# What are antiphotons?

1. Jun 26, 2008

### putongren

Are photons and antiphotons the same thing? Crackpot theorists say they are different, mainstream scientists say they are the same. Who should I believe? The mainstream scientists right?

2. Jun 26, 2008

### Xezlec

Yes. Antimatter has been created, observed, and studied. There is no evidence I know of for antiphotons. Antimatter creates and responds to ordinary photons just like matter does.

3. Jun 27, 2008

### pmb_phy

A photon is its own antiparticle. There is an article of that title as a matter of fact

Anti-photon, W.E. Lamb, Appl. Phys, B 60, 77-84. The abstract reads
The author is a Nobel Laureate. He won the Nobel Prize in Physics for his work in quantum electrodynamics as I recall.

Pete

4. Jun 27, 2008

### vociferous

Meaning that an antiphoton and a photon are exactly the same, correct? They both should have a spin of 1 and a charge of 0?

5. Jun 27, 2008

### BryanP

According to some threads I've read here in the past, anti-photons and photons are the same because the only difference is the charge when it comes to the anti-particles. The mass, spin, and energy should be identical.

Since photons are chargeless, photons and anti-photons are no different.

6. Jun 27, 2008

### vociferous

My understanding is that many particles, neutrons for instance, have the opposite spin of their antiparticles.

There are other differences than charge between particles and antiparticles. For instance, the types of quarks they have (if they are not leptons).

7. Jun 27, 2008

### D H

Staff Emeritus
Be careful. Antineutrons have no net charge and are not the same as a neutron. Neutrons are made of quarks while antineutrons are made of antiquarks.

8. Jun 27, 2008

### Xezlec

They are synonyms. Therefore, there is no reason to ever say "antiphoton".

9. Jun 27, 2008

### Staff: Mentor

Note that neutrinos and antineutrinos are different even though they are both neutral fundamental leptons. Neutrinos produce a negative lepton when they interact via W-boson exchange (e.g. an incoming $\nu_{\mu}$ produces a $\mu^{-}$) whereas an antineutrino produces a positive lepton (e.g. an incoming ${\overline \nu}_{\mu}$ produces a $\mu^{+}$).

10. Jun 27, 2008

### BryanP

Thanks for the correction, I'm also going through the subject right now so the clarification is helpful.

11. Jun 27, 2008

### malawi_glenn

putongren: NEVER trust a crackpot :-)

12. Jun 27, 2008

### BryanP

Crackpots aren't always wrong, crackpots just have a hard time listening to others when it comes to being told they are.

13. Jun 27, 2008

### malawi_glenn

No, but you shouldn't trust them on anything. You give them one finger, they take the whole hand! (swedish old telling)

14. Jun 27, 2008

### BryanP

I guess you're right with that one

15. Jun 27, 2008

### vanesch

Staff Emeritus
Whether a particle is the anti-particle of another one depends on its definition in a quantum field theory, if you look upon it on the theoretical side. In other words, on the theoretical side there's not much discussion: you look at the field theory, and at the names you've given to the emerging particle states, and you can clearly see what is going to be the anti-particle of another one. There's a transformation that turns particles into anti-particles: the so-called C transformation (from the CPT theorem).

This is on the level of the structure of a theory. Denying this would be like denying that -2 is the negative of 2 in the integer ring, and that 0 is its own negative. There's no discussion about that.

What can be discussed however, is how a certain quantum field theory is applicable to (is a good model for) a certain aspect of nature. One might come up with a theory (QED) where photons are their own anti-particles (in QED, they are, just as 0 is its own negative), and someone else might come up with another theory, where there are photons and anti-photons. In the standard model, photons are their own anti-particles.

There is an experimental way to see whether certain particles are the anti-particles of others: you have to try to find out whether they can "annihilate" eachother: that is, whether it is possible to have the particle and its anti-particle candidate to interact, and disappear in the interaction, while creating on the occasion a known particle-anti-particle pair, or vice versa. Of course, that's a bit circular, as you have to have initially something you know is a particle-anti-particle pair. The idea behind this is that a particle-anti-particle pair must have the "non-dynamic quantum numbers of the vacuum", and there is some potential ambiguity of what we call the vacuum.

Take it that you accept that an electron and a positron are a particle-anti-particle pair. Then, knowing that this can transform into a 2-photon state should indicate that one of the photons in the pair is the anti-particle of the other one. In the opposite way, having two photons interact to create an electron-positron pair should indicate also that they were eachother's anti-particle.

But probably the best indication is that QED describes the things we call photons pretty well, and that within QED, photons are theoretical constructs which are their own anti-particles.

16. Jun 27, 2008

### putongren

Thanks a lot everyone, I think I got the idea.

17. Jun 27, 2008

### putongren

Oh one last thing, I would like to know more about the CPT (Charge, Parity, Time) transformation. I looked at Wikipedia and I didn't understand their explanation. Can anyone explain the CPT transformation in simpler terms?

18. Jun 27, 2008

### humanino

• C exchange particles and antiparticles
• P takes the image in a mirror (actually it exchanges left and right... let's keep it simple)
• T is time reversal
. Local relativistic quantum field theories respect CPT. That means,
• In the standard model, violation of CP for instance is equivalent to a violation of T (since CPT must be respected)
• If you find a process for which CPT does not hold experimentally, that rules out not only the SM, but the QFT formalism altogether
To be commented by a theoretician