## Superfluid Quasicrystals

Condensed matter physics is so cool! Bounce 4 laser beams off mirrors to make an interference pattern with 8-fold symmetry. Put a Bose–Einstein condensate of potassium atoms into this “optical lattice” and you get a superfluid quasicrystal!

You see, no periodic pattern in the plane can have 8-fold symmetry, so the interference pattern of the light is ‘quasiperiodic’: it never repeats itself, thought it comes arbitrarily close, sort of like this pattern drawn by Greg Egan:

In the Bose–Einstein condensate all the particles have the same wavefunction, and the wavefunction itself, influenced by the light, also becomes quasiperiodic.

But that’s not all! As you increase the intensity of the lasers, the Bose-Einstein condensate suddenly collapses from a quasicrystal to a ‘localized’ state where all the atoms sit in the same place!

Below the gray curve is the potential V formed by the lasers, while the blue curve is the absolute value squared of the wavefunction of the Bose–Einstein condensate, |ψ0|2.

At top the lasers are off so V is zero and |ψ0|2 is constant. In the middle the lasers are on, but not too bright, so V and |ψ0| is quasiperiodic. At the bottom the lasers are brighter, so V is quasiperiodic and larger, and |ψ0|2 is localized.

It’s well known that when a crystal is sufficiently disordered, its electrons may localize: instead of having spread-out wavefunctions, they get trapped in specific regions as shown here:

This phenomenon is called ‘Anderson localization’, and it was discovered around 1958.

But when a Bose-Einstein condensate localizes, all the atoms get trapped in the same place—because they’re all in exactly the same state! This phenomenon was discovered experimentally at the University of Cambridge very recently:

• Matteo Sbroscia, Konrad Viebahn, Edward Carter, Jr-Chiun Yu, Alexander Gaunt and Ulrich Schneider, Observing localisation in a 2D quasicrystalline optical lattice.

The evidence for it is somewhat indirect, so I’m sure people will continue to study it. Localization of a Bose–Einstein condensate in a one-dimensional quasiperiodic potential was seen much earlier, in 2008:

• Giacomo Roati, Chiara D’Errico, Leonardo Fallani, Marco Fattori, Chiara Fort, Matteo Zaccanti, Giovanni Modugno, Michele Modugno and Massimo Inguscio, Anderson localization of a non-interacting Bose–Einstein condensate, Nature 453 (2008), 895–898.

The holy grail, a ‘Bose glass’, remains to be seen. It’s a Bose-Einstein condensate that’s also a glass: its wavefunctions is disordered rather than periodic or quasiperiodic.

New forms of matter with strange properties—I love ’em!

For more popularizations of these ideas, see:

• Julia C. Keller, Researchers create new form of matter—supersolid is crystalline and superfluid at the same time, Phys.org, 3 March 2018.

• University of Texas at Dallas, Solid research leads physicists to propose new state of matter, Phys.org, 9 April 2018.

The latter says “The term ‘superfluid quasicrystal’ sounds like something a comic-book villain might use to carry out his dastardly plans.”

### 14 Responses to Superfluid Quasicrystals

1. Ishi Crew says:

https://en.wikipedia.org/wiki/Herbert_Frohlich also wrote on bose-einstein condensates. (his wife was an artist philosopher and part of a group called ‘collective states’).

2. Here’s one of the things that are puzzling me: Potassium atoms are bosons, and you can condense them so that they occupy one tight location. But they are composed of fermions. Does it mean that the component fermions are also squeezed? But fermions can’t be squeezed. Do they lose their individualism when we compose them?

• John Baez says:

Potassium atoms are bosons, and you can condense them so that they occupy one tight location.

Tight… but not too tight!

This used to worry me. It also applies to helium-4 and Cooper pairs (pairs of electrons in a superconductor).

A simple calculation shows what’s going on. Suppose you have a fermion whose state lies in the Hilbert space spanned by vectors

$e_1, \dots, e_n$

Their creation operators obey

$a^\dagger_i a^\dagger_j = - a^\dagger_j a^\dagger_i$

for $1 \le i, j \le n.$

Imagine that these are position eigenstates of a fermion. Suppose fermions of this sort like to stick together in pairs for some reason. They can’t lie directly on top of each other, but they can sit side by side.

