# Is a photon technically a set of two particles?

When looking at the classification of massless particles, one finds that there is the (half-integer) quantum number "helicity" $h$. For every possible $h$ there is a certain particle kind. In the case of the $h=1$ representation it is the photon which we group together with the $h=-1$ rep (because of parity invariance of the electromagnetic interaction). So is the photon we use in the Standard Model actually a set of two distinct fundamental particles?

• The helicity (related to the spin) is another way to view the polarization of photons. A photon can have one helicity or the other, or even a combination of both (quantum superposition). In any case, all of these represent just one type of particle, a photon. – fffred Feb 18 '15 at 15:44

• This is another aspect that confuses me: if we take "your" position and equate particle = mode of free quantum field, then in the photon case we look at the field-operator $A^\mu(x)$ of the EM-field which does not tell us anything about the possible polarization states. In fact it is only by looking at the Wigner reps that we see that there are two ($h=\pm1$) polarizations and the two remaining components of the four-vector $A^\mu$ have to be eliminiated by gauge transformations. – quantumorsch Feb 18 '15 at 16:28
• @quantumorsch: That is not true. The BRST quantization procedure as well as Gupta-Bleuler quantization both eliminate the unphysical polarisations by imposing unitarity/positivity of the norm on the "naive" Hilbert space, or directly by defining the physical Hilbert space to be the cohomology of the BRST operator. You don't need to know Wigner's classification at all to conclude that a gauge field in $d$ dimensions has $d-2$ polarisations. – ACuriousMind Feb 18 '15 at 16:35