Found: Elusive particle!

[freyar]

It is and it isn't - I'm mixing my voice a bit, for the non-technical audience.

Sitting in the same quantum state is, for purposes of the layman, equivalent to being in the same place at the same time (it is also more than that, but for these purposes, that's not relevant). Two things cannot get closer than being in the same quantum state. It is the ultimate in "closeness". Sure as anything, if two particles were in the same state, if annihilation were *required*, it would happen.

Since bosons can be in the same state, annihilation is not strictly required, even if it is possible. That's all I mean.

I appreciate speaking for the technical audience, and I am also using "in the same place" to mean "in the same state." But that really has nothing to do with annihilation. It's that last paragraph that's a problem. It has nothing to do with boson vs fermion. An electron and positron with overlapping wavefunctions ("in the same place") are not *required* to annihilate any more than two photons or Higgs bosons are. In each of these cases, there is a probability that the two particles will annihilate (see below about terminology) in any given time.

The real problem I see with what you've said (and what the article says) is that it mixes up the Pauli exclusion principle with annihilation. They are not related in any way. Just think about what I said before. An electron and positron are not the same kind of particle, so the Pauli exclusion principle does not apply. Yet they are the poster children for particle annihilation. Two electrons do feel the Pauli exclusion principle -- they can never be "in the same place" --- but they can never annihilate. That's forbidden.

By my recollection (and, admittedly, I've not done the math in a long time), if you want the full cross section for "two photons in, and two of the same out" you must include the diagrams for the scattering process *and* for annihilation. I recall there being a singularity in there that needs renormalization.

In usual QED, it's just a single box diagram. It's just the terminology that 2 things in and 2 other things out is annihilation (or pair creation) while 2 things in and 2 of the same things out is scattering. But...

And, if you want the cross section for, "two photons in, and two of *something* out," then it isn't distinguishable from an annihilation process that reduces to a scattering under the energy required for electron pair production.

I'm taking the "looks, walks, and quacks like a duck" line here.

Fair enough.

 

[El Mahdi]

...Majorana Fermion: Particle and antiparticle are the same, they do annihilate on contact...

So does this mean that when the Quantum Laptop Computer is finally developed using these, overheating Lithium Batteries are going to be the least of our worries...?

 

[freyar]

So does this mean that when the Quantum Laptop Computer is finally developed using these, overheating Lithium Batteries are going to be the least of our worries...?

Fortunately, I don't think these annihilations would make the laptop any hotter than a standard processor gets anyway.

 

[Umbran]

It has nothing to do with boson vs fermion.

It does have a little to do with it...

An electron and positron with overlapping wavefunctions ("in the same place") are not *required* to annihilate any more than two photons or Higgs bosons are. In each of these cases, there is a probability that the two particles will annihilate (see below about terminology) in any given time.

I was using the (admittedly non-technical) approach - probability of interaction increases with proximity. Sharing the same state is the ultimate in proximity.

Correct me if I misremember - Effectively, at larger distances, scattering processes dominate. At shorter distances, annihilation processes dominate. This is a result of the fact that the forces that drive annihilation are stronger at short distances, and the electromagnetic forces that drive most scattering are stronger at long distances.

The fact that bosons can get into the ultimate state of proximity and not annihilate is, honestly, pretty remarkable. It is something very special - to you and me it plays out rather simply in the math, but that means we can forget how cool that really is. It's how we get the frikkin' lasers, man! While I'm usually a big critic of science writing, and that article is no star, I'm still cutting them some slack for that particular over-emphasis.

 

[freyar]

I was using the (admittedly non-technical) approach - probability of interaction increases with proximity. Sharing the same state is the ultimate in proximity.

Correct me if I misremember - Effectively, at larger distances, scattering processes dominate. At shorter distances, annihilation processes dominate. This is a result of the fact that the forces that drive annihilation are stronger at short distances, and the electromagnetic forces that drive most scattering are stronger at long distances.

Right so far...

The fact that bosons can get into the ultimate state of proximity and not annihilate is, honestly, pretty remarkable. It is something very special - to you and me it plays out rather simply in the math, but that means we can forget how cool that really is. It's how we get the frikkin' lasers, man! While I'm usually a big critic of science writing, and that article is no star, I'm still cutting them some slack for that particular over-emphasis.

It's not an over-emphasis, it's a completely incorrect definition. And that's my issue with this whole discussion.

The point is that the "non-annihilation" of photons in lasers (as you say, "the ultimate state of proximity") is not due to boson-ness. but due to the fact that they are massless. If electrons and positrons (fermions) were massless, then lined up photons would annihilate into electrons and positrons at precisely the same rate that lined up electrons and positrons annihilate into photons.

There is one case where boson vs fermion makes a difference in annihilations. That's when a particle is its own anti-particle (like a Majorana fermion). But the effect is the opposite of what the article states! That is, if a boson is its own antiparticle, then it and its antiparticle can "share the same space" easily and therefore annihilate easily. Two of the same kind of Majorana fermion are restricted by the Pauli exclusion principle --- they can't occupy the same space --- so it is harder for them to annihilate.

The problem isn't overemphasis. It's that the article is mixing up two separate things --- statistics vs antiparticles. And in the one case where there is a relationship, it states the opposite of what really happens! While it's neat that this story made it through several major news outlets, this article is just a poor, poor explanation of the relevant physics.

 

[Umbran]

The point is that the "non-annihilation" of photons in lasers (as you say, "the ultimate state of proximity") is not due to boson-ness. but due to the fact that they are massless. If electrons and positrons (fermions) were massless, then lined up photons would annihilate into electrons and positrons at precisely the same rate that lined up electrons and positrons annihilate into photons.

Okay, that's where my memory doesn't match what you're saying. My recollection was that, all other things being equal, if you used a heavier boson (a W or Z, or a composite boson, for example - let's leave unobserved particles out of this as speculative), the annihilation rates would be lower than for fermions of similar mass. My memory, however, is admittedly imperfect.

 

[freyar]

We have to be careful; the couplings for the W/Z and fermions are a bit different, and it depends what you want it to annihilate them into. You also have to worry about masses and such being different. So if you have two fermions that can annihilate into two bosons of the same mass as the fermions, those bosons will annihilate into those two fermions, and the cross sections will be the same.

 

[Umbran]

We have to be careful; the couplings for the W/Z and fermions are a bit different, and it depends what you want it to annihilate them into.

I thought there was weirdness there, so maybe I'm half-remembering correctly.

Fine, let's simplify. The gauge bosons are all weird anyway - either they have the W/Z odd couplings, or they are zero mass. So, let's use a meson - a composite boson, not a quasiparticle.

My recollection is that the math works out such that annihilation is still less likely for the meson than for fermions of similar mass and charge.

We can even compare here, those mesons which are their own antiparticle (like a charmed eta), with ones that aren't (say, a Pi+).

Do you know offhand how the math works out for those? I admit I don't.

 

[freyar]

Whew, really getting into the nitty-gritty. I guess it's ok since I suspect we've lost the rest of the audience for this thread anyway.

I might be able to figure out meson annihilations if I looked the right things up, but I'm not sure I've ever seen them discussed. Because they're all unstable, it's usually the decays that are worked out.

The main difficulty with using Standard Model particles in examples is making sure that physics that doesn't have to do with statistics is roughly held constant in the comparison (phase space for the annihilation, couplings, etc). So I'd tend to go for thought experiments.

 

[El Mahdi]

I'm still reading. I'm not understanding the majority of it, but I'm learning some new stuff. It's quite interesting.