In particular:

$$ W(\mathbf{\hat{x}}) W(\mathbf{\hat{y}}) = e^{i \frac{2 \pi}{k}} W(\mathbf{\hat{y}}) W(\mathbf{\hat{x}}) $$

It is worthwhile to mention that the operators $\alpha$ cannot be chosen as observables because they are not gauge invariant. Thus we seek a Hilbert space representation of $W(\mathbf{\hat{x}})$ and $W(\mathbf{\hat{y}})$. One way to find such a representation is to postulate the existence of a state (vacuum) $|0\rangle$ upon which we can generate the whole Hilbert space by the action of the operators $W(\mathbf{\hat{x}})$ and $W(\mathbf{\hat{y}})$.

First, we observe that, without loss of generality, we can choose $W(\mathbf{\hat{x}})$ as diagonal. Next we observe that

$$ W(\mathbf{\hat{x}}) W(\mathbf{\hat{y}})^k = W(\mathbf{\hat{y}})^k W(\mathbf{\hat{x}}) $$

Thus $ W(\mathbf{\hat{y}})^k $ commutes with the diagonal $W(\mathbf{\hat{x}})$, thus they must be diagonal, since it commutes also with $W(\mathbf{\hat{y}})$, therefore with any sequence of products of the two operators. Thus this operator must be represented by the unit operator: $$ W(\mathbf{\hat{y}})^k = \mathbf{I}$$

Since $ W(\mathbf{\hat{x}})$ is diagonal, it does not alter the vectors it acts on, thus, a basis of the Hilbert space must be given by:

$$\mathrm{Span} \{ |0\rangle$, $ W(\mathbf{\hat{y}}) |0\rangle, W(\mathbf{\hat{y}})^2 |0\rangle,..., W(\mathbf{\hat{y}})^{k-1} |0\rangle \}$$

This implies that the Hilbert space is finite dimensional, and if we denote:

$$ |n\rangle = W(\mathbf{\hat{y}})^n |0\rangle$$

We obtain the torus algebra action on the Hilbert space:

$$ W(\mathbf{\hat{x}}) |n\rangle = e^{i \frac{2 \pi n}{k}} |n\rangle $$

$$ W(\mathbf{\hat{y}}) |n\rangle = |n+1\rangle $$

This is not the only way to obtain this result. This Hilbert space can be obtained more naturally as the space of Jacobi theta functions in the coordinate representation. This would require to use the techniques of Berezin quantization or Coherent state quantization, please see, for example, section 6 of the following article by: Spradlin, and Volovich.

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Partial answer :

With a classical abelian connection, we would have $W(a)W(b) = W(b) W(a)$. In the quantum version of the theory, there is a deformed algebra (quantum algebra) : $W(a)W(b) = e^{ \large \frac{2i\pi}{k}} W(b) W(a)$.

The Hilbert space is $k$-dimensional, with states $|0 \rangle, ....|k-1 \rangle$, and we have to choose a corresponding representation of the deformed algebra.

We may choose $W(a)|0 \rangle = |0 \rangle$, and considering that the Hilbert space is spanning by the powers of $W(b)$. Because the Hilbert space is $k$-dimensional, we must have : $W^k(b)|0 \rangle = |0 \rangle$.

Now, it is not difficult to see that the relations :

$$W(b)|n \rangle = |n+1 \mod k \rangle \quad \quad,W(a)|n \rangle = ^{ \large \frac{2i\pi n}{k}}|n \rangle$$ satisfy the wished precedent conditions.

See this Ref.

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@ Trimok, it seems to me that even for Abelian connection (such as level-k U(1) Chern-Simons theory), you still have $$W(a)W(b) = e^{ \large \frac{2i\pi}{k}} W(b) W(a)$$ instead of $W(a)W(b) = W(b) W(a)$. Am I correct? It is NOT the same as what you said in the beginning. Please let me know. thanks. –  Idear Jan 9 at 3:55
    
@Idear : Yes, you are correct. It is the quantum version for an abelian connection, and this is clear in the ref I gave.... It is an error.... I modify the answer. Thanks. –  Trimok Jan 9 at 9:29

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