As I understand it, the reason why there is no Spin 0 Photon is because the polarisation of an EM field lives in two dimension. Hence we only have two basis vectors, yielding two pairs of ladder operators in the quantization.

Is this correct?

If I recall correctly, in a cavity the polarizations must not be perpendicular to the direction of propagation, so we should have three allowed basis vectors for our polarization. Does this mean that in this case we can have Spin 0 photons? Would this be equivalent to massive photons?


I have now found out that for massive photons in free space, we have a massive term $\frac{1}{2}M²A_\mu A^\mu$ in the Lagrangian, which thereby loses its gauge freedom (Lorenz gauge goes mandatory). Hence we now get three polarizations (without mass, the Coulomb gauge $\partial_i A_i = 0$ directly yields $\vec{k}\cdot\vec{A}=0$), and the (longitudinal) polarization $\vec{\varepsilon}_3 \propto \vec{k}$ generates the spin 0 photon.


You are correct that the reason the photon is not spin 0 is because there are multiple polarizations.

Saying that a particle is spin 0 means that under rotations of your spatial coordinate system the mathematical description of the internal state of the particle does not change. For the massive photon case there are three polarizations and when you rotate your coordinate system the polarizations rotate like a three-dimensional vector. In such a case we say it has spin 1. Different ways that the internal state can change under rotations are called different representations and are labeled with different half-integers of spin.

Note that if a particle has multiple internal states that have nothing to do with rotations of space it can still be spin 0. This is the case for the pion triplet which also has three states, but it doesn't change under rotations so it has spin 0 (but isospin 1).

In the massless case it is trickier to define spin this way because it is trickier to define what we mean by internal state when we can not go to a rest frame. It turns out that you always only get two internal helicity states, so just knowing that there are two polarizations does not tell us what the spin is. But it is still true that the helicity is quantized in half-integers, and the photon still happens to have spin 1.

This is related to the fact that you have to rotate the polarization vector by a full 360 to come back to same state. A gravitational wave also has two polarizations but you only have to rotate by 180 to get back to the same state (2 cycles in one full rotation) so it has spin 2.


Photons have a varying electric and a varying magnetic field, perpendicular to each other and perpendicular to the photons propagation (in vacuum). This fields have a direction, for example from plus to minus and from north to south (this is a convention). What ever you try, you will end up with exact two combinations of the fields directions:

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  • $\begingroup$ But my question is specifically not about free space. Is your point that even in TE or TM modes, even though one field is not perpendicular to k, the polarization will still be restricted to 2 dimensions? $\endgroup$ – Pascal Engeler Jun 18 '15 at 7:46
  • $\begingroup$ Exact. But my opinion is not representiv. $\endgroup$ – HolgerFiedler Jun 18 '15 at 11:07
  • $\begingroup$ I don't understand what that's supposed to mean. $\endgroup$ – Pascal Engeler Jun 18 '15 at 12:43
  • $\begingroup$ QED is good for processes inside the nucleus. But since they tried to explain photons as disturbances in space photons are claimed as not more real. That is why my explanation about spin and its visualization may be would not accepted. $\endgroup$ – HolgerFiedler Jun 18 '15 at 13:53

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