10
$\begingroup$

I need to compute the "topologically massive photon" propagator.

I've started with :

$$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{\mu}{4}\epsilon^{\mu\nu\lambda}A_\mu\partial_\nu A_\lambda $$ $$ =A_\mu\underbrace{[\frac{1}{2}g^{\mu\lambda}\partial^2+\frac{\mu}{2}\epsilon^{\mu\nu\lambda}\partial_\nu]}_{(\Delta^{-1})^{\mu\lambda}}A_{\lambda} $$

So how can I invert The under braced part which will yield the topologically massive photon propagator?

$\endgroup$
  • 4
    $\begingroup$ Comment to the question (v3): The mentioned Lagrangian density is not just CS, since it also contains a Maxwell term. $\endgroup$ – Qmechanic Feb 6 '14 at 13:34
  • 1
    $\begingroup$ I've edited the title. Its just "Chern-simmons abelian gauge theory" $\endgroup$ – Aftnix Feb 6 '14 at 13:42
  • $\begingroup$ Usually the factor $1/2$ is not included in the inverse propagator. And I think you also forgot a derivative term (the index $\nu$ is by itself). $\endgroup$ – Adam Feb 6 '14 at 14:26
3
$\begingroup$

The following is a rough calculation. If the operator,

$$\triangle^{\mu\lambda} = \eta^{\mu \lambda} \partial^2 + \mu \epsilon^{\mu\nu\lambda}\partial_{\nu},$$ is the one you wish to invert, then we must solve the differential equation

$$ \triangle^{\mu\lambda} G = \left[\eta^{\mu \lambda} \partial^2 + \mu \epsilon^{\mu\nu\lambda}\partial_{\nu} \right] G = -i\delta^{(4)}(x-y)$$

where $(-i)$ is by convention. I think the momentum space equivalent is

$$\left[ \eta^{\mu \lambda}p^2 + i\mu \epsilon^{\mu\nu\lambda}p_{\nu}\right]\hat{G} = -i\mathrm{e}^{-ip\cdot y}.$$

To obtain $\hat{G}$ in position/physical space, you must perform an inverse fourier transform. Let me know what you get, I'm curious!

$\endgroup$
1
$\begingroup$

You might try it by the standard way of calculating Green's function (which is what the propagator is) of linear differential operators. For this purpose you have to derive the equations of motion from the given Lagrangian, which will be some operator acting on the gauge field. The crucial observation is now that the Green's function can be calculated by solving the equation

$$L G(x,y)=\delta(x-y),$$

which schematically shows that the differential operator's action on the Green's function gives a delta function. This equation can be solved for example by transforming it to momentum-space.

Note that I have not tried it for your specific example, so there might me caveats to my approach I am not aware of.

$\endgroup$
1
$\begingroup$

Eq. 26 and then you can send gauge fixing term to be very large. Fourier transform back to real space in 2+1 D. http://cds.cern.ch/record/379553/files/9902115.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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