Q & A: Antiphotons?

Are there such things as anti-photons. And if there is, what would happen if it collided with a photon? Thank you
- Matthew Ervin (age 16)

Hi Matthew,

The short answer to "are there anti-photons" is "yes", but the disappointment here is that anti-photons and photons are the same particles. Some particles are their own antiparticles, notably the force carriers like photons, the Z boson, and gluons, which mediate the electromagnetic force, the weak nuclear force, and the strong force, respectively. Particles that are their own antiparticles must be electrically neutral, because an aniparticle has the opposite electrical charge as its partner particle. Other things must also be zero, like the number of quarks. A neutron cannot be its own antiparticle because it is made up of quarks and an antineutron is made up of antiquarks. A pi_0 is made up of a quark and an antiquark and is in fact its own antiparticle also.

You can find lots out about particles at , part of the Particle Data Group's .

You also ask, in a follow-up question:

I have just thought of some stuff to add to my other question. When the antiphoton and photon collide, would they fuse? And if so, would they form a particle that has mass, or one that is massless. And what kind of particle is it?

The answer is yes, photons may collide and produce other particles. One familiar reaction is the low-energy annihilation of an electron and an anti-electron (known as a positron)-- the result is usually a pair of photons (sometimes you get more than two). You need at least two, in order to conserve both energy and momentum. This reaction also works in reverse -- a pair of photons may collide to make an electron-positron pair. This happens all the time in particle physics experiments.

In high-energy electron-positron collisons, often what collides are not the electrons or the positrons, but the photons in the "entourage" around the beam electrons and positrons. These photons come together with enough energy to produce a pair of particles like an electron and a positron, a muon and an antimuon, or some quarks, depending on how much energy is available. The quarks may have lots of energy to pull on the strong force holding them together so that they may produce jets of subatomic particles.

These collisions also happen in proton-antiproton collisions at high energy. The protons and antiprotons also have a cloud of photons around them which may interact with the photons in the opposite beam to produce pairs of particles which may be observed in a detector. The energies of the photons has to be really high, however.

For low-energy photons (like visible light, radio waves, x-rays and just about anything outside of a high-energy physics laboratory), photons for the most part just go right past each other. This is because the equations of electricity and magnetism are "linear" -- the local field strentgths of two electromagnetic waves colliding is just the sum of the two, with no interaction. It works just like waves on a pond -- they will pass through each other without interacting. There is a very tiny effect, called "light-on-light scattering", where, with a very low probability, a photon will "bounce off" of another one. This proceeds by exchanging a virtual electron around in a loop. The resulting photons are massless just like the incoming ones. So if you ask if photons when they collide make massive or massless particles, the answer is: they can make either, but the mass of the whole system (that is, the total energy in a frame in which the total momentum is zero) is the same before and after the collision.

Back to high-energy collisions: There are ideas floating around the high-energy physics community to build a "photon collider" out of an electron-positron collider. This can be done by focusing laser light head on on a beam of very high-energy electrons. The photons which bounce backwards from the electrons will have very high energies, taking a large fraction of the electron's energy. The same can be done pointing in the opposite direction, and the beams of high-energy photons can be brought into collision. This is proposed to study the production of, for example, Higgs bosons, which can be made in this way via loops of W bosons and top quarks.

You can do a search on Google for "photon collider" -- there is quite a lot of information available. You can ask us more questions about anything that sounds weird.

Tom

(published on 10/22/2007)

Quote: "Particles that are their own antiparticles must be electrically neutral, because an antiparticle has the opposite electrical charge as its partner particle." like photon, Neutrino is electrically neutral (although they might not be force carriers like photons). but we have particles called anti-neutrinos. are neutrino and anti-neutrino one and the same?
- An (age 17)
India

Hello An,
This is a very interesting, and fundamental, question. In order for a particle to be its own anti-particle it must be invariant under the three simultaneous operations, Charge Conjugation, Parity Inversion, and Time Inversion.  The CPT theorem (that physics as a whole is invariant under this operation) is one of the tenets of modern field theory.  No experiment has proven it wrong.   Even though some particles may violate P or C the combination of CPT always turns out to be invariant.
 
 As you pointed out in a previous question any charged particle cannot be its own anti-particle since under the operation of charge conjugation the sign of the electric charge  is reversed.   Neutral bosons (integer spin particles) can be their own anti-particles, for example the photon and presumably the graviton, but not necessarily if there is an additional associated quantum number.   The neutral K meson is an example of the latter since it has a quantum number, called strangeness, that is not charge conjugate invariant. 

Back to neutrinos.   If the neutrino has mass, which we now believe it does, then it has the capability of having a magnetic moment.  If so, then it cannot be its own anti-particle because, like electric charge, the magnetic moment changes sign under the operation of charge conjugation.  If the mass of the neutrino were zero then the anti-neutrino could in principle be its own anti-particle.  There is a complete theory of this type of particle, called a Majorana neutrino (See:) that describes this possibility.   So far, no experimental evidence for these particles have been found, the absence of neutrino-less double beta decay being the best test.

