Fractional Quantum Hall Effect- degeneracy of ground state (Tong's notes)

In summary, the conversation discusses the ground state degeneracy on a manifold and the comment made in regards to it. The identity and dimension of the manifold are discussed, leading to the conclusion that the ground state must have a degeneracy of m b for some positive integer b. However, it is not proven that b=1 for genus 1. The use of Chern-Simons theory may provide a solution.
  • #1
binbagsss
1,254
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Hi , I'm looking at the argument in David Tongs notes (http://www.damtp.cam.ac.uk/user/tong/qhe/three.pdf) for ground state degeneracy on depending on the topology of the manifold (page 97, section 3.2.4).

I follow up to getting equation 3.31 but I'm stuck on the comment after : ' But such an algebra of operators can’t be realized on a single vacuum state.'

I have no idea where this comment comes from- (also I then don't understand the smallest representation comment, dimension m, and where the action comes from).

Any links to relevant background knowledge assumed to follow these comments of this section, or any help, would be greatly appreciated. many thanks.
 
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  • #2
In the end this is purely a math problem: we know that there is the following identity on the ground state manifold,
$$
T_1 T_2 T_1^{-1} T_2^{-1} | 0 \rangle = e^{2 \pi i/m} | 0 \rangle,
$$
where ##m## is a positive integer. Given that, what is the dimension of the manifold ##| 0 \rangle##? It turns out to just be equivalent to asking what dimension of matrices do you need to have the two matrices ##T_{1,2}## satisfy
$$
T_1 T_2 = e^{2 \pi i/m} T_2 T_1.
$$
Here I will admit that I don't remember my abstract algebra well enough to know how to solve this problem in complete generality, but you should be able to convince yourself pretty quickly that there are no one-dimensional solutions, so you need more than one ground state. With a little bit of trial-and-error for small values of ##m## I quickly found the ##m##-dimensional solution
$$
T_1 =
\begin{pmatrix}
1 & 0 & 0 & \cdots & 0 \\
0 & e^{2 \pi i/m} & 0& \cdots & 0 \\
0 & 0 & e^{4 \pi i/m} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & e^{2 \pi i (m-1)/m}
\end{pmatrix},
\qquad
T_2 =
\begin{pmatrix}
0 & 0 & 0 & \cdots & 0 & 1 \\
1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 1 & 0 & \cdots & 0 & 0 \\
\vdots &\vdots &\vdots &\ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 1 & 0
\end{pmatrix}
$$
Wen's textbook claims that there is only a single ##m##-dimensional irreducible representation of the above algebra, so this construction must be it. But I'd go over to the Algebra forum if you want a full proof of that statement. In any case, this proves that representations of the above algebra must be direct sums of the above representation, so the ground state must have degeneracy ##m b## for some positive integer ##b##. Unfortunately, I don't think this argument tells you that ##b=1## for genus 1, but if you're planning on completing Tong's notes I think you will eventually be able to show this from Chern-Simons theory.
 
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1. What is the Fractional Quantum Hall Effect (FQHE)?

The Fractional Quantum Hall Effect is a phenomenon that occurs in two-dimensional electron systems at extremely low temperatures and high magnetic fields. It is characterized by the formation of quantized energy levels and the appearance of fractional charges in the system.

2. How does the degeneracy of the ground state in FQHE differ from that of the integer quantum Hall effect?

In the integer quantum Hall effect, the degeneracy of the ground state is proportional to the strength of the magnetic field. However, in the FQHE, the degeneracy is independent of the magnetic field and is determined by the filling fraction of the system.

3. What is the significance of the degeneracy of the ground state in FQHE?

The degeneracy of the ground state in FQHE is important because it allows for the formation of collective excitations called quasiparticles. These quasiparticles have fractional charges and obey fractional statistics, which are key properties of the FQHE.

4. How is the degeneracy of the ground state calculated in FQHE?

The degeneracy of the ground state in FQHE can be calculated using the Laughlin wave function, which describes the ground state of the system at certain filling fractions. This wave function takes into account the interactions between electrons and predicts the degeneracy of the ground state.

5. What are the applications of understanding the degeneracy of the ground state in FQHE?

Understanding the degeneracy of the ground state in FQHE is crucial for studying and manipulating the behavior of electrons in low-dimensional systems. It has potential applications in quantum computing and the development of new materials with unique properties.

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