5 ms·

A faster way to make Bose-Einstein condensates

101 points by CarolineW 9y ago

mudil 9y ago
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 9y ago
  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.
  - mudil 9y ago
    Thanks!
  - pishpash 9y ago
    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/rayc.gif http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/ray..., 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 9y ago
      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_equation https://en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equat... ) 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 9y ago
        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 9y ago
        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).
- archgoon 9y ago
  This is a good question and should not have been downvoted.
- mickronome 9y ago
  After a bit of reading, this is how it appears to work: For a coherent state the uncertainty relation is certainly affecting the system behaviour. In this case, the standard deviation of the position times the devotion in speed/moment is h-bar/2 However, the effect in this case is evident in a Heisenberg limited momentum, as the spatial dimension/position of the condensate becomes well defined by the physical dimensions of the trap. Sources: * Wikipedia on Heisenberg uncertainty for Coherent states ( BECs exhibit a close to ideal coherent state ) * Observing Properties of an Interacting Homogeneous Bose--Einstein Condensate: Heisenberg-Limited Momentum Spread, Interaction Energy and Free-Expansion Dynamics https://arxiv.org/abs/1403.7081 https://arxiv.org/abs/1403.7081 * Guesstimations and extrapolations.
- k__ 9y ago
  I think, yes. If you think of a particle as waves on fields this makes sense. If the frequency of a wave goes down their wavelength goes up, as the frequency approaches zero, the wavelength goes to infinity (or at least to the size of the fields they is on)
- lippel82 9y ago
  It was elaborated upon in other comments, but to drive home the point of extending an atom over the whole universe: The quantum mechanical ground state of an atom is actually determined by the size of the box you put it in. Any finite size box leads to a finite ground state energy and indeed, only by increasing the box size to infinity can you reduce the ground state energy to zero.
avsteele 9y ago
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 9y ago
  And here's the arxiv version (no paywall): https://arxiv.org/abs/1705.03421 https://arxiv.org/abs/1705.03421
Triesault 9y ago
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&t=2m34s https://www.youtube.com/watch?v=OvgZqGxF3eo&feature=youtu.be...

101 points by CarolineW 9y ago

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 9y ago
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.
- mudil 9y ago
  Thanks!
- pishpash 9y ago
  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/rayc.gif http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/ray..., 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 9y ago
    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_equation https://en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equat... ) 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 9y ago
      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 9y ago
        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).
archgoon 9y ago
This is a good question and should not have been downvoted.
mickronome 9y ago
After a bit of reading, this is how it appears to work: For a coherent state the uncertainty relation is certainly affecting the system behaviour. In this case, the standard deviation of the position times the devotion in speed/moment is h-bar/2 However, the effect in this case is evident in a Heisenberg limited momentum, as the spatial dimension/position of the condensate becomes well defined by the physical dimensions of the trap. Sources: * Wikipedia on Heisenberg uncertainty for Coherent states ( BECs exhibit a close to ideal coherent state ) * Observing Properties of an Interacting Homogeneous Bose--Einstein Condensate: Heisenberg-Limited Momentum Spread, Interaction Energy and Free-Expansion Dynamics https://arxiv.org/abs/1403.7081 https://arxiv.org/abs/1403.7081 * Guesstimations and extrapolations.
k__ 9y ago
I think, yes. If you think of a particle as waves on fields this makes sense. If the frequency of a wave goes down their wavelength goes up, as the frequency approaches zero, the wavelength goes to infinity (or at least to the size of the fields they is on)
lippel82 9y ago
It was elaborated upon in other comments, but to drive home the point of extending an atom over the whole universe: The quantum mechanical ground state of an atom is actually determined by the size of the box you put it in. Any finite size box leads to a finite ground state energy and indeed, only by increasing the box size to infinity can you reduce the ground state energy to zero.

avsteele 9y ago

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 9y ago
And here's the arxiv version (no paywall): https://arxiv.org/abs/1705.03421 https://arxiv.org/abs/1705.03421

Triesault 9y ago

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&t=2m34s https://www.youtube.com/watch?v=OvgZqGxF3eo&feature=youtu.be...