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I Bose-Einstein Condensate Properties

  1. Jan 3, 2017 #1
    So a Bose Einstein condensate is another state of matter at temperatures below those where a solid state exists.

    Bose_EInstien_Condensate.png

    The temperature is reduced so much, that the quantum wave states overlap and become one single object.

    So what are the properties of this object? Is it more rigid than solids?

    Also, does it stay this way after the atoms have become one, even after the temperature goes up? If so, how? Conceptually, raising the temp should be different in solid to liquid to gas since those are simply aggregates of atoms, whereas a BEC is completely different. Are the identities of each component atom erased once it becomes part of the Bose Einstein condenstate?

    What about the quarks and electrons, do they merge as well?
     
    Last edited: Jan 3, 2017
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  3. Jan 3, 2017 #2

    DrClaude

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    Definitely not. It is a superfluid.

    In reality, you never get a pure condensate at T = 0K. There is a condensed fraction, with thermal atoms also present.

    You can't simply put a BEC in the s-l-g hierarchy, because it is not a stable state of matter. At the temperatures at which condensation occurs, the stable state would be solid. This is why dilute gases need to be used, to limit 3-body collisions that would lead to the formation of a solid.

    You still have individual atoms, but they are fundamentally indistinguishable, and they occupy the same state and behave coherently. There is no "merging" taking place.
     
  4. Jan 3, 2017 #3
    Interesting. So I'm guessing that for a solid to exists, more pressure needs to be applied, and more pressure means more temperature.




    Makes sense. Why would it be classified as another state of matter if it's just a special case of a liquid?


    So they are indistinguishable from each other? So they are all behaving in the same exact way, basically in unison?
     
  5. Jan 4, 2017 #4

    DrClaude

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    It has nothing to do with pressure. When two identical atoms collide, because of conservation of energy, the only possible outcome is that the two atoms fly away from each other. For them to stick together, energy needs to be removed, which usually happens through a three-body collision, with the third atom carrying the extra energy.

    In a sense, this is similar to the process of regular condensation, such as in the atmosphere, where a nucleation seed is needed for a rain drop or a snow flake to form.


    Because it is a special case? A superfluid state is so different from a normal liquid state that we consider it as separate. It also comes about through a phase transition, and thus the passage from normal liquid to superfluid is not fundamentally different from the passage from liquid to gas. If you look at the phase diagram for helium, you will clearly see also the distinction between the normal liquid and superfluid phases.

    (Note that you could have said the same thing about liquids as compared to gases: liquids are just a special case of gases, since both are fluids. Above the critical point, you even lose the distinction between liquid and gas.)

    That's a simplified picture, but correct.
     
  6. Jan 12, 2017 #5
    Interesting. Does pressure help facilitate the process of this? I can imagine higher pressures help increase the likelihood of 3 body collisions.

    Also, are BECs more dense than solids?
     
  7. Jan 13, 2017 #6
    Still can't wrap my head around this. How exactly do we characterize entities made of fermions as bosons? If we have a bunch of cooper pairs, and they become a condensate, what exactly is the state of system look like if we return to the characterization of a bunch of electrons?
     
  8. Jan 13, 2017 #7

    DrClaude

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    Anything that increases 3-body collisions will make it more difficult to form a BEC. We usually don't think of it in terms of pressure, as all this is done with trapped atoms in vacuum chamber, as collisions with background gases need to be avoided (it heats up the atoms).

    No, they are very dilute. If you tried to make it dense, it would again form a solid.
     
  9. Jan 13, 2017 #8

    DrClaude

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    What is important is the property of a system as a whole. Take a hydrogen atom, made up of a proton and an electron, two fermions. The Pauli principle tells you that the wave function must change sign under exchange of two identical fermions, and when you are exchanging two hydrogen atoms, you are actually exchanging two electrons and two protons. So one exchange (the atoms) corresponds to two minus sings that cancel each other out, so the total wave function doesn't change sign under the exchange, hence the composite particle is a boson.

    I am not familiar enough with superconductivity and the BCS model to answer that.
     
  10. Jan 17, 2017 #9

    radium

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    In mean field BCS theory you have Hamiltonian which has pairing terms that do not conserve particle number. These tell you that there is some attraction causing electrons to pair. Most of the time you consider a bound state of two electrons with zero net momentum, but if the pair has a finite momentum, you get what is called a pair density wave (PDW). If you diagonalize the Hamiltonian by introducing new fermionic creation and annihilation operators, you will get a spectrum of noninteracting Bogoliubov quasiparticles (superpositions of electrons and holes) which will be gapped. Excitations in this spectrum correspond to breaking apart Cooper pairs. So in this system below the critical temperature you will have a fraction of the electrons bound in Cooper pairs reflected in the trial BCS wavefunction by the Bogoliubov coefficients. The gap is related to these coefficients and is determined self consistently from mean field equations. When the gap is zero you've reached the critical temperature and the fraction of Cooper pairs goes to zero.
     
