Is the third spin vector of a photon always suppressed? I like to tell people interested in light polarization that the photon is a vector boson for which the third spin axis, the one in the direction of travel, is suppressed due to photons being massless and traveling at c.
It's an argument that kind of makes sense for classical physics.
However, mulling over quantum physics makes me a bit more cautious. In QED for example, a photon that travels over a sufficiently short distance can have non-zero amplitudes for both superluminal and subluminal velocities.
But that leads to an interesting question: Does a photon with a non-zero amplitude for subluminal travel also have a non-zero amplitude for exhibiting a real and measurable $\pm$spin axis pointed along its direction of travel? That is, does such a photon include an amplitude to behave like a true three-axis vector boson? And if so, what kind of experiment might detect that amplitude?
Also, what does all of this imply for a superluminal photon amplitude? Does that amplitude imply the existence of some sort of mirror-image vector boson version of the photon? What would that even mean?
And finally: How does any of this relate to fully classical circularly polarized photons?
While circularly polarized photons have same symmetry as particles with spin along the direction of travel, they behave like fully classical particles that can travel indefinite distances. But by the same arguments I just made, that means they cannot possibly be true vector boson states, which would only exist as small amplitudes over very short distances! Are circularly polarized photons perhaps more accurately understood as some kind of spin-cancelling superposition of the subluminal and superluminal true vector boson states?
 A: The photon polarization directions are only transversal when it is free in space . The polarization of an interacting photon or a photon with nonfree boundary conditions is not transversal in general. One example is that electromagnetic waves possess longitudinal polarizations in waveguides. 
Another relatively simple example where (a certain combination of) the transversal and the scalar polarizations becomes physical when the photon is a mediator of a Coulomb interaction. This can be seen as follows: 
The covariant (off-shell) photon propagator without gauge fixing has the form:
$$D^{\mu\nu}(k)                                           = \frac{1}{k^2}\epsilon^{\mu}_{\lambda}(\mathbf{k}) \epsilon^{\nu\lambda}(\mathbf{k}) = \frac{g^{\mu\nu}}{k^2}$$ 
Where the Minkowski summation convention is adopted:
$$a.b = a^{\lambda} b_{\lambda} = - a^0b^0+a^1b^1+a^2b^2 + a^3b^3$$
The four polarization vectors can be chosen to be orthonormal: 
$$\epsilon_{\mu}^{\lambda}(\mathbf{k}) \epsilon_{\nu\lambda}(\mathbf{k}) = g^{\mu\nu}$$ 
Therefore three of them should be spacelike and one of them timelike.
Without loss of generality they can be chosen as:
The transversal polarization vectors
$$ \epsilon_r^{\mu}= [0, \mathbf{\epsilon}_r], \   \ r=1,2$$
Which are defined to be orthogonal to the momentum
$$\mathbf{\epsilon}_r . \mathbf{k} = 0$$ 
The scalar polarization vector
$$\epsilon_0^{\mu} \equiv  n^{\mu} =  [1, 0, 0, 0]$$
The longitudinal polarization vector:
$$\epsilon_3^{\mu} =  [0,\frac{\mathbf{k}}{k}]$$
The longitudinal polarization vector can be written covariantly as:
$$\epsilon_3^{\mu} =  \frac{k^{\mu} - (k.n) n^{\mu}}{[ (k.n) ^2- k^2]^{\frac{1}{2}}}$$
Substituting into the propagator we get
$$D^{\mu\nu}(k)                                          = \frac{1}{k^2}\big [\sum_{r=1,2} \epsilon^{\mu}_{r}(\mathbf{k}) \epsilon^{\nu}_{r}(\mathbf{k})  
+ \frac{[k^{\mu} - (k.n) n^{\mu}][k^{\nu} - (k.n) n^{\nu}]}{[ (k.n) ^2-k^2]} + n^{\mu}n^{\nu} \big ]$$
The propagator can be arranged as:
$$D^{\mu\nu}(k) = D_{\mathrm{trans}}^{\mu\nu}  + D_{\mathrm{Coulomb}}^{\mu\nu}    + D_{\mathrm{resid}}^{\mu\nu} $$   
With
$$D_{\mathrm{trans}}^{\mu\nu}(k)         = \frac{1}{k^2}\sum_{r=1,2} \epsilon^{\mu}_{r}(\mathbf{k}) \epsilon^{\nu}_{r}(\mathbf{k}) $$
$$D_{\mathrm{Coulomb}}^{\mu\nu}(k)  = \frac{n^{\mu}n^{\nu}}{[ (k.n)^2-k^2]}$$
$$D_{\mathrm{resid}}^{\mu\nu}(k)        = \frac{k^{\mu} k^{\nu} - (k^{\mu} n^{\nu} + k^{\nu} n^{\mu})(k.n)}{k^2[ (k.n) ^2-k^2]}$$
The Coulomb and the residual parts are combinations of the scalar and the longitudinal parts of the propagator. 
The residual part always gives a vanishing magnitude when sandwitched  between two conserved currents ($k^{\mu}J_{\mu} = 0$):
$$D_{\mathrm{resid}}^{\mu\nu} J^{(1)}_{\mu}J^{(2)}_{\nu} = 0$$
Thus it does not contribute to physical observables. The Coulomb part is just the Fourier transform of the instantaneous Coulomb potential: 
$$D_{\mathrm{Coulomb}}^{\mu\nu}(x-y)  = \delta^{\mu0} \delta^{\nu0}\frac{\delta(x_0-y_0)}{4 \pi| \mathbf{x}-\mathbf{y}|}$$
Thus free photons can never result a Coulomb interaction.