These states are ‘composite bosons’. We can take the operator for creating one of these composite bosons to be

$A_i^\dagger = a_i^\dagger a_{i+1}^\dagger$

It’s easy to check that these creation operators commute, as you’d expect for true bosons:

$A_i^\dagger A_j^\dagger = A_j^\dagger A_i^\dagger$

However, you can’t create two of these composite bosons in the same position state, since you can also check that

$\displaystyle{ \left(A_i^\dagger\right)^2 = 0 }$

You can put two of these composite bosons in the same ‘maximally spread out state’ where they have the same amplitude to have any position, since

$\displaystyle{ \left(\sum_i A_i^\dagger \right)^2 \ne 0 }$

This would describe two composite bosons in such a smeared-out state. However, if you try to put n or more of these bosons in this state, it’s impossible since

$\displaystyle{ \left(\sum_i A_i^\dagger \right)^n = 0 }$

Further simple computations along these lines show that composite bosons act a lot like true bosons unless you try to pack too many of them too tight.

In a Cooper pair, by the way, the two electrons are not next to each other in position space. They’re next to each other in momentum space! This allows electrons to ‘stick together’ despite the fact that like charges repel.

3. So a composite of two fermions is a boson only approximately. Pack them too tightly and their fermionic nature starts showing. I wonder if this has been seen in an experiment.

I’ve found this paper that sheds some light on this problem: https://arxiv.org/abs/1310.8488 .

Also, I think our idea that a bound state is just a composition of individual states plus interaction is pretty naive.

• John Baez says:

I had a vague impression that this effect has been seen experimentally, but I’m not sure.

I’m not sure what you’re saying is naive. My model of a bound state as an antisymmetrized tensor product of two single-particle position eigenstates

$e_i \otimes e_{i+1} - e_{i+1} \otimes e_i$

is extremely naive; it’s the simplest model that illustrates what’s going on.

• My understanding is that this only works for non-interacting particles. The S-matrix gives you amplitudes between initial and final states, which are supposed to be asymptotically free. Bound states are S-matrix poles. I admit, this has always been confusing to me, because what’s a “free proton”? If you turn off strong interactions, doesn’t it fall apart?

• John Baez says:

This post was about condensed matter physics so I thought we were talking about composite bosons that can be understood well using nonrelativistic quantum mechanics, like Cooper pairs or helium-4 atoms (thought of as a bound state of a bosonic helium-4 nucleus and two electrons).

Everything gets more complicated in quantum field theory, and I don’t really want to think about bound states in quantum field theory today! It’s been a productive day so far.

4. Blake Stacey says:

them: you can’t just put science words together and call it science

physics: superfluid quasicrystal

them: no

physics: also it’s quantum

them: stop

physics: we made it with lasers

• John Baez says:

Yes, all this fits in perfectly with that quote:

The term ‘superfluid quasicrystal’ sounds like something a comic-book villain might use to carry out his dastardly plans.

5. Wolfgang says:

I wonder, since this localization seems to happen in the 2D plane (there are no 3D octagonal quasicrystals), do the trapped atoms at some point form a kind of jet in the direction perpendicular to the quasicrystalline plane, and can’t this be used to make a laser kind of machine for these atoms?

• John Baez says:

The potassium atoms lie in a 2d plane. They don’t form a “jet”, they’re just attracted to points where the potential is lowest. This potential is created by the standing waves formed by intersecting laser beams. That’s how an optical lattice works.

But while an ordinary optical lattice creates a periodic potential, so that the atoms arrange themselves in a kind of crystalline pattern, here the beams are cleverly arranged so that the atoms form a quasicrystalline pattern — that is, a pattern that never quite repeats itself. To make things even more exciting, the atoms are in a Bose–Einstein condensate, meaning that every atom is doing exactly the same thing as every other.

If you turn up the strength of the laser beams, the condensate ‘localizes’, meaning that the wavefunction of all the atoms becomes sharply peaked near one point.

All this is happening at extremely low temperatures, with merely 300,000 atoms, and it lasts for much less than a second. It’s absolutely useless for any sort of practical device, at least currently.

• Wolfgang says:

Yes, I think I understood this part, but there must be some sort of confinement laser field that prevents the atoms from moving outwards in a direction perpendicular to the quasiperiodic plane? I mean, if they might have a fifth laser in the z direction creating a periodic potential this could already explain it. Then I wonder, what happens if you first localize the atoms and then switch of this perpendicular laser?

• John Baez says:

There’s no fifth laser beam. I believe the atoms are attracted to places where the light is brightest. Since the light is brightest near a plane, the atoms stay near that plane.

The Wikipedia article on optical lattices says that neutral atoms are trapped in the light thanks to the Stark effect: an effect where an electric field changes the energies of atomic orbitals. The Wikipedia article doesn’t condescend to say whether the atoms’ energies are lower in the bright regions or the dim regions. But I’m guessing they must be lower in the bright regions. Otherwise they wouldn’t stay in the laser beams!

• Wolfgang says:

I see, ok… interesting that they are attracted. I thought that one has to force them into the plane first (or independently from the remaining laser experiment). Thanks for the explanation.

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