  11. Feb 17, 2017 #10
    What does it mean to diagonalize the Hamiltonian in physical terms if you are doing an experiment? Thanks.
     
  12. Feb 17, 2017 #11

    DrClaude

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    How do you observe Cooper pairs experimentally?
     
  13. Feb 18, 2017 #12
    Please briefly tell me how that measurement equates to "diagonalizing the Hamiltonian" which is a mathematical operation. I'm interested in how mathematical terms equate to experimental reality. Are you just saying that any measurement effectively represents a diagonalized Hamiltonian? Thanks.

    BTW, I've always wondered. When a BEC is used to slow light, what happens to the momentum of the light? Thanks.
     
    Last edited: Feb 18, 2017
  14. Feb 19, 2017 #13

    radium

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    You can measure the energy gap and also the critical temperature (which of course is related to the energy gap). You can also measure things like the London penetration depth and the critical magnetic field strength at which superconductivity vanishes.

    Diagonalizing a Hamiltonian tells you the spectrum of low energy excitations.

    Right now I am talking about BCS type I superconductors which are much simpler than high Tc superconductors. In BCS theory, superconductivity is mediated by an electron phonon interaction causing a distortion in the lattice and a net attraction between electrons. This happens because of something called Cooper instability at low temperatures (of order the Debye frequency) which causes the electron phonon interaction (and hence the ability to form bound states) to win over electron repulsion. Another important thing to note is that the pairing here is a bound state of electrons (of zero momentum for BCS) and that electrons aren't actually right next to each other (like a ballroom dancer analogy). They could be quite far away.
     
  15. Feb 22, 2017 #14
    Thanks, Radium!

    Any thoughts anyone about my photon momentum question in a BEC?
     
  16. Feb 22, 2017 #15

    DrClaude

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    It gets transfered to the atoms.
     
  17. Feb 22, 2017 #16
    The real question is if the momentum of photons is enhanced or not as the speed of light slows to a crawl.
    Since in a vacuum the momentum is E/c, if c is 1m/s in a BEC is the momentum enhanced or not? The reason I'm asking is this, if one could exchange the power of a beam of photons to momentum of the BEC at a near 1/1 ratio, one has an incredible rocket engine that uses very low amounts of fuel. Thanks.
     
  18. Feb 23, 2017 #17

    DrDu

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    Why should a BEC be special as far as momentum is concerned? In every solid or liquid, momentum of light is transferred ot the medium and, if you are interested in acceleration by a laser, the maximum transfer of momentum is obtained in simple reflection from a mirror, as the momentum of the photons is reversed, so that 2 times the momentum of the photon is transferred to the mirror.
     
  19. Feb 23, 2017 #18
    It's not so simple as the nature of momentum in a solid has been debated for a century. The main issue is that in the BEC, c can be slowed to virtually nothing. If one takes the energy/momentum relationship literally it implies as c goes to say 1 m/s, the momentum transferred equates to the energy, not E/(3x10^8). If the BEC could be engineered to carry away much larger momentums than typical photon momentums, a small amount of fuel, the BEC atoms, might provide a reasonable thrust in space, comparable to at least an ion engine and perhaps better. For example, if one could get Newtons of thrust per KW of power for micrograms of BEC atoms, one has a great engine.
     
  20. Feb 23, 2017 #19

    DrDu

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    From what I remember, the low velocity of light is due to extreme deviations of dispersion from linearity, so your argument is moot.
     
  21. Feb 24, 2017 #20

    Dale

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    The debate had been resolved https://www.researchgate.net/profile/Brandon_Kemp/publication/234850066_Resolution_of_the_Abraham-Minkowski_debate_Implications_for_the_electromagnetic_wave_theory_of_light_in_matter/links/0a85e536a487b8673b000000.pdf [Broken]

    For some reason, the controversy seems more appealing than the resolution, leaving the false impression that it is unresolved.

    You don't need BEC for that. Ordinary chemical rockets carry away larger momentum than typical photon momentum. Energy and momentum conservation still apply, but in principle you can analyze the trade off flexibly.
     
    Last edited by a moderator: May 8, 2017
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