Hamiltonian written in terms of field operators: The kinetic energy (KE) part of Hamiltonian is
$$
H=-\frac{\hbar^2}{2m}\int d\mathbf{r} \Psi^\dagger(\mathbf{r}) \nabla^2\Psi(\mathbf{r}) \tag{1}
$$
Hamiltonian written in tight-binding form:
One reaches to tight-binding form by transforming the field operators as
$$
\Psi(\mathbf{r}) = \sum_i \phi_i(\mathbf{r})c_i \quad ; \quad
\Psi(\mathbf{r})^\dagger = \sum_i \phi_i^*(\mathbf{r})c_i^\dagger \tag{2}
$$
here $i$ goes over all the lattice sites inside the system, and $\phi_i(\mathbf{r})$ is the wavefunction of $i$-the lattice site (assume only one state per lattice site). When we put this transformation in $(1)$, we get
$$
H=\sum_{ij}c_i^\dagger \left[-\frac{\hbar^2}{2m}\int d\mathbf{r} \phi_i^*(\mathbf{r}) \nabla^2 \phi_j(\mathbf{r})\right] c_j
$$
$$
H=\sum_{ij}c_i^\dagger t_{ij} c_j\tag{3}
$$
Question:
In Hamiltonian $(1)$, the field operators create and destroy particles at $\mathbf{r}$ position. This position vector $\mathbf{r}$ includes all the lattice sites (for a discrete system). In the tight-binding Hamiltonian, the same job is done by operators ($c_i^\dagger, c_i$). So, can we say that the discrete version of the field operator Hamiltonian $(1)$ is equal to the tight-binding Hamiltonian $(3)$? What I mean is to discretize the position vector and $\nabla$ in $(1)$
$$
H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i) \nabla^2\Psi(\mathbf{r}_i)\\
H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i)
\left(\frac{\Psi(\mathbf{r}_i+a) – \Psi(\mathbf{r}_i) + \Psi(\mathbf{r}_i-a)}{a^2} \right) \tag{4}
$$
here $a$ is lattice constant, and $\Psi^\dagger(\mathbf{r}_i)$ creates particle at site $i$. This is exactly what $c_i^\dagger$ does. So, can replace $\Psi^\dagger(\mathbf{r}_i)$ with $c_i^\dagger$:
$$
H=-\frac{\hbar^2}{2ma^2}\sum_i
\left(c_i^\dagger c_{i+a} – c_i^\dagger c_{i} +c_i^\dagger c_{i-a} \right) \tag{5}
$$
Is not $(5)$ similar to $(3)$? Why do we need tight-binding model when we can just discretize field operators?
Best Answer
The operator $\Psi^\dagger(\mathbf{r}_i)$ creates a particle at $\mathbf{r}_i$ which is infinitely localized (i.e. it has a delta-function wave function.) In contrast, $c_i^\dagger$ creates a particle which is localized at position $\mathbf{r}_i$, but has wavefunction $\phi_i(\mathbf{r})$. You should think of this as essentially an atomic wavefunction (e.g. 1s, 2s, 2p, etc.), not a delta-function.
The choice of $\phi_i(\mathbf{r})$ depends on the problem, and the chemistry involved in your materials. Take the simplest case: Assume you are dealing with a chain of hydrogen atoms. Here, $\phi_i(\mathbf{r})$ should be taken as the 1s orbitals localized on each hydrogen atom. Of course, you could in principle include in your tight-binding model the 2s,2p,3s,3p,3d,etc. orbitals, and this should make your model more accurate. However, such complications are often unnecessary to describe the essential physics of the problem.