Can bosons have anti-particles? In the past, I would have answered this question with a yes, primarily because I can imagine writing down a QFT for complex scalars that has a $U(1)$ symmetry that allows me to assign a conserved charge. That is, I expect to obtain a charged spin-0 boson with an additive quantum number. A $CP$-transformation would change these quantum numbers into their negatives and I would consider the corresponding particle an anti-particle.

Of course I know at the same time that Standard Model particles, such as the $Z$-boson and the Higgs boson, are considered not to have observable anti-particles (in the way that electrons have, for instance). On the other hand, mesons are considered (composite) bosons and are known to have anti-particles. I used to take the viewpoint that the mentioned elementary bosons are their own anti-particles, because they are charge-neutral.

After reading, by chance, an interview with Geoff Taylor (Melbourne) I am a bit confused, however. He says that bosons can not have anti-particles, because this property is restricted to Fermions and explicitly refutes the idea that they are their own anti-particles:

"Really fermions are the things where we have this idea of a particle and anti-particle pair," says Taylor, "anti-particles at the fundamental level are fermions with the opposite charge."

"The $W+$ and $W-$ bosons only differ by charge so it's an easy mistake to talk about it that way [as particle and anti-particle], but it's just a pair of different charges."

"While they behave in some sense like particle and anti-particle, we don't think of one as the anti-particle counterpart of the other because they're force carriers," says Taylor

"Fermions have conservation laws associated with them, so for example they are created in particle-anti-particle pairs, the sum of their quantum numbers cancelling to maintain the conservation laws," explains Taylor.

"Bosons operate under different laws and can be created singly. This is a crucial distinction and is in nature of being either matter particles or force carriers."

(It should perhaps be mentioned that he works in experimental HEP-data analysis and not theory, but still he could know more.)

Which, if any, of these viewpoints is correct?

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    $\begingroup$ Well, this crucially depends on your definition of anti-particle. $\endgroup$
    – ACuriousMind
    Commented May 15, 2015 at 14:09
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    $\begingroup$ For me, it is the C/CP-conjugate of the corresponding particle state. Of course, C and CP are not exact symmetries of the Standard Model but, if this is an issue, I would set the corresponding CP-violating phases to zero for that matter. Which other definition would be sensible? $\endgroup$
    – highsciguy
    Commented May 15, 2015 at 14:14
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    $\begingroup$ I can't think of any sensible definition of antiparticle under which it's a mistake to say bosons could have antiparticles (you could say anti-particles are only fermions by definition, but why would you say that?!) $\endgroup$
    – innisfree
    Commented May 15, 2015 at 14:21
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    $\begingroup$ I guess Geoff Taylor has a very non-standard viewpoint. Don't let him confuse you. $\endgroup$ Commented May 15, 2015 at 19:59
  • $\begingroup$ By definition charge conjugation $C$ is that operator which swaps particles and anti-particles. And I have to agree with @innisfree. $\endgroup$
    – Siva
    Commented May 16, 2015 at 20:19

3 Answers 3


This is the answer to the bounty call "Can anyone give a proper source, does Higgs boson has an antiparticle or not, it is the own antiparticle itself. Both Wikipedia and ChatGPT are sure it does not but these are not best sources ever."

I don't think you can easily pinpoint the statement in some textbook or scientific paper "Higgs boson is its own antiparticle" because it is a rather evident statement for the people who study this stuff in detail. Importantly, you should understand that there is no fundamental distinction between the particles and antiparticles. In fact if you have a particle state $|\phi\rangle$ and its antiparticle state $|\bar{\phi}\rangle$ the quantum mechanics allows you to consider their even and odd superpositions $\frac{1}{\sqrt{2}}(|\phi\rangle \pm |\bar{\phi}\rangle)$ that would transform under charge conjugation to themselves. They may be considered as the particles that are their own antiparticles. Depending on the interaction it may happen that these superpositions better describe what you may call a "physical particle" (e.g. that has a specific mass, not a superposition of a states with different masses like in case of $K^0$ and $\bar{K}^0$ mesons).

