
A faster way to make Bose-Einstein condensates - CarolineW
http://news.mit.edu/2017/faster-way-make-bose-einstein-condensates-1123
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mudil
Here's a question from a non-physicist. As Bose-Einstein condensate is formed,
it starts to occupy larger and larger volume, because atoms slow down, and
according to Heisenberg uncertainty principle, if momentum is slow, the
location expands. So, theoretically, as one atom approaches the standstill, it
can expand to occupy the entire universe?

~~~
mgibbs63
You can't actually suck all the energy out - the best you can do is get the
atom into its ground state, which is still non-zero.

The de Broglie wavelength of the atom is (h/p), where h is Planck's constant,
and p is the atom's momentum. This is the wavelength of the atom's probability
wave, so at the minimum value of p, the atom has some 'fixed maximum size', if
you want to call it that (but size isn't really an accurate descriptor, more
like 'the region in which you might find the atom').

The Bose-Einstein condensate is defined as the state where the de Broglie
wavelength for atoms in a cloud is larger than the spacing between atoms - the
probability waves overlap, and it is no longer possible to distinguish one
from another.

~~~
pishpash
It's not just that, right? You make it sound the same as hitting the
diffraction limit in optics: [http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/imgpho/ray...](http://hyperphysics.phy-
astr.gsu.edu/hbase/phyopt/imgpho/rayc.gif),

but at the Raleigh criterion and under, the waveforms can still be quite
different depending on the source distance, and furthermore you can definitely
tell the difference of those waveforms from that of a single source.

What I've read about Bose-Einstein condensates seems to imply that in the
condensate form, the probability waves not only become unresolvable but also
synchronized in phase, AND the energy behavior of the aggregate is markedly
different since they "all" (or at least according to Bose-Einstein statistics)
occupy the same quantum state:
[https://www.youtube.com/watch?v=shdLjIkRaS8](https://www.youtube.com/watch?v=shdLjIkRaS8)

Is the transition from Maxwell-Boltzmann statistics to Bose-Einstein
statistics a sharp transition or not? In other words, are condensates a
descriptive marker or a suddenly different state?

~~~
VorticesRcool
I am not sure what you mean by the probability waves (amplitude?) becomes
unresolvable for the condensate, but it is true that the condensate will have
a continuous phase with integer windings of 2 pi around vortices etc. The
wavefunctions of the atoms overlap below the critical temperature and you get
Bose-Einstein condensation for atoms with integer spin.

But those atoms themselves are still made up of electrons, protons and
neutrons which have half integer spin, and at even smaller scales of quarks
and gluons. If you probe the condensate with high enough frequency without
thermalizing it you would be able to resolve those details, but at the
macroscopic level of the condensate those details are not resolvable (is that
what you were getting at?).

When you cool an atomic cloud below a critical temperature there will be a
condensate fraction and non condensate fraction. If you are just looking at
the condensate fraction then you can use Bose-Einstein statistics.

At zero temperature with 100% of the atomic cloud as condensate ( in reality
we can never get to zero temperature, but we can get pretty damn close), the
Gross–Pitaevskii equation (
[https://en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equat...](https://en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equation)
) is a good model for the dynamics of the condensate. If you want to go above
zero temperature and include interaction with the thermal cloud (the non-
condensate fraction), then you can use the SPGPE, the stochastic projected
Gross–Pitaevskii equation.

~~~
pishpash
I mean "unresolvable" in the same sense that two nearby point sources through
a diffraction-limited lens are unresolvable. If you watch the video, it
appears as if there is only one wavefunction for the BEC of many particles
(presuming that is what the video is depicting). Is that in any way an
accurate depiction of what's going on?

~~~
gji
Indeed, that's accurate (apart from finite temperature and interaction
effects). A BEC is in some ways pretty similar to a laser, where all the
photons are in the same state, even quantum mechanically.

But it's important to distinguish between a BEC and a superfluid. Superfluids
are the substances with strange collective properties, and these properties
come from being cold, bosonic, and interacting. BECs with very low inter-
particle interactions do not behave like superfluids, but will exhibit e.g.
interference (just like a laser, which is kinda like a non-interacting BEC).

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avsteele
Here's a link to the actual article in Science
[http://science.sciencemag.org/content/358/6366/1078](http://science.sciencemag.org/content/358/6366/1078)

~~~
rubidium
And here's the arxiv version (no paywall):
[https://arxiv.org/abs/1705.03421](https://arxiv.org/abs/1705.03421)

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Triesault
There is a good PBS Space Time episode discussing Absolute Cold. This portion
of the episode discusses the Bose-Einstein condensate:
[https://www.youtube.com/watch?v=OvgZqGxF3eo&feature=youtu.be...](https://www.youtube.com/watch?v=OvgZqGxF3eo&feature=youtu.be&t=2m34s)

