# The Higgs bosons is its own anti-particle...

### #1

Posted 5 July 2012 - 01:26 PM

### #2

Posted 5 July 2012 - 03:58 PM

Mathematical Ramblings.

### #3

Posted 5 July 2012 - 04:40 PM

The photon is another particle that is its own antiparticle.

What does that means?

### #4

Posted 5 July 2012 - 07:39 PM

When a particle and its anti-particle collide, they annihilate. A photon is annihilated when it collides with another photon.What does that means?

### #5

Posted 5 July 2012 - 11:21 PM

**anti**meaning?

Mathematical particle model concept?

electron<-->positron ;same mass(+-?), same property, different electrical charge.

proton <--> anti-proton ; same mass(+-?), same property, different electrical charge.

neutron <--> anti-neutron; same mass(+-?), same property, different( ? )

photon <--> anti-photon ; no mass , same property, different( ? )

photon=anti-photon

Higgs Boson <-->anti-Higgs Boson ??? Mathematically possible?

### #6

Posted 6 July 2012 - 07:37 AM

Mathematical Ramblings.

### #7

Posted 6 July 2012 - 01:31 PM

Neutrons are neutral, but they are not their own antiparticle. You also need to look at more than electrical charge. There's also color charge to consider. Some gluons are their own antiparticle, some are not.Basically an antiparticle has the same mass as, but opposite electric charge of the corresponding particle. Neautral particles are then considered to be their own antiparticle.

It's the neutral (electrical charge and color charge) gauge bosons and the Higgs boson that are their own antiparticles.

This is modern software; there's no useful manual. After all, changing how everything works every six months or so is more important than helping people use the features you have.

### #8

Posted 6 July 2012 - 03:06 PM

To answer alpha2cen's question. The anti means (mathematically) just take the complex conjugate. So if the particle is represented by a real number, it is its own antiparticle. If it is represented by a complex number, it is not.

**Edited by Severian, 6 July 2012 - 03:08 PM.**

### #9

Posted 7 July 2012 - 10:16 AM

Neutrons are neutral, but they are not their own antiparticle. You also need to look at more than electrical charge. There's also color charge to consider. Some gluons are their own antiparticle, some are not.

Right, I am specifically talking about fundamental particles. You are right, there is such a thing as an antineutron.

As Severian has hinted at, you mathematically need to consider how fields transform under the charge conjugation operator C. This acts on all additive quantum numbers: Charge, Baryon number, Lepton number, Isospin, Strangeness, Charm, etc. For simple scalar fields this is equivalent to the complex conjugate.

Calling it charge conjugation is a bit of a misnomer. It acts not only on the electric charge.

The important thing is that it acts an additive quantum numbers by switching the sign. It does not act on "space-time properties" like mass or spin. It also does not act on parity which is not additive.

Mathematical Ramblings.

### #10

Posted 7 July 2012 - 12:48 PM

I read that the Higgs boson is its own anti-particle. What does that mean? Also, are there any other particles that are their own anti-particles?

It means that particle and the antiparticle are indistinguishable. E.g. an electron has negative charge, but an antielectron has positive charge and you can differentiate both.

As ajb pointed, the photon is the typical example of particle which is indistinguishable from its antiparticle. The photon has zero charge and the anti-photon has zero charge (0 = -0). As a consequence, photon and antiphoton are not distinguishable.

**Edited by juanrga, 7 July 2012 - 12:53 PM.**

### #11

Posted 7 July 2012 - 08:13 PM

It means that particle and the antiparticle are indistinguishable. E.g. an electron has negative charge, but an antielectron has positive charge and you can differentiate both.

As ajb pointed, the photon is the typical example of particle which is indistinguishable from its antiparticle. The photon has zero charge and the anti-photon has zero charge (0 = -0). As a consequence, photon and antiphoton are not distinguishable.

I would raise one point about the photon being considered its own antiparticle. When two particles like the electron and positron collide, they "annihilate" each other in the sense that the product of their collision is energy (photons). All of the particles' mass is converted to energy. But the photon is already pure energy (no rest mass),so when two photons collide, the product of the collison is what exactly? More photons? This is just a guess on my part, but aren't there Feynman diagrams in which two photons merge (collide) to produce particles with rest mass like the electron?

### #12

Posted 7 July 2012 - 08:40 PM

Nothing. The Photons don't interact at all.I would raise one point about the photon being considered its own antiparticle. When two particles like the electron and positron collide, they "annihilate" each other in the sense that the product of their collision is energy (photons). All of the particles' mass is converted to energy. But the photon is already pure energy (no rest mass),so when two photons collide, the product of the collison is what exactly? More photons?

This is just a guess on my part, but aren't there Feynman diagrams in which two photons merge (collide) to produce particles with rest mass like the electron?

This is known as pair production and the result is a particle and antiparticle (electron and positron for example). This can only happen if the combined energy of the photons is great enough to account for the mass of the particles

*and*when a nucleus is involved. You need the nucleus to act as sort of a mediator for the pair production.

### #13

Posted 8 July 2012 - 10:38 AM

I would raise one point about the photon being considered its own antiparticle. When two particles like the electron and positron collide, they "annihilate" each other in the sense that the product of their collision is energy (photons). All of the particles' mass is converted to energy. But the photon is already pure energy (no rest mass),so when two photons collide, the product of the collison is what exactly? More photons? This is just a guess on my part, but aren't there Feynman diagrams in which two photons merge (collide) to produce particles with rest mass like the electron?

The product of the collision of two photons depends of the situation. For instance two photons can annihilate to give a pair electron positron, as you notice

but to higher energies two photons can collide and scatter

Here you have a basic tutorial on two-photon physics

http://www.hep.ucl.a...g-tutorial.html

### #14

Posted 10 July 2012 - 09:22 AM

Example (Massive Mass Increase): A missile shooting another missile; I hope I am on the right track with what everyone was saying.

### #15

Posted 26 July 2012 - 06:03 PM

When a particle and its anti-particle collide, they annihilate. A photon is annihilated when it collides with another photon.

I find this hard to believe. Can someone confirm or deny?

### #16

Posted 27 July 2012 - 12:47 AM

I find this hard to believe. Can someone confirm or deny?

Sometimes true, but not in general. The energy has to go somewhere so there needs to be another particle to carry it off. For example, if an electron and positron meet, they do indeed annihilate, but convert into a photon which carries the energy away. A photon and another photon can't do this, since there is no 3-photon vertex (because photons are neutral). A gluon though could annihilate with another gluon to turn into a third gluon (the third gluon would need to be virtual though).

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