34
$\begingroup$

What is the record for the largest number of bosons placed in a Bose-Einstein condensate?

What are the prospects for how high this might get in the future?

EDIT: These guys reported 20 million atoms in 2008. "Large atom number Bose-Einstein Condensate machines" (pdf).

EDIT 2: van der Straten's group reported 100-200 million.

https://doi.org/10.1063/1.2424439
https://doi.org/10.1103/PhysRevA.80.043606
https://doi.org/10.1103/PhysRevA.80.043605

But these guys don't seem to claim the largest BEC, so I wonder if there's larger.

EDIT 3: ~ 1 billion hydrogen atoms back in 2000 by Greytak et al.

https://doi.org/10.1016/S0921-4526(99)01415-5

Of course, hydrogen has only 1 nucleon while sodium (used by van der Straten et al.) has 23, so these are quite comparable in terms of mass.

$\endgroup$
9
  • 2
    $\begingroup$ Hmm, are BEC coherent states? Do they have definite particle number? Coherent states are eigenstates of creation operator, a linear combination of states with all numbers of excitations... $\endgroup$
    – innisfree
    Commented Nov 25, 2014 at 17:13
  • 4
    $\begingroup$ I am pretty sure they are in number eigenstates for massive bosons like atoms. That's not even super important though, as coherent States are strongly peaked around a certain number for large N. $\endgroup$ Commented Nov 25, 2014 at 17:26
  • 2
    $\begingroup$ OK, yes, $\delta n / n \propto 1/\sqrt{n}$ for a coherent state. What $n$ has been achieved... $\endgroup$
    – innisfree
    Commented Nov 25, 2014 at 17:34
  • 1
    $\begingroup$ It should be the highest one for atomic BEC, but if you also count for other quasi-particle BEC, it should be even higher. $\endgroup$
    – unsym
    Commented Nov 25, 2014 at 18:21
  • 2
    $\begingroup$ No, I want the number of boson that have been placed in an identical quantum state. I'd be interested to know how many helium atoms within a macroscopic sample of liquid helium are in the exact same state; it's clearly not most of them and I haven't seen a paper that makes a serious attempt to answer that question. $\endgroup$ Commented Jul 4, 2019 at 21:19

2 Answers 2

2
$\begingroup$

Superfluid helium 4 is the winner here, and by a vast margin. There's a body of literature from the 90s that attempts to estimate the fraction of helium atoms that are Bose condensed in superfluid helium 4. To quote this paper, "These works... estimate the condensate fraction to be ≈7% at saturated vapor pressure (SVP)." At a mass density of 125 g/L, this would give about $10^{24}$ condensed atoms per liter!

Note that while helium liquifies below 4K, you need to go below ~2K to get superfluid helium. The transition from liquid to superfluid is continuous, and so the amount of superfluid helium right at the transition temperature is basically zero. However below about 1K, virtually all helium atoms are in the superfluid state.

That an Avogadro's number or more of atoms can occupy the zero-momentum state of an arbitrarily large box is a pretty remarkable demonstration of Bose-Einstein condensation.

Honorable mentions

  • Regarding more conventional BECs, the largest I've come across is only slightly larger than the 1 billion you quote in your question; ­ $1.1 \pm 0.6 \times 10^9$ atoms in an earlier paper by the same authors. Most research these days works with only millions of atoms (and rarely hydrogen), so I'm not aware of any efforts to push above billions.

  • Superconductors have a lot in common with Bose Einstein condensates, since in this effect pairs of fermions also condense into a macroscopic 'supercurrent.' No two fermions can occupy the same state, but if you'll accept the similarities, then there are some theoretical proposals that entire stars can form a "BEC". See e.g. Bose-Einstein Condensate general relativistic stars, though I can't speak to the correctness of this paper.

Edit: Wait, really???

