# Gauge fixing in derivation of fractional QHE action

I'll copy the text from a relevant question:

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

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

$$\mathbf{A} = \mathbf{A}_\text{ext} + \mathbf{a}$$, 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

$$$$\Psi(\mathbf{x}_1,...) \rightarrow \Psi(\mathbf{x}_1,...) \exp \big(-2is \sum_{i

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

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

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

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

I don't understand why we integrate over gauge potentials in the Coulomb gauge only? Altland and Simons claims this is because the gauge potential

\begin{align} \mathbf{a}=-2s\int {\rm d}^2{x'} \frac{\hat{z} \times (x-x')}{|x-x'|^2}\rho(x') \end{align}

is transverse i.e. $$\nabla \cdot \textbf{a} = 0$$. However the condition $$b+4\pi s\rho=0$$ that is used to promote the vector potential to an integration variable is gauge-invariant, so it should not matter what gauge I am working in, right?

Isn't there a direct way to get to the gauge-invariant expression $$\mathcal{L}_{CS}=\epsilon^{\mu \nu \rho} a_\mu \partial_\nu a_\rho$$? for the Chern-Simons term?

1. When performing the gauge transformation $$U = \exp \big(-2is \sum_{i you can show that $$U^{-1}\Big(-i\mathbf{\nabla} + \mathbf{A_{\text{ext}}} \Big)U = \Big(-i\mathbf{\nabla} + \mathbf{A_{\text{ext}}} + \mathbf{a} \Big),$$ with $$\mathbf{a}=-2s\int {\rm d}^2{x'} \frac{\hat{z} \times (x-x')}{|x-x'|^2}\rho(x')$$. Hence your statistical field $$\mathbf{a}$$ is fixed in this case. The gauge redundancy would come from amending the gauge transformation with an irrelevant phase factor: $$\tilde U = U e^{i\phi(\mathbf{x})}$$. So when now implementing the $$\mathbf{a}$$ field in the path integral, we need to remember that it is only uniquely defined when considering the two equations: $$\mathbf{\nabla} \wedge \mathbf{a} + 4\pi s \rho = 0,$$ and $$\mathbf{\nabla} \cdot \mathbf{a} = 0.$$ Thus, you multiply the path integral with a delta function enforcing the first equation and say you only integrate over transverse fields $$\mathbf{a_{\perp}}$$.
2. Actually, we have already obtained the reasonable path integral where we get rid of the gauge redundancy above. If we ask ourselves, what is the gauge-invariant term in the Lagrangian that still gives us the first defining equation from above, the answer is indeed the Chern-Simons term. Bare in mind though that the full Chern-Simons term $$ada$$ will have an additional factor of $$\frac{1}{2}$$ compared to the $$a_0 \mathbf{\nabla} \wedge \mathbf{a}$$ term because there are two such terms in the full Chern-Simons term.