# Anti photon

1. Aug 10, 2010

### binbots

I was recenlty watching a lecture by Feynman where is talks about particles, and how all particles have anti particles. A photon is a particle but I can't find any discussion about a anti photon. Reason?

2. Aug 10, 2010

### mgb_phys

Photons are bosons, which don't have anti-particles (or are their own anti-particle if you prefer)

edit - Although thinking about it you can have +/- W bosons ?

3. Aug 10, 2010

### diazona

In Standard Model-lingo (if I remember my stuff correctly), the quantum state that represents the photon is an eigenstate of the C parity operator, which is the technical reason that it is its own antiparticle.

In normal-person-lingo, that means that all its quantum numbers are zero: zero charge, zero baryon number and lepton number, zero flavor (upness downness strangeness etc.), and zero isospin.

In real normal-person-lingo, it just is

4. Aug 15, 2010

### PipBoy

I forget, is it because the Photons travel back in time, so to speak, that we observe them to be their own anti-particles or am I getting this horribly confused with Electrons?

5. Aug 15, 2010

### bcrowell

Staff Emeritus
This is a general thing in quantum field theory. Any particle traveling backward in time can be interpreted as its antiparticle. For a photon, the version traveling backward in time is the same as the version traveling forward in time.

6. Sep 26, 2010

### Bararontok

The polarity of the photon’s electric charge is positive because it represents its energy content which is a scalar quantity. The reason why the electromagnetic fields of charged Fermion particles can be positive or negative because it is a vector quantity that represents the direction of force exerted by the photons comprising the field. It is negative for inwardly directed electromagnetic fields and positive for outwardly directed fields. The existence of antiparticle counterparts for Fermions only applies to elementary particles that have electromagnetic fields which are comprised of multiple photons. The Fermion elementary particles that have no electromagnetic fields which are the electron, muon, and tau neutrinos have no anti-particles because they have electric charges of 0e.

Last edited: Sep 27, 2010
7. Sep 26, 2010

### Xela

If I'm not wrong being a fermion/boson is irrelevant to having an antiparticle. What I remember is that the field has to be a complex one to have charge and antiparticles. Photon field is a real one. Charge conservation for for complex fields comes out of the invariance of the complex field Lagrangian to the global phase transformation. And you can't have phase in a real field. This is only a mathematical explanation, but I guess I don't have a more intuitive one.

8. Sep 28, 2010

### piareround

Last edited by a moderator: Apr 25, 2017
9. Sep 28, 2010

### Staff: Mentor

Photons do not carry charge.

10. Sep 28, 2010

### Bararontok

Actually, according to the particle data group, each photon can carry charge if it is traveling through space and that charge only disappears when the photon is absorbed by a charged fermion particle that converts the absorbed energy into thermal excitation. But when the photon is traveling freely, it still retains its mass and charge which are scaled values derived from the energy carried by the photon. The charge of a freely traveling photon is an absolute value and not a vector quantity such as the direction at which electromagnetic force is applied which is only applicable to an electromagnetic force field of photons surrounding a charged fermion.

Here is the PDF showing the maximum amount of mass and charge that a photon can carry when traveling in a vacuum:

http://pdg.lbl.gov/2009/tables/rpp2009-sum-gauge-higgs-bosons.pdf

Last edited: Sep 28, 2010
11. Sep 28, 2010

### daschaich

It is actually still an open question whether or not neutrinos and anti-neutrinos are identical. If they are identical, then neutrinos and anti-neutrinos would be "Majorana fermions". If they are distinct, then neutrinos and anti-neutrinos would be "Dirac fermions".

12. Sep 28, 2010

### daschaich

Please note that the maximum photon mass and charge are incredibly tiny. They are, in fact, zero as best we can measure them. No photon mass or charge has ever been measured, but experiments are not yet perfect, and never will be.

13. Sep 28, 2010

### Bararontok

But because there is no measurable electromagnetic field emanating from these neutrino particle groups, whether they are classified as Majorana or Dirac Fermions does not change the fact that they are uncharged particles.

The mass and charge in this case are only a scaling of the energy content of the photon because mass and charge is only applicable to the fermions that carry force fields and use them for action at a distance interactions. An individual gauge boson can only truly be measured by the energy it carries because it is incapable of this type of interaction and can only exert influence if it directly comes in contact with other photons or a force field.

