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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.