How does Helium-4 exhibit bosonic properties despite being composed of fermions?

[Iamu]

The wikipedia article on Bose-Einstein condensates mentions that helium-4 is, or can be, a boson. It says that a condensate is made by putting many bosons, such as helium-4, into the lowest energy state.

How can an atom be a boson? I figured that an atom, composed of fermions, would have to be a fermion as well. How do the electrons and protons of helium-4 occupy the same space?

 

[f95toli]

Add the spins of all the particles in the nucleus.
He-4 2 has protons and 2 neutrons -> 4*1/2 =2 ->Integer spin meaning it is a boson
He-3 2 has protons and 1 neutron -> 3*1/2 =2 ->half-integer spin meaning it is a fermion

 

[Meir Achuz]

Boson is not defined as occupying the same space.
The wave function of two identical bosons is unchanged if the two are interchanged,
while the WF of two fermions changes sign.
If two He atoms are interchanged , the WF has four sign changes, which results in no overall sign change.

 

[michael879]

wow, I made an almost identical post to this and I thought this was mine. haha sorry for posting you can delete this if you want. Thanks for the answer the meir.

 

[akhmeteli]

I am afraid the situation is not so simple. The statement "Helium-4 is a boson" is just approximately correct (i.e. only correct for relatively low density of Helium-4, when wave functions of the fermions in the Helium-4 atoms essentially do not overlap). The commutation relations for operators of creation/annihilation of Helium-4 atoms are derived from the anticommutation relations for the operators of creation/annihilation of protons, neutrons, and electrons that are parts of Helium-4 atoms, and those commutation relations approximately coincide with the commutation relations for boson creation/annihilation operators in the limit of low density. See the details (for the example of deuterons) in the book by Lipkin called Quantum Mechanics, or something of the kind. So Iamu actually asked a good question.

 

[Meir Achuz]

I am afraid the situation is not so simple. The statement "Helium-4 is a boson" is just approximately correct (i.e. only correct for relatively low density of Helium-4, when wave functions of the fermions in the Helium-4 atoms essentially do not overlap). The commutation relations for operators of creation/annihilation of Helium-4 atoms are derived from the anticommutation relations for the operators of creation/annihilation of protons, neutrons, and electrons that are parts of Helium-4 atoms, and those commutation relations approximately coincide with the commutation relations for boson creation/annihilation operators in the limit of low density. See the details (for the example of deuterons) in the book by Lipkin called Quantum Mechanics, or something of the kind. So Iamu actually asked a good question.

I think you (and Lipkin?) are describing a dynamical complication for composite bosons, but they still satisfy the usual definition of "boson".

 

[akhmeteli]

I agree, if you use the definition of bosons as particles with integer spin. However, e.g., Dirac in his book "The principles of quantum mechanics" defines bosons as particles for which only symmetric states exist in nature (my quote may be a bit imprecise as my book is a translation from English). If the usual Bose commutation relationships are not satisfied, as is the case for composite particles, the particles are not bosons under this definition. On the one hand, arguing about choice of definitions does not make much sense, on the other hand, the usual conclusions about Bose-Einstein condensation and so on are not exactly correct for composite integer spin particles. For example, one cannot have too many Helium-4 atoms with limited energy within limited space, because they do consist of fermions. This important fact is not widely appreciated, and it seemed highly relevant to the question raised by the original poster. That's why I mentioned it.