# When physicists say they have discovered a “particle” that is it's own anti-matter what does this mean?

Not being a physicist but liking to keep up with the news I read that new particles were discovered that are their own anti-particles. I can see that perhaps we can discover a particle that has an anti particle as in the case of the electrons and when they come in contact we get all energy from the "reaction". I believe they are called positrons , being the anti particle to the electron. But how can ANY particle be its own antiparticle, if that was the case would they not annihilate each other and you would never even know they ever existed since as soon as one was in the process of being created, and it is its own antiparticle it would go into non existence as fast as it came into existence and you would never know which particle it was that was its own antiparticle.

Just because two particles are antiparticles of each other does not mean they have to annihilate. For example, an electron and a positron can form an atom called positronium without immediately annihilating. That's because in order to collide, they need to have the right combination of momenta and position.

So what makes a particle an antiparticle? An antiparticle is simply a particle with the same properties but some of its quantum numbers are opposite its normal particle equivalent. For example, an electron has a mass of $511~\mathrm{keV}/\mathrm{c}^2$, a spin of $1/2~\hbar$, and a charge of $-\mathrm{e}$ while the positron has the same mass and spin, but a charge of $+\mathrm{e}$.

How can a particle be its own antiparticle? Simple, all the quantum numbers that change from particle to antiparticle are 0. The example you would be most familiar with is the photon. Photons are massless, have a spin of $1~\hbar$, but a charge of $0$. If you tried to create an anti-photon, it would just be a photon.

So why don't particles that are their own antiparticles decay automatically? Because in order for a particle-antiparticle pair to annihilate, you need two particles. If only one particle is created, it can't annihilate with itself.

Now, you mentioned "new particles" that are their own antiparticle. Any new particles being discovered right now are not fundamental particles like electrons or photons (unless you're talking about the Higgs, in which case you're a few years behind), they're composite particles, made up of combinations of fundamental particles. The best examples of such particles are the neutron and the proton. They are part of a class of particle called baryons, made up of three quarks. There is another class of particles, called mesons, which are composed of a quark and an antiquark. The quark and antiquark that make up a meson don't have to be of the same type (e.g. the $\pi^+$ meson is composed of an up quark and a down antiquark), so they won't annihilate each other. But some mesons are (e.g. the $\pi^0$ meson is a mix of up-antiup quarks and down-antidown quarks). Like the positronium example above, they don't immediately annihilate each other because they are in a bound state which require specific types of interactions to annihilate.

Edit:

After looking it up, I've identified the new particle you were talking about (press release, Science paper, arXiv).

What's happening here is a completely different phenomenon. In condensed matter physics, they study a variety of materials. When they study those materials, they will see local excitations that behave like other objects in physics. I've seen condensed matter systems recreate black holes, create magnetic monopoles, and mimic subatomic particles. When these excitations mimic particles, they are called quasiparticles.

Among those materials are superconductors and topological insulators. A superconductor is a material that has absolutely no electrical resistance. A topological insulator is a material that either is an insulator in its volume and a conductor on its surface (for 3D materials) or an insulator on its surface and conductor on its edges (for 2D materials). What they did was take a sheet of superconductor and a sheet of topological insulator, modified the topological insulator a bit so it showed some magnetic effects, and then sandwiched the two together. This created a system where quasiparticles moved along the edges in a way that simulated fermions. These fermions were their own antiparticles.

• I was talking about "Stanford News" apparently they have evidence of the "majorana fermion", but as you pointed out it's not really a new particle , it's a fermion ? The Standard Model predicts 24 type of fermions so syas wiki but there are three classes , massive , massless and having its own antiparticle , but is this one of those types? and which types are quarks. Gosh I need a flow chart....but you answered my question...the "a particle cannot annihilate with itself. " Must be a conservation law thing? – Sedumjoy Jul 24 '17 at 3:24
• Fermions are any particle with half-integer spin. Quarks, leptons (like electrons and neutrinos), baryons are all fermions. – Johnathan Gross Jul 24 '17 at 3:56
• I just looked up the article and it's not really a particle. It's a quasiparticle. What they did was they took two sheets of material with different properties and stuck them together. Then they ran a magnet over it and that created excitations in the material that behaved like Majorana fermions. The excitations have no charge and have a spin of 1/2. – Johnathan Gross Jul 24 '17 at 4:09
• To any condensed matter physicists out there, feel free to correct any mistakes I made in my explanation in my edit. – Johnathan Gross Jul 24 '17 at 5:36
• Particles are real objects that exist. Quasiparticles are objects that mimic particles in materials. The best example would be electron holes in a conductor. If you ionize an atom in a conductor, it creates a free electron and a hole where the electron used to be. The hole behaves almost exactly like a positron. – Johnathan Gross Jul 24 '17 at 14:36

The 'charge conjugation operator' $C$ is defined so that it converts elementary particles into their antiparticles, so the charge conjugation operator acting on a state representing an electron give a state representing a positron: $$C |e^-\rangle = |e^+\rangle \;,$$ and the operator acting on a anti-strange quark give a strange quark $$C |\bar{s}\rangle = |s\rangle \;,$$ and so on.

But not all particles are elementary: the proton, for instance, has a valance make-up of two up quarks and a down quark $| p \rangle = | uud \rangle$. 1

What does charge conjugation do to a particle like that? It changes all the elementary particles making up the compound particle to their anti-particles.

Now consider the group of particles known as 'mesons'. They have a valance structure consisting of one quark and one anti-quark. For instance a positive pion is $$|\pi^+\rangle = | u\bar{d} \rangle \;,$$ and the charge conjugation operator acting on one give you \begin{align} C|\pi^+\rangle &= C| u\bar{d} \rangle\\ &= | \bar{u}d \rangle\\ &= |\pi^-\rangle \;, \end{align} a negative pion.

Finally, there are some meson's who valence content is made up of a quark and an anti-quark of the same basic kind. Something like $|\pi_u^0 \rangle = | u \bar{u} \rangle$.2 When we let charge conjugation act on that we get \begin{align} C|\pi_u^0\rangle &= C| u\bar{u} \rangle\\ &= | \bar{u}u \rangle\\ &= |\pi_u^0\rangle \;, \end{align} because the ordering of the quark labels doesn't matter here. We see that the anti-particle of the $\pi_u^0$ is the same as the original particle.

1 I am going to steadfastly ignore the quark-gluon sea in this discussion.

2 Because of the nearly correct symmetry known as iso-spin a real neutral pion is actually $$| \pi^0 \rangle = \frac{1}{\sqrt{2}}\left( | u \bar{u} \rangle - | d \bar{d} \rangle\right) \;,$$ but this complication doesn't affect the discussion or the conclusion other than making that math longer to write out.

• Does the math show that a particle cannot be an antiparticle of itself??? – Sedumjoy Jul 25 '17 at 1:20
• No. It gives an explicit example of a particle that is it's own antiparticle. Well, the example in the text is hypothetical, but the proper $\pi^0$ from the footnote also has the same property. Though it seems there has been a copy-n-paste-o in there for nine hours which I have fixed now. – dmckee Jul 25 '17 at 1:28