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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

\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}

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?

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1 Answer 1

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Great question!

  1. When performing the gauge transformation $ U = \exp \big(-2is \sum_{i<j} \text{arg}(\mathbf{x}_i - \mathbf{x}_j) \big) $ 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.

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