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I'm not sure whether this question is redundant, but I think it is probably not due to the focus of the question.

We might say that photon does have antiparticle and it is photon it self. I know for a charged particle, it must have an antiparticle due to the constraint imposed by quantum field theory. However, photon is neutral, I want to know whether there is any constraint imposed by any theory that photon MUST have an antiparticle, even though it is just photon itself?

Therefore, what I want to know is that is there any constraint, and then where does it come from. In addition, I want to know whether the statement "every particle in the Standard Model has an antiparticle" is precise or not.

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    $\begingroup$ Every particle in the Standard Model has an antiparticle when you allow the particle to be its own antiparticle. The quantum state of an antiparticle is obtained from the quantum state of a particle by applying charge conjugation, parity and time inversion. These transformations are well defined for the state of any particle. $\endgroup$ – Adomas Baliuka Sep 19 '18 at 9:09
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    $\begingroup$ Yes, it's a general fact that in relativistic QFT every particle has an antiparticle, and that can be proven rigorously. $\endgroup$ – knzhou Sep 19 '18 at 9:51
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    $\begingroup$ @AdomasBaliuka That seems like an answer rather than a comment. $\endgroup$ – rob Sep 19 '18 at 10:44
  • $\begingroup$ @knzhou You also: That seems like an answer rather than a comment. $\endgroup$ – rob Sep 19 '18 at 10:44
  • $\begingroup$ So photon must have an antiparticle? What happen in not? $\endgroup$ – Kevin Kwok Sep 19 '18 at 11:03
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Quantum Field Theory, QFT, requires every complex field to have an antiparticle field. When I say field, I actually mean particle, but in QFT it's the same thing. Antiparticle are defined as the electric charge conjugated, for example, the electron has an eletric charge equal to $e$, thus the positron, electron's antiparticle, will have an eletric charge equal to $-e$. Let's now analize the case where the particle has no charge. If this particle is not fundamental, which means is made up of other particles, such as the neutron, its antiparticle still have no charge, but his made up of the corresponding antiparticles. The neutron is made up of 3 quarks, which have eletric charge, thus the antineutron is made up of 3 antiquarks. If the particle is fundamental, it is said to be antiparticle of itself. This is the photon case.

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    $\begingroup$ For the complex scalar field, from the diagonalized Hamiltonian we see that there are two types of particles, lets call them a and b particles, and then add (remove) either a or b particles will increase (decrease) the energy of the system. From the Noether conserved charge, we see that they have opposite effect when we consider the number operator, add one a (b) particle get +1 (-1) and vice versa. After coupled with the A field, we can identify this conserved charge as electric charge. That's the idea of antiparticle of charged particle. $\endgroup$ – Kevin Kwok Sep 19 '18 at 21:20
  • $\begingroup$ Actually, I've allways thought about the relationship between complex scalar fields and antiparticles due to the cluster decomposition principle. In order to be true for a complex scalar field, and we know it is, there must be two "twin" partiacles, with opposite eletric charge. This is obviously a mathematical demonstration, but I'd never lookep up to the problem in the way you described. $\endgroup$ – Matteo Campagnoli Sep 19 '18 at 21:27
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    $\begingroup$ Turning this complex KG field back into the real one by requiring self-antiparticle, we know that the issue of [ φ(x) , φ(y) ] = D(x-y) - D(y-x) ≠ 0 in spacelike region will no longer be an issue for spin-0 particle. I am thinking whether the same thing will happen when we quantize the EM field or Proca field--non-vanishing commutator difference and then use the same idea to cure the situation. $\endgroup$ – Kevin Kwok Sep 19 '18 at 21:27
  • $\begingroup$ I don't quite get it. How does the CDP requires the existence of "twin" particles in order for the complex scalar field to work? $\endgroup$ – Kevin Kwok Sep 19 '18 at 21:45
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    $\begingroup$ Read Steven Weinber, The Theory of Quantum Fields, chapter 4. $\endgroup$ – Matteo Campagnoli Sep 20 '18 at 14:06

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