Antiferromagnetism one or two types of Holstein-Primakoff Bosons? In these lecture notes (pg10) the author defines two types of Holstein-Primakoff bosons - one for each sublattice of the Bipartite lattice. E.g.:
$$ S^Z_{A_j}=S-a_j^\dagger a_j, \quad S_{Bl}^Z=-S +b_l^\dagger b_l$$
Contrary to this (Altand and Simon, 2010; pg78) define first flip the spin on the $B$ sublattice: 
$$S_z\rightarrow -S_Z, \quad S_y \rightarrow -S_y, \quad S_x\rightarrow S_x$$
then procced to only define one type of Holstein-Primakoff boson. I have also seen sources (not publicly available) both flip the spin and define two Holstein-Primakoff bosons. 
Are these two approaches equivalent? If so how and if not why not?
 A: What Ryan says is completely correct. Let me just add a bit more from a quantum number of point of view. The original Heisenberg model has translation symmetry $T_{x,y}$ (I will work in the 2D case here for notational convenience). This will be spontaneously broken by the ground state, but we will still have translation-symmetry of a two-site unit cell. So we have the residual translation symmetry $T_{1,1}$ and $T_{1,-1}$. Hence, we can use this to associate well-defined quantum numbers (i.e. momenta!) to our eigenstates. Another symmetry preserved by the Hamiltonian, is the $S^z_\textrm{tot} = \sum_{\boldsymbol n} S^z_{\boldsymbol n}$ (at least if I take the Neel state to order along the $z$-direction). One way of doing spin-wave theory is using these two (commuting!) operators to uniquely label the magnons. Note that we then get two bands, each living on half the Brillouin zone of the original periodic lattice, with the bands labeled by $S^z_\textrm{tot} = \pm1$. Also note that by symmetry, the two bands will exactly coincide (to convince oneself, note that $\prod_{\boldsymbol n} \exp{\left(i \pi S^x_{\boldsymbol n}\right)}$ is a symmetry of the Hamiltonian which switches the sign of $S^z_\textrm{tot}$). Moreover, these two bands are entirely decoupled (due to having distinct quantum numbers), so they cannot hybridize!
Alternatively, we can observe that even though the fundamental translation symmetries $T_{1,0}$ and $T_{0,1}$ are broken by the ground state, we can define the modified translation symmetries $\tilde T_{1,0} = \prod_{\boldsymbol n} \exp{\left(i \pi S^x_{\boldsymbol n}\right)} \; T_{1,0}$ and $\tilde T_{0,1} = \prod_{\boldsymbol n} \exp{\left(i \pi S^x_{\boldsymbol n}\right)} \; T_{0,1}$. These commute with the Hamiltonian, but moreover, the ground state preserves this symmetry! This is conceptually similar to working in the `local' frame. Anyway, we can use the quantum numbers of these two commuting transation symmetries to uniquely label the single-magnon states. We now get only one band, but this lives on the original Brillouin zone (which is twice as large as the aforementioned reduced zone, so the state counting is consistent). Note that these modified translation symmetries do not commute with $S^z_\textrm{tot}$, such that in this notation we can no longer consistent associate that quantum number to these states.
In conclusion: both descriptions are entirely equivalent, and it is a matter of choice. The latter has the benefit of only giving rise to one magnon band, but it has the downside of not being able to associate a well-defined notion of $\pm 1$ `spin flip' value to the magnons.
A: You can map the two pictures to each other by flipping the sign of the $b$ boson creation/annihilation operators. I was just staring at this in Affleck's notes from Les Houches '88, which explain things beautifully.
