I have a question regarding Eq. (3.5) in Moore & Read's paper. They said \begin{equation} \Psi_{\text{Laughlin}}=\left\langle\prod_{i=1}^{N}e^{i\sqrt{q}\phi(z_i)}\exp\left[-i\int \mathrm d^2z^{\prime}\sqrt{q}\rho_0\phi(z^{\prime})\right]\right\rangle \end{equation} is equivalent to Laughlin wave function described below \begin{equation} \Psi_{\text{Laughlin}}(z_1,\cdots,z_N)=\prod_{i<j}(z_i-z_j)^q\exp\left[-\frac{1}{4}\sum|z_i|^2\right] \end{equation} $\phi(z)$ is a free massless scalar filed satisfying $\langle\phi(z)\phi(w)\rangle=-\log(z-w)$. My question is how do I get the second equation from the first one. Especially, I don't know how do I obtain nonholomorphic part, that is, $\exp\left[-\frac{1}{4}\sum|z_i|^2\right]$ starting from the first equation. I don't understand at all the explanation below Eq. (3.7) in their paper.

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    $\begingroup$ Especially considering the paper you link is not freely available, please include all relevant context into the question. What does the notation means, what are you trying to compute, and what specifically about the given explanation is unclear to you? $\endgroup$
    – ACuriousMind
    Commented Apr 7, 2016 at 10:44
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    $\begingroup$ The non-holomorphic part $\exp\{-|z|^2/4\}$is not included I think. $\endgroup$
    – mike stone
    Commented Jan 22, 2021 at 23:09
  • $\begingroup$ Related physics.stackexchange.com/q/805997/226902 $\endgroup$
    – Quillo
    Commented Mar 12 at 8:18

1 Answer 1


Perhaps you could have a look at David Tong's lectures on QHE (http://www.damtp.cam.ac.uk/user/tong/qhe.html) you might find Section 6.2 helpful. The paper in question is available here (http://www.physics.rutgers.edu/~gmoore/MooreReadNonabelions.pdf)


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