One for Karen…
I understand that the condensate can be formed because Helium nuclei are bosons, with integral spin. But Helium nuclei are made from protons and neutrons, which are in turn made from quarks, all of which have half-integral spin. How are they able to overlap, just because they are in Helium nuclei?
It is too clear, and so it is hard to see.
Wouldn’t a limit be governed by the Pauli exclusion priciple?
Well that’s what my question is. How can a BEC be formed from Helium nuclei, when the constituants of the nuclei are subject to the Pauli exclusion principle?
I thought that fermions can group together so that the overall spin is boson.
I have just the basest understanding of physics, so I don’t have a clue what you’re talking about. I just have to say that it sounds like a Monty Python bit. “Since the kluvions eclipsed the Curiesque fractal, it would seem to me that far-nodal shift only occurs in semi-doovardian liquids and hyperfreens.” 
Lumpy is on the right track here. It’s the net spin that matters, not the individual spin on the particles. Helium has two electrons, one with spin +1/2 and the other with spin -1/2, thus the net spin is 0. Hydrogen, on the other hand, has one electron with spin 1/2 (sign doesn’t matter here) and so Hydrogen acts like a fermion.
The spin of the particles in the nucleus doesn’t contribute.
– Sylence
I don’t have an evil side. Just a really, really apathetic one.
Here is what they are talking about. I won’t claim to be able to answer the question, but the phenomenon is a big deal in the physics community lately.
FIRST OBSERVATION OF A NEW QUANTUM GAS
<P ALIGN=“CENTER”>Tris</P>
Lumpy and Sylence are glossing over the question I’m asking.
The first two electrons added to a nucleus go into the lowest S orbital. They can both go in because they have opposite spin. The combined spin of the two is zero, however they don’t cancel each other, and allow a third electron to also enter the lowest S orbital. Even though the first two electrons have a combined spin 0, other electrons are still excluded from the orbital.
With a BEC, even though the nuclei are spin zero, their constituents are spin 1/2, and so must obey the Pauli exclusion principle in their own states. I’m wondering why this doesn’t come up in discussions of the BEC. Is it just because these BEC only have a few particles (hundreds or thousands, not 10^20 or so)? How many particles would have to be added before the fermion states were all filled, even though the nuclei were bosons? Or is there something else going on here?
It is too clear, and so it is hard to see.
The helium atoms must be bosons. The atom includes the two spin 1/2 protons, the one (He-3) or two (He-4) spin 1/2 neutrons, and the two spin 1/2 electrons. Angular momentum is a vector, so the possible spin angular momenta for the atom are the possible vector sums of the particle momenta. If the result is integer, the particle is a boson, if half-integer, a fermion. The easiest way to see the result for any given composite particle is to figure the maximum value of the spin, which is just the number of particles times 1/2.
Hence: He-3 has 5 particles, so max spin is 5/2, it is a fermion. He-4 has 6 particles, max spin is 6/2 = 3, it is a boson. An H atom has two (proton + electron), so max spinis 2/2 = 1, a boson.
The protons and neutrons in a nucleus do not overlap, because of the Pauli principle. Thus the nucleus has a finite size, roughly the size of all the nucleons gathered into a liquid lump. For the same reason neutron stars have a finite size.
There appears to be confusion about the nature of the condensate: it is not that the atoms are all in the same place at the sametime. This could not except under very weird conditions be an eigenstate of the Hamiltonian. Rather it is roughly that theparticles all have the same momentum (and it is zero) at the same time.