# How to calculate a momentum space of a semi-finite lattice?

If we have a 2D square lattice of lattice constant a whose $$x$$ axis has only $$N_x$$ cells each with one atom and no with spin degeneracy, and periodic boundary conditions on $$y$$ with $$N_y$$ cells along it, how can we calculate the $$k$$-space of this lattice? I know if we don't have a PBC we can't have a Bloch wave. But sometimes, we refer to a 1D lattice with a finite number of sites as having a first Brillouin zone, there is a many uniform distribution of many sites and the number of k-space sites is equal that in real-space.

Consider for simplicity a particle hopping from a site to a nearest neighbor on a 1d lattice. If the lattice has a finite number of sites $$L$$, then, depending on the specific Hamiltonian that describes the system, we can have two situations:
• Periodic boundary conditions (PBC) which means that the nearest neighbors of site $$i=L$$ are $$i=L-1$$ and $$i=1$$ (the particle can hop from site $$L$$ to site $$1$$): you can imagine this as a ring geometry. In this case, the system is invariant under spatial translations, and hence the translation operator and the Hamiltonian have common eigenstates. These states (known as Bloch waves) are labeled by a number $$k$$ that can assume $$L$$ distinct values and is interpreted as the crystal momentum. The set of independent values of $$k$$ is the first Brillouin zone.
• Open boundary conditions (OBC) which means that the site $$i=L$$ has a single nearest neighbor $$i=L-1$$, and so does the site $$i=1$$, whose only neighbor is $$i=2$$. You can imagine this as an open chain. In this case, the system is no longer translation invariant and hence the eigenstates of the Hamiltonian are not eigenstates of the translation operator. The whole picture above breaks down in this case, and Brillouin zone is not properly defined here.
In a 2d square lattice with $$L_x \times L_y$$ sites, with PBC on the y direction and OBC on the x direction, a hybrid situation happens. There is actually translation invariance along $$y$$, which means that the eigenstates of the Hamiltonian are also eigenstates of the translation operator along the y direction and can be labeled by a number $$k_y$$ which takes $$L_y$$ distinct (independent) values. However there is no translation invariance along x. You can thus build a (sort of) Brillouin zone as the set of independent values of $$k_y$$, but this is not a 2d properly called Brillouin zone.