14. Sep 28, 2010

### piareround

So later checked with my professor about this an interestingly optical phase conjugation, or time reversal, and concept of going backward in time are indeed to separate phenomena.
Thus binbots the best answer to you question may be this one:

Last edited by a moderator: Apr 25, 2017
15. Sep 28, 2010

### Parlyne

The fact that neutrinos carry no electric charge does not tell you whether or not they are their own anti-particle. Neutrinos carry both weak isospin and hypercharge. These change under charge conjugation. The only way that neutrinos can be their own antiparticles is if some additional new physics causes a certain type of mixing between the fundamental neutrino and antineutrino states, which can't occur in the standard model alone (and, in fact, must be related to the mechanism by which neutrino masses are generated).

The W bosons, which are two of the three carriers of the weak force, have both mass and electric charge. In fact, the W bosons have the same magnitude of charge as the electron and the mass of about 86 protons. The photon, by contrast, has never been measured to have any mass or charge. The limits in the PDG file you linked simply state the smallest values of each that would be detectable by experiments that have been performed to date. It is not, in any sense, a statement that these value are not zero. It is simply a statement that we could not, at this point, know if their values were different from zero but didn't violate those limits.

16. Sep 29, 2010

### Bararontok

Interestingly enough, when I check the data sheets for both the neutrinos and anti-neutrinos, the pages link only to the neutrinos which makes the properties of both particles exactly identical. What then would differentiate the neutrinos from their anti-particle counterparts?

I was referring to mass and charge in terms of energy and I did not say that other bosons did not have any fixed mass and charge values. I was referring to the photon's mass and charge as simply another way of denoting the energy that it carries but that when the photon is incorporated into the electromagnetic field of a charged fermion, that energy is used to thermally excite the charged fermion.

17. Sep 30, 2010

### Parlyne

And, if you looked up the data sheet for positrons, you'd find that is sent you to the sheet for electrons. This is because the properties of an antiparticle follow directly from those of its particle; so, there's no need to have separate sheets.

If neutrinos and anti-neutrinos are distinct, they can be differentiated by the processes they can be involved in. For example, in a standard beta decay, a neutron decays to a proton, an electron and an electron anti-neutrino. If the anti-neutrino subsequently scatters off of a nucleus, there is a (small, but non-zero) chance that it could be transmuted into a charged lepton. If the neutrino is its own anti-particle, this could produce either an electron or a positron; but, if the neutrino is not its own anti-particle, this could only lead to a positron.

In practice, this is a difficult experiment to perform because the neutrino interacts so weakly. The test most commonly discussed to determine whether the neutrino is its own anti-particle is neutrinoless double beta decay. Essentially, there are some nuclei that undergo double beta decay - in other words, their decays involved two neutrons simultaneously decaying to protons and emitting electrons and anti-neutrinos (or the analogous process with protons decaying to neutrons, depending on the nature of the nucleus involved). If neutrinos are their own antiparticle, such decays should occasionally occur in a way that one nucleon absorbs the neutrino from the other instead of each emitting their own. This would lead to double beta decays with no emitted neutrinos.

Mass and charge are not, in any context, a denotation of the energy carried by a particle. Also, I don't see what thermal excitation has to do with the fundamental properties of the photon.

18. Sep 30, 2010

### Staff: Mentor

At particle accelerators, one can produce beams of muon-neutrinos or muon-antineutrinos by selecting for decays of negative or positive muons respectively. In neutrino detectors these beams give you interactions which accordingly produce negative or positive muons.

19. Sep 30, 2010

### Bararontok

So this means that having an electromagnetic field has nothing to do with whether a particle will have an anti-particle or not because even the uncharged neutrino group has anti-particles. What then causes a particle to be an anti-particle if it does not have a charge opposite to its identical particle? Because when I looked at data sheets and compared the charged particles to their charged anti-particles the anti-particles were the same in every respect except for the polarity of their electromagnetic fields which are opposite to their counterparts.

Read this definition for antiparticle and you will see that having an opposite electric charge is a prerequisite to being an anti-particle:

http://en.wikipedia.org/wiki/Antiparticle

Additionally, you previously mentioned that weak iso-spin and weak hyper-charge can be used to tell the particle and anti-particle pairs apart and I saw the values RH and LH for the weak iso-spin and weak hyper-charge of the neutrinos so does this mean that RH and LH are used to differentiate the particles and anti-particles as opposed to only using the polarity of the electric charge to make the differentiation?

Last edited: Sep 30, 2010
20. Sep 30, 2010

### Parlyne

Shockingly enough, Wikipedia isn't always right, particularly about technical subjects. For instance, the article you linked claimed that all self-conjugate particles (i.e. particles that are their own anti-particle) are "Majorana particles," when, in fact, Majorana particles are specifically self-conjugate fermions.

