How can Helium 4 be a Boson and Helium 3 be a Fermion?

Calcifur

Hi there,

Just a quick question which I'm sure I'm over complicating in my head.

I've read that Helium 4 is a Boson because it has 0 spin and that Helium 3 is a Fermion because it has 1/2 spin. Is this right? I don't see how whole atoms can be associated with fundamental particle groups.

Why do each have those particular spins and why does that constitute them to be either Fermion or a Boson.

Also one other thing: Is it "integer spin" or "integral spin"? Do they mean the same thing because I keep seeing them being interchanged when talking about Fermions and Bosons?

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tom.stoer

In order to associate a spin to a composite system one has to couple the individual spins (and orbital angular momenta); in the case of He-n one couples n nucleons and n electrons. Assuming that higher spin i.e. higher angular momentum is associated with higher energy, the ground state of such a system is always the state with lowest angular momentum.

Lowest total spin for n=4 i.e. 8 fermions is S=0, lowest total spin for n=3 i.e. 6 fermions is S=1/2.

Calcifur

Hi Tom,

Thanks for your response but I think I need the response in simpler terms. So are you saying that due to He 3 lacking a neutron its total spin cannot cancel to zero?

Is it just a fundamental definition that Fermions have half spin and Boson's have integer spin?

And does that mean that ANY atom can be either a Fermion or a Boson depending on its resultant total spin?

I only say this because I always assumed that F's and B's were fundamental particles.(i.e. NOT associated with atoms but more their components.)

Furthermore I've also read that 2 Bosons can occupy the same space but Fermions cannot. Does this mean that two He 4 atoms can be in the same place at the same time?

mfb

Mentor
Thanks for your response but I think I need the response in simpler terms. So are you saying that due to He 3 lacking a neutron its total spin cannot cancel to zero?
Right.
Is it just a fundamental definition that Fermions have half spin and Boson's have integer spin?
The fundamental definition is the statistics they follow, but for all known particles (and in 3 spatial dimensions), the spin value can be used as definition, too.

And does that mean that ANY atom can be either a Fermion or a Boson depending on its resultant total spin?
Right, and it has to be one of those. You can even see this with Rubidium (with >100 components) at Bose-Einstein condensation, for example.

Furthermore I've also read that 2 Bosons can occupy the same space but Fermions cannot. Does this mean that two He 4 atoms can be in the same place at the same time?
This happens in Bose-Einstein condensates.

By the way: Protons and neutrons are not elementary particles either, and they have a well-defined spin of 1/2.

Calcifur

Hi mfb,

Thank you so much for answering each question individually. You have really helped me to get things straight in my head so I really appreciate it!

Calcifur

akhmeteli

Hi there,

I've read that Helium 4 is a Boson because it has 0 spin and that Helium 3 is a Fermion because it has 1/2 spin. Is this right? I don't see how whole atoms can be associated with fundamental particle groups.

Why do each have those particular spins and why does that constitute them to be either Fermion or a Boson.
Actually, there are two different and inequivalent definitions of, say, bosons. On the one hand, they are often defined as particles with integer spin, on the other hand, sometimes they are defined as particles for which only symmetric states exist in nature (see, e.g., Dirac's "Principles of Quantum Mechanics"). Composite particles, such as He 4 atoms, can be bosons under the first definition, but not under the second one.The commutation relations for operators of creation/annihilation of He 4 atoms are derived from the anticommutation relations for the operators of creation/annihilation of protons, neutrons, and electrons that are parts of the atoms (and an even more precise treatment would require starting with quarks for neutrons and protons), and those commutation relations approximately coincide with the commutation relations for boson creation/annihilation operators in the limit of low density. Please see the details (for the example of deuterons) in the book by Lipkin called "Quantum Mechanics. New Approaches to Selected Topics".

Calcifur

Hi akhmeteli,
Thank you for your response. I think its safe to assume that the text I was reading was referring to the first definition for Bosons then. Thanks for your references too. I will investigate further.

Bill_K

Without access to Lipkin's book, I'm wondering if the discussion in this paper on the commutation relations obeyed by "quasibosons" is equivalent. They do have a reference to Lipkin. (Just skip over the paper's speculative parts!)

akhmeteli

Without access to Lipkin's book, I'm wondering if the discussion in this paper on the commutation relations obeyed by "quasibosons" is equivalent. They do have a reference to Lipkin. (Just skip over the paper's speculative parts!)
Looks similar

Pseudo Epsilon

so are you saying that he-4 atoms can occupy the same space at the same time?

akhmeteli

so are you saying that he-4 atoms can occupy the same space at the same time?
I don't know if you're asking me, but the commutation relations for He 4 creation/annihilation operators, strictly speaking, differ from the canonical commutation relations for boson operators, so you cannot place an unlimited number of limited energy He 4 atoms in a limited space due to the Pauli principle for the constituent particles of He 4 atoms.

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