# Integral spin and half integral spin

I am reading a book (Laudau and Lifshitz, Vol. 4, page 94) and it derived why spin-0 should obey Bose quantization and spin-1/2 should obey Fermi Quantization.

Then it says, all integral spin particles should obey Bose quantization and half-integral spin particles should obey Fermi quantization. The reason is integral spin is regarded as "composed" of even number of spin-1/2 particles while half-integral spin is "composed" of odd number of spin-1/2 particles. This is where I don't understand. Why "regarded as composed of even number of spin-1/2 particles" necessitates the use of Bose quantization?

• – Qmechanic Feb 4 '17 at 22:04

An element of answer is that the tensor product of an even number of copies of the fundamental irrep of $su(2)$, which describes spin-$1/2$ particles (thus fermions), decomposes into a sum of irreps containing only integer values of $j$. For instance, $1/2\otimes 1/2=1\oplus 0$ etc.
Since the fermionic wave function picks up a sign under transposition, the wave function for an even number -say $2n$ - of fermions would pick up a sign $(-1)^{2n}=+1$ under any transposition.
I don't know what Landau had in mind and I don't have that book handy right now, but when it comes to permutations it is possible to think of a system composed of an even number of fermion as bosonic, somewhat like above. He$^4$ is an example of a system of 4 fermions that behaves like a boson. Of course there are fundamental (non-composite) particles that are bosons, so it's suspicious to think of those as a kind of composite of an even number of fermions, but from a representation theory perspective this may be a useful way of getting insight in this issue.