What Wikipedia does get right is that having non-zero electric charge is not a prerequisite for having a distinct anti-particle. There may be other charges that distinguish the two states.

In the standard model of particle physics there are three forces, each of which is related to a type of charge. The strong force involves color charge, the SU(2) part of the electroweak force involves weak isospin, and the U(1) part involves hypercharge. While color charge is conserved in all interactions (much like electric charge), weak isospin and hypercharge are not, due to the spontaneous breaking of electroweak symmetry by the vacuum state of the Higgs field. However, since it is only the vacuum state of the system that breaks the symmetry, the actual dynamics of the fundamental fields must be the same as they would be were the symmetry exact. In other words, each field in the standard model has a well defined isospin and hypercharge (or, at worst, is a linear superposition of fields that each have well defined isospin and hypercharge).

The RH and LH you've seen discussed are chiral states, which is one basis we use to discuss spin. As it happens, the SU(2) part of the EW force discriminates by chirality. LH states have non-zero weak isospin, while RH states have 0. This means that normal, everyday fermions actually take their two spin states from different fundamental fields (which is a little weird, but works).

Weak isospin itself is actually a spinor-valued charge. (It doesn't have anything to do with the spin of the particles involved. It just uses the same kind of math that's involved in the description of spin.) As such, there will be two distinct isospin states, which we can think of as isospin up and isospin down. Fundamental fields that carry weak isospin, then, must come in doublets - one field with isospin up and one with isospin down. (This is generally characterized by a quantity called $I_3$, which takes the value 1/2 for isospin up and -1/2 for isospin down.) So, for instance, the (LH) electron neutrino is the isospin up component of a doublet which has the LH electron as the isospin down component. The LH up and down quarks form another such doublet (with the caveat that symmetry breaking generate mixing among the quarks in a way such that the physical down quark is really a superposition of the fundamental down, strange, and bottom quarks).

Hypercharge (denoted $Y$) is a scalar valued charge (like electric charge). For the dynamics to be consistent, the fields in an isospin doublet must have the same hypercharge as each other. And, after symmetry breaking, electric charge (in units of e) is given by $Q = I_3 + Y$. So, we can immediately see that, for example, the electron neutrino/LH electron doublet must have hypercharge -1/2, while the RH electron has hypercharge -1.

Under charge conjugation, $I_3$ and hypercharge each change sign. So, for a particle to be identical to its antiparticle requires at least one more condition than its electric charge being 0, at least in the case of a fundamental field with definite weak isospin and hypercharge.

Models in which the neutrinos are self-conjugate get around this by making the physical neutrino a superposition of a fundamental state with $I_3 = 1/2$ and $Y=-1/2$ and one with $I_3=-1/2$ and $Y=1/2$.

21. Oct 1, 2010

### Bararontok

In conclusion, there are three states that determine whether a particle will be matter or anti-matter and that would be the electric charge, isospin and hypercharge. The LH values apply to the neutrinos while the RH values apply to the anti-neutrinos.

I have summarized your data in this table to differentiate the LH and RH values:

Neutrino LH Isospin Up = 1/2 || Anti-Neutrino RH Isospin Down = -1/2

Neutrino LH Hypercharge = -1/2 || Anti-Neutrino RH Hypercharge = -1

But I was wondering if these values apply to all three neutrinos. Could you give me a data table for this because wikipedia does not have one?

22. Oct 1, 2010

### Parlyne

You can calculate electric charge from isospin and hypercharge; so, it's not really an independent value. Also, a correction, the RH anti-neutrino has hypercharge 1/2, not -1.

You can find this information in books that cover the standard model of particle physics; but, you generally won't find it in experimental data sheets, as hypercharge and isospin are not conserved quantities in the SM (making their values difficult to determine experimentally other than from symmetry considerations).

23. Oct 3, 2010

### Bararontok

Sorry, but I cannot make the correction for the value of the RH anti-neutrino because the edit function is somehow missing. Can a technical supporter please fix this problem?

24. Oct 16, 2010

### Xia Ligang

Neutrinos have no charge. What happen if C operates neutrinos? I think it deserves thinking over. Maybe we should extend the perception of anti-pariticles.

25. Oct 16, 2010

### Parlyne

That's exactly what I was saying. It's not known whether the physical neutrino states are invariants under C. It is known that the neutrino states in the SM (before EWSB) have weak isospin and hypercharge that means they are not invariant under C; but, the relation of these to the physical states depends on the mass generation mechanism.