I have been reading a little on the Maxwell-Chern-Simons Lagrangian, an attemp to create a massive photon in $2+1$ dimensions. \begin{align}\mathcal{L} &= -\frac14 F_{\mu\nu}F^{\mu\nu} + \frac{m}{4} \epsilon_{\sigma\mu\nu}F^{\mu\nu}A^{\sigma}\\ &= -\frac14 F_{\mu\nu}F^{\mu\nu} + \frac{m}{2} \epsilon_{\sigma\mu\nu} \partial^{~\mu}A^{\nu}A^{\sigma} \end{align} The action but not the Lagrangian is gauge invariant (changes by a total derivative). I derived the equations of motion, \begin{align} \partial_{\mu}F^{\nu\mu}-m\epsilon^{\nu\mu\sigma}\partial_{\mu}A_{\sigma}=0 \end{align} If we combine this with the Bianchi identity and define a new vector $~f^{\mu} = \frac12\epsilon^{\mu\nu\sigma}F_{\nu\sigma}$, it takes a bit of algebra to show, $$ (\partial^{~\mu}\partial_{\mu}+m^2)f^{\nu} = 0$$

I'm really interested in the number of polarizations this "photon" can have. The little group should be $SO(2)$ right? So a $J_z$ should generate this? So one polarization? I wonder if there is a nice argument for this? Maybe something similar to how it's done in the normal case from using the Lorenz gauge? Note that the Lorenz gauge condition holds trivially from the Bianchi identity.


1 Answer 1


This system has a primary constraint: $$\Pi_0=0$$ and a secondary constraint: $$-\partial_i \Pi_i+m\epsilon_{0ij} \partial^i A^j=0$$ The Poisson bracket of these constraints is zero, so they are first class constraints. Using the formula for counting the degrees of freedom of a constrained system:$$DOF=\frac{N-2M-S}{2}$$

see, for example:

Counting degrees of freedom in presence of constraints

Counting number of degrees of freedom in constrained system

where $N$ is the dimension of the phase space, $M$ is the number of first class constraints and $S$ is the number of second class constraints, the degree of freedom is $\frac{6-2\times2-0}{2}=1$.

The little group for a massive particle is indeed $SO(2)$,but the number of independent polarizations is not equal to the number of generators of the little group; it equals to the dimension of the irreducible representation of the little group that the particle carries.

  • $\begingroup$ So they just happened to coincide this time? Could you give an example of where they wouldn't? I'm just learning all this stuff $\endgroup$ Commented Sep 30, 2016 at 4:47
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    $\begingroup$ @ClassicStyle In 4-dimensions, the little group for a massive particle is $SO(3)$, whose irreducible representations according to which particles transforms are labeled by the spin of the particles, which can be any non-negative integer or half integer. Only a vector particle (spin 1) has the number of degrees of freedom that equals to the number of generators. $\endgroup$
    – Xavier
    Commented Sep 30, 2016 at 5:11
  • $\begingroup$ A massive particle in four dimensions would have three polarizations no? $\endgroup$ Commented Sep 30, 2016 at 5:18
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    $\begingroup$ @ClassicStyle No, electrons, neutrinos, etc. all have spin 1/2, i.e. two polarizations. $\endgroup$
    – Xavier
    Commented Sep 30, 2016 at 5:22
  • $\begingroup$ Duh...I was really confused there with the definition of polarization I had in my head Jesus! Ok thank you! $\endgroup$ Commented Sep 30, 2016 at 5:32

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