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It is well known that in ordinary $4$ dimension, the photon has on shell only two physical degrees of freedom. Physically this means its elicity is either $\lambda=+1$ or $\lambda=-1$ but cannot assume the value $0$, or that out of the four components of $A^{\mu}$ only two are really independent.

Now, is it possible to work how many degrees of freedom does the photon have (on shell) in an arbitrary number $n$ of dimensions?

Federico Carta
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  • Note: question is related to this http://physics.stackexchange.com/q/31143/ however, what I ask explicitly is quite different. – Federico Carta Jan 14 '14 at 11:38
  • Don't the arguments in http://physics.stackexchange.com/q/46643/ apply to any number of dimensions? – John Rennie Jan 14 '14 at 12:02
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    On shell, we have $k^\mu = (k^0, k^1, 0,0....0)$, with $k^o k_0 - k^1 k_1=0$. Gauge-fixing by applying the Lorentz gauge $k^\mu.\epsilon_\mu(k) = 0$ leads to $\epsilon_\mu(k)= (k_0 \phi(k), k_1\phi(k), \epsilon_2(k),\epsilon_3(k),.....\epsilon_{d-1}(k))$. However, by a gauge transformation $\epsilon_\mu(k) \to \epsilon_\mu - k_\mu \phi(k) $ (compatible with the Lorentz gauge, and which does not change the electromagnetic field), we may eliminate the non-physical $k_\mu\phi$ terms. So, finally, we have only $d-2$ degrees of freedom $\epsilon_2(k),\epsilon_3(k),.....\epsilon_{d-1}(k)$ – Trimok Jan 14 '14 at 12:17
  • There is a related discussion in my answer to this Phys.SE post. – Qmechanic Jan 14 '14 at 13:43
  • @FedericoCarta : Correction typo : It is $k^o k_0 + k^1 k_1=0$ – Trimok Jan 14 '14 at 16:26

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