Some scientists say that for every particle there exist an anti-particle. If this is true, does it mean that there would be an anti-particle for the Higgs boson also? Could it be that the massless particles would interact with this anti-particle for the Higgs boson?


1 Answer 1


Your first statement is true. However, there is a subtlety - some particles are their own antiparticles.

My understanding is that antiparticles are one particle states (irreducible representations in quantum state space of the Poincaré group see my answer here of what this means) that are mapped into one another by the CPT operator (or better written $C\circ P\circ T$) - time reversal followed by parity inversion followed by charge conjugation.

Some such one particle states are invariant under the action of the CPT operator. It follows by the above definition then that such particles are their own antiparticles. The Higgs Boson is one such particle.

"Antiparticle" and "particle" are partly just words and definitions. Essentially we say that a particle is its own antiparticle if it can emerge alone (i.e. not belonging to a pair) from any quantum state with the requisite momentum, energy and angular momentum. There are no other quantities, such as electric charge, which need to be conserved and therefore pairs of opposite charges or the like are not needed to conserve such quantities.

See my other related answer on this topic for other details.

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    $\begingroup$ Pretty sure you've just defined all SM particles as their own antiparticles. Charge conjugation (not CPT conjugation) is the act of replacing a particle with its antiparticle. The part about 'emerging alone' doesn't make any sense to me. $\endgroup$
    – dukwon
    Nov 26, 2016 at 15:15
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    $\begingroup$ see hyperphysics.phy-astr.gsu.edu/hbase/Particles/antimatter.html "Going beyond the basics, we can say that an antiparticle is related to the particle by charge conjugation. This includes more than just electric charge; it inverts all internal quantum numbers such as baryon number, lepton number, and strangeness. " $\endgroup$
    – anna v
    Jan 14, 2018 at 13:49
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    $\begingroup$ @dukwon I somewhat disagree: the antiparticle is the CPT conjugate, as a property of the Dirac equation. Plain C is violently (maximally) violated in the SM, and P as well, and even T by a little. The vacuum is CPT invariant, and, since the Higgs is shifted by a real number into it in the eponymous mechanism, so is the Higgs itself. The full Higgs complex doublet field, of course, is not. Do you want an answer explicitly demonstrating the field's flip and the conjugate rep gig that achieves that? $\endgroup$ Jan 14, 2018 at 17:56
  • $\begingroup$ @duckwon. I assume "alone" means "no need for associated production" . So, e.g., in a strong process, $K^0$ cannot emerge "alone", but in association with , e.g., $\bar{K}^0$, because, in this case, strangeness is a conserved quantity "such as electric charge.... or the like", all right. But, of course, it might be for Rod to clarify. My sense is the OP may have developed qualms by the complex Higgs doublet not being its CPT conjugate, but as I indicated, the Higgs and the goldston of $\tau_3$ are self-conjugate. $\endgroup$ Jan 15, 2018 at 0:51

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