What is the spin of photon [duplicate]

I am reading a book. It says that spin $s$ describes the symmetry property of a particle. Basically, the number of different wavefunctions which are transformed into linear combination of one another is $2s+1$, under rotation.

Since photon wavefunction is a vector, therefore it has spin $s=1$. So it has $2s+1=3$ components under transformation.

But photon is massless. So it doesn't have rest frame. It always moves with light speed in any frame. As a result, it has a preferred direction which is direction of the momentum of the photon.

What I don't understand is since photon can never be at rest, it doesn't have the rotation symmetry of spin = 1 (seems to me). Indeed, when writing the polarization in density matrix, it is always 2 components. And I thought this corresponds to spin = 1/2 (i.e. 2s+1=2).

So, it seems photon behaves as spin=1/2. What's my misconception here? Is it possible to have longitudinal photon? Thanks a lot!

marked as duplicate by ACuriousMind♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 24 '16 at 16:51

It is true that the little group of massive particles, $SO(3)$, has $2s+1$ states associated with a spin $s$ particle.
The little group of massless participles is $ISO(2)$, which is different. Massless spin-0 particles have 1 degree of freedom. All higher spin-$s$ particles have 2 degrees of freedom. (There's a tricky topological argument for this given for example in Weinberg Vol 1, chapter 2). A massless spin $s>0$ particle has helicity $h=\pm s$ states. Helicity is defined by transformation under rotation about the 3-momentum. Under rotation by an angle $\theta$ about the 3-momentum, a helicity $h$ state transforms as $|h\rangle\rightarrow e^{i h \theta} |h\rangle$.