This may be a stupid question

[Chirios]

For energy to be released in an antiparticle-particle collision, does the antiparticle have to be specific to the particle? For example, if an electron hits an anti-proton would energy still be released?

[Feil]

Quantum numbers are conserved. For particle-antiparticle annihilation (put more usefully, spontaneous decay of massive particles into massless photons following mass-energy equivalence), the totality of the quantum numbers before the decay must equal the totality of the quantum numbers after the decay.

In the case of your hypothetical electron-antiproton annihilation, we can follow a simple rubric:

Suppose an electron and an antiproton can annihilate
Then the sum of each additive quantum number numbers must be zero
But the charge quantum numbers for an electron and an antiproton are both Q=-1, thus their sum is -2
-2!=0
Ergo the premise is false.



An interesting exercise might be to consider the time-independent Schrodinger equation for a system of particles consisting of an electron and an antiproton in an infinite-square well, and prove that they have 0 probability of occupying the same position.



A macroscopic freshman physics approach to the same problem would be to consider the electron-antiproton system as simple point-charges in a system governed by Coulomb's Law. Clearly, two charges of the same sign cannot occupy the same location. No matter with how much energy they are projected towards one another, they will always deflect.



I think a more interesting question - and one for which I don't have a ready answer - would be to consider the interaction between an antiproton (composed of two anti-up-quarks and one anti-down-quark, I should expect) and an up quark. Their charges are of different sign, so they should attract, and it is quite possible for quantum numbers to be conserved in the reaction. Would the system decay into a bunch of photons, an anti-up-quark, and an anti-down-quark? I leave that question to better educated people than I.



If I have made a mistake somewhere, please point it out. I am positively awful at all things quantum, though I'm improving... slowly....

[Eframepilot]

Energy can be released by any particle collision. For instance, the LHC collides bunches of protons together, because the center of mass energy of the protons is much higher than the rest mass of the proton (and antiproton), so little would be gained and much would be lost by trying to make an equally energetic antiproton beam.

[Narkis]

I think a more interesting question - and one for which I don't have a ready answer - would be to consider the interaction between an antiproton (composed of two anti-up-quarks and one anti-down-quark, I should expect) and an up quark. Their charges are of different sign, so they should attract, and it is quite possible for quantum numbers to be conserved in the reaction. Would the system decay into a bunch of photons, an anti-up-quark, and an anti-down-quark? I leave that question to better educated people than I.

There are a couple problems with that. Namely, it's impossible to have free quarks due to their "colour" forces confining them into forming a particle. But even if you could have a free quark, Up has a charge of +2/3, while an antiproton has a charge of -1. Annihilation would be a little problematic.

[Feil]

There are a couple problems with that. Namely, it's impossible to have free quarks due to their "colour" forces confining them into forming a particle.

That makes sense. No quark-proton interactions because no free quarks to begin with.

But even if you could have a free quark, Up has a charge of +2/3, while an antiproton has a charge of -1. Annihilation would be a little problematic.

Well, protons are not elementary particles. My thinking was that the up quark might annihilate one of its opposite numbers in the antiproton. Of course, this leaves the original problem - where did the free quark come from in the first place.

[Simon_Jester]

Well, protons are not elementary particles. My thinking was that the up quark might annihilate one of its opposite numbers in the antiproton. Of course, this leaves the original problem - where did the free quark come from in the first place.

Inside a nucleus, I can at least imagine it being possible- say, one of the strange baryons decaying inside a nucleus and resulting in a spray of close range quark-quark interactions until the smoke clears and you have some (indefinite) number of bound quark-combos.

Unfortunately, I cannot do the math on quantum chromodynamics, so I may be flat wrong here.

[Kuroneko]

An interesting exercise might be to consider the time-independent Schrodinger equation for a system of particles consisting of an electron and an antiproton in an infinite-square well, and prove that they have 0 probability of occupying the same position.

It wouldn't be a square well unless they're noninteracting, in which case there would be no problem in them occupying the same position. Being non-indentical, they would not have any exchange forces acting on them (i.e., no Pauli/spin-exclusion). But if they are interacting, then their contributions to the potential well are infinite at their locations, which means that the probability is indeed zero by the contrainst that the wavefunction be C^1.

Just intergrate the time-independent Schrödinger equation in one dimension around a potential divergence (say, at zero), in the limit of the interval of integration going to zero:
[1] -(ℏ²/2m)[limx→0⁺{ψ'(x)} - limx→0⁻{ψ'(x)}] + limε→0⁺{Int-ε+ε Vψ dx} = 0,
so that continuity of ψ' implies that the integral of Vψ is zero in the limit, i.e., ψ must go to zero faster than V diverges.

I think a more interesting question - and one for which I don't have a ready answer - would be to consider the interaction between an antiproton (composed of two anti-up-quarks and one anti-down-quark, I should expect) and an up quark.

If your goal is to have a quark interacting with an antiquark, then that sort of thing happens all the time with the residual color force, which is what bind the nucleons in an atom. For example, a positive pion has quark composition ud, so:
[2] p → n + π⁺, n + π⁺ → p
A proton can become into a neutron, which another neutron absorbing the π⁺ and becoming a proton. The reverse can happen with a negative pion exchange. Thus, in nucleus, the protons and neutrons don't have monolithic identities, but in a sense flip between being one another (neutral pion exchange doesn't do this, though).

[Chirios]

Quantum numbers are conserved. For particle-antiparticle annihilation (put more usefully, spontaneous decay of massive particles into massless photons following mass-energy equivalence), the totality of the quantum numbers before the decay must equal the totality of the quantum numbers after the decay.

so would an electron and an antineutron annihilate then?

[Feil]

No. 0!=-1, again considering only the charge quantum number. You can do these yourself: look up the quantum characteristics of the particles under consideration, and see if they sum to zero. You will find that, since distinguishable particles types may be identified uniquely by their quantum numbers, that for fundamental particles, only the exact antiparticle may annihilate it.