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Integrating out Chern-Simons fields

We will derive a much used identity used when 'integrating out Chern-Simons fields'. We mean we have a partition function, which is a path integral over multiple fields. We then perform one of the integrations and yield an action depending on the rest of the fields. We do this because the field we 'integrate out' is not important to the effect under study.

$\displaystyle Z=\int \pmaat{a}\pmaat{b} e^{iS[a,b]} = C \int \pmaat{b} e^{iS\prime[b]}$ (A.1)

We have a $ U(1)$ gauge field with Chern-Simons term coupled to a current,

$\displaystyle S=\intm{x}{3} \frac{1}{2} \big\{ a_{\lambda} \epsilon^{\lambda\mu\nu} \partial_{\mu} a_{\nu} + j^{\nu}a_{\nu} \big\}.$ (A.2)

We can check that the operator $ \epsilon\partial^{\lambda\nu} \equiv \epsilon^{\lambda\mu\nu}\partial_{\mu}$ is symmetric,

\begin{displaymath}\begin{split}\intm{x}{3} a_{\lambda} \epsilon^{\lambda\mu\nu}...
...epsilon^{\nu\mu\lambda} \partial_{\mu} a_{\lambda}. \end{split}\end{displaymath} (A.3)

And we can find an inverse,

\begin{displaymath}\begin{split}\epsilon\partial^{\rho}_{\phantom{\rho}\lambda}\...
...\partial_{\sigma} a^{\rho} = -\partial^{2} a^{\rho} \end{split}\end{displaymath} (A.4)

In going to last line we chose a gauge, where $ \partial\dot a=0$. Now the inverse can be written as.

$\displaystyle (\epsilon\partial^{-1})^{\mu\nu} = -\frac{\epsilon\partial^{\mu\nu}}{\partial^{2}} .$ (A.5)

In (A.2) we now shift the field $ a$.

$\displaystyle a_{\mu} \to a_{\mu} - (\epsilon\partial^{-1})_{\mu}^{\phantom{\mu}\lambda} a_{\lambda}$ (A.6)

When we assume the integral over $ a$ is invariant under such transformations we find,

\begin{displaymath}\begin{split}S&=\frac{1}{2} \intm{x}{3} \big\{ a_{\lambda} \e...
...epsilon\partial^{\lambda\nu}}{\partial^{2}} j_{\nu} \end{split}\end{displaymath} (A.7)

In the second line the term containing $ a$ has a Gaussian form, hence we can 'integrated out' the gauge field $ a$.

If we take the following form the current,

$\displaystyle j_{\lambda}=\epsilon^{\lambda\mu\nu}\partial_{\mu}b_{\nu}=\epsilon\partial^{\lambda\nu}b_{\nu},$ (A.8)

we get the following useful identity,

$\displaystyle \intm{x}{3} \frac{1}{2} \big\{ a_{\lambda} \epsilon^{\lambda\mu\n...
...c{1}{2} \intm{x}{3} b_{\lambda} \epsilon^{\lambda\mu\nu}\partial_{\mu}. b_{\nu}$ (A.9)



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Tim Dijkstra 2002-05-08