# Why do electron-positron pair annihilate upon contact?

I'll a appreciate a layman's explanation, if there exists one, to this question that arose when reading an popular-science level article on Einstein and the $E=MC^2$ equation.

What I mean is that, why is the product of the reaction not an electrically neutral doublet of mass $2M_e$ (= 1.022 MeV) and which the opposite charges keep the two particles attracted to each-other?

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You in fact can get that product: en.wikipedia.org/wiki/Bhabha_scattering. If the energy of the collision is enough, you can also have products like $e^{+}e^{-}\rightarrow \mu^{+}\mu^{-}$ –  Jerry Schirmer Sep 9 '11 at 23:46

An explanation for a layman.

A hydrogen atom exists because of quantum mechanics, and not classical mechanics. In quantum mechanics there exist energy levels where the electron of the hydrogen atom, which has a tiny mass with respect to the proton, can only occupy fixed energy levels around the proton. The lowest one is the ground state and is non zero, allowing for the existence of the hydrogen atom. Otherwise, if we only had the classical theory, the electron would fall onto the proton and no hydrogen atom would be possible. In the quantum ground state, the electron does not have enough energy to change a proton into a neutron, energy conservation prevents the disappearance.

Still, in quantum mechanics, which is a probabilistic theory, there may exist a tiny probability for the electron wave function to meet a quark within the proton nucleus and interact, with the effect of destroying the hydrogen atom. The probability is so small that it can be ignored.

In contrast, the positronium "atom" has energy levels like hydrogen , and that is why we observe it, but the probabilities for the electron and the positron at ground state to overlap is huge, because they have equal masses and interact mainly with the electromagnetic field, and once the positron reaches the ground state (corresponding to the ground state of the hydrogen electron mathematically) it overlaps with the electron and annihilates because that is the most probable outcome.

It is all a matter of sizes of constants describing the interactions, and conservation laws, whose details you need to study to comprehend.

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I'm not really happy with this answer, because ground state hydrogen atoms have a non-zero expectation for the electron to be inside the proton (it is an s-wave after all), what prevents the reaction $e + p \to n + \nu$ is insufficient energy. –  dmckee Sep 10 '11 at 19:48
@dmckee I thought I covered that with my tiny probability comment. The S wave probability with the proton is not enough since the proton is composite, and I was thinking on the lines of proton decay like combinations. I will edit in your energy comment,thanks. –  anna v Sep 11 '11 at 3:51

Since you asked for a layman's explanation, I'll try to give you one that sticks to the basics (e.g. about what I learned in high school) and hopefully it's not too basic. I'm also going to anthropomorphize the sub-atomic particles since it makes the explanation quicker :)

The basic idea is, the two particles annihilate because they can, and any time something can decay that's probably going to happen. This is largely due to the combination of two physics concepts: minimum energy states, and conservation of quantum states.

The first part says, in essence, that particles "prefer" to be at the lowest mass and energy state they can. Whenever they have "too much" mass or energy, they try to fix it by decaying. This is why high-mass particles, like the ones created in huge particle colliders, decay so quickly. Although mass an energy are equivalent, for various reasons, energy is "preferred" over mass when a particle is deciding what form to take.

The lowest-mass particle we know of is the photon - it has zero mass, and is entirely made of energy. So other particles are pretty quick to give off and/or turn into photons if they can. This is why light bulbs glow: the electrons are giving off some of their own energy as photons so they can be in the lowest energy state possible.

The second idea, though, says that certain quantities about a system cannot ever change. You've probably heard of "conservation of energy" or "conservation of momentum", but lots of physical quantities are conserved, including charge. In order for an electron to decay into something, that thing must otherwise have all of the same conserved quantities, but less mass, and there simply is no such particle. A photon is less massive, but it has no charge, so that isn't an option.

Now, we introduce a positron. The positron on its own has the same problem as the electron: it's charged so it cannot decay into a photon. But it has a positive charge, which is the exact opposite value as the negative electron charge. I'm skipping over lots of details here, including other conserved quantities, but you should get the idea. The key is, other than mass, a positron has exactly the opposite value for each of those conserved quantities as an electron.

When the two collide, all of those other conserved quantities cancel out and become zero. We're left with just an object with the combined mass of the two original particles. There is now no reason not to decay into photons, so that's what happens.

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$2M_e$ would be for a couple of very distant particles where the contribution of the negative potential energy into the total mass is negligible. Positronium has a smaller mass and an annihilated positronium has the mass zero, if you like.

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Thanks. I am not sure I understand. What you are saying that when approaching the two particles, the potential energy has to be preserved somehow, and this is why a simple mass can't form? So why not release the excess energy as light and keep just the 2Me mass? –  ysap Sep 9 '11 at 21:38
"Positronium has a smaller mass" - than what? –  ysap Sep 9 '11 at 21:39
Because there are energy levels, like in Hydrogen atom but that end up at zero total mass in case of positronium. –  Vladimir Kalitvianski Sep 9 '11 at 21:40
Positronium has smaller mass that $2M_e$ because the energy levels are negative: $E_n<0$. –  Vladimir Kalitvianski Sep 9 '11 at 22:08
While technically correct, the binding energy of positronium is a perturbation to the rest mass of the electron/positron pair. Functionally, the mass of positronium is $2m_{e}$. –  Jerry Schirmer Sep 10 '11 at 0:48

The product of the reaction is an electrically neutral doublet of mass 2M (up to negligible potential energy corrections) in which the opposite charges keep the two particles attracted to each other. This electron-positron system is called "positronium". Positronium then decays into photons, because the electron and positron can annihilate in a relativistic Compton wavelength volume when they get close enough. The annihilation process is an elementary interaction, I don't know how to describe it in simpler terms.

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