# On derivation of Chern-Simons action for fractional quantum Hall effect (cf. Altland and Simons 9.5)

This follows the discussion in Altland and Simons Condensed Matter Field Theory -- section 9.5 on deriving the Chern-Simons action for FQHE.

Starting with the real-time field integral representation: $\mathcal{Z} = \mathcal{N} \int D(\bar{\psi},\psi)e^{iS[\bar{\psi},\psi]}$ where

\begin{equation} S[\bar{\psi},\psi] = \int dt \, d^2x \, \bar{\psi} \bigg(i\partial_t + \mu - \frac{1}{2m} (-i\partial_x + \mathbf{A}[\bar{\psi},\psi])^2 - V(\mathbf{x}) \bigg)\psi \end{equation}

$\mathbf{A} = \mathbf{A}_\text{ext} + \mathbf{a}$, with $\mathbf{A}_\text{ext}$ is the vector potential of the magnetic field responsible for the QHE and $\mathbf{a}$ is the vector potential from the phases factor of the singular gauge transformation

\begin{equation} \Psi(\mathbf{x}_1,...) \rightarrow \Psi(\mathbf{x}_1,...) \exp \big(-2is \sum_{i<j} \text{arg}(\mathbf{x}_i - \mathbf{x}_j) \big) \end{equation}

As stated in the book, $\mathbf{A}$ present a complication that can be avoided by promoting the vector potential to an integration variable whose value is set so as to generate the flux pattern. This is done by multiplying $\mathcal{Z}$ by

\begin{equation} (1) \qquad \qquad 1=\mathcal{N}\int D\mathbf{a_\perp} \prod_{\mathbf{x},t} \, \delta\big(b(\mathbf{x},t)+4\pi s \rho(\mathbf{x},t)\big) \end{equation}

where $b=\epsilon_{ij} \partial_i (a_\perp)_j$ and the subscript "$\perp$" indicates that the integration extends only over transversal configuration of the vector potential (i.e. $\partial_i a_i =0$). This results to

\begin{equation} (2) \qquad \mathcal{Z} = \mathcal{N} \int D(\bar{\psi},\psi) D\mathbf{a_\perp} \prod_{\mathbf{x},t} \, \delta\big(b(\mathbf{x},t)+4\pi s \rho(\mathbf{x},t)\big) \exp\big[-S[\bar{\psi},\psi,\mathbf{a}_\perp] \big] \end{equation} \begin{equation} (3) \qquad \mathcal{Z} = \mathcal{N} \int D(\bar{\psi},\psi) D\mathbf{a_\perp} D\phi \exp \bigg( iS[\bar{\psi},\psi,\mathbf{a}_\perp] - i\int dt\, d^2x \, \phi (b/4\pi s + \rho ) \bigg) \end{equation}

My questions are as follow:

• How does eq. (1) achieved the statement in bold preceding it?
• How does one goes from eq. (2) to eq. (3)? The general method involved as I am unfamiliar with what could have possibly happened.

I would also appreciate any supplemental texts, preferably good for a graduate student starting to study this rich field.

• Concerning the second question, they used $\int d \lambda e^{i \lambda x}=\delta(x)$, promoted to the case of functional integrals.
$$\begin{equation} S[\bar{\psi},\psi] = \int dt \, d^2x \, \bar{\psi} \bigg(i\partial_t + \mu - \frac{1}{2m} (-i\partial_x + \mathbf{A}[\bar{\psi},\psi])^2 - V(\mathbf{x}) \bigg)\psi \end{equation}$$ $$\mathbf{A}=\mathbf{A_{ext}+a}$$
$$\mathbf{a}$$ is not a dynamic variable. Instead, it is determined by $$\rho(x)$$ as Altland has mentioned before. \begin{align} \mathbf{a}=-2s\int {\rm d}^2{x'} \frac{\hat{z} \times (x-x')}{|x-x'|^2}\rho(x') \end{align} However, this means that $$\overline{\psi} (-i\partial_x+\mathbf{a})^2\psi$$ contain 4 and 6 fermion operators we do not know how to deal with. The formula (1) which you mentioned itself is just an identity since integration of delta function yields constant, but when you put this into the expression of action, the delta function allow you to make the substitution from $$-2s\int {\rm d}^2{x'} \frac{\hat{z} \times (x-x')}{|x-x'|^2}\rho(x')$$ to $$a_\perp$$(where now $$a_\perp$$ is a dynamic variable do not dependent on $$\rho(x))$$ , since the solution of $$\begin{equation} \left\{\begin{array}{c}\epsilon_{ij}\partial_i a_{\perp}(\mathbf{x},t)+4\pi s \rho(\mathbf{x},t)=0\\\partial_i (a_{\perp})_i=0\end{array}\right. \end{equation}$$ is just the expression of $$a_{\perp}$$.
just $$\int {\rm d}x e^{i\lambda x}\propto \delta(\lambda)$$