Consider the hopping terms for a lattice Hamiltonian with bosons:

$$\sum_{i,j\neq i} t_{i,j} b^\dagger_i b_j$$

when writing this term in the basis of Hubbard operators $$X^{a,b}_i =| a,i \rangle \langle b,i | $$ becomes:

$$\sum_{i,j\neq i} t_{i,j} \sum_{p} \gamma_p \gamma_q X^{p+1,p}_i X^{q,q+1}_j$$ being $$\gamma_q = \sqrt{q+1}$$

My question refers to the Fourier transform of this term. I would expect that due to the translation symmetry of the hopping, this term can be diagonalized in momentum space and therefore it should be possible to contract the two Hubbard operators to:

$$\sum_k t_k X^{p+1,p}_k X^{q,q+1}_k = \sum_k t_k X^{p+1,p+1}_k$$

but I see that for example in ref. "Cluster Perturbation Theory in Hubbard Model Exactly Taking into Account the ShortRange Magnetic Order in 2 × 2 Cluster" with DOI: 10.1134/S1063776110100146, Eq. 13 does not simplify the product of these two Hubbard operators. Could someone explain to me why?

Thank you in advance