# Behaviour of quantum spins

I am reading the Jordan-Wigner transformation in the book "Introduction to many-body physics" by Piers Coleman. When I read the introduction of this chapter, it is stated that:

Quantum spins are notoriously difficult objects to deal with in many-body physics, because they do not behave as canonical fermions or bosons. In one dimension, however, it turns out that spins with $$S=1/2$$ actually behave like fermions.

As far as I understand, in quantum mechanics, you can identify any spin half particle to be fermion and any spin one particle to be boson. Even if we have many of them, why do they not behave just like one of this (fermion/boson)?

• Quick comment: Commutation relationship between $spin-\frac{1}{2}$ systems on any lattice are fermionic like locally on sites but are bosonic like for spins on different sites. But for 1D case with nearest neighbour interactions, they can be mapped on to free fermions using Jordan-Wigner transformation. Otherwise, spins can be handled using Faddev-Fedotov fermion representation or Schwinger boson representations for example. – Sunyam Feb 25 '19 at 0:09

If I understand your question correctly, there seems to be a misunderstanding in your statement

As far as I understand, in quantum mechanics, you can identify any spin half particle to be fermion and any spin one particle to be boson. Even if we have many of them, why do they not behave just like one of this (fermion/boson)?

The truth is that many particle states of fermions can act like bosons. However, many particle states of bosons cannot act as fermions. The reason for this comes down to the fact that angular momentum of spins $$s_1, s_2$$ adds in such a way that the possible values of total spin $$s_{tot}$$ are

$$s_{tot} = |s_1 + s_2 |, | s_1 + s_2 -1|, \ldots |s_1 - s_2|.$$

That is, you can add two half integers to get an integer but you can't add two integers and get a half integer.

Example: Suppose we have two spin 1/2 particles. Then, the possible spin states are

the states with $$s_{tot} = 1$$ given by

\begin{align*} & |1, 1\rangle = | \uparrow \uparrow\rangle\\ & | 1, 0 \rangle = \frac{1}{\sqrt{2}}(| \uparrow\downarrow\rangle + |\downarrow\uparrow \rangle)\\ &|1, -1 \rangle = | \downarrow\downarrow\rangle \end{align*}

and the $$s_{tot}= 0$$ state

$$|0,0\rangle = \frac{1}{\sqrt{2}}(| \uparrow\downarrow\rangle - |\downarrow\uparrow \rangle).$$

That is, if you "zoomed out" you would just see either a spin 1 or spin 0 object behaving like a boson. This is the basic idea behind "cooper pairs".

• If we have many lattice sites and every site is identical, should we not be able to say with certainty that the behaviour between sites must be either fermions or bosons? For example, if each site consists of two electrons, then every site on the lattice must be boson. Or if each site consists of an odd number of electrons, they should behave as a fermion. – TangBear Feb 25 '19 at 8:29