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A well-known result of Wigner's classification of relativistic particles is that massless particles transform with helicity $h \oplus -h$ under $ISO(2)$. Thus, such particles have two helicity states.

Wigner's classification in this form, while explaining why all massless particles have two helicity states, does not guarantee that all quantized massless fields have only two degrees of freedom (d.o.f.). The true d.o.f. are usually determined in a case-by-case basis. For instance, we achieve two on-shell degrees of freedom for a gauged vector field in 3+1 dimensions through (1) the equations of motion and (2) the choice of gauge.

Question: Does there exist a constructive proof for why all massless quantum fields have only two degrees of freedom?

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  • $\begingroup$ Massless fermions (Weyl fermions) have two degrees of freedom because they are spin 1/2 particles and gauge bosons, which are massless, have only two degrees of freedom due to the reasons you give. Perhaps it is just a coincidence that in both cases they have ony two degrees of freedom. Are there other examples of massless particles? $\endgroup$ – flippiefanus Aug 16 '17 at 5:30
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    $\begingroup$ What sort of proof are you looking for, if not Wigner's classification? In the end, it is an accident of 4 dimensions probably best explained through Wigner's classification rather than brute-force computation of the d.o.f. of any given field, note that e.g. a massless symmetric traceless 2-tensor (like the graviton) has $\frac{1}{2}(d-2)(d-1)$ d.o.f. in $d$ dimensions. $\endgroup$ – ACuriousMind Aug 16 '17 at 9:27

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