I'm considering that, the [tex](1/2,1/2)[/tex] representation of Lorentz group is the four-vector representation, and this four-vector representation has spin 0,1.(adsbygoogle = window.adsbygoogle || []).push({});

As we know, this four-vector representation can serve for the gauge field [tex]A_\mu[/tex], i.e. photon. However, the photon has spin 1. So we face a problem---how to get rid of the spin 0 part of (1/2,1/2) representation.

I found in a QFT book by Maggiore, after quantizing the EM field, she computed the angular momentum operator from Noether theorem, then extracting out the intrinsic spin part (throwing away the orbital angular momentum part); finally, computing the eigenvalues of intrinsic spin operator on one particle state, she showed that the circular polarized one particle states are eigenstates of the intrinsic spin operator and the eigenvalues are 1 and -1. This shows that the spin of photon is 1.

My question is, is there a concrete way to realize that the spin-0 part of (1/2,1/2) field representation is ruled out by some mechanism? (gauge symmetry or...) Because by explicit quantizing the theory and compute the spin eigenvalues of one particle state is a tough work. Moreover, in other theory, we also take the four-vector field [tex]A_\mu[/tex], i.e. (1/2,1/2) representation as a spin-1 particle, so probably there is a way to understand that spin-0 part is ruled out?

Any ideas or instructions will be appreciated.

Sincerely

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# (1/2,1/2) representation and photon

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