Even if the ordinary Dirac-type neutrinos have mass, but no magnetic moment, then they can be their own anti-neutrino.  Lowest order calculations using minimal extensions to the Standard Model give a moment value of the order of 10^-20 times that of the electron's Bohr magneton.   Several experiments have set upper limits of around 10^-11 or so.  We have to improve our experimental limits by quite a bit before we will know the answer.
See:


LeeH

(published on 02/05/2011)

When shooting a photon through a double-slit experiment, is it that same photon that actually hits the detector? I'm curious about the interaction of that photon and the quantum fluctuations that occur in a vacuum. Assuming Pilot Wave theory, where the uncertainty isn't a fundamental property of reality, is it possible that your initial photon interacts with virtual particles? Possibly getting annihilated by one, allowing the surviving anti-photon (see http://van.physics.illinois.edu/qa/listing.php?id=1153) to continue on? Could these fluctuations then be part of the Pilot Wave? My thought process. If Hawking Radiation can take advantage of virtual particles, can we apply this to other observable phenomena?
- Joe Esposito (age 28)
Pittsburgh, PA, USA

The sea of virtual photons in the quantum vacuum doesn't affect the real photon propagation. Perhaps photons are affected by the sea of virtual particles of some deeper type, sort of as quarks are affected by the Higgs field. But in that case the thing we call a photon would already be an object that included the interactions.

You ask if the initial photon could annihilate and be replaced with a different one. I believe that that process would have no symptoms whatsoever, even in principle. Therefore I think that the question has no meaning. We all often stumble into questions like that when we take try to picture quantum processes in classical terms. Classically, no matter how similar two particles are they are in some subtle way not identical. Yet quantum particles of the same type truly are identical, so it doesn't mean anything to say that one is "new".

I'm not positive what change a pilot-wave (Bohm) interpretation would make for the discussion.  

Mike W.

(published on 01/02/2014)

Thank you for answering, Mike! > Perhaps photons are affected by the sea of virtual particles of some deeper type, sort of as quarks are affected by the Higgs field. But in that case the thing we call a photon would already be an object that included the interactions. Gotcha. That's an interesting thing to think about. > I believe that that process would have no symptoms whatsoever, even in principle. I ask because I wonder if this could give insight to the _mechanism_ of the interference pattern we see. We know there's a sea of virtual photons, so perhaps that could be a hint of something more fundamental rather than pure mathematics. In other words, I was wondering because if the math does allow for this kind of interaction, it could be a clue that these virtual particles and what we're calling the "pilot wave" may be more closely related. Specifically, the virtual particles could have been what is deterministically guiding the photon to its seemingly-random final destination on the detector, similar to what the description of the pilot wave. > I'm not positive what change a pilot-wave (Bohm) interpretation would make for the discussion. It's because when you instead assert that particles still exist even when you aren't observing them, you can start to talk about what happens between the initial and final locations. "In de Broglie–Bohm theory, the wavefunction travels through both slits, but each particle has a well-defined trajectory that passes through exactly one of the slits." If we assumed the Copenhagen interpretation, then you're right, my question would be meaningless. Yet you said "The sea of virtual photons in the quantum vacuum doesn't affect the real photon propagation," so my question is false. So in the case of Hawking Radiation, what happens to the virtual particle whose pair gets lots in the black hole? Does it get promoted to a real photon and can then interact with other real photons? Or does it instead drift on forever without ever interacting? (Or perhaps get annihilated by other virtual photon pairs, ad infinitum?)
- Joe Esposito (age 28)
Pittsburgh, PA

Aha, I see what you're wondering about- if some mechanism can be given for the pilot wave force on the particle coordinate in the Bohm picture. maybe some sort of little collisions with local particle fluctations, etc. It's a nice thought, but I think that if anything comes out of an attempt like that it will be even weirder than the view in which the wave function is the sole ingredient. The reason is that all such processes can violate the Bell Inequalities. That means that there is no local picture at all (other than universal conspiracies) that can reproduce the observations. So there's little motivation to pursue yet another local picture.

The emitted photo in a Hawking picture is a real photon, and, like Pinocchio, can do all the things a real photon can do.

Mike W.

