Negative kinetic energy in Bose Hubbard mode?

[Questionasker]

Too familiar with Bose Hubbard model, but suddenly got stuck by a simple question: Why is the kinetic term (to be precise should be single body part) negative?

 

[fzero]

Too familiar with Bose Hubbard model, but suddenly got stuck by a simple question: Why is the kinetic term (to be precise should be single body part) negative?

The hopping term is not just the kinetic energy, but also includes the interaction of the particles with the lattice. If the interaction between the particles and the lattice sites is attractive, then often the energy spectrum will contain negative energy bound states (ignoring the other interactions, these are just the analogue of electrons being bound in atoms). Note that the hopping parameter (usually called $t$) must still be specified. The - sign is usually chosen in the expectation that $t>0$, but for a specific system, it might not be so.

 

[Questionasker]

The hopping term is not just the kinetic energy, but also includes the interaction of the particles with the lattice. If the interaction between the particles and the lattice sites is attractive, then often the energy spectrum will contain negative energy bound states (ignoring the other interactions, these are just the analogue of electrons being bound in atoms). Note that the hopping parameter (usually called $t$) must still be specified. The - sign is usually chosen in the expectation that $t>0$, but for a specific system, it might not be so.

Thank you for your response! Can you show one example, in which $t>0$?