Propagator, mass and electrostatic potential of an modified 2+1 dimensional Maxwell action Consider the modified Maxwell action in 2 + 1 dimensions
$$S=\int d^3x[-\frac{1}{4} F^{\mu\nu} F_{\mu\nu}+\frac{\theta}{2}\epsilon^{\alpha\mu\nu}A_\alpha F_{\mu\nu}]  .$$
The action invariant under gauge transformation, I want to find the propagator and the mass term of the gauge field and then understand what is the electrostatic potential between two charged particles coupled to $A_\mu$ at large distances.
I think the mass term is $\theta$ only by dimensional analysis but I'm still confused for the fact that the last term look like $\frac{\theta}{2}A_\alpha\partial_\mu A_\nu$ with the anti simmetrisazion come from Levi-Civita symbol and not very like $\frac{m^2}{2}A_\mu A^\mu.$
I remember that for the Maxwell term in 3+1 dimensions the action is $S=\int d^4x[-\frac{1}{4} F^{\mu\nu} F_{\mu\nu}]$ and after integration by parts and using Fourier transform we find that the propagator via the quadratic form $\eta^{\mu\nu}q^2-q^\mu q^\nu$' and imposing Lorenz gauge condition: $q^\nu A_\nu=0$  so the final propagator would be:
$\Delta_{\mu\nu}=\frac{\eta_{\mu\nu}}{q^2}$. but in this case i'm not quiet sure if the propagator is the same or I need to do some manipulation with the second term in the action to get some "modified propagator"
After I'll get the propagator to find the electrostatic potential would be automatic.
 A: I am recently learning Chern-Simons theory. The E.O.M is
$$\partial_\mu F^{\mu\nu}+
\frac{\theta}{2}\epsilon^{\nu\rho\sigma}F_{\rho\sigma}=0.$$
As we are working in 3d, there are only 3 $F$'s. So it is easier to work with $F$ rather than $A$. Define
$$F^{\nu}=\frac{1}{2}\epsilon^{\nu\rho\sigma}F_{\rho\sigma},$$
or explicitly
$$F^2=F_{01},\quad F^1=-F_{02},\quad F^0=F_{12}.$$
Then the E.O.M looks like (If I did the computation correctly)
$$\begin{align}
\partial_1 F^2 - \partial_2 F^1 + \theta F^0 &= 0 \tag{1}\\
\partial_0 F^2 + \partial_2 F^0 - \theta F^1 &= 0 \tag{2}\\
\partial_0 F^1 + \partial_1 F^0 + \theta F^2 &= 0. \tag{3}\\
\end{align}$$
Then do some organization. By $-\partial_1(1)+\partial_0(2)+(3)$, we get
$$\partial_0^2 F^2-\partial_1^2 F^2+\partial_2(\partial_0F^0+\partial_1F^1)+\theta^2F^2=0.$$
Finally, notice that $\partial_0F^0+\partial_1F^1=-\partial_2F^2$, we arrived at a wave equation with mass $\theta$
$$\partial_0^2 F^2 - \partial_1^2 F^2 - \partial_2^2 F^2 + \theta^2F^2=0.$$
You can also using vector calculus to simplify the EOM. Suppose that we have done a Wick rotation so that there will not be $-1$ around the $t$ index bothering us, the EOM reads
$$\vec{\nabla}\times \vec{F}-\theta \vec{F}=0$$
Then
$$\vec{\nabla}\times\vec{\nabla}\times \vec{F}-\theta \vec{\nabla}\times\vec{F}=
\nabla(\vec{\nabla}\cdot\vec{F})-\nabla^2\vec{F}-\theta^2\vec{F}=0.$$
Then by $\vec{\nabla}\cdot\vec{F}=0$, you can get the same wave equation.
The $\theta$ here need not be an integer classically. However, when you quantize it, it has to be in $\mathbb{Z}$ (I saw only examples in non-Abelian case).
