Do periodic boundary conditions affect the k-space Hamiltonian for the Hubbard model? In real space, the Fermi-Hubbard model can be written as:
$$ H = -t \sum_{\langle i,j \rangle\sigma} (c^\dagger_{i\sigma} c_{j\sigma} + c.c.) + U \sum_in_{i\uparrow}n_{i\downarrow}$$
The only difference between having periodic boundary conditions and not having them is that the nearest-neighbor pair, $\langle i, j \rangle$ will also include the pair $(N, 1)$ for periodic boundary conditions.
If we apply a transform to move into k-space, the real space summation really comes into effect when simplifying the exponential (in the form: $ e^{i(k-k')\cdot r_i}$). (I'm assuming that the sites are equally spaced.) I'm wondering if the k-space Hamiltonian is the same, whether or not we have periodic boundary conditions in real space. It seems that the only difference is that our sum over all real sites, $i$, is restricted from 1 to $N$ in the periodic boundary condition case (in order to include $(N, 1)$ at the end, but goes from 1 to $N-1$ in the non-periodic boundary condition case.
The delta function comes (for example in the kinetic term) from:
$$ -\frac{t}{N^2}\sum_{i \sigma}\sum_{kk'}e^{-i(k-k')r_i}e^{ik'\delta}a^\dagger_{k\sigma}a_{k'\sigma} + ...$$
where the $a$'s are my k-space creation/annihilation operators. I'm assuming $\sum_ie^{-i(k-k')r_i} = N\delta_{kk'}$ regardless of whether we sum $i$ to $N$, or $i$ to $N-1$.
Is this true? Is there also a way to physically think about this, to justify whether the k-space Hamiltonian should change?
 A: Short answer: yes and no.
Please keep in mind that $\sum_i e^{-i (\vec{k}-\vec{k}\prime) \cdot \vec{r}_i} = N \delta_{\vec{k} \vec{k}\prime}$ is not always valid in periodic boundary conditions, i.e., it implies that $\langle \vec{k} | i \rangle = e^{i \vec{k} \cdot \vec{r}_i}$, which is not true, since the electrons described by the Hubbard model are definitely not free electrons that can be described by plane waves!
Instead, you need to involve the Bloch theorem $\psi_i(\vec{r}+\vec{G})=e^{i \vec{k} \cdot \vec{G}} \psi_i(\vec{r})$ where $\vec{G}$ is an integer multiple of the lattice vectors, and $\psi_i(\vec{r}) = \langle \vec{r} | i \rangle$ is the electron wavefunction in real space for an electron occupying site $i$.
$\langle \vec{k} | i \rangle$ actually represents a basis transformation from $\vec{k}$-space to the local basis {i} space. The annihilation and creation operators are also transformed according to $\hat{a}_{\vec{k}\sigma}=\sum_i \langle \vec{k} | i \rangle \hat{c}_{i \sigma}$ and $\hat{a}^\dagger_{\vec{k}\sigma}=\sum_i \langle i | \vec{k} \rangle \hat{c}^\dagger_{i \sigma}$. As expected, these operators anticommute in both bases, $[\hat{c}_{i\sigma},\hat{c}^\dagger_{j\sigma}]_{+} = \delta_{ij}$ and $[\hat{a}_{\vec{k}\sigma},\hat{a}^\dagger_{\vec{k}\prime\sigma}]_{+} = \delta_{\vec{k}\vec{k}\prime}$
In fact, in $\vec{k}$-space, the kinetic term of the Hubbard model takes the form of $\sum_\sigma \epsilon(\vec{k}) \hat{a}^\dagger_{\vec{k}\sigma} \hat{a}_{\vec{k}\sigma}$, and $\epsilon(\vec{k})$ depends on the geometry of the system. For instance, with the 1-D Hubbard model (assume monoatomic chain), $\epsilon(\vec{k})=-2t \cos(|\vec{k}|a)$, where $a$ is the lattice constant. Thus, it indeed changes going from the local basis {i} to momentum basis $\vec{k}$.
(Side note: I wish I can send you my professor's lecture notes on this topic, where he lays out the algebra for the basis transformation.)
