Superfluidity, helium-3 and helium-4

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SUMMARY

Helium-3 and helium-4 exhibit different lambda points due to their distinct atomic properties; helium-4 is a boson, allowing it to directly form a Bose-Einstein condensate, while helium-3, a fermion, requires the formation of Cooper pairs to achieve superfluidity. The superfluid transition temperature for helium-4 is significantly higher than that of helium-3, with the latter being sensitive to atomic interactions and occurring around 1 mK. The mechanisms of superfluidity in these isotopes are fundamentally different, highlighting the unique characteristics of their atomic structures and interactions.

PREREQUISITES
  • Understanding of Bose-Einstein condensation
  • Knowledge of fermions and bosons
  • Familiarity with Cooper pairs and the Pauli exclusion principle
  • Basic concepts of quasiparticles and dispersion relations
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  • Research the properties of Bose-Einstein condensates in helium-4
  • Study the formation and implications of Cooper pairs in helium-3
  • Explore the role of quasiparticles in superfluidity
  • Investigate Landau's theory of superfluidity and Galilean invariance
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Physicists, researchers in condensed matter physics, and students studying superfluidity and quantum mechanics will benefit from this discussion.

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Why do helium-3 and helium-4 have different lambda points?
 
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Helium 3 also has drastically different density and boiling point.

The big reason for difference in lambda points is that while a helium 4 atom is a boson, and they can all go to ground state, helium 3 atoms are fermions. Which means they do not all fit to ground state.

Furthermore, superfluidity is something that can only happen to Bose condensate of bosons. Since helium 4 atoms already are bosons, they can directly form Bose condensate. Whereas fermions can only form Bose condensate if they somehow are turned into bosons by forming Cooper pairs. Quite different mechanism.
 


So you are asking why the two helium isotopes have different superfluid transition temperatures (only the T_c of ^4He is called the lambda point). There is actually quite a large difference in the temperatures, three orders of magnitude. Fundamentally the reason is that ^4He is a boson and ^3He is a fermion. As a boson, ^4He can directly form a "Bose-Einstein condensate", which is the superfluid state. One could say that the wave functions of the helium atoms begin to overlap and they lose their identity. The fermionic ^3He, on the other hand, must form pairs of atoms, called Cooper pairs to form the condensate. This is because fermions do not like to be too close to each other due to the Pauli exclusion principle. This fermion transition temperature is very sensitive to the interactions between the atoms, and for ^3He is quite low being of the order of 1 mK.

So the short answer is that the mechanisms through which the two isotopes form the superfluid state are very different.
 


j.gal said:
Why do helium-3 and helium-4 have different lambda points?
It is more wonderful that both heliums can be superfluid!
It means that some structures in helium3 and helium4 are similar!

Minich explained me, that such structure is zero phonon modes in he4. Those phonon zero modes in he4 are similar to he3 atoms and are similar to fermion-like particles. The number of those modes in he4 is 3*(number of he4 atoms).

Superfluidity is due to dispersion relations of quasiparticles in helium. At some temperatures some number of quasiparticles become diode-like waves (superfluid modes) near Fermi surface.
 


M@2 said:
Minich explained me, that such structure is zero phonon modes in he4. Those phonon zero modes in he4 are similar to he3 atoms and are similar to fermion-like particles. The number of those modes in he4 is 3*(number of he4 atoms).

Yes, while necessary, Bose Einstein condensation is not sufficient to get superfluid behaviour. The linear dispersion of the phonons at low k values allows for superfluidity and determines the maximum velocity. This was nicely shown by Landau using Galilean invariance.
 

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