(published on 01/06/2014)

Follow up question regarding the double slit experiment. If a photon is it's own anti-particle, does it have something to do with the particle-wave conundrum? I find it very interesting that in the 2 slit experiment, when ever systems are in place to observe witch slit the photon passes through the wave form disappears and becomes only 2 lines, making it impossible to solve. Now with new discoveries in dark matter and dark energy...and also negative energy...dose a dark matter particle have an anti-particle, and can dark energy have a negative form?
- Travis (age 39)
Phoenix, AZ, USA

The basic two-slit behavior works the same for particles that are their own antiparticles (e.g. photons) and ones that aren't (e.g. electrons, buckyballs). The particle-like aspect of quantum waves is that they have a "number operator" that gives them something that has discrete integer counts. That also holds for each type of particle. It has a little different behavior for bosons (e.g. photons,  ⁴He) than for fermions (e.g. electrons, ³He), but that is a different distinction than that between ones that are their own antiparticles and ones that aren't.

Mike W.

(published on 05/20/2014)

QED states that probability of a photon hitting an electron is a square of the sum of amplitudes for all physically possible independent paths a photon could take to reach the detector electron. Amplitude itself for a given energy photon is a vector whose direction essentially depends on just the length of each such path or some imaginary timing required to travel that path. Experiments confirm that it is not possible to tell which path a photon actually takes. Assuming that photon does not actually split into an infinite number of sub-particles each travelling every possible path and reassembles at the detector, here are some conclusions that I would like to falsify by previously done experiments and theories: 1. Every emitted photon knows exactly the location of each electron (and any other particle it interacts with) in the universe and it just hits randomly one of them with some probability according to its relative location with respect to all such particles existing in the universe. 2. Any attempt to detect which path a photon takes does not succeed because detection implies interaction with some particle but the photon already knows the location of the detector particle hence the photon's behaviour changes accordingly. 3. Assuming the above i.e., that each photon somehow knows the location of every particle in the universe, when does it acquire such knowledge: a) at the point of emission? b) or its knowledge about the world is constantly updated as it travels? 4. What about the following version of the double-slit experiment – calculate/observe what interference pattern should be by emitting photons one-by-one through the double-slit barrier and onto a detector screen behind it. Then perform the same experiment but keep shifting the double-slit barrier slightly after each photon has been emitted but before it is supposed to go through the slits. Will the interference pattern be exactly the same and in the same location as in the original static version of the experiment or will it be different? 5. If the above experiment produces the same expected interference pattern in both cases it could mean that photon acquires knowledge about all existing particles it interacts with at the point of emission and it doesn’t even travel at all. There are just two discrete events: a photon is emitted and then it gets absorbed by a randomly selected particle after some time. The selection of which particle will absorb the photon is performed at the point of emission and the fact that there is a delay between emission and absorption creates an illusion of space travelling which in reality does not happen. 6. If the above experiment produces a shifted interference pattern, at what point does the photon update its knowledge about the world? What if the double-slit barrier is shifted immediately before photon can reach the barrier? I think, it would have been nice if photon did in fact behave like described in point 5, since this could potentially explain many quantum world properties.
- David (age 38)
Reading, UK

Hi David,

It is true that photon amplitudes take all possible paths in an interaction. For example, if you make an interferometer, half of the photon's amplitude goes down each path, and the two amplitudes interfere at the beamsplitter. Most physicists don't try to make statements about the photon's past (beyond saying that it has amplitudes for every possibility), and instead just discuss the outcomes of detector measurements.

That said, you can make some statements, which I believe are true:

1. The photon does NOT have a huge nonlocal database telling it everything about all the other particles in the universe. However, it can have properties which are (nonlocally) correlated to ("entangled with") the other particles that it has interacted with, or that are connected with the chain of events leading to its creation.

2. Attempts to determine which path a particle took can certainly be successful. However, even if such a measurement doesn't destroy the photon, it always collapses the photon amplitudes from "all possible paths" to the measured path, and so limits any further wavelike behavior. (For example, if you put a nondemolition measuring device in one arm of an interferometer, then you won't get the usual interference at the output.) The physical (i.e. classical-sounding) explanation of this effect is that any position measurement transfers random momentum to the particle, which washes out any fringes. (A less classical but more robust description can be made in terms of information: correlations and density matrices.)

3. Since the photon doesn't know about all the other particles, this question doesn't make sense. However, you can ask something similar: if a photon is entangled with a partner particle, and something affects the partner particle, then when does the photon find out? Really, it never "finds out." The relationship between distant entangled particles is not causal, it is just a correlation. The correlation can change without the photon "having any knowledge" about its partner. (See https://van.physics.illinois.edu/qa/listing.php?id=24896.)

4, 5, and 6: The interference pattern you see in any such dynamic experiment will depend only on what configuration the photon saw locally at each point in time. For example, if you emit a photon towards two slits, but yank them out a picosecond before the photon reaches the slits, then you won't get any interference pattern. There isn't anything surprising here, as far as I can think of: the photon really does travel from the source to the detection screen, and the pattern doesn't change if you change something before the photon gets there or after the photon has left. The photon interacts with objects at the same spacetime coordinate, that's all.

Hope that made sense. Let me know if you want further details or explanations, of which there are many.

David Schmid

(published on 06/09/2014)