Fock space with mixed anti-commutation/commutation relations?

Question

Let's say we have two modes, with the following labeling of occupation number states:

$ \lvert \Psi \rangle = \begin{pmatrix} 0,0 \\ 0,1 \\ 1,0 \\ 1,1 \end{pmatrix} $

An example of (what I assume to be) fermionic creation operators for the two modes is

$\hat a_1^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \quad \hat a_2^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$

These operators obey full anti-commutation relations.

$\{\hat a_1,\hat a_1^\dagger\} = \{\hat a_2,\hat a_2^\dagger\} = 1$

$a^\dagger_1 a^\dagger_1 = a^\dagger_2 a^\dagger_2 = 0$

$\{\hat a_1,\hat a_2^\dagger\} = \{\hat a_1,\hat a_2\} = 0$

If we don't include the ($-$) sign, then operators corresponding to the same mode still anti-commute, but those corresponding to different modes commute.

$\hat b_1^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \quad \hat b_2^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$

$\{\hat b_1,\hat b_1^\dagger\} = \{\hat b_2,\hat b_2^\dagger\} = 1$

$b^\dagger_1 b^\dagger_1 = b^\dagger_2 b^\dagger_2 = 0$

$[\hat b_1^\dagger,\hat b_2^\dagger] = [\hat b_1^\dagger,\hat b_2] = 0$

It looks like we started constructing a boson Fock space, but only included states for which the occupation numbers are 0 or 1. Is there some reason these operators aren't suitable, other than the observation that all elementary particles are either fermions or bosons? Are there any quasi-particles in condensed matter physics that behave like this?

Answer 1

The operators $b_i$ defined by the OP correspond to the algebra of hardcore bosons, that is, bosons that cannot be put at the same place.

Hardcore bosons correspond to the limit of infinite interaction ($U\to\infty$) of the Bose-Hubbard model $$ H=-t\sum_{\langle i,j\rangle}b^\dagger_i b_j-\mu\sum_i n_i+\frac U2 \sum_i n_i(n_i-1) , $$ with $n_i=b^\dagger_i b_i$.

Hardcore bosons are also related to $\frac12$-spins, with the mapping $b=\sigma^-$, $b^\dagger=\sigma^+$ and $b^\dagger b-\frac12=\sigma^z$. In particular, the Bose-Hubbard model at infinite interaction can be mapped onto the XY model in transverse field (up to a constant) $$ H_{XY}=-J\sum_{\langle i,j\rangle } (\sigma^x_i\sigma^x_j+\sigma^y_i\sigma^y_j)-h\sum_i \sigma^z_i. $$