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.


1 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 $y = y+1$.


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 :

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

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

  • $\begingroup$ I can follow your argument except for the first line. Are W(a) and W(b) obtained from first principle calculation from the original definition of the Wilson loop? Also isn't $\left|n\right\rangle = e^{i2\pi n y}$ (just a Fourier compenent of \psi)? Does this then correctly reproduce the action of W(a) and W(b) on $\left|n\right\rangle$? $\endgroup$
    – MrLee
    Commented Oct 9, 2013 at 20:14
  • $\begingroup$ @MrLee : I edited the answer to add precisions for Wilson loops, and a basis $|n\rangle$ $\endgroup$
    – Trimok
    Commented Oct 10, 2013 at 7:41
  • $\begingroup$ Thanks a lot Trimok. Your answer is very detailed and nice. Please allow me one more question: In your derivation of the Wilson loop (which as you have stated) you omit the "Path ordering" that acts on the exponential. Why can you ignore the path ordering in this particular case? $\endgroup$
    – MrLee
    Commented Oct 10, 2013 at 7:47
  • 2
    $\begingroup$ Path-ordering is referring to a particular order, relatively to a parameter, which is usually the time $t$. The paths have been defined at $t=$ constant, so, it seems that there is no problem. $\endgroup$
    – Trimok
    Commented Oct 10, 2013 at 8:12

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