Now, for a complex scalar field $\Phi$ the particles correspond to the positive energy solutions of the wave equation on $\Phi$. The antiparticles correspond to the positive energy solutions of the wave equation on its complex conjugate $\Phi^\ast$. Now the Higgs fields is a complex doublet, \begin{equation} \Phi=\begin{pmatrix}\Phi_u\\\Phi_d\end{pmatrix}\neq \Phi^\ast \end{equation} Therefore it has distinct solutions that are associated with particles and what you can call their antiparticles. However when we go to the low energies we look at the fluctuations of this field near its minimum, \begin{equation} \Phi=\begin{pmatrix}0 \\ \frac{1}{\sqrt{2}}v\end{pmatrix} + \begin{pmatrix}G^{+} \\ \frac{1}{\sqrt{2}}h + \frac{i}{\sqrt{2}} G^0\end{pmatrix} \end{equation} The fluctuation $G^{+}$ is complex and describes the scalar particle that would be positively charged under the electromagnetic field and would have a negatively charged antiparticle described by $G^{-}\equiv (G^{+})^\ast$.

In contrast both $h$ and $G^0$ are real and produce states that happen to be eigenstates of the charge conjugation operators. I.e. particle for $h$ is its own antiparticle ($\mathcal{C}$-even). Likewise $G^0$ produces also a particle that is its own antiparticle but its wavefunction changes the sign when you exchange all particles to their antiparticles ($\mathcal{C}$-odd). Those are basically the even and odd superpositions of $\Phi_d$ and $\bar{\Phi}_d$

So naively the Higgs field should produce a charged particle $G^{+}$, its antiparticle $G^{-}$ and two neutral particles $h$ and $G^0$ that are their own antiparticles. However, because of the interactions with the gauge bosons $G^{\pm}$ and $G^0$ actually produce not spin 0 particles but the longitudinal polarizations for spin 1 $W^{\pm}$ and $Z^0$ bosons. As result you have only one particle $h$ that we call Higgs particle. As I've explained above, it is indeed its own antiparticle.

In summary, fundamentally the Higgs field may produce both particles and antiparticles but its only low-energy spin-0 excitation, the famous Higgs particle happens to be its own antiparticle.

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    $\begingroup$ I obviously upvote this and thanks indeed for clarification. $\endgroup$
    – Nightrider
    Commented Oct 10, 2023 at 12:33

In the standard model, there is no elementary spin 0 boson being electrically charged (but there are many charged spin 0 composite particles). However, in many extensions such as supersymmetry, there are such particles: the scalar partner of the electron, the selectron carries the same charge as the electron. The anti-selectron is the spin 0 partner of the positron. Thus the answer to your question is yes.

  • $\begingroup$ Would you call the Higgs (doublet; not just the radial mode) an elementary scalar that is electrically carged? $\endgroup$
    – Siva
    Commented May 16, 2015 at 15:20
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    $\begingroup$ @Siva: before the spontaneous symmetry breaking (S.S.B.), yes, I would say that the upper component of the Higgs doublet is indeed a charged scalar whose antiparticle is the lower component of the charge conjugate of the doublet. However, these charged scalars become degrees of freedom of the $W^\pm$ after S.S.B. Hence, the physical scalar of the Standard Model is only the neutral scalar (radial excitation), whose antiparticle is itself (and spin 1 massive $W^+$ is the anti-particle of $W^-$). $\endgroup$
    – Paganini
    Commented May 16, 2015 at 16:56
  • $\begingroup$ How do i learn particle physics like you, Paganini? $\endgroup$ Commented Jan 30, 2019 at 21:56

In the Rishon Model (still non-mainstream, but how knows what the future brings; the model has a lot of advantages), made up by the great mind of Haim Harari, the $Z^0$ particle is considered to be a composite particle which consists of three T-Rishons and three anti-T-Rishons. Each T-rishon is considered to have a charge of $+\frac 1 3$ and two other kinds of charges (the color force and the hyper color force). So all these charges are opposite for the anti-$Z^0$ particle. In other words, the $Z^0$ and the $\bar{Z}^0$ are the same.

In the same model, the $W^+$ is considered to be a combination of three T-rishons and three V-rishons (which have a zero electric charge). In the model, it is considered not to be the fundamental weak force transmitting particle, just like the pion, which was once thought to be the fundamental transmitter of the strong force, turned out not to be the fundamental transmitter, but the gluon. In the model this is the hyper color gluon, transmitting the hyper color force, a particle similar to the gluon, which reduces the weak force to a residue force (like the strong force transmitted by the pion turned out to be a residue force). The $W^-$ in this model is the antiparticle of the $W^+$.

The gluons and similar hypercolor (see the article) gluons (both bosons) are force transmitting particles, though their antiparticles are clearly different particles.

Maybe you can think for yourself how the Higgs is represented in the model.

It's true though that the antiparticles of all quarks and leptons (fermions) are different particles (especially the neutrino, of which only the left handed one exists).


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