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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?

quantumorsch
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  • 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

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Well, it depends how you define "distinct fundamental particle".

If you insist that Wigner's classification is what defines a particle, i.e. "particle = irreducible unitary representation of the Lorentz/Poincare group", then the photon is two particles, as you say.

But, more commonly, we do not look at the particles like this - particles arise as the states created by the modes of a free quantum field, and to every field there corresponds a particle. The photon (with either helicity) is the quantum of the electromagnetic gauge field.

Since nothing distinguishes the "two photons" except for helicity, to say that there are two distinct photons is probably more confusing than helpful.

ACuriousMind
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  • 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
  • I've struggled to find a concise definition of a particle that allows us to count 17 particles in the standard model: saying "things with distinct mass" lumps the gluons in with the photon; "things with distinct couplings" makes the left-handed and right-handed fermions distinct; "things with distinct mass, and charges that can't be transformed onto each other though a global symmetry" isn't really very concise... – Shep Feb 18 '15 at 17:15
  • I don't think Wigner's classification suggests that the photon is two particles. The photon corresponds to the irreducible $m=0$, $(j_1,j_2)=(\frac{1}{2},\frac{1}{2})$ Poincare rep, and its helicity states correspond to the eigenvectors of the diagonal generator of the little group. There is only one representation, so there is only one particle. On the other hand, the electron, being the reducible $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ rep, does correspond to two particles in Wigner's classification. – Kris Walker Mar 19 '23 at 06:31
  • @KrisWalker There is no such thing as "$(j_1,j_2) = (1/2,1/2)$" for a massless particle - the irreducible representations for $m=0$ are the one-dimensional helicity representations and the acausal continuous spin representations; please review the correct classification of representations of the Poincaré group. See also this and this answer of mine – ACuriousMind Mar 19 '23 at 10:35