Some of the comments are understandably skeptical of the claim in this answer. As OP points out (essentially), a helium atom confined to a 1 mm $\times$ 1 mm $\times$ 1 mm box would have a ground state energy of $E=3\hbar^2 \pi^2/2mL^2 = 2.5\times 10^{-35}$ Joules, which corresponds to a temperature of

$$T_\text{ground state} = E/k_b = \frac{3\hbar^2\pi^2}{2m L^2 k_B} = 1.8 \times 10^{-12}~K$$

How then can it be that liquid helium atoms would occupy the ground state of an arbitrarily large box at 2 $K$ or more?

But the magic of Bose-Einstein condensation is that a dense collection of particles will start to pile up in the ground state of a potential well at much higher temperatures than would an isolated particle of the same kind. The critical temperature for a Bose-Einstein condensate in 3D is

$$T_c = \frac{\hbar^2 \rho_n^{2/3}}{m k_b} = \frac{\hbar^2 N^{2/3}}{mL^2 k_B}$$

where $\rho_n$ is the particle density and $N$ is the total number of particles. Notably these two expressions differ by the factor of $N^{2/3}$ in the numerator, which makes $T_c$ much much larger than $T_\text{ground state}$. If we plug in the number density for liquid helium (125 $kg/m^3 \rightarrow 1.8\times 10^{28}~ \text{particles}/m^3$), then we get a critical temperature of 0.8 $K$. This agrees quite well with the actual critical temperature of 2 $K$, despite the fact that we incorrectly treated the helium atoms as non-interacting particles.

$\endgroup$
6
  • $\begingroup$ As mentioned in one of my comments above, I'm interested in the number of bosons that have been places in the same, identical quantum state. The condensate fraction of 7% is not actually for the same literal quantum state, but rather represents something like the fraction of atoms in each characteristic region that are in the same local state, right? In other words, the condensed atoms are distributed over many individual modes? $\endgroup$ Commented Feb 9, 2023 at 20:11
  • $\begingroup$ What do you mean by 'characteristic region'? In the textbook example of a BEC, the modes are just the free-particle modes of the confining potential, which can be arbitrarily large. We then wind up with a large fraction in the unique motional ground state of that potential, which fills the whole region. My understanding of these papers is that you get the same result for the condensed fraction of the superfluid. Nothing else sets a length scale, so it really is 7% in the same, container-sized quantum state. $\endgroup$
    – user34722
    Commented Feb 10, 2023 at 19:20
  • 1
    $\begingroup$ If the real-life experiments were like the textbook example, then I'd agree with you that the 7% component would all be in a single mode. However, if you have a macroscopic amount of superfluid helium-4, say in a 1 mm square well, then the ground state temperature would be 10^-18 Kelvin. In contrast, the coldest lab temp ever is 10^-11 Kelvin (PRL) $\endgroup$ Commented Feb 11, 2023 at 0:29
  • 1
    $\begingroup$ What I think is going on is that some fluid properties of helium-4 (which I don't understand) sets a characteristic length scale much smaller than 1 mm, and that within each region of that size roughly 7% of the atoms are in the same mode of that size. Indeed, the "characteristic lambda transition temperature" for superfluid helium-4 is a relatively balmy 2.17 Kelvin. Need an expert to comment. $\endgroup$ Commented Feb 11, 2023 at 0:39
  • $\begingroup$ No superfluid helium expert comments here yet, but I updated the answer explain the flaw in your logic. Hopefully you'll agree now. $\endgroup$
    – user34722
    Commented Jun 2 at 5:29
-3
$\begingroup$

I would suspect the What is the largest number of bosons placed in a BEC would be the largest vat of liquid below boiling point 4He. I think 4He must be in the ground state to be a non boiling liquid and this is effectively a BEC.

$\endgroup$
1
  • 2
    $\begingroup$ The liquid phase of $^4\mathrm{He}$ is a distinct phase from BEC. $\endgroup$
    – wyphan
    Commented Dec 4, 2020 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.