Timeline for How many degrees of freedom does a photon have in 2+1D?
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| Sep 8, 2023 at 18:17 | comment | added | knzhou | @NormalsNotFar No, I'm agreeing with your comment. OP's real misconception is that they think a scalar particle should have $0$ degrees of freedom. | |
| Sep 8, 2023 at 9:34 | comment | added | NormalsNotFar | I'm not sure if you're replying to my comment @Knzhou, but if you take a look at Binegar's paper "Relativistic field theories in three dimensions", he mentions at the end that there is a single degree of freedom for the massless vector field (with EoMs coming from Maxwell action), and it is a scalar excitation. | |
| Sep 8, 2023 at 7:00 | comment | added | knzhou | I think this is just a trivial counting issue. In 3+1 dimensions, the gauge redundancy argument says photons have $4 - 2 = 2$ internal degrees of freedom, i.e. there are $2$ photon modes for each photon momentum. A scalar has $1$ internal degree of freedom, i.e. there is $1$ mode for each momentum. That's what we expect for the photon in 2+1 dimensions, by your first argument, and indeed it's true because $3 - 2 = 1$. If you had gotten $0$ instead, that would not be a scalar field, that would be a totally nondynamical field. | |
| Sep 8, 2023 at 6:47 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
deleted 65 characters in body; edited tags
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| Sep 8, 2023 at 4:47 | comment | added | NormalsNotFar | You could also ask what the dynamical content of the Chern-Simons action for a vector field is, i.e. $\mathcal{L} \propto \varepsilon^{abc}A_a\partial_bA_c$, with gauge symmetry $\delta A_a = \partial_a \lambda$. The answer is that in this case there is no degrees of freedom. | |
| Sep 8, 2023 at 4:43 | comment | added | NormalsNotFar | The massless little group in D = 2+1 is $\mathbb{Z}_2$ (actually it's $\mathbb{R}\times \mathbb{Z}_2$, but the $\mathbb{R}$ is usually represented trivially, i.e. continuous spin reps are ignored). There are only two inequivalent irreps of $\mathbb{Z}_2$, and this is interpreted as meaning that there are only two types of spin: 'spin-0 or scalar' (the trivial rep) and spin-1/2 (the other one). So there is no spin 1 in D = 2+1. Nevertheless, we can still ask whats dynamical content of the Maxwell action. You would find that there is a single degree of freedom, and it is the scalar rep. | |
| Sep 8, 2023 at 3:57 | history | asked | Panopticon | CC BY-SA 4.0 |