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If gravitons exists, then would there be anti-gravitons as well? 1)If not, why? 2) If yes, what are their expected properties?

Qmechanic
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sanyam sharma
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2 Answers2

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If by graviton you mean the spin 2 boson we get from attempting to describe gravity using quantum field theory, then the graviton is its own antiparticle just like the photon.

John Rennie
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    Oh, what a let down. There I was, briefly hopeful that anti-grav beams could allow spaceships the size of asteroids to hover just above our cities. Theoretical physics yet again ruins science fiction. ;-) – Chappo Hasn't Forgotten Aug 12 '16 at 08:27
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    @Chappo "Might there be anti-gravitons" is a different question from "might there be anti-gravity". Consider that the electromagnetic force can be both attractive and repulsive, and yet its force carrier (the photon) is its own antiparticle. For antigravity, one needs a negative gravitational charge; which is to say, a negative mass. – zwol Aug 12 '16 at 13:34
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    @Chappo : don't you worry. All you need is an artificial gravitational field above the ship to counterbalance the Earth's gravitational field. It hovers. – John Duffield Aug 18 '16 at 14:51
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I would have guessed the answer John Rennie gives, but I actually suddenly realize I do not understand why, so I sought to fix this lack of knowledge by some reading. You should wait for upvotes / my deletion of the answer to tell whether to trust it.

So, now, why is the postulated graviton its own antiparticle?

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.

So now we need a sensible definition of when we call these two states "distinct" particle and anti-particle, or whether we say a particle is its own antiparticle as we do for the photon - charge conjugation is the identity map here and $P\circ T$ swaps left and right hand polarized one photon states.

The crucial difference between particles identified as their own antiparticles and the others then seems to be that self-anti particles can be created alone given some other quantum state with the requisite energy. momentum and angular momentum, whereas non-self-anti particles cannot be created alone with the requisite energy (because such a creation would violate the conservation of charge, or some other conserved quantum number) and must be created in $\psi$ and $C\circ P\circ T\,\psi$ pairs so as conserve electric charge or, I'm guessing more generally, generally conserved quantum numbers.

A graviton is postulated to be chargeless particle state. Moreover, it doesn't have any other quantum numbers (aside from spin) that would need to be conserved. Therefore, a quantum state with the requisite energy, momentum and angular momentum can evolve into a lone graviton without breaking known conservation laws. We would therefore call the graviton its own antiparticle, just as we do the photon.

This answer and the definitions assumed in it would seem to be supported by This Question and Answer Here and the links and discussions it cites, although there is not a great deal of discussion to be found there.

Community
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Selene Routley
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    Put another way, a graviton and an anti-graviton would be indistinguishable, as all the relevant numbers are 0, and -0 == 0. – OrangeDog Aug 12 '16 at 09:28
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    Dose this mean spin 2 == spin -2? – Joshua Aug 12 '16 at 14:22
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    @Joshua Kind of. The spin "class" of a particle simply names the set of values the angular momentum projection operator can yield as measurements. A massive spin $n$ particle in an AM eigenstate can have spin projections of $-n,,-n+1,,\cdots,,+n-1,,n$ times $\hbar$ onto any given reference axis. A photon can have an AM projection of +1 or -1 onto an axis. I'm guessing a graviton (I'm out of my depth here) in an AM eigenstate could have all possible projections of $\pm2,,\pm1$ but not the zero spin state (as for the photon). – Selene Routley Aug 12 '16 at 15:08
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    "I'm guessing a graviton (I'm out of my depth here) in an AM eigenstate could have all possible projections of ±2,±1 but not the zero spin state" No, only ±2. Massless particles can have only extreme values of spin projection. – BartekChom Aug 12 '16 at 21:52
  • @BartekChom Thanks, Bartek. I might ask a question myself about that one. – Selene Routley Aug 12 '16 at 23:15
  • @WetSavannaAnimalakaRodVance Hey! I enjoyed reading this, because I was wondering about a really similar issue a while back, and came to the same conclusion. So I wasn't being crazy! I'm linking my thing here in case people reading your answer or my question want to see another example of the reasoning. – knzhou Aug 19 '16 at 01:44
  • I don't think this works. The $W^+$ and $W^-$ are mutual antiparticles but can be created alone. The $K^0$ and $\bar K^0$ are mutual antiparticles, while the $K_L$ and $K_S$ are self-antiparticles, but they are literally the same particles. A Majorana fermion would be its own antiparticle but couldn't be produced alone because that violates angular momentum conservation (unless pair-production with a particle of a different type counts as alone, in which case all SM fermions can be produced alone in weak interactions). – benrg Jan 24 '21 at 02:05