2

We know that we do have composite particles (for example Atoms) made of fermions or bosons or mixture of them with fermionic or bosonic statistics. So why can't a gas of $2N$ fermions become a gas of $N$ bosons and condense to the lowest state at low temperature (just like what happens in superfluidity and superconductivity)?

richard
  • 4,194

1 Answers1

5

It can. This is exactly what happens when Helium-3 becomes superfluid. It's also what happens in superconductivity, which you mention in your question, when electrons combine into Cooper pairs.

Well, it's not exactly what you ask since neither liquid Helium-3 nor electrons are a gas. It's very unlikely a gas of fermions could pair up to form a gas of bosons because the energy required to disrupt the pair is typically very low, and requires temperatures at which the gas would liquify. The nearest to a gas would be a fermionic condensate, which was first made in 2004.

John Rennie
  • 355,118
  • But we never consider this possibility in statistical physics of fermionic gases even at zero temperature! – richard Oct 10 '13 at 06:53
  • 1
    @richard: Deborah Jin's experiment required temperatures of $10^{-8}$K, and even then it took the assistance of a magnetic field to increase the pairing energy. The error in your stat mech calculations by ignoring the process is pretty small :-) – John Rennie Oct 10 '13 at 06:55
  • But this process leads to a completely different ground state :-)! – richard Oct 10 '13 at 06:59
  • 1
    @richard In basic statistics/condensed matter courses you won't consider this possibility because often the pairing energy between electrons due to their common interaction with the lattice is neglected. This is still a useful approach because many properties of metals at room temperature can be inferred from the zero temperature behaviour, since the Fermi energy is so much larger than typical thermal energies. If you really want to study very low temperature behaviour of metals then you need to take into account the effect of the ionic lattice, leading to the BCS theory of superconductivity. – Mark Mitchison Oct 10 '13 at 10:26
  • @MarkMitchison, What if there was no lattice? What prevents pair formation? – richard Oct 10 '13 at 10:35
  • @richard Well, in the case of metals obviously that doesn't make any sense. In general, in order to see Bose condensation you need some attractive force between the fermions so that they form pairs. In cold matter experiments usually you need to use external magnetic fields to do this. I guess the fundamental reason why fermions don't just "pair up" without an attractive force is because it's entropically unfavourable. You need some additional force so the loss in energy due to pairing counteracts the loss in entropy. – Mark Mitchison Oct 10 '13 at 10:39
  • 1
    If the attractive interaction between the fermion is strong enough, there might be a bound-state, and the fermions form a bosonic molecule. These molecules can then Bose-Einstein condense. One can continuously go from the BCS regime (loosely bound Cooper pair) to the BEC regime (strongly bound molecules), this is the BCS-BEC cross-over. If the interaction is strictly repulsive, the system stays a Fermi liquid down to $T=0$. – Adam Oct 10 '13 at 13:29
  • As Mark said, in metals, the typical temperature at which the Fermi liquid description breaks down is of the order of the Kelvin, whereas the Fermi temperature is of the order of 10000K. So for most of the temperature range of interest, the Fermi gas description is correct (Fermi liquid is roughly a Fermi gas with some renormalized parameters). Interestingly, for cold atoms, $T_c/T_F\simeq 0.1$, and in some sense, cold atoms form some kind of high-$T_c$ superconductors. – Adam Oct 10 '13 at 13:34