Follow up question on "Wilson Loops as Raising Operators"

Question

This is a follow-up question on the topic that I opened a few days ago, Wilson Loops as raising operators.

The paper

Topological Degeneracy of Quantum Hall Fluids. X.G. Wen, A. Zee. Phys. Rev. B 58 no. 23 (1998), pp. 15717-15728. arXiv:cond-mat/9711223.

gives a nice derivation of the explicit ground states of the $U(1)$ Chern-Simons Theory on a torus in Section 2 on Abelian Quantum Hall States.

In particular Eq. (12) gives the generic form of a ground state $\psi(y) = \sum_{n=-\infty}^{\infty} c_{n} \ e^{i\ 2\pi ny}$. Due to the fact that the theory lives on a torus the ground state manifold is found to be $k$-fold degenerate.

My question: Is it possible (by direct calculation) to obtain the relations \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} from the previous question?

I don't have a particularly strong background in field theory so I am feeling somewhat uneasy when it comes to the explicit evaluation of the Wilson Loop (with its exponentiated gauge field and the path ordering) acting on the constructed state.

I am looking forward to your responses.

Answer 1

The correspondence is : $ W(a)= e^{2i \pi y}, \quad W(b) = e^{-2i \pi x}$.

From $[x,y] = \frac{i}{2\pi k}$ $(9)$, we get : $W(a)W(b) = e^{ \large \frac{2i\pi}{k}} W(b) W(a)$ (see precedent answer).

As it is explained in the text between formulae $(19)$ and $(20)$, $x$ and $y$ may only take discrete values $\frac{1}{k},\frac{2}{k},....., \frac{k-1}{k},1$.

A direct consequence of that is : $W^k(a) = W^k(b) = \mathbb{Id}$, as wished, because of the identifications $x= x+1$ and $y = y+1$.

[EDIT]

The Wilson loops

Here is a non rigorous argument, but this is my feeling. If we look at equation $(4)$ of your paper, we see that :

$$a_0(x_1,x_2,t)=0, \quad a_1(x_1,x_2,t)= 2 \pi \frac{x(t)}{L_1}, \quad a_2(x_1,x_2,t)= 2 \pi \frac{y(t)}{L_2}\tag{4}$$

Now, choose for the path $(a)$ : $x_2$ increasing , $0 \le x_2 \le L_2$, and $x_1,t$ constant (so $dx_1=0$) we would have :

$$\oint_a (a_i dx^i) = \oint_a (a_2 dx^2) = \int_0^{L_2} 2 \pi \frac{y(t)}{L_2} = 2 \pi y(t) \tag{5}$$

So, we would have :

$$W(a) = e^{ \large i\oint_a (a_i dx^i)} = e^{2 i\pi y} \tag{6}$$

In the same spirit choose the path $(b)$ : $x_1$ decreasing , $0 \le x_1 \le L_1$, and $x_2,t$ constant (so $dx_2=0$) we would have :

$$\oint_b (a_i dx^i) = \oint_a (a_1 dx^1) = \int_{L_1}^0 2 \pi \frac{x(t)}{L_1} = -2 \pi x(t) \tag{7}$$

So, we would have :

$$W(b) = e^{\large i\oint_b (a_i dx^i)} = e^{-2 i\pi x} \tag{8}$$

The basis

Because the momentum $p$ is $p=2\pi kx = -i\frac{\partial}{\partial y}$, we may rewrite $W(b)$ as $W(b) = e^{\large -\frac{1}{k}\frac{\partial}{\partial y}}$.

So, we have : $W(b) \psi(y) = \psi(y-\frac{1}{k})$

It is natural, then to postulate the following basis :

$$\psi_n(y) = \langle n|y\rangle = \delta (ky - n) \tag{9}$$

We see that :

$$W(a) \psi_n(y) = e^{2i \pi y} \delta (ky - n)= e^{2i \pi \large \frac{n}{k}}\delta (ky - n) = e^{2i \pi \large \frac{n}{k}} \psi_n(y) \tag{10}$$

$$W(b) \psi_n(y) = \psi_n(y - \frac{1}{k}) = \delta(k (y -\frac{1}{k}) - n) = \delta (ky - (n+1)) = \psi_{n+1}(y)\tag{11}$$