
Evidence for Anyons, a Third Kingdom of Particles - beefman
https://www.quantamagazine.org/milestone-evidence-for-anyons-a-third-kingdom-of-particles-20200512/
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ivalm
Just to be clear for a general reader. These are quasiparticles. They are
effective mathematical descriptions of certain effects in the real world, but
the physical particles responsible for the effects are fermions and bosons,
not anyons.

~~~
knzhou
Yeah. In this case it's fine, but in general it can lead to confusion.

For some weird reason, condensed matter physicists keep giving their
quasiparticles the same names particle physicists give their particle
particles, even though the mathematics are quite different. If you hear
someone proclaiming the discovery of Majorana fermions, axions, or monopoles,
it's essentially guaranteed to be a quasiparticle in a lump of metal or
something. Sometimes even the articles themselves get confused about this.

~~~
twic
Is this really so different from particle physicists' "resonances", where they
see a peak in a graph and pronounce a new particle:

[https://en.wikipedia.org/wiki/Resonance_(particle_physics)](https://en.wikipedia.org/wiki/Resonance_\(particle_physics\))

~~~
knzhou
That's fine because basically everything's a resonance. The only particles
that aren't technically resonances are those that have infinite lifetime, i.e.
just the proton, electron, and photon.

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rhn_mk1
What does the "looping around" mean here? There's no definition given, and the
expression can be interpreted in several ways.

Does it mean "being moved on a circular path around something"? That would be
simply a couple of translations, which change exactly nothing. Or does it
happen in a sort of a field?

Taking the interpretation of "engulfing", like "a rope _looped around_ a post
engulfs it" shows a clear difference between 2D and 3D, and seems
topologically relevant, but what does it mean "to go back again"?

Not knowing the topological model for particles beforehand, I am unable to
learn anything from the article.

~~~
fennecfoxen
> Does it mean "being moved on a circular path around something"? That would
> be simply a couple of translations, which change exactly nothing.

I can't exactly tell you how the topological part works but I can tell you
that this statement isn't as simple as you think.

You're not reasoning about the location of a particle, but you should actually
be reasoning about the Feynman path integral, the sum of all the probabilities
of the potential paths that the particle could travel through in the future —
because this is physics and our particles are all waves.

Say that translating the particle does not change its state. Imagine for a
moment a macroscopic two dimensional system in which you have a globe
depicting the Earth. Initially, this globe is set with the north pole facing
up and the Greenwich meridian facing north. Draw an circle on the floor and
place another sphere in the middle. Roll this globe around the circle. When
you it returns to the initial location, does the globe look the same? It might
not!

The globe has state and that state changes as it is moved around. Your globe
changes direction because it is rotating on the floor. Particles don't have a
floor to roll on, but they can have internal state.

If your globe ends up the same place after a circle, then your globe is a
boson (phase factor 1). If the globe is upside down with Antarctica on top but
otherwise the same, it is a fermion (phase factor -1). If Florida or the
Ukraine or God-knows-what end up on top instead, it is an anyon.

Now, you're going to rig this up to a camera and do a time lapse (this is us
taking the integral). Every time the globe passes the camera, take a photo.
Stack all the photos after each other in a video. If you have a boson globe,
any picture looks like one facet of the globe. If you have a fermion globe,
you'll have a double-image, with "north" pointing two directions at once. If
you have an anyon globe, well, it could look like a bunch of different things,
and that's the point.

In physics, of course, we actually want to run this sort of integral on all
sorts of circles simultaneously, instead of just one, and some of these
circles are more probable than others, and everything gets all quantized
(which is _why_ the path integrals usually add up in a way that exclude anyons
instead of that being the default.)

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amluto
This is a surprising error from Quanta, which is otherwise pretty good at
getting the details right:

> Imagine two indistinguishable particles, like electrons. Take one, then loop
> it around the other so that it ends up back where it started. Nothing seems
> to have changed. And indeed, in the mathematical language of quantum
> mechanics, the two wave functions describing the initial and final states
> must be either equal or off by a factor of −1.

As explained later in the article, in three or more dimensions, if you loop a
particle around another and back where it started, you did nothing at all. The
factor is 1, not -1.

The actual thought experiment is that you _swap_ two indistinguishable
particles. The resulting state must be the same up to a scalar factor _k_ ,
since, if it wasn’t, you could tell the particles apart. Moreover, if you swap
two particles twice, you end up exactly where you started, so _k_ ^2=1.

In two dimensions, this all breaks down a bit, and it gets very complicated
very quickly.

~~~
MooMooMilkParty
I'm a little bit curious here. My understanding is that the wave function has
complex values and the "state" is described by the wave function times its
complex conjugate. Couldn't they have meant the factor of -1 to be on the
imaginary portion and then things still work out to have the same probability
distribution?

~~~
amluto
Suppose you have a very simple system with two indistinguishable particles in
one dimension and no spin or any other complications. The wavefunction is just
a complex valued function of two variables:

f(x0, x1)

where x0 and x1 are coordinates of the two particles.

The fact that the particles are indistinguishable means that no possible
experiment can tell them apart. After a bit of math, this means that f(x0, x1)
= k * f(x1, x0) where k is a constant. All wavefunctions are normalized, so
|k|^2=1, which constrains k to be on the unit circle.

But you can go farther: it's possible, at least in principle, to come up with
an experiment that measures k. So you start with the particles in their
initial state, do your experiment, and learn k. (Maybe you need to repeat it a
bunch of times -- no big deal.) Now you repeat the same experiment but start
with the two particles having their roles swapped. Since the particles are
indistinguishable, you had better measure the same value for k. This means:

f(x0, x1) = k * f(x1, x0)

and also

f(x1, x0) = k * f(x0, x1) = k^2 * f(x1, x0)

So k^2 = 1, and that rules out everything except k=1 or k=-1.

