The Laughlin wave function at filling fraction $\nu=\frac{1}{m}$ is

\begin{equation} \Psi_m=\prod_{i<j}(z_i-z_j)^m e^{-\sum|z_i|^2/4l_B^2} \end{equation}

It is claimed in section 7.2.3 of Wen's book that the wave function of a quasi-hole excitation on the top of this state is

\begin{equation} \Psi^h(\xi,\xi^*)=\sqrt{C(\xi,\xi^*)}\prod_i(\xi-z_i)\Psi_m \end{equation}

I am wondering why this is a wave function of a quasi-hole, it looks more like the wave function of a quasi-particle to me. If this is indeed the wave function for a quasi-hole instead of a quasi-particle, what is the wave function of a quasi-particle?

Later in section 7.2.4, the author discusses a way to construct generalization of Laughlin states by adding quasi-holes or quasi-particles on the top of Laughlin states, and when the density of quasi-holes or quasi-particles reaches certain value, they form a Laughlin state by itself. And there are two examples: by condensing quasi-holes of $\nu=1/3$ Laughlin state, we get a $\nu=2/7$ FQH state with wave function:

\begin{equation} \Psi=\int\prod_id^2\xi_i\Psi_3\prod(\xi_i-z_j)(\xi_i^*-\xi_j^*)e^{-\frac{1}{4l_B^2}\frac{1}{3}|\xi_i|^2} \end{equation}

and by condensing quasi-particles of $\nu=1/3$ Laughlin state, we get a $\nu=2/5$ FQH state with wave function:

\begin{equation} \Psi=\int\prod_id^2\xi_i\prod_{i<j}(\xi_i-\xi_j)^2(\xi_i^*-2\partial_{z_i})\Psi_3 \end{equation}

My understanding is, the original electrons and added excitations form a Laughlin state with a common filling fraction, then by charge conservation, we should have

\begin{equation} {\rm original\ filling}+{\rm resulting\ filling}\times{\rm charge\ of\ excitation}={\rm resulting\ filling} \end{equation}

However, these two examples are telling me (notice the factor $\frac{1}{2}$):

\begin{equation} {\rm original\ filling}+{\rm resulting\ filling}\times{\rm charge\ of\ excitation}\times\frac{1}{2}={\rm resulting\ filling} \end{equation}

(The first example fits as \begin{equation} \frac{1}{3}-\frac{1}{3}\times\frac{2}{7}\times\frac{1}{2}=\frac{2}{7} \end{equation} and the second example fits as \begin{equation} \frac{1}{3}+\frac{1}{3}\times\frac{2}{5}\times\frac{1}{2}=\frac{2}{5}.) \end{equation}

My questions are:

1) Why should there by a $\frac{1}{2}$ factor?

2) Is it related to the power of $(\xi_i-\xi_j)^2$, namely $2$?

3) If it is related to that $2$, can I change that $2$ into another integer to get another FQH state?


1 Answer 1


To see that the expression for the quasi-hole is consistent, one notes that the order of the polynomial part of this expression is one higher than that of the original Laughlin state. The order of the polynomial corresponds to the number of flux and thus the quasi-particle expression can be seen to have one additional flux with the same number of particles, which is what one expects of a quasi-hole state.

I don't know of a simple expression for a Laughlin state with a quasi-particle excitation. An expression with a product of derivatives acting on the Laughlin state as is hinted at in the expression for the $\nu=\frac{2}{5}$ state could work. This would at least give the correct filling factor.

I believe you could be miss-understanding slightly what the filling factor is and what quasi-particle and quasi-hole excitations are. When considering the filling factor, one does not consider the quasi-particles or quasi-holes. The filling factor is simply the ratio of electrons (or in some cases Bosons) to flux quanta. The effect of adding a quasi-particle excitation to a state is to increase the flux by one while leaving the number of underlying electrons (or Bosons) the same. It is simply a localised absence of charge. Likewise adding a quasi-particle excitation reduces the number of flux by one while leaving the number of underlying electrons (or Bosons) the same. It is a localised excess of charge.

Now in the case of the expression for the $\nu=\frac{2}{7}$ hierarchy state, quasi-hole excitations are added over the $\nu=\frac{1}{3}$ Laughlin state until they condense into a $\nu = \frac{1}{3}$ Laughlin state of their own (question is missing the 2 exponent). The original Laughlin has filling $\nu=\frac{1}{3}=\frac{N}{3N}$, where $N$ is the number of electrons. If one adds $\frac{N}{2}$ quasi-holes the resulting filling factor is:

$$ \nu = \frac{N}{3N+\frac{N}{2}} = \frac{2}{7} $$

The situation with the $\nu = \frac{2}{5}$ case is similar, except with quasi-particles. When $\frac{N}{2}$ quasi-particles are added to the system one gets:

$$ \nu = \frac{N}{3N-\frac{N}{2}} = \frac{2}{5} $$

The first question has been addressed. As for the others. The exponent on the $(\xi_i-\xi_j)$ term does not affect the overall filling factor but can change the filling of the state the quasi-holes condense into.

  • $\begingroup$ How do you know $N/2$ quasi-holes has been added to the system? Is this piece of information encoded in the wave function? $\endgroup$ Jun 27, 2016 at 5:26
  • $\begingroup$ The wave-function with $\frac{N}{2}$ additional quasi-holes is not written explicitly in either the question or my answer. Such an expression can be written in a similar way to the expression for the wave-function with a single quasi-hole, but with a product of additional sets of Jastrow factors with different positions $\{\xi_j\}$. One arrives at the number $\frac{N}{2}$ as this is the number that gives the desired filling of $\nu=\frac{2}{7}$. $\endgroup$ Jun 27, 2016 at 9:31
  • $\begingroup$ What exactly do we mean by saying quasihole-excitation over Laughlin state?. Are those states excited states over degenerate Laughlin ground states in the spectrum of the Fqh Hamiltonian (like freeze out kinetic term, project contact interaction to LLL and diagonalize it; quasi hole excitations are excited states above ground state?) OR we sort of mean that the Laughlin ground state is kind of a vacuum and it harbors quasi-particle (hole or electron) excitations? $\endgroup$
    – Quarkonia
    Mar 12, 2021 at 11:38

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